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# 2020 AMC 8 Problems/Problem 15
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Suppose $15\%$ of $x$ equals $20\%$ of $y.$ What percentage of $x$ is $y?$
$\textbf{(A) }5 \qquad \textbf{(B) }35 \qquad \textbf{(C) }75 \qquad \textbf{(D) }133 \frac13 \qquad \textbf{(E) }300$
## Solution 1
Since $20\% = \frac{1}{5}$, multiplying the given condition by $5$ shows that $y$ is $15 \cdot 5 = \boxed{\textbf{(C) }75}$ percent of $x$.
## Solution 2
Letting $x=100$ (without loss of generality), the condition becomes $0.15\cdot 100 = 0.2\cdot y \Rightarrow 15 = \frac{y}{5} \Rightarrow y=75$. Clearly, it follows that $y$ is $75\%$ of $x$, so the answer is $\boxed{\textbf{(C) }75}$.
## Solution 3
We have $15\%=\frac{3}{20}$ and $20\%=\frac{1}{5}$, so $\frac{3}{20}x=\frac{1}{5}y$. Solving for $y$, we multiply by $5$ to give $y = \frac{15}{20}x = \frac{3}{4}x$, so the answer is $\boxed{\textbf{(C) }75}$.
## Solution 4
We are given $0.15x = 0.20y$, so we may assume without loss of generality that $x=20$ and $y=15$. This means $\frac{y}{x}=\frac{15}{20}=\frac{75}{100}$, and thus the answer is $\boxed{\textbf{(C) }75}$.
## Solution 5
$15\%$ of $x$ is $0.15x$, and $20\%$ of $y$ is $0.20y$. We put $0.15x$ and $0.20y$ into an equation, creating $0.15x = 0.20y$ because $0.15x$ equals $0.20y$. Solving for $y$, dividing $0.2$ to both sides, we get $y = \frac{15}{20}x = \frac{3}{4}x$, so the answer is $\boxed{\textbf{(C) }75}$.
## Solution 6
$15\%$ of $x$ can be written as $\frac{15}{100}x$, or $\frac{15x}{100}$. $20\%$ of $y$ can similarly be written as $\frac{20}{100}y$, or $\frac{20y}{100}$. So now, $\frac{15x}{100} = \frac{20y}{100}$. Using cross-multiplication, we can simplify the equation as: $1500x = 2000y$. Dividing both sides by $500$, we get: $3x = 4y$. $\frac{3}{4}$ is the same thing as $75\%$, so the answer is $\boxed{\textbf{(C) }75}$.
~NiuniuMaths
~Math-X
## Video Solution (🚀Very Fast🚀)
~Education, the Study of Everything
~savannahsolver
~Interstigation
## See also
2020 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 16 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
## Pages
Welcome to William's Math Analysis Blog
## Monday, June 2, 2014
### I/D#2 - Unit O Concept 7-8: Deriving Special Right Triangles
How can we derive the 30-60-90 triangle from an equilateral triangle with a side length of 1?
First we have the triangle as shown above.
Then we divide the triangle in half making one of the 60 degree angles become two 30 degree angles. By doing this we also create two 90 degree angles. Now we have to find the missing side being the dotted line. we can do this with the pythagorean theorem. With the bottom being 1/2 and the side being 1. The answer should be y = rad3 / 2. Then we multiply all the values of the triangle by two so that we don't have any fractions.
This should be the final result of the triangle.
How can we derive the 45-45-90 triangle from an square with a side length of 1?
First you draw the square and label each one of the sides one.
Then you make a diagonal dividing the square into two triangles. You may notice that these are the two triangles are the triangles that you're looking for (45, 45, 90)
Then you use the pythagorean theorem to find the length of r. this resulting to be rad2.
Then when you get that you multiply it by n so that you get the derivation of the 45 - 45 - 90 triangle, as shown below
Something I never noticed before about special right triangles is…
That they are derived from another shape and the angles come from there.
Being able to derive these patterns myself aids in my learning because…
now I where the formulas and the angles come form. if one day I don't remember how the concept works I can always come back here and know the basics of the special right triangles. |
### click - Uplift Education
```FORCES APPLIED AT AN ANGLE &
INCLINED PLANES
FORCES APPLIED AT AN ANGLE
When forces are applied at angles other than 90o, we
need to resolve the force into its component vectors.
Then, we find the net force in each direction, write our
force equations (Fnet = ma), and solve.
FORCES AT AN ANGLE – WE DO
Luke Skywalker starts to pull a sled with Princess Leia across a
large ice pond with the force of 100 N at an angle of 30.0° with
the horizontal. Find normal force and initial acceleration if the
weight of sled and Princess Leia is 800 N and the friction force
is 40 N.
What do we do first?
FORCES AT AN ANGLE – WE DO
Luke Skywalker starts to pull a sled with Princess Leia across a
large ice pond with the force of 100 N at an angle of 30.0° with
the horizontal. Find normal force and initial acceleration if the
weight of sled and Princess Leia is 800 N and the friction force
is 40 N.
What do we do first?
Draw a free-body diagram.
FORCES AT AN ANGLE – WE DO
Luke Skywalker starts to pull a sled with Princess Leia across a
large ice pond with the force of 100 N at an angle of 30.0° with
the horizontal. Find normal force and initial acceleration if the
weight of sled and Princess Leia is 800 N and the friction force
is 40 N.
Now what?
FORCES AT AN ANGLE – WE DO
Luke Skywalker starts to pull a sled with Princess Leia across a
large ice pond with the force of 100 N at an angle of 30.0° with
the horizontal. Find normal force and initial acceleration if the
weight of sled and Princess Leia is 800 N and the friction force
is 40 N.
Now what?
Resolve F into components
FORCES AT AN ANGLE – WE DO
Luke Skywalker starts to pull a sled with Princess Leia across a
large ice pond with the force of 100 N at an angle of 30.0° with
the horizontal. Find normal force and initial acceleration if the
weight of sled and Princess Leia is 800 N and the friction force
is 40 N.
What next?
FORCES AT AN ANGLE – WE DO
Luke Skywalker starts to pull a sled with Princess Leia across a
large ice pond with the force of 100 N at an angle of 30.0° with
the horizontal. Find normal force and initial acceleration if the
weight of sled and Princess Leia is 800 N and the friction force
is 40 N.
Add the forces in each direction. What should the forces equal in
the horizontal? In the vertical?
FORCES AT AN ANGLE – WE DO
Luke Skywalker starts to pull a sled with Princess Leia across a
large ice pond with the force of 100 N at an angle of 30.0° with
the horizontal. Find normal force and initial acceleration if the
weight of sled and Princess Leia is 800 N and the friction force
is 40 N.
Add the forces in each direction. What should the forces equal in
the horizontal? Unbalanced Fnet = ma In the vertical? Fnet = 0.
FORCES AT AN ANGLE – WE DO
Luke Skywalker starts to pull a sled with Princess Leia across a
large ice pond with the force of 100 N at an angle of 30.0° with
the horizontal. Find normal force and initial acceleration if the
weight of sled and Princess Leia is 800 N and the friction force
is 40 N.
mg = 800 N m = 80 kg F = 100 N
Horizontal direction:
F cos θ – Ffr = ma
86.6 – 40 = 80 a
a = 0.58 m/s2
Ffr = 40 N
vertical direction :
F sin θ + Fn - mg = 0
50 + Fn = 800
Fn = 750 N
FORCES AT AN ANGLE – YOU DO
1. A box of books weighing 325 N moves with a
constant velocity across the floor when it is pushed
with a force of 425 N exerted downward at an angle
of 35.2 degrees below the horizontal. Find the
coefficient of friction between the box and the floor.
2. Two forces act on a 4.5-kg block resting on a
frictionless surface as shown. What is the
magnitude of the horizontal acceleration of the
block?
FORCES AT AN ANGLE – YOU DO
1. A box of books weighing 325 N moves with a constant velocity
across the floor when it is pushed with a force of 425 N exerted
downward at an angle of 35.2 degrees below the horizontal. Find
the coefficient of friction between the box and the floor.
Constant velocity means no acceleration, which means that the
forces in ALL directions are balanced.
Ff = 425 cos (35.2) = 361.7 N
F f = μs F n
Fn = mg + 425 sin (35.2) = 325 N + 425 sin(35.2)
μs = 0.66
Fn = 548.2 N
FORCES AT AN ANGLE – YOU DO
2. Two forces act on a 4.5-kg block resting on a frictionless
surface as shown. What is the magnitude of the horizontal
acceleration of the block?
Resolve into components, Add the forces, then find Fnet = ma.
3.7 N + 5.9 N cos 43 = Fnet = 8.3 N
a = Fnet / m = 8 .3 N / 4.5 kg = 1.8 m/s2
Draw a free-body diagram for a
block accelerating down a ramp.
FORCES ON AN INCLINED PLANE
Fn
Ffr
a
Oh no!
This will be ugly!
We will have to resolve Fn,
Ffr, Fnet, and a into
components!!!
Or … will we?
FORCES ON AN INCLINED PLANE
Fn
y
Ffr
x
a
Solution:
Choose a more convenient
coordinate system!
Make x be parallel to incline and y
be perpendicular.
Now, only mg has to be resolved into
components.
Motion and all the other forces will
be in the x or y direction.
FORCES ON AN INCLINED PLANE
The only force
that we have
to resolve into
components is
weight
Fn
Ffr
a
y
x
FORCES ON AN INCLINED PLANE
The only force
that we have
to resolve into
components is
weight
Notice that the
angle for resolving
mg is the same as
the angle of the
incline
a
y
x
Fn
Ffr
Resolve vector mg into two components. Now
instead of three forces, we have four forces
Fn
Ffr
direction perpendicular to the
incline:
Fnet = ma = 0
Fn = mg cos θ
Resolve vector mg into two components. Now
instead of three forces, we have four forces
Fn
Ffr
Write the Fnet equations
direction perpendicular to the
incline:
Fnet = ma = 0
Fn = mg cos θ
force pressing the object into the surface is not full weight mg,
but only part of it,
So the normal force acting on the object is only part of full
weight mg: Fn = mg cos θ
Resolve vector mg into two components. Now
instead of three forces, we have four forces
Fn
Ffr
direction parallel to the incline:
Fnet = ma
Fnet = Ff – mg sin θ = ma
The force that causes acceleration downward is only part of the
full force of gravity.
Greater acceleration the steeper the slope.
If the incline = 0, then there is no horizontal movement due to
gravity.
INCLINED PLANE – WE DO
A cute bear, m = 60 kg, is sliding down an iced incline 300. The ice
can support up to 550 N. Will bear fall through the ice?
If the coefficient of the friction is 0.115, what is the acceleration of
the bear?
What’s my strategy??
INCLINED PLANE – WE DO
A cute bear, m = 60 kg, is sliding down an iced incline 300. The ice
can support up to 550 N. Will bear fall through the ice?
If the coefficient of the friction is 0.115, what is the acceleration of
the bear?
What’s my strategy??
1) Draw the free-body diagram
2) Choose a coordinate system with x parallel to incline
3) Resolve mg into components
4) Add vectors perpendicular to plane and set Fnet = ma = 0.
5) Add vectors parallel to plane and set Fnet = ma.
INCLINED PLANE – WE DO
A cute bear, m = 60 kg, is sliding down an iced incline 300. The ice
can support up to 550 N. Will bear fall through the ice?
If the coefficient of the friction is 0.115, what is the acceleration of
the bear?
m = 60 kg
θ = 300
μ = 0.115
g = 10 m/s2
Perpendicular direction:
Fnet = ma
a=0
Fn - mg cos θ = 0
Fn = 520 N < 550 N
ice can support him, but he
should not eat too much
Parallel direction:
Fnet = ma
mg sin θ – Ffr = ma
300 – 60 = 60 a
a = 4 m/s2
Ffr = μ Fn = 60 N
cute bear is speeding up!!!!
INCLINED PLANE – YOU DO
3. A block weighing 15.0 newtons is on a ramp inclined at
40.0° to the horizontal. A 3.0 Newton force of friction, Ff ,
acts on the block as it is pulled up the ramp at constant
velocity with force F, which is parallel to the ramp.
Find F.
4. A 75 kg box slides down a ramp inclined at 25O with an
acceleration of 3.60 m/s2.
a) Find the coefficient of friction.
b) What acceleration would a 175 kg box have on this
ramp?
INCLINED PLANE – YOU DO
3. A block weighing 15.0 newtons is on a ramp inclined at 40.0°
to the horizontal. A 3.0 Newton force of friction, Ff , acts on the
block as it is pulled up the ramp at constant velocity with force F,
which is parallel to the ramp.
Find F.
Constant velocity = no acceleration. This means that forces in
parallel to the inclined plane are also balanced.
Ff = F
F = 12.6 N
INCLINED PLANE – YOU DO
4. A 75 kg box slides down a ramp inclined at 25O with an
acceleration of 3.60 m/s2.
a) Find the coefficient of friction.
b) What acceleration would a 175 kg box have on this
ramp?
Fnet = mg sin 25 – Ff = ma
Ff = mg sin 25 – ma = 40.6 N
F f = μs F n
μs = Ff / Fn = 40.6 / mgcos25 = 0.06
a = (mg sin 25 – Ff ) / m =
EXIT TICKET
Using pictures, words, and equations, describe
how to solve an inclined plane force problem.
``` |
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# Class 12 NCERT Solutions – Mathematics Part I – Chapter 5 Continuity And Differentiability – Exercise 5.5 | Set 2
• Last Updated : 05 Apr, 2021
### (x cos x)x + (x sin x)1/x
Solution:
Given: (x cos x)x + (x sin x)1/x
Let us considered y = u + v
Where, u = (x cos x)x and v = (x sin x)1/x
So, dy/dx = du/dx + dv/dx ………(1)
So first we take u = (x cos x)
On taking log on both sides, we get
log u = log(x cos x)
log u = xlog(x cos x)
Now, on differentiating w.r.t x, we get
………(2)
Now we take u =(x sin x)1/x
On taking log on both sides, we get
log v = log (x sin x)1/x
log v = 1/x log (x sin x)
log v = 1/x(log x + log sin x)
Now, on differentiating w.r.t x, we get
………(3)
Now put all the values from eq(2) and (3) into eq(1)
### Question 12. xy + yx = 1
Solution:
Given: xy + yx = 1
Let us considered
u = xy and v = yx
So,
………(1)
So first we take u = xy
On taking log on both sides, we get
log u = log(xy)
log u = y log x
Now, on differentiating w.r.t x, we get
………(2)
Now we take v = yx
On taking log on both sides, we get
log v = log(y)x
log v = x log y
Now, on differentiating w.r.t x, we get
………(3)
Now put all the values from eq(2) and (3) into eq(1)
### Question 13. yx = xy
Solution:
Given: yx = xy
On taking log on both sides, we get
log(yx) = log(xy)
xlog y = y log x
Now, on differentiating w.r.t x, we get
### Question 14. (cos x)y = (cos y)x
Solution:
Given: (cos x)y = (cos y)x
On taking log on both sides, we get
y log(cos x) = x log (cos y)
Now, on differentiating w.r.t x, we get
### Question 15. xy = e(x – y)
Solution:
Given: xy = e(x – y)
On taking log on both sides, we get
log(xy) = log ex – y
log x + log y = x – y
Now, on differentiating w.r.t x, we get
### Question 16. Find the derivative of the function given by f(x) = (x + 1)(x + x2)(1 + x4)(1 + x8) and hence find f'(1).
Solution:
Given: f(x) = (x + 1)(x + x2)(1 + x4)(1 + x8)
Find: f'(1)
On taking log on both sides, we get
log(f(x)) = log(1 + x) + log(1 + x2) + log(1 + x4) + log(1 + x8)
Now, on differentiating w.r.t x, we get
∴ f'(1) = 2.2.2.2.
f'(1) = 120
### Do they all give the same answer?
Solution:
(i) By using product rule
dy/dx = (3x4 – 15x3 + 24x2 + 7x2 – 35x + 56) + (2x4 + 14x2 + 18x – 5x3 – 35x – 45)
dy/dx = 5x4 – 20x3 + 45x2 – 52x + 11
(ii) By expansion
y = (x2 – 5x + 8)(x3 + 7x + 9)
y = x5 + 7x3 + 9x2 – 5x4 – 35x2 – 45x + 8x3 + 56x + 72
y = x5 – 5x4 + 15x3 – 26x2 + 11x + 72
dy/dx = 5x4 – 20x3 + 45x2 – 52x + 11
(iii) By logarithmic expansion
Taking log on both sides
log y = log(x2 – 5x + 8) + log(x3 + 7x + 9)
Now on differentiating w.r.t. x, we get
dy/dx = 2x4 + 14x2 + 18x – 5x3 – 35x – 45 + 3x4 – 15x3 + 24x2 + 7x2 – 35x + 56
dy/dx = 5x4 – 20x3 + 45x2 – 52x + 11
Answer is always same what-so-ever method we use.
### Question 18. If u, v and w are function of x, then show that
Solution:
Let y = u.v.w.
Method 1: Using product Rule
Method 2: Using logarithmic differentiation
Taking log on both sides
log y = log u + log v + log w
Now, Differentiating w.r.t. x
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# 10.1 Add and subtract polynomials (Page 2/10)
Page 3 / 10
Evaluate $3{x}^{2}-9x+7$ when
1. $\phantom{\rule{0.2em}{0ex}}x=3$
2. $\phantom{\rule{0.2em}{0ex}}x=-1$
## Solution
ⓐ $x=3$ $3{x}^{2}-9x+7$ Substitute 3 for $x$ $3{\left(3\right)}^{2}-9\left(3\right)+7$ Simplify the expression with the exponent. $3·9-9\left(3\right)+7$ Multiply. $27-27+7$ Simplify. $7$
ⓑ $x=-1$ $3{x}^{2}-9x+7$ Substitute −1 for $x$ $3{\left(-1\right)}^{2}-9\left(-1\right)+7$ Simplify the expression with the exponent. $3·1-9\left(-1\right)+7$ Multiply. $3+9+7$ Simplify. $19$
Evaluate: $2{x}^{2}+4x-3$ when
1. $\phantom{\rule{0.2em}{0ex}}x=2$
2. $\phantom{\rule{0.2em}{0ex}}x=-3$
1. 13
2. 3
Evaluate: $7{y}^{2}-y-2$ when
1. $\phantom{\rule{0.2em}{0ex}}y=-4$
2. $\phantom{\rule{0.2em}{0ex}}y=0$
1. 114
2. −2
The polynomial $-16{t}^{2}+300$ gives the height of an object $t$ seconds after it is dropped from a $300$ foot tall bridge. Find the height after $t=3$ seconds.
## Solution
Substitute 3 for $t$ Simplify the expression with the exponent. Multiply. Simplify.
The polynomial $-8{t}^{2}+24t+4$ gives the height, in feet, of a ball $t$ seconds after it is tossed into the air, from an initial height of $4$ feet. Find the height after $t=3$ seconds.
4 feet
The polynomial $-8{t}^{2}+24t+4$ gives the height, in feet, of a ball $x$ seconds after it is tossed into the air, from an initial height of $4$ feet. Find the height after $t=2$ seconds.
20 feet
## Practice makes perfect
Identify Polynomials, Monomials, Binomials and Trinomials
In the following exercises, determine if each of the polynomials is a monomial, binomial, trinomial, or other polynomial.
$5x+2$
binomial
${z}^{2}-5z-6$
${a}^{2}+9a+18$
trinomial
$-12{p}^{4}$
${y}^{3}-8{y}^{2}+2y-16$
polynomial
$10-9x$
$23{y}^{2}$
monomial
${m}^{4}+4{m}^{3}+6{m}^{2}+4m+1$
Determine the Degree of Polynomials
In the following exercises, determine the degree of each polynomial.
$8{a}^{5}-2{a}^{3}+1$
5
$5{c}^{3}+11{c}^{2}-c-8$
$3x-12$
1
$4y+17$
$-13$
0
$-22$
In the following exercises, add or subtract the monomials.
${\text{6x}}^{2}+\phantom{\rule{0.2em}{0ex}}9{x}^{2}$
15 x 2
${\text{4y}}^{3}+\phantom{\rule{0.2em}{0ex}}6{y}^{3}$
$-12u\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}4u$
−8 u
$-3m\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}9m$
$5a\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}7b$
5 a + 7 b
$8y\phantom{\rule{0.2em}{0ex}}+\phantom{\rule{0.2em}{0ex}}6z$
Add: $\text{}4a\phantom{\rule{0.2em}{0ex}},\phantom{\rule{0.2em}{0ex}}-3b,\phantom{\rule{0.2em}{0ex}}-8a$
−4 a −3 b
Add: $4x\phantom{\rule{0.2em}{0ex}},\phantom{\rule{0.2em}{0ex}}3y\phantom{\rule{0.2em}{0ex}},\phantom{\rule{0.2em}{0ex}}-3x$
$18x-2x$
16 x
$13a-3a$
Subtract $5{x}^{6}\phantom{\rule{0.2em}{0ex}}\text{from}\phantom{\rule{0.2em}{0ex}}-12{x}^{6}$
−17 x 6
Subtract $2{p}^{4}\phantom{\rule{0.2em}{0ex}}\text{from}\phantom{\rule{0.2em}{0ex}}-7{p}^{4}$
In the following exercises, add or subtract the polynomials.
$\left(4{y}^{2}+10y+3\right)+\left(8{y}^{2}-6y+5\right)$
12 y 2 + 4 y + 8
$\left(7{x}^{2}-9x+2\right)+\left(6{x}^{2}-4x+3\right)$
$\left({x}^{2}+6x+8\right)+\left(-4{x}^{2}+11x-9\right)$
−3 x 2 + 17 x − 1
$\left({y}^{2}+9y+4\right)+\left(-2{y}^{2}-5y-1\right)$
$\left(3{a}^{2}+7\right)+\left({a}^{2}-7a-18\right)$
4 a 2 − 7 a − 11
$\left({p}^{2}-5p-11\right)+\left(3{p}^{2}+9\right)$
$\left(6{m}^{2}-9m-3\right)-\left(2{m}^{2}+m-5\right)$
4 m 2 − 10 m + 2
$\left(3{n}^{2}-4n+1\right)-\left(4{n}^{2}-n-2\right)$
$\left({z}^{2}+8z+9\right)-\left({z}^{2}-3z+1\right)$
11 z + 8
$\left({z}^{2}-7z+5\right)-\left({z}^{2}-8z+6\right)$
$\left(12{s}^{2}-15s\right)-\left(s-9\right)$
12 s 2 − 16 s + 9
$\left(10{r}^{2}-20r\right)-\left(r-8\right)$
Find the sum of $\left(2{p}^{3}-8\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left({p}^{2}+9p+18\right)$
2 p 3 + p 2 + 9 p + 10
Find the sum of $\left({q}^{2}+4q+13\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(7{q}^{3}-3\right)$
Subtract $\left(7{x}^{2}-4x+2\right)\phantom{\rule{0.2em}{0ex}}\text{from}\phantom{\rule{0.2em}{0ex}}\left(8{x}^{2}-x+6\right)$
x 2 + 3 x + 4
Subtract $\left(5{x}^{2}-x+12\right)\phantom{\rule{0.2em}{0ex}}\text{from}\phantom{\rule{0.2em}{0ex}}\left(9{x}^{2}-6x-20\right)$
Find the difference of $\left({w}^{2}+w-42\right)$ and $\left({w}^{2}-10w+24\right)$
11 w − 66
Find the difference of $\left({z}^{2}-3z-18\right)$ and $\left({z}^{2}+5z-20\right)$
Evaluate a Polynomial for a Given Value
In the following exercises, evaluate each polynomial for the given value.
$\text{Evaluate}\phantom{\rule{0.2em}{0ex}}8{y}^{2}-3y+2$
1. $\phantom{\rule{0.2em}{0ex}}y=5$
2. $\phantom{\rule{0.2em}{0ex}}y=-2$
3. $\phantom{\rule{0.2em}{0ex}}y=0$
1. 187
2. 40
3. 2
$\text{Evaluate}\phantom{\rule{0.2em}{0ex}}5{y}^{2}-y-7\phantom{\rule{0.2em}{0ex}}\text{when:}$
1. $\phantom{\rule{0.2em}{0ex}}y=-4\phantom{\rule{0.2em}{0ex}}$
2. $\phantom{\rule{0.2em}{0ex}}y=1$
3. $y=0$
$\text{Evaluate}\phantom{\rule{0.2em}{0ex}}4-36x\phantom{\rule{0.2em}{0ex}}\text{when:}$
1. $\phantom{\rule{0.2em}{0ex}}x=3$
2. $\phantom{\rule{0.2em}{0ex}}x=0$
3. $x=-1$
1. −104
2. 4
3. 40
$\text{Evaluate}\phantom{\rule{0.2em}{0ex}}16-36{x}^{2}\phantom{\rule{0.2em}{0ex}}\text{when:}$
1. $\phantom{\rule{0.2em}{0ex}}x=-1\phantom{\rule{0.2em}{0ex}}$
2. $\phantom{\rule{0.2em}{0ex}}x=0$
3. $x=2$
A window washer drops a squeegee from a platform $275$ feet high. The polynomial $-16{t}^{2}+275$ gives the height of the squeegee $t$ seconds after it was dropped. Find the height after $t=4$ seconds.
19 feet
A manufacturer of microwave ovens has found that the revenue received from selling microwaves at a cost of p dollars each is given by the polynomial $-5{p}^{2}+350p.$ Find the revenue received when $p=50$ dollars.
## Everyday math
Fuel Efficiency The fuel efficiency (in miles per gallon) of a bus going at a speed of $x$ miles per hour is given by the polynomial $-\frac{1}{160}\phantom{\rule{0.1em}{0ex}}{x}^{2}+\frac{1}{2}\phantom{\rule{0.1em}{0ex}}x.$ Find the fuel efficiency when $x=40\phantom{\rule{0.2em}{0ex}}\text{mph.}$
10 mpg
Stopping Distance The number of feet it takes for a car traveling at $x$ miles per hour to stop on dry, level concrete is given by the polynomial $0.06{x}^{2}+1.1x.$ Find the stopping distance when $x=60\phantom{\rule{0.2em}{0ex}}\text{mph.}$
## Writing exercises
Using your own words, explain the difference between a monomial, a binomial, and a trinomial.
Eloise thinks the sum $5{x}^{2}+3{x}^{4}$ is $8{x}^{6}.$ What is wrong with her reasoning?
## Self check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math, every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no—I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
how do I set up the problem?
what is a solution set?
Harshika
find the subring of gaussian integers?
Rofiqul
hello, I am happy to help!
Abdullahi
hi mam
Mark
find the value of 2x=32
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
yes i wantt to review
Mark
use the y -intercept and slope to sketch the graph of the equation y=6x
how do we prove the quadratic formular
Darius
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
An investment account was opened with an initial deposit of $9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years? Kala Reply lim x to infinity e^1-e^-1/log(1+x) given eccentricity and a point find the equiation Moses Reply A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place. Kimberly Reply Jeannette has$5 and \$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? |
# If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is
Question:
If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is
(a) n(n − 2)
(b) n(n + 2)
(c) n(n + 1)
(d) n(n − 1)
Solution:
Here, we are given an A.P. whose $n^{\text {th }}$ term is given by the following expression, $a_{n}=2 n+1$. We need to find the sum of first $n$ terms.
So, here we can find the sum of the $n$ terms of the given A.P., using the formula, $S_{n}=\left(\frac{n}{2}\right)(a+l)$
Where, a = the first term
l = the last term
So, for the given A.P,
The first term (a) will be calculated using in the given equation for nth term of A.P.
$a=2(1)+1$
$=2+1$
$=3$
Now, the last term (l) or the nth term is given
$l=a_{n}=2 n+1$
So, on substituting the values in the formula for the sum of n terms of an A.P., we get,
$S_{n}=\left(\frac{n}{2}\right)[(3)+2 n+1]$
$=\left(\frac{n}{2}\right)[4+2 n]$
$=\left(\frac{n}{2}\right)(2)(2+n)$
$=n(2+n)$
Therefore, the sum of the $n$ terms of the given A.P. is $S_{n}=n(2+n)$. So the correct option is (b). |
EDUCATION UPDATE BLOGS
Dr. Alfred Posamentier: August 2010 Archives
# August 2010 Archives
## A Power Loop
| Comments |
Can you imagine that a number is equal to the sum of the cubes of its digits? Take the time to explain exactly what this means. This should begin to "set them up" for this most unusual phenomenon. By the way, this is true for only five numbers. Below are these five most unusual numbers.
Students should take a moment to appreciate these spectacular results and take note that these are the only such numbers for which this is true.
Taking sums of the powers of the digits of a number leads to interesting results. We can extend this procedure to get a lovely (and not to mention, surprising) technique you can use to have students familiarize themselves with powers of numbers and at the same time try to get to a startling conclusion.
Have them select any number and then find the sum of the cubes of the digits, just as we did above. Of course, for any other number than those above, they will have reached a new number. They should then repeat this process with each succeeding sum until they get into a "loop." A loop can be easily recognized. When they reach a number that they already reached earlier, then they are in a loop. This will become clearer with an example.
Let's begin with the number 352 and find the sum of the cubes of the digits.
The sum of the cubes of the digits of 352 is:
33 + 53 + 23 = 27 + 125 + 8 = 160.
Now we use this sum, 160, and repeat the process:
The sum of the cubes of the digits of 160 is:
13 + 63 + 03 = 1 + 216 + 0 = 217.
Again repeat the process with 217:
The sum of the cubes of the digits of 217 is:
23 + 13 + 73 = 8 + 1 + 343 = 352.
Surprise! This is the same number (352) we started with.
You might think it would have been easier to begin by taking squares. You are in for a surprise. Let's try this with the number 123.
Beginning with 123, the sum of the squares of the digits is:
12 + 22 + 32 = 1 + 4 + 9 = 14.
1. Now using 14, the sum of the squares of the digits is: 12 + 42 = 1 + 16 = 17.
2. Now using 17, the sum of the squares of the digits is:
12 + 72 = 1 + 49 = 50.
3. Now using 50, the sum of the squares of the digits is:
52 + 02 = 25.
4. Now using 25, the sum of the squares of the digits is:
22 + 52 = 4 + 25 = 29.
5. Now using 29, the sum of the squares of the digits is:
22 + 92 = 85.
6. Now using 85, the sum of the squares of the digits is:
82 + 52 = 64 + 25 = 89.
7. Now using 89, the sum of the squares of the digits is:
82 + 92 = 64 + 81 = 145.
8. Now using 145, the sum of the squares of the digits is:
12 + 42 + 52 = 1 + 16 + 25 = 42.
9. Now using 42, the sum of the squares of the digits is:
42 + 22 = 16 + 4 = 20.
10. Now using 20, the sum of the squares of the digits is:
22 + 02 = 4.
11. Now using 4, the sum of the squares of the digits is:
42 = 16.
12. Now using 16, the sum of the squares of the digits is:
12 + 62 = 1 + 36 = 37.
13. Now using 37, the sum of the squares of the digits is:
32 + 72 = 9 + 49 = 58.
14. Now using 58, the sum of the squares of the digits is:
52 + 82 = 25 + 64 = 89.
Notice that the sum, 89, that we just got in step 14 is the same as in step 6, and so a repetition will now begin after step 14. This indicates that we would continue in a loop.
Students may want to experiment with the sums of the powers of the digits of any number and see what interesting results it may lead to. They should be encouraged to look for patterns of loops, and perhaps determine the extent of a loop based on the nature of the original number.
In any case, this intriguing unit can be fun just as it is presented here or it can be a source for further investigation by interested students. We need to bring more of these "spectacular" math wonders to our students!
Education Update, Inc. All material is copyrighted and may not be printed without express consent of the publisher. © 2011. |
# If $A$ has a maximum, prove that it only has one.
Let $A\subseteq \mathbb{R}$. We say that a real number $M\in\mathbb{R}$ is a maximum of $A$ if $M$ is an upper bound for $A$ and $M\in A$.
If $A$ has a maximum, prove that it only has one; and prove that if $\max A$ exists, then so does $\sup A$, and $\sup A=\max A$.
To prove that $A$ has only one maximum, I'm pretty sure I have to assume that there are two maximums but then show that they are the same. I'm just confused as to how to do this.
Any help is greatly appreciated. Thank you!
Let $M$ and $M'$ two maximum of a set $A$ then $M,M'\in A$ and
$$\forall a\in A,\; a\le M$$ so in particular and since $M'\in A$ then we have $M'\le M$. Similarly we get $M\le M'$ hence $M=M'$.
• Similarly, since $\sup A$ is the least upper bound for $A$ and $M$ is an upper bound for $A$, we have $\sup A \leq M$. But $\sup A$ is an upper bound for $A$ and $M \in A$, so $M \leq \sup A$. – Alex Wertheim Jan 20 '15 at 21:48
Let $M_1$ and $M_2$ both be maxima of $A$, and suppose $M_1 > M_2$. Then since $M_2$ is an upper bound of $A$, $M_1 \not \in A$, contradicting the definition of a maximum which requires $M \in A$.
Then suppose $M$ is a maximum of $A$. Then $\forall a \in A: a \leq M$, so that part of the definition of $\sup$ is met by $M$. Now suppose there were some other value $M' < M$ also meeting that definition. Since $M\in A$, $M$ cannot be less than or equal to than $M'$. Amd that contradicts the statement about $M'$ also meeting that part of the definition of $\sup A$. So $M$ is the lowest number such that all members of $A$ are less than or equal to $M$, completing the requirements for $M = \sup A$. |
# How do you evaluate 2^10 please solve and give an example and explanation? I really need help please.
May 18, 2018
2^10=2·2·2·2·2·2·2·2·2·2=1024
#### Explanation:
The expresion ${2}^{10}$ means that you have to repeat 2 as factor 10 times.
As you did long time ago, when you repeat a number adding a number of times you said for example 3+3+3+3=4·3=12
Now the number you are repeating is as factor and the notation is a^n=a·a·a·a····a (n times)
Hope this helps
May 18, 2018
${2}^{10} = 1024$
#### Explanation:
The powers of $10$ are well known ...start with $1$ and keep multiplying by $10$ (just add another $0$ each time. You get
$1 , \text{ "10," "100," "1000," "10,000," } 100 , 000 \ldots .$
For the powers of $2$, start with $1$ and keep multiplying by $2$ which is the same as doubling each time. You get:
$1 , \text{ "2," "4," "8," "16," "32," } 64 \ldots . .$
$2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 1024$
When you have multiplied by $2$ a total of $10$ times you have ${2}^{10}$
#
The answer will be $1024$
May 18, 2018
${2}^{10} = 1024$
#### Explanation:
${2}^{10} = {2}^{4} \cdot {2}^{4} \cdot {2}^{2}$
$\therefore \text{when multiplying add the exponents together}$
$\therefore {2}^{10} = 16 \cdot 16 \cdot 4 = 1024$
example:-
$\therefore {3}^{16} = {3}^{4} \cdot {3}^{4} \cdot {3}^{4} \cdot {3}^{2} \cdot {3}^{2}$
$\therefore {3}^{16} = 81 \cdot 81 \cdot 81 \cdot 9 \cdot 9 = 43046721$
For ${2}^{10} : -$
"on your calculator(hp9s):- $p r e s s$2 "then press ${x}^{y}$ then press $10$ then press enter display$= 1024$ |
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# Evaluate the expression $\underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{\sqrt{7+2x}-\left(\sqrt{5}+\sqrt{2} \right)}{{{x}^{2}}-10}$
Last updated date: 15th Apr 2024
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Hint: You could use either of the two here, you could use L’ Hopital’s rule after confirming this limit has
an indeterminate form, or you could directly work by rationalising the numerator, to do away with the
radicals, and then simply substituting $x$ with its limiting value in the expression you get thereafter.
We’ll use the simple way of rationalising to find out the limit over here, since it’s easier.
Now the given equation is;
$\underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{\sqrt{7+2x}-\left( \sqrt{5}+\sqrt{2} \right)}{{{x}^{2}}- 10}$ ………………..(i)
Now, let’s try rationalising equation (i).
For that, let’s multiply the numerator and the denominator with the conjugate of whatever part of the
fraction has the radicals. Since we have radicals in the numerator, we’ll multiply the numerator and
denominator by its conjugate, which $=\sqrt{7+2x}+(\sqrt{5}+\sqrt{2})$
Doing so, we get :
$\Rightarrow \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{\sqrt{7+2x}-\left( \sqrt{5}+\sqrt{2} \right)}{{{x}^{2}}-10}\times \dfrac{\sqrt{7+2x}+\left( \sqrt{5}+\sqrt{2} \right)}{\sqrt{7+2x}+\left( \sqrt{5}+\sqrt{2} \right)}$
Now, above equation numerator looks similar to the identity,
$\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$, where $a=\sqrt{7+2x},b=\sqrt{5}+\sqrt{2}$
So what we could do here is, apply the identity mentioned above. Doing so, we get :
$\Rightarrow \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{{{\left( \sqrt{7+2x} \right)}^{2}}-{{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}}{({{x}^{2}}-10)\left( \sqrt{7+2x}+\left( \sqrt{5}+\sqrt{2} \right) \right)}$
Now, let’s try further simplifying the above equation. After solving the above equation further, we get;
$\to \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{\left( 7+2x \right)-\left( 5+2+2\sqrt{10} \right)}{\left( {{x}^{2}}-10 \right)\left( \sqrt{7+2x}+\sqrt{5}+\sqrt{2} \right)}$ (Using property:
$\underset{{}}{\mathop{{{\left( a+b \right)}^{2}}}}\,={{a}^{2}}+{{b}^{2}}+2ab$ )
Now, on opening the brackets in the numerator and solving it, we get :
$\Rightarrow \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{{7}+2x-{7}- 2\sqrt{10}}{\left( {{x}^{2}}-10 \right)\left( \sqrt{7+2x}+\sqrt{5}+\sqrt{2} \right)}$
Taking $2$ common from the terms in the numerator, we get :
$\Rightarrow \underset{x\to \sqrt{10}}{\mathop{\lim }}\,\dfrac{2{\left( x-\sqrt{10} \right)}}{{\left( x-\sqrt{10} \right)}\left( x+\sqrt{10} \right)\left( \sqrt{7+2x}+\sqrt{5}+\sqrt{2} \right)}$……………….(ii)
Now, let’s put the $\underset{x\to \sqrt{10}}{\mathop{\lim }}\,$ in the equation (ii). Doing so, we get :
$\Rightarrow \dfrac{2}{\left( \sqrt{10}+\sqrt{10} \right)\left( \sqrt{7+2\sqrt{10}}+\sqrt{5}+\sqrt{2} \right)}$
$\Rightarrow \dfrac{{{2}}}{{2}\sqrt{10}\left( \left( \sqrt{7+2\sqrt{10}} \right)+\sqrt{5}+\sqrt{2} \right)}$
$\Rightarrow \dfrac{1}{\sqrt{10}\left( \sqrt{7+2\sqrt{10}}+\sqrt{5}+\sqrt{2} \right)}$
Hence, the solution is $\dfrac{1}{\sqrt{10}\left( \sqrt{7+2\sqrt{10}}+\sqrt{5}+\sqrt{2} \right)}$
Note: We, here have used rationalisation followed by substitution to solve the limit. We could also use
L’ Hopital Rule, because the limit becomes a $\dfrac{0}{0}$ form on putting $x=\sqrt{10}$ in the limit.
Note that,
$\sqrt{7+2\sqrt{10}}=\sqrt{{{(\sqrt{5})}^{2}}+{{(\sqrt{2})}^{2}}+2.\sqrt{2}.\sqrt{5}}=\sqrt{{{\left( \sqrt{5}+\sqrt{2} \right)}^{2}}}=\sqrt{5}+\sqrt{2}$, and this is what makes the numerator zero on putting
$x=\sqrt{10}$. Thus, you could’ve used L’ Hopital Rule over here as well. |
# Quick Answer: How To Find Square Root Using Prime Factorization?
## What is the square root of 576 by prime factorization?
What is the square root of 576 in prime factorization form? The square root of 576 is √ 576 = 24.
## What is the square root of 784 by prime factorization method?
Thus, by prime factorisation method, we obtain the square root of 784 to be 28. The correct answer is 28.
## What is the square root of 2025 by prime factorization method?
Let us look at the prime factorization of the given number. So we can say that 2025 =45×45. Therefore, the square root of 2025 is 45.
## How do you find the square root of 7744 by prime factorization?
The numbers that we multiply are factors of the product. Hence, a factor can be defined as a number which completely divides a number without leaving any remainder. For example, factors of 12 are 1, 2, 3, 4, 6 and 12. Hence the square root of 7744 is 88 by using the prime factorization method.
You might be interested: Often asked: Why Is The Square Root Of -1 I?
## What is the square number of 576?
The value of the square root of 576 is equal to 24. In radical form, it is denoted as √ 576 = 24.
## What is the square of 4225?
The answer is 65. To solve this, find a number when multiplied twice will give you an answer of 4,225. In this case, 65 x 65 = 4,225.
## How do you find the square root of 1156?
Square Root of Solved Examples
1. Write 1156 as 11 56 in pairs. 11 is the dividend now.
2. Find a (number x number) that gives ≤ 11. We find 3 × 3 = 9.
3. Bring down the next pair of 56.
4. Double the quotient.
5. Find a (number + 60) × number that gives the result ≤ 2 56.
6. Subtract this from 256.
7. Thus, √ 1156 = 34.
## How do you find the square root of 729?
To find the square root of 729: We can use prime factorization method to obtain one number from each pair of the same numbers and multiply them. The prime factorization of 729 is 3 × 3 × 3 × 3 × 3 × 3 which consists of 3 pairs of the same number, 3 × 3 x 3. Thus, the square root of 729 = 9 × 3 = 27.
## What is the square of 1296?
So, the square root of number 1296 is 36.
## How do you solve under root 1521?
Square root of 1521 =13×13⇒ Square root of 1521 =39⇒√ 1521 =39.
## What is the prime factorization of 2025?
Positive Integer factors of 2025 = 3, 9, 27, 81, 5, 405, 2025 divided by 3, 3, 3, 3, 5, 5, gives no remainder. They are integers and prime numbers of 2025, they are also called composite number.
You might be interested: Readers ask: Square Root Of 2 How To Calculate?
## What is a square root of 4096?
∴ The square root of 4096 is 64.
## What is the square root of 5776?
Hence, the square root of 5776 is 76.
## What is the positive square root of 6400?
Therefore, the square root of 6400 is 80. It can be represented as √ 6400 =80.
## What is the square root of 529 by prime factorization method?
Square Root of 529 by Prime Factorization Hence, the square root of 529 is 23. Thus, the square root of 529 is 23. |
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Adding and Subtracting Scientific Notation Lesson
How to Add and Subtract Scientific Notation
When numbers are expressed in scientific notation, we must match the powers of the base 10's before adding or subtracting them.
For example, 3.5 x 104 and 6.1 x 106 have different base 10 powers (the first number's base 10 is raised to the 4th power and the second number's base 10 is raised to the 6th power).
To add or subtract these numbers in scientific notation we will need to match the powers of the base 10's. We have three options for this, all of which will reach the same solution. We can:
1. Make the smaller power bigger to match the bigger power.
2. Make the bigger power smaller to match the smaller power.
3. Do a combination of #1 and #2 so the powers meet somewhere in the middle.
In this lesson we will show how to change scientific notation base 10 powers up or down. At the end of the lesson there are example problems for adding and subtracting in scientific notation.
INTRODUCING
How to Change the Power of the Base 10
Changing the power of the base 10 in scientific notation is simple. We can increase or decrease the power at will.
For every increase in the index of the power, we move the decimal to the left one place.
For example:
3.55 x 104 = 0.0355 x 106
We moved the base 10 power's index up by two and the decimal to the left by two places. Both scientific notation expressions equate to 35,500.
Likewise, for every decrease in the index of the power, we move the decimal to the right one place.
For example:
8.10 x 105 = 8100.00 x 102
We moved the base 10 power's index down by three and the decimal to the right by three places. Both scientific notation expressions equate to 810,000.
Example Problems
Let's go through a couple of example problems together to reinforce our ability to add and subtract scientific notation.
Find the sum of 2.50 x 104 and 4.14 x 103.
Solution:
1. Let's start by moving the second number's base 10 power up from 3 to 4 so we can then add the numbers.
2. 4.14 x 103 = 0.414 x 104
3. 2.50 x 104 + 0.414 x 104 = 2.914 x 104
Subtracting Scientific Notation Example Problem
Find the difference 1.25 x 107 - 5.00 x 105.
Solution:
1. Let's try making the base 10 powers meet in the middle this time. We will do this by moving the first number's power down to 6 and second number's power up to 6.
2. 1.25 x 107 = 12.50 x 106
3. 5.00 x 105 = 0.50 x 106
4. 12.50 x 106 - 0.50 x 106 = 12.00 x 106.
5. We currently have two digits left of the decimal. Let's reformat it to proper scientific notation format.
6. 12.00 x 106 = 1.20 x 107.
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# Sequencing Questions and Answers – Set 3
This set of Verbal Reasoning Questions and Answers (MCQs) focuses on “Sequencing – Set 3”.
1. Five girls are standing in a row to be photographed. Komal is to the left of Aakriti. Disha is to the right of Aakriti. Ankita is between Aakriti and Soumya. Who would be second from the right in the photograph?
a) Soumya
b) Aakriti
c) Ankita
d) Disha
Explanation: Here, the sitting order is Komal→Aakriti→Ankita→Soumya→Disha.
2. Identify the wrong number in the series. 0, 1, 1, 2, 3, 5, 8, 13, 20, 34?
a) 8
b) 13
c) 20
d) 34
Explanation: Here, the series is a Fibonacci series.
i.e. a number in the series is the sum of previous two numbers.
So, in this 8+13 = 21. So the wrong number is 20.
3. Find the next number in the series: 80, 99, 120, 143, 168, …. ?
a) 182
b) 123
c) 195
d) 249
Explanation: Here, each number is one less than square of a number.
i.e. 92=81-1=80, 102=100-1=99, 112=121-1=120, 122=144-1=143, 132=169-1=168.
So, the next number will be 142=196-1=195.
4. A person is fourth in the queue from starting. If he shifts 2 position forward, now he is seventh from the end. How many people are there in the queue?
a) 9
b) 11
c) 7
d) 8
Explanation: Here, that person is in fourth position from starting of the queue i.e. 3 people stands in front of him and when he shifts 2 positions forward he is in seventh position from the end i.e. 6 people stand behind him and 1 person stands in front of him.
Therefore, total number of people will be 8.
5. A cage had 13 birds. All but three manages to escape. How many birds actually escape ?
a) 7
b) 3
c) 10
d) 9
Explanation: Here, it is given that all birds escapes except 3 birds i.e. out of 13, 10 birds escape.
6. Complete the series 23, 45, 67, ?, 54, 32.
a) 28
b) 62
c) 76
d) 63
Explanation: Here, in the series after first three numbers, remaining three numbers are reversed number of previous numbers.
7. Complete the following series 4, 12, 30, 56, …?
a) 68
b) 80
c) 110
d) 40
Explanation: Here, the pattern followed is:-
2×2=4, 3×4=12, 5×6=30, 7×8=56. Similarly, 11×10=110.
8. Complete the following series 4, 20, 29, 45, 70…?
a) 86
b) 94
c) 106
d) 89
Explanation: Here, the difference between the digits is the square of digits i.e.
20-4 = 16 = 42, 29-20 = 9 = 32, 45-29 = 16 =42, 70-45 = 25 = 52.
Similarly, 106-70 = 36 = 62
9. Complete the following series 6, 12, 36, 144, 720…?
a) 8249
b) 6825
c) 4320
d) 2843
Explanation: Here, the series follows the following pattern.
6×2=12, 12×3=36, 36×4=144, 144×5=720.
Similarly, 720×6=4320.
10. In a row of 26 boys, when Sourav shifted five places towards the left, he become 10th from the left end. What was his earlier position form the right end of the row?
a) 10th
b) 11th
c) 12th
d) 13th
Explanation: Here, there are 26 boys and Sourav is in 10th position from left end after shifting. i.e. Sourav is in 15th position from left before shifting. So 26 – 15 = 11 i.e. there are 11 boys before Sourav, so his position was 12th from the right end.
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# Solving ##y'' - 5 y' - 6y = e^{3x}## using Laplace Transform
• Hall
In summary: Sorry, I made a mistake in my previous post. Here's the corrected version:In summary, we have to solve the differential equation with initial conditions of $y'' - 5y' - 6y = e^{3x}$, $y(0) = 2$, and $y'(0) = 1$. By applying the Laplace transform, we get $s^2 Y(s) - (sy(0) + y'(0)) - 5s Y(s) + y(0) - 6Y(s) = 1/(s-3)$. This simplifies to \$Y(s) = (2s^2 - 15s + 27)/(s-3)(
#### Hall
Homework Statement
Nil
Relevant Equations
Nil
We have to solve
\begin{align*} y'' - 5y' - 6y = e^{3x} \\ y(0) = 2,~~ y'(0) = 1 \\ \end{align*}
Applying Laplace Transform the equation
\begin{align*} L [ y''] - 5 L[y'] - 6 L[y] = L [ e^{3x} ] \\ s^2 Y(s) - \left( s y(0) + y'(0) \right) - 5s Y(s) + y(0) - 6 Y(s) = \frac{1}{s-3} \\ Y(s) \{ s^2 - 5s - 6\} = \frac{1}{s-3} + 2s - 9 \\ Y(s) = \frac{2s^2 - 15s +27}{(s-3) (s+1) (s-6)} \\ \textrm{On Partial Fraction Decomposition}\\ \frac{A}{s-3} + \frac{B}{s+1} + \frac{C}{s-6} = \frac{2s^2 -15s +27}{(s-3) (s+1) (s-6)} \\ A (s+1)(s-6) + B(s-3)(s-6) + C (s-3)(s+1) = 2s^2 -15s +27 \\ \textrm{putting s =6} \\ 21C = 9 \implies C = \frac{3}{7} \\ \textrm{similarly,} \\ s = -1 ~\text{gives}~ B = 11/7 \\ s = 3 ~ \text{gives} ~ A = 0\\ \textrm{Thus,}~~ y_p = 11/7 L^{-1} [1/(s+1)] + 3/7 L^{-1} [1/(s-6)] \\ y_p = 11/7 e^{-x} + 3/7 e^{6x} \end{align*}
But this doesn't match with the answer given in the book. Where is my mistake?
For starters, $$\frac{1}{s-3} + 2s - 9 = \frac{1 + (2s-9)(s-3)}{s-3} = \frac{2s^2 - 15s + 28}{s-3}.$$
Hall and FactChecker
At the end of your third line, I think I get a -1 instead of a -9.
In general, it would be good for you to double-check your arithmetic.
May I suggest using Wolfram alpha or similar (Mathematica is great if you have access to it) to double check arithmetic?
FactChecker
Hall said:
Homework Statement:: Nil
Relevant Equations:: Nil
We have to solve
\begin{align*} y'' - 5y' - 6y = e^{3x} \\ y(0) = 2,~~ y'(0) = 1 \\ \end{align*}
Applying Laplace Transform the equation
\begin{align*} L [ y''] - 5 L[y'] - 6 L[y] = L [ e^{3x} ] \\ s^2 Y(s) - \left( s y(0) + y'(0) \right) - 5s Y(s) + y(0) - 6 Y(s) = \frac{1}{s-3} \\ Y(s) \{ s^2 - 5s - 6\} = \frac{1}{s-3} + 2s - 9 \\ Y(s) = \frac{2s^2 - 15s +27}{(s-3) (s+1) (s-6)} \\ \textrm{On Partial Fraction Decomposition}\\ \frac{A}{s-3} + \frac{B}{s+1} + \frac{C}{s-6} = \frac{2s^2 -15s +27}{(s-3) (s+1) (s-6)} \\ A (s+1)(s-6) + B(s-3)(s-6) + C (s-3)(s+1) = 2s^2 -15s +27 \\ \textrm{putting s =6} \\ 21C = 9 \implies C = \frac{3}{7} \\ \textrm{similarly,} \\ s = -1 ~\text{gives}~ B = 11/7 \\ s = 3 ~ \text{gives} ~ A = 0\\ \textrm{Thus,}~~ y_p = 11/7 L^{-1} [1/(s+1)] + 3/7 L^{-1} [1/(s-6)] \\ y_p = 11/7 e^{-x} + 3/7 e^{6x} \end{align*}
But this doesn't match with the answer given in the book. Where is my mistake?
I find the method of characteristic to be straightforward ...i.e
Let
##m^2-5m-6=0##
##m_1=-1## ##m_2=6## therefore,
##y = Ae^{-x} + Be^{6x}## on applying boundary condition; ##y(0)=2## we shall have
##2=A+B##
and on applying the second boundary condition; ##y^{'}(0)=1## we shall have
##1=-A+6B##
solving the simultaneous equation;
##2=A+B##
##1=-A+6B##
yields;
##A=\dfrac{11}{7}## and ##B=\dfrac{3}{7}## therefore our complementary solution is;
##y_c(x) =\dfrac{11}{7} e^{-x} + \dfrac{3}{7} e^{6x}## now on the inhomogenous part;
Let ##y_p(x)= Ce^{3x}##
##y^{'}_p(x)= 3Ce^{3x}##
##y^{''}_p(x)= 9Ce^{3x}## therefore;
##9Ce^{3x}-15Ce^{3x}-6Ce^{3x}= e^{3x}##
##-12Ce^{3x}=e^{3x}##
##C=-\dfrac{1}{12}##
Therefore ##y(x)= y_c(x) + y_p(x)= \dfrac{11}{7} e^{-x} + \dfrac{3}{7} e^{6x}-\dfrac{1}{12}e^{3x}##
Hall
Hall said:
Homework Statement:: Nil
Relevant Equations:: Nil
We have to solve
\begin{align*} y'' - 5y' - 6y = e^{3x} \\ y(0) = 2,~~ y'(0) = 1 \\ \end{align*}
Applying Laplace Transform the equation
\begin{align*} L [ y''] - 5 L[y'] - 6 L[y] = L [ e^{3x} ] \\ s^2 Y(s) - \left( s y(0) + y'(0) \right) - 5s Y(s) + y(0) - 6 Y(s) = \frac{1}{s-3} \\ Y(s) \{ s^2 - 5s - 6\} = \frac{1}{s-3} + 2s - 9 \\ Y(s) = \frac{2s^2 - 15s +27}{(s-3) (s+1) (s-6)} \\ \textrm{On Partial Fraction Decomposition}\\ \frac{A}{s-3} + \frac{B}{s+1} + \frac{C}{s-6} = \frac{2s^2 -15s +27}{(s-3) (s+1) (s-6)} \\ A (s+1)(s-6) + B(s-3)(s-6) + C (s-3)(s+1) = 2s^2 -15s +27 \\ \textrm{putting s =6} \\ 21C = 9 \implies C = \frac{3}{7} \\ \textrm{similarly,} \\ s = -1 ~\text{gives}~ B = 11/7 \\ s = 3 ~ \text{gives} ~ A = 0\\ \textrm{Thus,}~~ y_p = 11/7 L^{-1} [1/(s+1)] + 3/7 L^{-1} [1/(s-6)] \\ y_p = 11/7 e^{-x} + 3/7 e^{6x} \end{align*}
But this doesn't match with the answer given in the book. Where is my mistake?
Where is your solution for the inhomogenous part?...your answer does not look complete to me! just check if my solution is correct. Cheers.
Hall said:
Homework Statement:: Nil
Relevant Equations:: Nil
We have to solve
\begin{align*} y'' - 5y' - 6y = e^{3x} \\ y(0) = 2,~~ y'(0) = 1 \\ \end{align*}
Applying Laplace Transform the equation
\begin{align*} L [ y''] - 5 L[y'] - 6 L[y] = L [ e^{3x} ] \\ s^2 Y(s) - \left( s y(0) + y'(0) \right) - 5s Y(s) + y(0) - 6 Y(s) = \frac{1}{s-3} \\ Y(s) \{ s^2 - 5s - 6\} = \frac{1}{s-3} + 2s - 9 \\ Y(s) = \frac{2s^2 - 15s +27}{(s-3) (s+1) (s-6)} \\ \textrm{On Partial Fraction Decomposition}\\ \frac{A}{s-3} + \frac{B}{s+1} + \frac{C}{s-6} = \frac{2s^2 -15s +27}{(s-3) (s+1) (s-6)} \\ A (s+1)(s-6) + B(s-3)(s-6) + C (s-3)(s+1) = 2s^2 -15s +27 \\ \textrm{putting s =6} \\ 21C = 9 \implies C = \frac{3}{7} \\ \textrm{similarly,} \\ s = -1 ~\text{gives}~ B = 11/7 \\ s = 3 ~ \text{gives} ~ A = 0\\ \textrm{Thus,}~~ y_p = 11/7 L^{-1} [1/(s+1)] + 3/7 L^{-1} [1/(s-6)] \\ y_p = 11/7 e^{-x} + 3/7 e^{6x} \end{align*}
But this doesn't match with the answer given in the book. Where is my mistake?
Taking the Laplace transform of ##5y’## you have transformed this to ##5sY(s) - y(0)##…
chwala said:
Where is your solution for the inhomogenous part?...your answer does not look complete to me! just check if my solution is correct. Cheers.
The Laplace transform gets both the homogeneous and inhomogeneous parts since it solves the differential equation including the initial conditions.
Yes, you can use the characteristic equation, but the OP specifically asked about the Laplace transform method.
Mark44 and chwala
## 1. How do I use Laplace Transform to solve a differential equation?
To use Laplace Transform, you first need to take the Laplace Transform of both sides of the differential equation. This will give you an algebraic equation in terms of the Laplace Transform of the function. Then, you can use algebraic methods to solve for the Laplace Transform of the function. Finally, you can take the inverse Laplace Transform to get the solution to the original differential equation.
## 2. What is the Laplace Transform of a derivative?
The Laplace Transform of a derivative is equal to the product of the Laplace Transform of the function and the initial value of the function. In other words, if the Laplace Transform of a function is F(s), then the Laplace Transform of its derivative is sF(s) - f(0).
## 3. How do I handle the e^{3x} term in the given differential equation?
To handle the e^{3x} term, you can use the property of Laplace Transform that states e^{at} transforms to 1/(s-a). In this case, the e^{3x} term will transform to 1/(s-3).
## 4. What do I do with the constants in the differential equation?
The constants in the differential equation, such as the -5 and -6 in the given equation, will also transform to constants in the algebraic equation. You can then solve for these constants using algebraic methods.
## 5. How do I know if I have the correct solution using Laplace Transform?
To check if you have the correct solution, you can take the Laplace Transform of your solution and see if it matches the algebraic equation you obtained by taking the Laplace Transform of the given differential equation. You can also plug in your solution to the original differential equation and see if it satisfies the equation. |
# Sets – Class XI – Exercise 1.6
1. If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩Y).
Solution:
It is given that: n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38
n(X ∩ Y) = ?
We know that, n(X ∩Y) = n(X) + n(Y) – n(X∩Y)
38 = 17 + 23 – n(X ∩Y)
n(X ∩Y) = 40 – 38 = 2
Therefore n(X ∩Y) = 2
1. If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩Y have?
Solution:
It is given that, n(X U Y) = 18, n(X) = 8, n(Y) = 15
We know that,
n(X ∩Y) = n(X) + n(Y) – n(X∩Y)
18 = 8 + 15 – n(X∩Y)
n(X∩Y) = 23 – 18 = 5
n(X∩Y) = 5
1. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Solution:
Let H be the set of people who speak Hindi, and E be the set of people who speak English
∴ n(H ∪ E) = 400, n(H) = 250, n(E) = 200
n(H ∩ E) = ?
We know that, n(Hz n(H ∩ E)
⇒ 400 = 450 – n(H ∩ E)
⇒ n(H ∩ E) = 450 – 400
∴ n(H ∩ E) = 50
Thus, 50 people can speak both Hindi and English.
1. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?
Solution:
It is given that, n(S) = 21, n(T) = 32, n(S ∩ T) = 11
We know that, n (S ∪ T) = n (S) + n (T) – n (S ∩ T)
∴ n (S ∪ T) = 21 + 32 – 11 = 42
Thus, the set (S ∪ T) has 42 elements.
1. If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?
Solution:
It is given that, n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10
We know that, n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
∴ 60 = 40 + n(Y) – 10
∴ n(Y) = 60 – (40 – 10) = 30
Thus, the set Y has 30 elements.
6: In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?
Solution:
Let C denote the set of people who like coffee, and T denote the set of people who like tea
n(C ∪ T) = 70, n(C) = 37, n(T) = 52
We know that, n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
∴ 70 = 37 + 52 – n(C ∩ T)
⇒ 70 = 89 – n(C ∩ T)
⇒ n(C ∩ T) = 89 – 70 = 19
Thus, 19 people like both coffee and tea.
1. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solution:
Let C denote the set of people who like cricket, and T denote the set of people who like tennis
∴ n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10
We know that, n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
∴ 65 = 40 + n(T) – 10
⇒ 65 = 30 + n(T)
⇒ n(T) = 65 – 30 = 35
Therefore, 35 people like tennis.
Now, (T – C) ∪ (T ∩ C) = T and (T – C) ∩ (T ∩ C) = Φ
∴ n (T) = n (T – C) + n (T ∩ C)
⇒ 35 = n (T – C) + 10
⇒ n (T – C) = 35 – 10 = 25
Thus, 25 people like only tennis.
## 1 thought on “Sets – Class XI – Exercise 1.6”
1. Reblogged this on HarsH ReaLiTy and commented:
What if my answer is a letter. That’s bad huh… 😉 I have huge respect for people that hack at numbers all day! I look at computer code instead. 😄
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Liked by 1 person |
# Subtracting 5 Worksheets || Grade 1 Math
In this post we will try to understand subtraction of number with 5. First we will try to understand the subtraction concept with the help of number line then we will move on to solve some questions related to it.
The worksheet provided below are in the form of plain text and images. We have tried to post questions which are entertaining and engaging to solve while keeping the requirements of USA Math curriculum.
## Subtracting with 5 || Concept Explanation
When we subtract any number with five, we have to take 5 jumps towards left on number line to get the required answer.
Let us understand the concept with the help of examples:
Example 01
7 – 5 = ?
Solution
Here we are asked to subtract 7 with 5.
Step 01: Start at number 07 on the number line
Step 02: Take 5 jumps towards the left
Step 03: Stop at number 02
Please look at the following number line illustration to understand the above steps
Hence, 7 – 5 = 2 is the required answer
Example 02
12 – 5 = ?
Solution
Here we are asked to subtract 12 with 5.
Step 01: Start at number 12 on the number line
Step 02: Take 5 jumps towards the left
Step 03: Stop at number 07
Hence, 12 – 5 = 7 is the required answer
Example 03
10 – 5 = ?
Solution
Here we are asked to subtract 10 with 5.
Step 01: Start at number 10 on the number line
Step 02: Take 5 jumps towards the left
Step 03: Stop at number 05
Please look at the following number line illustration to understand the above steps
Hence, 10 – 5 = 5 is the required answer
Now we have seen some examples of subtraction of number with 5, let us look at the below table which shows subtraction of numbers from 5 to 20.
## Subtracting 5 Worksheet
All the numbers below are subtracted with 5. Do the calculation and find the right answer
(a) 16 – 5 = ?
Solution
11
(b) 10 – 5 = ?
Solution
05
(c) 5 – 5 = ?
Solution
0
(d) 19 – 5 = ?
Solution
14
(e) 23 – 5 = ?
Solution
18
(f) 47 – 5 = ?
Solution
42
(g) 52 – 5 = ?
Solution
47
(h) 11 – 5 = ?
Solution
6
(i) 29 – 5 = ?
Solution
24
(j) 77 – 5 = ?
Solution
72
(k) 6 – 5 = ?
Solution
1
(l) 31 – 5 = ?
Solution
26
(m) 7 – 5 = ?
Solution
2
(n) 9 – 5 = ?
Solution
4
(o) 66 – 5 = ?
Solution
61
(p) 22 – 5 = ?
Solution
17
(q) 98 – 5 = ?
Solution
93
(r) 13 – 5 = ?
Solution
8
(s) 106 – 5 = ?
Solution
101
(t) 78 – 5 = ?
Solution
73
## Subtracting Five – Image worksheet
Given below are collection of subtraction questions where one image is shown, your job is to do subtraction as shown in the image and select the right answer.
(01) How many stars are there after subtraction?
(a) 1
(b) 2
(c) 3
(d) 4
Solution
The above image can be expressed as:
7 – 5 = 2
Option (b) is the right answer
(02) Count the number of triangles after subtraction and select the right answer
(a) 3
(b) 2
(c) 1
(d) 0
Solution
The above image can be expressed as:
5 – 5 = 0
Option (d) is the right answer
(03) Count the number of boxes after subtraction and select the right answer
(a) 1
(b) 2
(c) 3
(d) 5
Solution
The above image can be expressed as:
6 – 5 = 1
Option (a) is the right answer
(04) Do the subtraction as shown in the below image and select the right answer
(a) 3
(b) 7
(c) 5
(d) 6
Solution
The above image can be expressed as:
10 – 5 = 5
Option (c) is the right answer
(05) Select the correct answer for the below subtraction
(a) 2
(b) 4
(c) 6
(d) 8
Solution
The above image can be expressed as:
7 – 5 = 2
Option (a) is the right answer
(06) Subtract the number of stars as shown in below image and select the right answer
(a) 11
(b) 8
(c) 10
(d) 6
Solution
The above image can be expressed as:
11 – 5 = 6
Option (d) is the right answer
(07) Find the number of hearts after subtraction
(a) 14
(b) 10
(c) 13
(d) 12
Solution
The above image can be expressed as:
15 – 5 = 10
Option (b) is the right answer
(08) Find the number of objects after subtraction
(a) 4
(b) 1
(c) 3
(d) 2
Solution
The above image can be expressed as:
8 – 5 = 3
Option (c) is the right answer
(09) Count the number of triangles ater subtraction
(a) 3
(b) 2
(c) 5
(d) 1
Solution
The above image can be expressed as:
7 – 5 = 2
Option (b) is the right answer
(10) Count the number of circles after subtraction
(a) 8
(b) 9
(c) 10
(d) 11 |
# Multiplying fractions and whole numbers - Fractions
## What is a whole number
You’ve encountered lots of whole numbers before now. Whole numbers are numbers that aren’t fractions—they are integers. For example, $2, 12$, and $50$ would all be whole numbers.
On the other hand, numbers that aren’t whole numbers would look something like $1.25$ or $\frac{4}{5}$. Although a fraction is a rational number, it is not a whole number. Knowing the difference will be important in this lesson
## How to multiply fractions with whole numbers
When you’re given a question that requires you to deal with multiplying fractions with whole numbers, there’s $4$ main steps you’ll have to carry out.
Firstly, rewrite the question so that the whole number is turned into a fraction. As you probably already know, when you have a whole number, turning it into a fraction just requires you to put it over $1$. So for example, if you wanted to convert $8$ into a fraction, it’ll be rewritten as $\frac{8}{1}$.
Secondly, multiply the two numerators in the two respective fractions. This just means taking the two numbers on top of each of the fractions and then multiplying them with one another.
For the third step, do the same as step two but now you’re using the two numbers in the denominators in the fractions. You’ll end up with a new fraction after doing steps two and three!
Lastly, you’ll just have to simplify the fraction you’ve gotten after solving the problem. You have to show your answer in the lowest terms possible, or you may get marks deducted for not having completely finished the question. Let’s take a look at some examples and put the four steps into use to help you with multiplying fractions and whole numbers.
## Practice problems
Question 1:
Calculate
$2 \times \frac{1}{5}$
Solution:
First, we can express $2$ as a fraction:
$\frac{2}{1}$
Our question will then be converted to something that looks like this:
$\frac{2}{1} \times \frac{1}{5}$
We multiply these fractions, first tackling the top numbers ($2 \times 1$) and then doing the bottom ones ($1 \times 5$). Then we’ll get our final number, which is a new fraction.
$\frac{2}{1} \times \frac{1}{5} = \frac{2}{5}$
Since $\frac{2}{5}$ is already the most simplified form of the fraction, this will be your final answer.
Question 2:
A pizza had $12$ slices, and $\frac{3}{4}$ of it was eaten in a party. How many slices of pizza were eaten during the party?
Solution:
There were $12$ slices and $\frac{3}{4}$ were eaten. So we multiply $12$ and $\frac{3}{4}$ to get the answer. Let’s express $12$ as a fraction.
$\frac{12}{1} \times \frac{3}{4}$
Before doing multiplication, we can simplify the question first and get this:
$\frac{3}{1} \times \frac{3}{1}$
Now, do the multiplication. We got $9$ as the final answer
$\frac{3}{1} \times \frac{3}{1} = \frac{9}{1} = 9$
Good lessons to review and move forward with to help you understand this lesson better includes finding common factors, comparing rational numbers, and solving problems with rational numbers in fraction form. These will eventually lead you into the more advanced problems of solving two-step linear equations.
### Multiplying fractions and whole numbers
We learned previously that whole numbers can be written We learned previously that whole numbers can be written as fractions with 1 as the denominator and the whole number as the numerator. To make the calculation easier, we can first make the whole numbers into fraction when we multiply whole numbers with fractions. By doing so, we turn the questions into multiplying fractions only. |
## Puzzling Probability Part 3
This is part 3 of a series in which I am exploring some interesting probability puzzles.
As someone who loves to walk, probability puzzles that involve considering different routes to the same destination are interesting to me. However, this puzzle requires a warm up. Consider the following:
How many ways are there to walk from A to B (assuming no back tracking)?
At first, I tried to draw all the possible paths:
Then I realized that it was going to take far too long and I could never be sure that I didn’t miss a path. I needed another approach. Instead of trying to figure out how many ways there were to get from A to B, I asked a simpler question: How many ways to get from A to C?
Clearly, we can only get to C in only 1 way. We place a number 1 at the intersection. Similarly, we only have 1 way to get to each intersection point along the top and along the left side. Next, we need to determine how many ways to get to point D.
Since there is 1 way to get to the point above D and 1 way to get to the point left of D, we add these and obtain 2 ways to get to D. Continuing like this with each intersection point on the grid, we have:
Thus, there are 20 different ways to walk from A to B.
Now we can answer the actual question. What is the probability that a random walk from A to B (assuming no back tracking) passes through the middle of the grid below?
Using the above method, we find that there are 70 ways of walking from A to B. To count the number of ways through the middle, we simply redraw the gird to force us through the middle as follows:
Again, using the above method, we find that there are 36 ways of walking through middle.
Hence, the probability is 36/70. If you try this exercise for a 6 by 6 grid, you will find that the probability is 4900/12870.
This was counter intuitive to me. In a very large gird, if you were to randomly walk from one end to the other, it seemed to me very unlikely you would pass exactly through the center. The above solution shows that even for a 6 by 6 grid, the probability of walking through the exact center is 38%. As an extension, what do you think happens to this probably as the grid grows larger? Does it approach a specific value? Or does the probability eventually tend to zero?
Filed under Uncategorized
## Puzzling Probability Part 2
This is part 2 of a series, in which I am exploring some interesting probability puzzles.
Imagine a typical dice game. You roll 2 dice and compute their sum. For example, the first die is a 4 and the second die is a 3. The sum is 4 + 3 = 7. From our everyday experience with rolling dice, we know that 7 comes up most often, and numbers like 3 or 11 are rarer.
Now imagine a pair of strange dice, with the following numbers:
Die 1: 1, 3, 4, 5, 6, 8
Die 2: 1, 2, 2, 3, 3, 4
Is rolling these two dice is the same as rolling 2 standard dice? Or better yet, how could we determine if the strange dice are equivalent to standard dice?
The first idea I had was to add up all the numbers on both dice:
(1 + 3 + 4 + 5 + 6 + 8) + (1 + 2 + 2 + 3 + 3 + 4) = 42
If you add up all the numbers on two standard dice, you also get a sum of 42. At first, I thought this might be enough to confirm that the strange dice were equivalent. However, consider the following two dice:
Die 3: 6, 6, 6, 6, 6, 6
Die 4: 1, 1, 1, 1, 1, 1
The sum of the all the numbers on the above two dice is clearly 42. But the only number that comes up when rolling both is 7. For an alternate pair of dice to be equivalent to the standard dice, there should be a way of rolling a 2, or a 12, or the other numbers typically rolled from standard dice.
Creating a table is very helpful for eliciting the answer in this particular situation. Here are all the possible sums for two standard dice:
The numbers along the outside are the standard dice numbers and the numbers on the inside of the grid are the sums. If we count up the occurrences of each sum, we have the following:
Now, we can repeat the same process with the strange dice:
How interesting! The strange dice have the same sums with the same frequency as the standard dice. However, I doubt I could get away with using these dice at a casino, even if they are mathematically equivalent.
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## Puzzling Probability Part 1
In this series, I am going to explore some interesting probability puzzles. First, I want to consider a question about selecting balls from an urn.
“An urn contains four colored balls: two orange and two blue. Two balls are selected at random without replacement, and you are told that at least one of them is orange. What is the probability that the other ball is also orange?”
What an interesting problem. Take a guess at what the answer might be.
At first, I was partial to the idea that the probability would be 1/3. After you have been told the first ball is orange, there is a 1 orange ball left out of the total of 3 remaining balls.
Then, I thought the answer must be 1/6. The probability of drawing an orange on the first selection is 2/4 and the probability of drawing an orange on the second selection is 1/3. Using multiplication, we get 1/6.
Unfortunately, neither of these approaches is correct. In fact, the answer is 1/5. Why 1/5 you might ask? Let’s diagram out the situation.
I labelled the balls: O1, O2, B1, B2. We are told that of the 2 balls selected, 1 of the balls is orange. Thus, we must have one of the following situations:
O1, O2
O1, B1
O1, B2
O2, B1
O2, B2
As you can see, there are only 5 possible selections where at least 1 of the balls selected is orange. Further, only 1 of these possibilities has 2 orange balls. Thus, the probability that the other ball is also orange is 1/5.
This was very shocking to me. I did not expect such a strange answer from a seemingly simple problem.
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## Card Probability
Suppose that a deck of 52 cards has been shuffled and placed face down. You draw 1 card but do not look at it. You place it face down and off to the side. What is the probability that you drew the Ace of Hearts?
Since there is 1 Ace of Hearts in the deck and 52 cards total, the probability is 1/52.
Now you prepare to draw a second card. What is the probably that the new card is an Ace of Hearts? Is it more likely than before? Less likely? The same? Impossible to know?
To determine the answer, we need to use a probability tree. Here are all the different options that could occur:
If we first drew the Ace of Hearts and placed it face down, there is no chance of drawing the Ace of Hearts for our second card. However, if we didn’t draw the Ace of Hearts, then our second card could or could not be the Ace of hearts.
To compute probabilities, we multiply. We want to determine the probability of drawing an Ace of Hearts on the second card. Hence, we multiply the probability of “Not Ace” by the probability of “Ace.”
The above fraction reduces to 1/52. This is the same probability as before! We could apply this logic to drawing a third card (assuming we didn’t peak at the second card). We would find that the probability of drawing the Ace of Hearts is still only 1 in 52. No matter how many cards you remove, the probability doesn’t change. Why is this?
One way of understanding this enigma is to consider the information you have. At the beginning, you know that there are 52 cards and 1 specific card you are looking for. Once you start drawing cards, although you keep removing the cards from the deck, you don’t know which cards you are removing. You could be removing the Ace of Hearts, or you could be removing random cards. You don’t gain any new information with each card you remove. So each time you have the same probability as in the beginning.
Or consider another scenario. Suppose I draw a card from the bottom of a full deck. What is the probability that it is an Ace of Hearts? Since the deck is shuffled, the probability should be 1/52. But to get to this card, I could have physically removed all 51 cards above it (without looking at them), placed them in a pile, and then selected the bottom card. Having a pile of unknown cards does not change the probability of the remaining card being an Ace of Hearts.
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## Noodles Part 1
A dish contains 100 strands of cooked spaghetti. See the picture below:
Two ends of spaghetti are chosen at random and tied together. For example:
If this process continues until there are no more loose ends, what is the probability that all strands will form one big loop? In other words, what are the chances that this big mess is actually just a circle?
Seriously, take a guess. When I first read a version of this problem,1 I thought the answer must surely be less than 1%. Watch this:
Now you may think that I took a particularly well behaved group of noodles for the above animation. However, mathematically speaking, the probability of getting a single big loop is 9%, much higher that you probably guessed.
I want to walk you through some of the interesting twists and turns that it took for me to get the answer. First, I had to find a formula for calculating the above probability. My first method was to try less noodles. With 2 noodles, it seemed like I could figure out all the possibilities.
For the first noodle, I started with the left end. I could tie it to itself:
Then I would have to tie the other noodle to itself and I would not have a big loop:
Another option would be to tie the left end to the end below it:
Then I would tie the remaining 2 ends together to make a loop!
My last option would be to tie the top left end to the right bottom end:
Then I would have to tie the remaining 2 ends together:
Do some untwisting, and make a loop!
To recap, with 2 noodles, we have a 2/3 probability of getting a one big loop. How about 3 noodles?
Instead of going through all the probabilities, we can use a trick. If I tie the top left end of the noodle to the right end of the same noodle, I create an isolated loop like before. That is not good. So instead, consider tying the top left end to any other noodle end:
By doing this, we create one long noodle, which we can move around to form:
Voila! We are back to the 2-noodle case. Since there are 4 places I could tie the top left end (green), and 1 place I cannot (red, since it would make an isolated loop) I have a 4/5 chance.
We multiply this by the 2-noodle case to get
$\frac{2}{3} \times \frac{4}{5}$
You can continue this logic and find the probability for 5 noodles is:
$\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times \frac{8}{9}$
Now I noticed a pattern. Combine the top and bottom numbers:
$\frac{2 \times 4 \times 6 \times 8}{3 \times 5 \times 7 \times 9}$
We are actually just multiplying even numbers and dividing them by odd numbers! I was so excited! I went into excel, punched in the formula, and was thoroughly disappointed.
Apparently, multiplying 100 numbers got excessively big for my computer to handle. I needed a new approach. The noodle probability formula can be written as follows:
$f(n) = \cfrac{2 \cdot 4 \cdot 6 \cdot 8 \dots (2n-2)}{ 3 \cdot 5 \cdot 7 \cdot 9 \dots (2n-1)}$
Using double factorial notation, we can rewrite it as:
$f(n) = \cfrac{(2n-2)!!}{(2n-1)!!}$
Using basic double factorial identities2, we can convert the double factorials into regular factorials:
$f(n) = \cfrac{2^n n!}{2n} \times \cfrac{2^n n!}{(2n)!}$
And use some algebra to simplify:
$f(n) = \cfrac{2^{2n} n! n!}{2n (2n)!}$
Enter, the Stirling Approximation!3
$n! \approx \sqrt{2n\pi} \left( \cfrac{n^n}{e^n} \right)$
$f(n) \approx \cfrac{2^{2n}}{2n} \times \cfrac{\sqrt{2n\pi} \left( \frac{n^n}{e^n} \right) \times \sqrt{2n\pi} \left( \frac{n^n}{e^n} \right)}{\sqrt{4n\pi} \left( \cfrac{(2n)^{2n}}{e^{2n}} \right)}$
We can cancel a bunch of terms to get:
$f(n) \approx \cfrac{\sqrt{\pi n}}{2n} = \sqrt{\cfrac{\pi }{4n}}$
Finally, a formula my computer can handle! This formula will work for any number of noodles. Maybe you were wondering what number of noodles you would need for the probability to be below 1%. The formula states that we would need 7854 noodles! Personally, if you would have told me that before analyzing the problem, I would have thought you were out to lunch 😛
1The Mathematics Teacher, Volume 109, Number 3, October 2015, page 201.
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## Matching Tests Proof
Warning, this post uses some fierce math to prove why the pattern continues in the last post. If you want to know why the pattern continues, then welcome to the explanation.
In the last post we counted how many ways to hand student back their tests so that no student had his or her own test. This type of matching has a special name, a derangement. For a group of n students we denote the derangement as
!n
To determine this quantity we need a formula. From before, we know that !3 = 2 and !4 = 9. Now consider the general case. How many ways are there of handing back a test to the first student? We certainly can’t give him his own test. However, we could give him any of the other n – 1 tests. Let’s say we give him test i.
We could hand out the next test to anyone. Let’s consider handing it out to student i (the student with test i). We have two options. First, student i gets student 1’s test. Here is a picture:
Once these 2 students have been assigned their tests, we have n – 2 students who still need tests. However, assigned them tests is analogous to our original problem, and we can denote this as !(n – 2).
Second, student i does not get student 1’s test. Thus, student i has 1 forbidden test. All the other students also have 1 forbidden test (namely their own test). Again, this is analogous to our original problem with n – 1 students (since we have not given student i a test yet). We can denote this as !(n – 1).
To summarize, to hand out n tests, we have the following recursive formula:
$!n = (n-1)(!(n-1) + !(n-2))$
We can use this formula to calculate derangements as high as we like, provided we proceed in order.
!5 = 4(!4 + !3) = 4(9 + 2) = 44
!6 = 5(!5 + !4) = 5(44 + 9) = 265
These are exactly the numerators we found when computing the fractions in the previous post!
To determine what would happen in a large class, we need a closed form for derangements. To simplify the notation for the next section, we will denote !n as D(n).
Observe the following pattern:
D(n) D(n)-nD(n-1) D(4) = 9 9 – 4*2 = 1 D(5) = 44 44 – 5*9 = -1 D(6) = 265 265 – 6*44 = 1 D(7) = 1854 1854 – 7*265 = -1
We can prove the above pattern continues using induction. The base case has been dealt with above. Our inductive hypothesis is:
$D(n) - nD(n-1) = (-1)^n$
Consider n+1:
$D(n+1) - (n+1)D(n) = n(D(n)+D(n-1)) - (n+1)D(n)$ using the recurrence relation
$= nD(n) + nD(n-1) - nD(n) - D(n)$
$= nD(n-1) - D(n)$
$= -(D(n) - nD(n-1))$
$= -(-1)^n$ by the inductive hypothesis
$= (-1)^{n+1}$
Using this information, we have a new recurrence relation for D(n):
$D(n) = nD(n-1) + (-1)^n$
Unfortunately, this is still not a closed form solution. However, we will use induction again to get us the rest of the way. The following is the closed form solution for D(n):
$D(n)=n!\sum_{k=0}^n\frac{(-1)^k}{k!} = n! \left( \frac{1}{0!} -\frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!} \right)$
Checking the OEIS, we find that D(0) = 1 and D(1) = 0. Using the formula for D(3), we can see that the base case will hold:
$D(3) = 3! \left( \frac{1}{0!} -\frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} \right)$
$= 6 \left( 1 -1 + \frac{1}{2} - \frac{1}{6} \right)$
$= 6 \left(\frac{3}{6} - \frac{1}{6} \right)$
$= 6 \left(\frac{2}{6} \right)$
$D(3) = 2$ as expected.
Suppose the formula is true for D(n). Consider D(n+1):
$D(n+1)= (n+1)D(n) + (-1)^{n+1}$ from the new recursive formula
$= (n+1)n!\sum_{k=0}^n\frac{(-1)^k}{k!} + (-1)^{n+1}$ using the inductive hypothesis
$= (n+1)!\sum_{k=0}^n\frac{(-1)^k}{k!} + (-1)^{n+1}$ definition of factorial
$= (n+1)!\sum_{k=0}^n\frac{(-1)^k}{k!} + (-1)^{n+1} \frac{(n+1)!}{(n+1)!}$
$= (n+1)! \left[ \sum_{k=0}^n\frac{(-1)^k}{k!} + \frac{(-1)^{n+1}}{(n+1)!} \right]$
$= (n+1)! \sum_{k=0}^{n+1}\frac{(-1)^k}{k!}$
Thus, the formula is true for any n. This formula is closed, but complicated. The last step in the process is to relate derangements to our original problem. Since we were looking for a probability, we were really interested in $\frac{D(n)}{n!}$. Using the Taylor series approximation for e-1 we get the following:
$\frac{D(n)}{n!} = n! \left( \frac{1}{0!} -\frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!} \right) \frac{1}{n!}$
$= \left( \frac{1}{0!} -\frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!} \right)$
$\approx e^{-1}$
Finally! We have our answer! Since the number of students determines how many terms of the sequence for e we use, the approximation is valid for even small classes of only 10 or 15 students. The probability of handing out tests properly is approximately $e^{-1} = \frac{1}{e} = \frac{1}{2.71828} \approx 37\%$
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## Matching Tests
An interesting coincidence happened today in class. My co-teacher was handing the students back their tests. She wanted each student to mark another’s test. However, she accidentally handed a student back her own test. She casually said to me “weird, what are the chances of that?” Her question implied that this was a rare event. The answer is quite the opposite.
My co-teacher had a 63%, or $\frac{e-1}{e}$ to be precise, chance of handing a student back his or her own test. If you are wondering, e stands for a special number:
e = 2.71828…
This percentage is not good. It means that, if you randomly distribute tests to students to mark, about two thirds of the time at least 1 student will end up with his or her own test!
How in the world did I come up with that answer? Let me back up a bit. Suppose we had 3 students and we wanted to give them back their tests to mark so that no student had his or her own test. What options do we have? One option is:
Student A: test B
Student B: test C
Student C: test A
Another option is:
Student A: test C
Student B: test A
Student C: test B
Hence, we have 2 options where we will hand back the tests with no matches. How many options do we have in total?
3! = 3*2*1 = 6
Thus, there is a 2/6 = 33.3% chance that everyone has a different test. In other words, there is a 66.6% chance that at least one student receives his or her own test to mark (which would not be good for academic standards).
What if we had 4 students? You can do the math, and you will find that there are 9 options where the students have different tests. For example:
Student A: test B
Student B: test C
Student C: test D
Student D: test A
Again, basic counting principles dictate that there will be 4! = 24 different possible combinations. Hence, there is a 9/24 = 38% chance each student will have a proper test to mark. We could continue on like this; painstakingly grinding out the probabilities for each situation until we reached a class size of 25. However, there is a pattern we can exploit. Consider the following table:
Number of Students Fraction of proper distributions 3 2/6 = 33% 4 9/24 = 38% 5 44/120 = 37% 6 265/720 = 37% 7 1854/5040 = 37% 8 14833/40320 = 37%
Do you see the pattern? Once the class size gets to 5, the percentage becomes stable. Based on the above data, we could predict that a class of 25 students will have roughly the same percentage, 37%. Indeed, one can prove mathematically that this percentage will be true for any size class, large or small.1 Thus, my advice to teachers is as follows. Be careful distributing tests to students to mark. You are playing a game with a 37% chance of winning. And any gambler knows, those are not good odds.
1See next blog post.
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# Solving linear equations LO To solve linear equations
• Slides: 18
Solving linear equations LO: To solve linear equations www. mathssupport. org
The Equation Snake Do the operation to both sides of the equation 3 x + 1 = 16 +5 3 x + 6 = 21 – 6 – 4 3 x = 15 3 x – 4 = 11 6 x - 8 = 22 +10 2 6 x + 2 = 32 +2 x 2 8 x + 2 = 2 x + 32 4 x + 1 = x + 16 – 6 x 7 x + 1 = 4 x + 16 ÷ 3 x + 1 = -2 x + 16 +2 x www. mathssupport. org +3 x 3 x = 15 3 x + 1 = 16 – 1 x=5
Solving linear equations An equation means that two things are equal. A linear equation is an equation which contains a variable which is not raised to any power other than 1. The solutions of an equation are the values of the variable which make the equation true The Left-hand side of an equation (LHS) is on the left of the = sign The Right-hand side of an equation (RHS) is on the right of the = sign www. mathssupport. org
Equations – Keep it balanced An equation means that two things are equal. We can compare equations to a set of scales. If the scales balance, then the value of each side must be equal. www. mathssupport. org
Equations – Keep it balanced Two bags of x candies plus five extra candies equals 15 candies. If we remove a candy from one side, we must remove a candy from the other side so that the scales still balance. 2 x + 5 = 15 Take off 1 2 x +4 = 14 x x www. mathssupport. org
Equations – Keep it balanced The scales now show that two bags of x candies plus four extra candies equals 14 candies. We can continue to remove candies, as long as we do the same thing to each side. 2 x + 5 = 15 11 12 10 14 2 x + 14230 = 13 x x www. mathssupport. org
Equations – Keep it balanced The scales now show that two bags of x candies equals 10 candies. We can’t remove any more candies, but we can halve the values on each side. 2 x + 5 = 15 ÷ 2 x x www. mathssupport. org ÷ 2 2 x = 10 x=5
Equations – Keep it balanced Let’s try one more Two bags of x candies plus four extra candies equals one bag of x candies plus ten extra candies. 2 x + 4 = x + 10 Remove 4 candies. 2 x = x + 6 x x www. mathssupport. org x Remove a bag of candies. x=6
When solving equations, try to imagine the scales. To make sure the equation balances … Whatever we do to one side of the equation we must do to the other, in order to isolate the unknown. We consider how the expression has been built up and then isolate the unknown by using inverse operations in reverse order, For example For the equation 2 xx + 6 = 10 The LHS is built up by starting with x 2 2 x 2 x + 6 So, to isolate x, we first subtract 6 from both sides Then divide both sides by 2 2 x + 6 = 10 +6 2 – 6 2 x + 6 – 6 = 10 – 6 2 x = 4 2 www. mathssupport. org 2 x =2
Solving linear equations Whatever we do to one side of the equation we must do to the other. Example 1: 3 x + 7 = 22 – 7 3 x = 15 3 Subtracting 7 from both sides 3 x=5 www. mathssupport. org Dividing both sides by 3
Solving linear equations Whatever we do to one side of the equation we must do to the other. Example 2: Subtracting 11 from both sides – 11 – 5 www. mathssupport. org Dividing both sides by -5
Solving linear equations Whatever we do to one side of the equation we must do to the other. Example 3: Subtracting 2 from both sides – 2 3 www. mathssupport. org 3 Multiplying both sides by 3
Solving linear equations Whatever we do to one side of the equation we must do to the other. Example 4: 5 5 4 x + 3 = – 10 – 3 4 x = – 13 4 www. mathssupport. org 4 Multiplying both sides by 5 Subtracting 3 from both sides Dividing both sides by 4
Equations with unknown on both sides Equations where the unknown appears more than once need to be solved systematically. If necessary, expand any brackets and collect like terms. If necessary, remove the unknown from one side of the equation. Aim to do this so the unknown is left with a positive coefficient. Use inverse operations to isolate the unknown and maintain balance. Check that your solution satisfies the equation. www. mathssupport. org
Equations with unknown on both sides The equation can then be solved by performing the same operations on both sides until the solution is found. 5 x – 4 = 2 x + 14 +4 add 4 to both sides: +4 5 x = 2 x + 18 subtract 2 x from both sides: – 2 x 3 x = 18 divide both sides by 3: ÷ 3 x=6 Check by substituting x = 6 into the expressions in the original equation. Both sides are equal to 2, so the solution is correct. www. mathssupport. org
Equations with brackets Solve the equation: 5 – 3(– 1 + x) = x Expand the brackets: simplifying 5 + 3 – 3 x = x 8 – 3 x = x Subtracting 8 from both sides: – 3 x = x – 8 subtracting x from both sides: – 4 x = – 8 dividing both sides by -4: www. mathssupport. org x=2
Equations with brackets Solve the equation: ( x – 3)2 = (4 + x) (2 + x) Expand each side: Subtracting x 2 from both sides x 2 – 6 x +9 = 8 + 4 x + 2 x + x 2 – 6 x +9 = 8 + 4 x + 2 x – 6 x +9 = 8 + 6 x adding 6 x to both sides: 9 = 8 + 12 x subtracting 8 from both sides: 1 = 12 x dividing both sides by 12: 1= x 12 www. mathssupport. org
Thank you for using resources from For more resources visit our website https: //www. mathssupport. org If you have a special request, drop us an email [email protected] org Get 20% off in your next purchase from our website, just use this code when checkout: MSUPPORT_20 www. mathssupport. org |
# SKETCH THE GRAPH AND VERIFY THE CONTINUITY OF THE FUNCTION
## About "Sketch the Graph and Verify the Continuity of the Function"
Sketch the Graph and Verify the Continuity of the Function :
Here we are going to see some example problems to understand the concept of sketching the graph and verifying the continuity of the function.
## Sketch the Graph and Verify the Continuity of the Function - Practice questions
Question 1 :
Let f(x)
Graph the function. Show that f(x) continuous on (- ∞, ∞).
Solution :
There are three partitions in the piecewise function.
To check the continuity at the point 0, we should prove the following.
lim x-> 0- f(x) = lim x->0+ f(x) = lim x->0 f(0)
lim x-> 0- f(x) = 0 ----(1)
lim x->0+ f(x) = 0 ----(2)
f(0) = 0 ----(3)
(1) = (2) = (3)
Hence the function is continuous at x = 0.
Now let us check the continuity at the point 2.
lim x-> 2- f(x) = lim x->2+ f(x) = lim x->2 f(0)
lim x-> 2- f(x) = 22 = 4 ----(1)
lim x->2+ f(x) = 4 ----(2)
f(2) = 4 ----(3)
(1) = (2) = (3)
Hence the function is continuous at x = 2.
So the given piecewise function is continuous on (- ∞, ∞).
Question 2 :
If f and g are continuous functions with f(3) = 5 and lim x->3 [2 f(x) - g(x)] = 4, find g(3).
Solution :
lim x->3 [2 f(x) - g(x)] = 4
Let us apply 3 in the function instead of x.
[2 f(3) - g(3)] = 4
2(5) - g(3) = 4
10 - g(3) = 4
g(3) = 10 - 4
g(3) = 6
Hence the value of g(3) is 6.
Question 3 :
Find the points at which f is discontinuous. At which of these points f is continuous from the right, from the left, or neither? Sketch the graph of f.
Solution :
First let us check the continuity at the point x = -1
lim x-> -1- f(x) = lim x-> -1- 2x + 1
By applying the limit, we get
= 2(-1) + 1
= -2 + 1
= -1 -----(1)
lim x-> -1+ f(x) = lim x-> -1+ 3x
By applying the limit, we get
= 3(-1)
= -3 -----(2)
lim x-> -1- f(x) lim x-> -1+
So, the function is not continuous at x = -1.
Now let us check the continuity at the point x = 1
lim x-> 1- f(x) = lim x-> 1- 3x
By applying the limit, we get
= 3(1)
= 3 -----(1)
lim x-> -1+ f(x) = lim x-> -1+ 2x - 1
By applying the limit, we get
= 2(1) - 1
= 1 -----(2)
lim x-> 1- f(x) lim x-> 1+
So, the function is not continuous at x = 1.
To find at which of these points f is continuous from the right, from the left, or neither, we have to draw the number line.
let x0 ∈ (-∞, -1]
lim x-> x0 f(x) = lim x-> x0 2x + 1
Applying the limit, we get
= 2x0 + 1 ------(1)
f(x0) = 2x0 + 1 ------(2)
(1) = (2)
It is continuous in (-∞, -1].
let x0 ∈ (-1, -1)
lim x-> x0 f(x) = lim x-> x0 3x
Applying the limit, we get
= 3x0 ------(1)
f(x0) = 3x0 ------(2)
(1) = (2)
It is continuous in (-1, 1).
let x0 ∈ [1)
lim x-> x0 f(x) = lim x-> x0 2x - 1
Applying the limit, we get
= 2x0 - 1 ------(1)
f(x0) = 2x- 1 ------(2)
(1) = (2)
It is continuous in [1).
Graph of f(x) = 2x + 1 :
x = -1f(-1) = -1 x = -2f(-2) = -3 x = -3f(-3) = -5
Graph of f(x) = 3x :
-1 < x < 1
x = -0.5f(-0.5) = -1.5 x = -0.7f(-0.7) = -2.1 x = 0.5f(0.5) = 1.5
Graph of f(x) = 2x - 1:
x > = 1
x = 1f(1) = 1 x = 2f(2) = 3 x = 3f(3) = 5
After having gone through the stuff given above, we hope that the students would have understood, "Sketch the Graph and Verify the Continuity of the Function"
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# Solving quadratics by completing the square
For example, solve x²+6x=-2 by manipulating it into (x+3)²=7 and then taking the square root.
## What you will learn in this lesson
So far, you've either solved quadratic equations by taking the square root or by factoring. These methods are relatively simple and efficient, when applicable. Unfortunately, they are not always applicable.
In this lesson, you will learn a method for solving any kind of quadratic equation.
## Solving quadratic equations by completing the square
Consider the equation ${x}^{2}+6x=-2$. The square root and factoring methods are not applicable here.
But hope is not lost! We can use a method called completing the square. Let's start with the solution and then review it more closely.
$\begin{array}{rlrl}\left(1\right)& & {x}^{2}+6x& =-2\\ \\ \left(2\right)& & {x}^{2}+6x+9& =7& & \text{Add 9, completing the square.}\\ \\ \left(3\right)& & \left(x+3{\right)}^{2}& =7& & \text{Factor the expression on the left.}\\ \\ \left(4\right)& & \sqrt{\left(x+3{\right)}^{2}}& =±\sqrt{7}& & \text{Take the square root.}\\ \\ \left(5\right)& & x+3& =±\sqrt{7}\\ \\ \left(6\right)& & x& =±\sqrt{7}-3& & \text{Subtract 3.}\end{array}$
In conclusion, the solutions are $x=\sqrt{7}-3$ and $x=-\sqrt{7}-3$.
### What happened here?
Adding $9$ to ${x}^{2}+6x$ in row $\left(2\right)$ had the fortunate result of making the expression a perfect square that can be factored as $\left(x+3{\right)}^{2}$. This allowed us to solve the equation by taking the square root.
This was no coincidence, of course. The number $9$ was carefully chosen so the resulting expression would be a perfect square.
### How to complete the square
To understand how $9$ was chosen, we should ask ourselves the following question: If ${x}^{2}+6x$ is the beginning of a perfect square expression, what should be the constant term?
Let's assume that the expression can be factored as the perfect square $\left(x+a{\right)}^{2}$ where the value of constant $a$ is still unknown. This expression is expanded as ${x}^{2}+2ax+{a}^{2}$, which tells us two things:
1. The coefficient of $x$, which we know to be $6$, should be equal to $2a$. This means that $a=3$.
2. The constant number we need to add is equal to ${a}^{2}$, which is ${3}^{2}=9$.
Try to complete a few squares on your own.
Problem 1
What is the missing constant term in the perfect square that starts with ${x}^{2}+10x$ ?
Problem 2
What is the missing constant term in the perfect square that starts with ${x}^{2}-2x$ ?
Problem 3
What is the missing constant term in the perfect square that starts with ${x}^{2}+\frac{1}{2}x$ ?
Challenge problem
What is the missing constant term in the perfect square that starts with ${x}^{2}+b\phantom{\rule{-0.167em}{0ex}}\cdot \phantom{\rule{-0.167em}{0ex}}\phantom{\rule{-0.167em}{0ex}}x$?
This challenge question gives us a shortcut to completing the square, for those that like shortcuts and don't mind memorizing things. It shows us that in order to complete ${x}^{2}+bx$ into a perfect square, where $b$ is any number, we need to add ${\left(\frac{b}{2}\right)}^{2}$ to it.
For example, in order to complete ${x}^{2}+6x$ into a perfect square, we added ${\left(\frac{6}{2}\right)}^{2}=9$ to it.
### Solving equations one more time
All right! Now that you're a certified square-completer, let's go back to the process of solving equations using our method.
Let's look at a new example, the equation ${x}^{2}-10x=-12$.
$\begin{array}{rlrl}\left(1\right)& & {x}^{2}-10x& =-12\\ \\ \left(2\right)& & {x}^{2}-10x+25& =13& & \text{Add 25, completing the square.}\\ \\ \left(3\right)& & \left(x-5{\right)}^{2}& =13& & \text{Factor the expression on the left.}\\ \\ \left(4\right)& & \sqrt{\left(x-5{\right)}^{2}}& =±\sqrt{13}& & \text{Take the square root.}\\ \\ \left(5\right)& & x-5& =±\sqrt{13}\\ \\ \left(6\right)& & x& =±\sqrt{13}+5& & \text{Add 5.}\end{array}$
In order to make the original left-hand expression ${x}^{2}-10x$ a perfect square, we added $25$ in row $\left(2\right)$. As always with equations, we did the same for the right-hand side, which made it increase from $-12$ to $13$.
In general, the choice of the number to add in order to complete the square doesn't depend on the right-hand side, but we should always add the number to both sides.
Now it's your turn to solve some equations.
Problem 4
Solve ${x}^{2}-8x=5$.
Problem 5
Solve ${x}^{2}+3x=-\frac{1}{4}$.
## Arranging the equation before completing the square
### Rule 1: Separate the variable terms from the constant term
This is how the solution of the equation ${x}^{2}+5x-6=x+1$ goes:
Completing the square on one of the equation's sides is not helpful if we have an $x$-term on the other side. This is why we subtracted $x$ in row $\left(2\right)$, placing all the variable terms on the left-hand side.
Furthermore, to complete ${x}^{2}+4x$ into a perfect square, we need to add $4$ to it. But before we do that, we need to make sure that all the constant terms are on the other side of the equation. This is why we added $6$ in row $\left(3\right)$, leaving ${x}^{2}+4x$ on its own.
### Rule 2: Make sure the coefficient of ${x}^{2}$ is equal to $1$ .
This is how the solution of the equation $3{x}^{2}-36x=-42$ goes:
$\begin{array}{rlrl}\left(1\right)& & 3{x}^{2}-36x& =-42\\ \\ \left(2\right)& & {x}^{2}-12x& =-14& & \text{Divide by 3.}\\ \\ \left(3\right)& & {x}^{2}-12x+36& =22& & \text{Add 36,completing the square.}\\ \\ \left(4\right)& & \left(x-6{\right)}^{2}& =22& & \text{Factor.}\\ \\ \left(5\right)& & \sqrt{\left(x-6{\right)}^{2}}& =±\sqrt{22}& & \text{Take the square root.}\\ \\ \left(6\right)& & x-6& =±\sqrt{22}\\ \\ \left(7\right)& & x& =±\sqrt{22}+6& & \text{Add 6.}\end{array}$
The completing the square method only works if the coefficient of ${x}^{2}$ is $1$.
This is why in row $\left(2\right)$ we divided by the coefficient of ${x}^{2}$, which is $3$.
Sometimes, dividing by the coefficient of ${x}^{2}$ will result in other coefficients becoming fractions. This doesn't mean you did something wrong, it just means you will have to work with fractions in order to solve.
Now it's your turn to solve an equation like this.
Problem 6
Solve $4{x}^{2}+20x-3=0$.
## Want to join the conversation?
• how did you get 25/4 and how did you equal it to 7
• The 25/4 and 7 is the result of completing the square method.
To factor the equation, you need to first follow this equation: x^2 + 2ax + a^2.
In x^2 +5x = 3/4, The a^2 is missing.
To figure out the a, you need to take the 5 and divide it by 2 (because 2ax), which becomes 5/2. a=5/2.
Then you need to square it, (because a^2) which becomes 5^2/2^2.
5x5 is 25, and 2x2 is 4, so the a^2 is 25/4.
When you add 25/4 on the left side, you have to add 25/4 on the right side as well.
Remember, the 3/4 is still there, so add them: x^2 + 5x + 25/4 = 25/4 + 3/4
25/4 + 3/4 = 28/4. 28 divided by 4 is 7.
Hope this helps! :3
• How was 7 added at the 6 paragraph? Where it shows the steps on how to complete the square?
• When Sal adds 9 to the left side, he must also add 9 to the right side. The right side was "-2".
So, when he add the 9, he gets: "- 2 + 9 = 7"
Hope this helps.
• what does what does it mean when there is a letter i In the solution of the equation?
• An "i" means the answer is the square root of a negative number. Since that doesn't work in the normal everyday world - but does have uses elsewhere - the "i" is used to make it easier to simplify the answers (and confuse the people ~_^). "i" is defined as the square root of negative 1, and can be factored out. Like an "x" or other variable, terms with "i" can only be added to or subtracted from other terms containing "i". However, "i" squared = -1. It becomes a regular number and can be added to regular numbers. This is important to remember when checking your answers.
• How do you solve 5x^2 -10x =23
• I'm going to assume you want to solve by completing the square.
1) Divide the entire equation by 5: x^2 - 2x = 23/5
2) Complete the square: -2/2 = -1. (-1)^2 = +1. Add +1 to both sides: x^2 - 2x + 1 = 23/5 + 1
3) Rewrite the left side as a binomial squared, and add the fractions on the right: (x-1)^2 = 28/5
4) Take square root of both sides: sqrt(x-1)^2 = +/- sqrt(28/5)
x-1 = +/- sqrt(4*7/5)
x-1 = +/- 2 sqrt(7/5) * sqrt(5/5)
x-1 = +/- 2 sqrt(35) / 5
6) Add 1 to both sides:
x = 1 +/- 2 sqrt(35) / 5
x = 5/5 +/- 2 sqrt(35) / 5
x = [5 +/- 2 sqrt(35)] / 5
Hope this helps.
• Does each quadratic equation always have a square and those missing parts of the square; or is it (this method) just for the perfect square equations?
• Not every quadratic equation always has a square. It may have a square, missing parts for a square, or even both, in which case you could use the completing the square method. But no, for the most part, each quadratic function won't necessarily have squares or missing parts. It's possible, but not common. But since every number is a square and has a square root, you can still do it, though it would be much more painful.
• How do I do a quadratic equation using completing the square if a or x^2 have a number in front of it ( ie. 4x^2) because I have tried many things with it and it doesn't add up or subtract out.
(1 vote)
• This would be the same as rule 2 (and everything after that) in the article above.
You are correct that you cannot get rid of it by adding or subtracting it out. As shown in rule 2, you have to divide by the value of a (which is 4 in your case). In the example following rule 2 that we were supposed to try, the coefficient of x² is 4. So, we have to divide the x² AND the x terms by 4 to bring the coefficient of x² down to 1.
Let's use the example they gave us:
4x² + 20x -3 = 0
move the constant term
4x² + 20x = 3
divide through the x² term and x term by 4 to factor it out
4(4x²/4 + 20x/4 ) = 3
This leaves
4( x² + 5x ) = 3
Or, you can divide EVERY term by 4 to get
x² + 5x = 3/4 → I prefer this way of doing it
Now we complete the square by dividing the x-term by 2 and adding the square of that to both sides of the equation. That is 5/2 which is 25/4 when it is squared
x² + 5x +25/4 = 3/4 + 25/4 → simplify the right side
x² + 5x +25/4 = 28/4 → Hey, that is equal to 7
Now rewrite the perfect square trinomial as the square of the two binomial factors
We especially designed this trinomial to be a perfect square so that this step would work:
x² + 5x +25/4 = (x + 5/2)²
If you get stuck on the fractions, the right-hand term in the parentheses will be half of the x-term.
(x + 5/2)² = 7
Take the square root of both sides, remembering to take both the positive and negative square root of the number on the right
(x + 5/2)² = ± √ 7
x + 5/2 = ± √ 7
Isolate the x by subtracting away the constant that is still with it
x = - 5/2 ± √ 7
That gives us our two correct solutions for x
x = - 5/2 + √ 7
x = - 5/2 - √ 7
• Often when coming across quadratics with non-zero leading co-efficients I'm not sure whether to pull out a common factor, or to divide by the leading co-efficient.
For example in the above question in my first attempt I factored out a 4 to get 4(x^2 + 5x)-3=0.
However, when attempting to solve from there, I end up at a dead end, or with a different result to if I'd divided out the leading co-efficient.
Are these 2 methods interchangeable, and I'm just doing something wrong? Or, is there a preferred approach for some reason?
• I wouldn't say that both methods are exactly interchangable; however, it's best to factor out. Factoring it out preserves the equation whilst dividing the equation by, for example it's GCF, removes that aspect of the equation. As a result from that, there might be some loss of solutions.
Your completing the square method is exactly on point with the first half. Just remember when you find (b/2)², you must add that result in the parenthesis, and subtract it out and multiply it by the 4 by the number outside. Less abstractly:
4(x²+5x+(25/4)) - 3(4)(25/4) = 0
And solve by then. Hopefully that gives some insight
• This method is only available if factoring does not help correct?
• No, completing the square can be used to solve any quadratic equation whether or not factoring works. |
# How do you use the chain rule to differentiate y=cos^3(4x)?
Oct 28, 2017
$\frac{\mathrm{dy}}{\mathrm{dx}} = - 12 {\left[\cos \left(4 x\right)\right]}^{2} \sin \left(4 x\right)$
#### Explanation:
What the chain rule is that it takes a composition of functions, and "peels" it back layer by layer. In the context of a calculation, you'd take the derivative of your outermost function leaving the inside as is, and then multiply by the derivative of the next outermost function, till you reach the end.
It's hard to describe properly in words, so let me show you with your example.
So, you have 3 different functions built into each other here. You have $f \left(x\right) = {x}^{3}$, $g \left(x\right) = \cos \left(x\right)$, and $h \left(x\right) = 4 x$.
Our outermost function is ${x}^{3}$. So we'd take the derivative of that while preserving the contents inside . This leaves:
$\implies 3 {\left[\cos \left(4 x\right)\right]}^{2}$
as our first term. Now for the next function, $\cos \left(x\right)$. Following the same principle:
$\implies 3 {\left[\cos \left(4 x\right)\right]}^{2} \cdot - \sin \left(4 x\right)$
Lastly, $4 x$:
$\implies 3 {\left[\cos \left(4 x\right)\right]}^{2} \cdot - \sin \left(4 x\right) \cdot 4$
Cleaning this up gives:
$\frac{\mathrm{dy}}{\mathrm{dx}} = - 12 {\left[\cos \left(4 x\right)\right]}^{2} \sin \left(4 x\right)$ |
# GSEB Solutions Class 12 Maths Chapter 4 Determinants Ex 4.2
Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.2 Textbook Questions and Answers.
## Gujarat Board Textbook Solutions Class 12 Maths Chapter 4 Determinants Ex 4.2
Question 1.
$$\left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right|$$ = 0
Solution:
Operating C1 → C1 + C2, we get
$$\left|\begin{array}{lll} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{array}\right|$$ = $$\left|\begin{array}{lll} x+a & a & x+a \\ y+b & b & y+b \\ z+c & c & z+c \end{array}\right|$$ = 0,
since 1st and 3rd columns are identical.
Question 2.
$$\left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|$$ = 0
Solution:
Operating C1 → C1 + C2 + C3 we get
$$\left|\begin{array}{lll} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|$$ = $$\left|\begin{array}{ccc} a-b+b-c+c-a & b-c & c-a \\ b-c+c-a+a-b & c-a & a-b \\ c-a+a-b+b-c & a-b & b-c \end{array}\right|$$
= $$\left|\begin{array}{lll} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \end{array}\right|$$ = 0,
since all elements of first column are zero.
Question 3.
$$\left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right|$$ = 0
Solution:
Operating C3 → C3 – C1 – 9C2 we get:
$$\left|\begin{array}{lll} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{array}\right|$$ = $$\left|\begin{array}{lll} 2 & 7 & 0 \\ 3 & 8 & 0 \\ 5 & 9 & 0 \end{array}\right|$$ = 0
since all the elements of 3rd column are zero.
Question 4.
$$\left|\begin{array}{lll} 1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b) \end{array}\right|$$ = 0.
Solution:
Operating C3 → C3 + C2, we get:
$$\left|\begin{array}{lll} 1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b) \end{array}\right|$$ = $$\left|\begin{array}{lll} 1 & b c & a b+b c+c a \\ 1 & c a & a b+b c+c a \\ 1 & a b & a b+b c+c a \end{array}\right|$$
Taking out ab + bc + ca common from 3rd column, we get
= (ab + bc + ca)$$\left|\begin{array}{lll} 1 & b c & 1 \\ 1 & c a & 1 \\ 1 & a b & 1 \end{array}\right|$$ = 0.
since 1st and 3rd columns are identical.
Question 5.
$$\left|\begin{array}{ccc} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right|$$ = 2 $$\left|\begin{array}{lll} a & p & x \\ b & q & y \\ c & r & z \end{array}\right|$$
Solution:
Question 6.
$$\left|\begin{array}{ccc} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{array}\right|$$ = 0
Solution:
Question 7.
$$\left|\begin{array}{ccc} -a^{2} & a b & a c \\ b a & -b^{2} & b c \\ a c & c b & -c^{2} \end{array}\right|$$ = 4a²b²c².
Solution:
Taking a, b, c common from I, II and III rows respectively,
we get ∆ = $$\left|\begin{array}{ccc} -a & b & c \\ a & -b & c \\ a & b & -c \end{array}\right|$$
Again a, b and c are taking out common from I, II and III columns respectively. we get
∴ ∆ = a²b²c²$$\left|\begin{array}{ccc} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right|$$
Now operating R1 → R1 + R2, we get:
∆ = a²b²c²$$\left|\begin{array}{ccc} 0 & 0 & 2 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right|$$ = a²b²c².2(1+1) = 4a²b²c².
Question 8.
(i) $$\left|\begin{array}{lll} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}\right|$$ = (a – b)(b – c)(c – a)
(ii) $$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|$$ = (a – b)(b – c)(c – a)(a + b + c)
Solution:
(i)
Taking out a – b and b – c common from R1 and R2 respectively,
∆ = (a-b)(b-c)$$\left|\begin{array}{ccc} 0 & 1 & a+b \\ 0 & 1 & b+c \\ 1 & c & c^{2} \end{array}\right|$$
Expanding with the help of elements of first common, we get
∆ = (a-b)(b-c)$$\left|\begin{array}{ll} 1 & a+b \\ 1 & b+c \end{array}\right|$$
⇒ ∆ = (a – b)(b – c)[(b + c) – (a + b)]
⇒ ∆ = (a – b)(b – c)(c – a)
(ii)
Taking out a – b and b – c common from C1 and C2 respectively,
∆ = (a-b)(b-c)$$\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & c \\ a^{2}+a b+b^{2} & b^{2}+b c+c^{2} & c^{3} \end{array}\right|$$
= (a-b)(b-c) x [(b2 + bc + c²) – (a² + ab + b²)]
= (a – b)(b – c)[(bc – ab) + (c² – a²)]
= (a- b)(b – c)[b(c – a) + (c – a)(c + a)]
= (a- b)(b-c)(c-a)(a + b + c).
Question 9.
$$\left|\begin{array}{lll} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array}\right|$$ = (x – y)(y – z)(z – x)(xy + yz + zx)
Solution:
L.H.S. ∆ = $$\left|\begin{array}{lll} x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y \end{array}\right|$$.
Operating R1 → R1 – R2, R2 – R3, we get:
Question 10.
(i) $$\left|\begin{array}{ccc} x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4 \end{array}\right|$$ = (5x + 4)(4 – x)²
(ii) $$\left|\begin{array}{ccc} y+k & y & y \\ y & y+k & y \\ y & y & y+k \end{array}\right|$$ = k²(3y + k)
Solution:
(i)
Expanding with the help of elements of R1, we get
∆ = (5ac + 4)(x – 4)²[(1 – 0)]
= (5x + 4)(x – 4)2 = R.H.S.
(ii)
Question 11.
Solution:
Question 12.
$$\left|\begin{array}{lll} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|$$ = (1 – x³)²
Solution:
L.H.S = ∆ = $$\left|\begin{array}{lll} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|$$
Operating C1 → C1 + C2 + C3, we get:
Question 13.
$$\left|\begin{array}{ccc} 1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2} \end{array}\right|$$ = (1 + a² + b)³.
Solution:
Expanding with the help of elements of first row, we get
∆ = (1 + a² + b²)²[(1 – a² – b²) + 2a² + 2b²]
= (1 + a² + b²)² (1 + a² + b²)
= (1 + a² + b²)³ = R.H.S.
Question 14.
$$\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|$$ = 1 + a² + b² + c².
Solution:
L.H.S = ∆ = $$\left|\begin{array}{ccc} a^{2}+1 & a b & a c \\ a b & b^{2}+1 & b c \\ c a & c b & c^{2}+1 \end{array}\right|$$
= $$\left|\begin{array}{lll} a^{2}+1 & a b+0 & a c+0 \\ a b+0 & b^{2}+1 & b c+0 \\ c a+0 & c b+0 & c^{2}+1 \end{array}\right|$$
This may be expressed as the sum of 8 determinants as shown below:
Question 15.
Let A be a square matrix of order 3 x 3, then | kA | is equal to
(A) k|A|
(B) k²|A|
(C) k³|A|
(D) 3k|A|
Solution:
Taking out k common from each row, we get
|kA| = k³$$\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|$$ = k³|A|.
⇒ Part (C) is the correct answer.
Question 16.
Which of the following is correct:
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of them.
Solution:
Determinant is a number associated to a square matrix.
⇒ Part (C) is the correct. |
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# Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics chapter 8 - Quadrilateral : Constructions and Types [Latest edition]
Textbook page
## Chapter 8: Quadrilateral : Constructions and Types
Practice Set 8.1Practice Set 8.2Practice Set 8.3
### Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics Chapter 8 Quadrilateral : Constructions and Types Practice Set 8.1 [Page 43]
Practice Set 8.1 | Q 1 | Page 43
Construct the following quadrilateral of given measures.
In ☐ MORE, l(MO) = 5.8 cm, l(OR) = 4.4 cm, m∠M = 58°, m∠O = 105°, m∠R = 90° .
Practice Set 8.1 | Q 2 | Page 43
Construct the following quadrilateral of given measures.
Construct ☐ DEFG such that l(DE) = 4.5 cm, l(EF) = 6.5 cm, l(DG) = 5.5 cm, l(DF) = 7.2 cm, l(EG) = 7.8 cm.
Practice Set 8.1 | Q 3 | Page 43
Construct the following quadrilateral of given measures.
In ☐ ABCD l(AB) = 6.4 cm, l(BC) = 4.8 cm, m∠A = 70° , m∠B = 50° , m∠C = 140°.
Practice Set 8.1 | Q 4 | Page 43
Construct the following quadrilateral of given measures.
Construct ☐ LMNO such that l(LM) = l(LO) = 6 cm, l(ON) = l(NM) = 4.5 cm, l(OM) = 7.5 cm.
### Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics Chapter 8 Quadrilateral : Constructions and Types Practice Set 8.2 [Pages 46 - 47]
Practice Set 8.2 | Q 1 | Page 46
Draw a rectangle ABCD such that l(AB) = 6.0 cm and l (BC) = 4.5 cm.
Practice Set 8.2 | Q 2 | Page 46
Draw a square WXYZ with side 5.2 cm.
Practice Set 8.2 | Q 3 | Page 46
Draw a rhombus KLMN such that its side is 4 cm and m∠K = 75° .
Practice Set 8.2 | Q 4 | Page 46
If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.
Practice Set 8.2 | Q 5 | Page 47
Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.
Practice Set 8.2 | Q 6 | Page 47
Find the length of diagonal of a square with side 8 cm .
Practice Set 8.2 | Q 7 | Page 47
Measure of one angle of a rhombus is 50° .find the measures of remaining three angles.
### Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics Chapter 8 Quadrilateral : Constructions and Types Practice Set 8.3 [Pages 49 - 50]
Practice Set 8.3 | Q 1 | Page 49
Measures of opposite angles of a parallelogram are (3x − 2)° and (50 − x)° . Find the measure of its each angle.
Practice Set 8.3 | Q 2 | Page 50
Referring the adjacent figure of a parallelogram, write the answer of questions given below.
(1) If l(WZ) = 4.5 cm then l(XY) = ?
(2) If l(YZ) = 8.2 cm then l(XW) = ?
(3) If l(OX) = 2.5 cm then l(OZ) = ?
(4) If l(WO) = 3.3 cm then l(WY) = ?
(5) If m∠WZY = 120° then m∠WXY = ? and m∠XWZ = ?
Practice Set 8.3 | Q 3 | Page 50
Construct a parallelogram ABCD such that l(BC) = 7 cm, m∠ABC = 40° , l(AB) = 3 cm.
Practice Set 8.3 | Q 4 | Page 50
Ratio of consecutive angles of a quadrilateral is 1:2:3:4. Find the measure of its each angle. Write, with reason, what type of a quadrilateral it is.
Practice Set 8.3 | Q 5 | Page 50
Construct ☐ BARC such that l(BA) = l(BC) = 4.2 cm, l(AC) = 6.0 cm, l(AR) = l(CR) = 5.6 cm
Practice Set 8.3 | Q 6 | Page 50
Construct ☐ PQRS, such that l(PQ) = 3.5 cm, l(QR) = 5.6 cm, l(RS) = 3.5 cm, m∠Q = 110° , m∠R = 70° . If it is given that ☐ PQRS is a parallelogram, which of the given information is unnecessary?
## Chapter 8: Quadrilateral : Constructions and Types
Practice Set 8.1Practice Set 8.2Practice Set 8.3
## Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics chapter 8 - Quadrilateral : Constructions and Types
Balbharati solutions for Maharashtra state board (SSC) Class 8 Mathematics chapter 8 (Quadrilateral : Constructions and Types) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the Maharashtra State Board Maharashtra state board (SSC) Class 8 Mathematics solutions in a manner that help students grasp basic concepts better and faster.
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Concepts covered in Maharashtra state board (SSC) Class 8 Mathematics chapter 8 Quadrilateral : Constructions and Types are Construction of a Quadrilateral, Concept of a Rectangle, Concept of a Square, Concept of Rhombus, Concept of Parallelogram, Concept of Trapezium, Concept of Kite.
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A lot of students, over the years, keep making the same exponent mistakes. I’m going to go through some of the exponent rules so you don’t make the same mistakes I see kids make over and over again. Let’s get these rules straight once and for all. Let’s go to the board.
### Same coefficient rules
One of the exponents rules students confuse is, if you have X2 x X3, you have the same coefficient here, and you’re just multiplying X2 x X3; students will multiply the exponents. They’ll say X2 x X3 = X6. That’s wrong. You don’t do that. Let’s get this straight once and for all. It’s X2 x X3. You add the exponents when you multiply two of the same coefficients. X2 x X3 = X2+3; X5. You add the exponents when you multiply exponents with the same bases.
Another exponent rule students confuse is if you take (X3)4. What they often do wrong is they will add these exponents; they’ll just say that’s X7. That’s wrong. Let’s get this straight once and for all. If you have an exponent and you’re raising it to another exponent, that’s when you multiply the exponents. It’s the same as X3×4, or X12, final answer.
Sometimes, students make this mistake: If you have division with exponents. If you have the same base for a numerator and denominator, but it’s X6/ X3. Sometimes, students will say, “I’m just going to divide those exponents.” What they say wrong is they’ll say it’s X6/3 = X2. That is completely wrong, do not do that. What you want to do when you divide exponents, you subtract the exponents from one another. X6/ X3 is the same thing as X6-3; X3, final answer.
One last error I want to show you, that students often make, is if you have(2X)3. What students often do wrong is they will only apply the exponent to the X. They’ll say “That is 2X3, final answer.” That is completely false. Do not do that. What you’re going to do is apply the exponent to each entity in the parentheses. The answer to (2X)3 is the same thing is 23 x X3= 8X3, final answer.
Just go over those 4 rules I taught you, and you shouldn’t make any careless mistakes when you see an exponent problem on the SAT. |
# Square Triangles
I recently came across this youtube video, where Matt Parker asks whether 36 is the only triangle-square number or not. Immediately after seeing this, I decided to tackle the problem, and after some early results, I decided to chronicle my efforts thus far 1.
## The Problem
Before detailing my attempt to tackle this problem, it’s probably a good idea to specify what it is. We are trying to find all triangle-square numbers. A triangle-square number is a number that is both a triangle number and a square number 2. So, it is a number $x$ such that $x=n^2$ and $x=m(m+1)/2$ where $m,n$ are natural numbers. Therefore, we will find such $x$ by investigating solutions to the following equation:
$$$$n^2=\frac{m(m+1)}2$$$$
We are looking for pairs of natural number $m, n$ that make the above true.
## First Steps
Before I started trying to solve this problem, I thought about what I expected the solution to be. Initially, I figured 36 was probably the only triangle-square number 3. At the very least, I could not immediately think of any other such number, or an obvious connection between triangle and square numbers. When I started working on the problem, I expected a nice solution to appear “in plain sight” after a bit of algebra that would show that the only possible solution was 36 4.
Finally beginning to work, the very first thing I did was rewrite the equation. Division is much messier than multiplication, so I had to get rid of that divide by 2. I also thought it might be a good idea to distribute the $m$.
$$2n^2=m^2 + m$$
Before I continued, I thought it might be helpful to write down the one solution I had so far. A quick calculation showed that the triangle-square 36 corresponds to the solution $(m,n)=(8,6)$. Now that I could return to the problem at hand, one approach to solving diophantine equations 5 that is often helpful is looking for restrictions on the variables. This basically involves using algebra and properties of integers to show that the variables must behave a certain way or follow a certain pattern. Since this problem deals with squares, the first trick that came to mind was to consider the equation $\pmod 4$. Perfect squares are always congruent to 0 or 1 $\pmod 4$ 6, so since there are two squares in the equation, this can be used to find 4 (hopefully not distinct) possible values of $m\pmod 4$.
$\begin{matrix} n^2\pmod 4& &m^2\pmod 4& &\text{equation }\pmod 4& &m\pmod 4 \\ \hline 0& &0& &0\equiv0+m& &0\\ 0& &1& &0\equiv1+m& &3\\ 1& &0& &2\equiv0+m& &2\\ 1& &1& &2\equiv1+m& &1 \end{matrix}$
As you can see from looking at the rightmost column $m$ could potentially be congruent to anything $\pmod 4$. In other words, that was a waste of effort because it told us nothing.
## A Breakthrough
After my initial attempt proved less than fruitful, I decided I needed to tackle this problem from a different angle. My first thought was to refactor the right hand side of the equation, because multiplication is nice.
$$2n^2 = m(m+1)$$
After staring at this equation for some time, and idea popped into my head: I should consider the prime factorizations of these numbers. The left hand side is twice a square. In the prime factorization of a square number, every prime has an even exponent. This means that if $n^2=p_1^{a_1}p_2^{a_2}\dots p_k^{a_k}$ where each $p_i$ is a prime number, then every $a_i$ is even. Therefore, in the prime factorization of the left hand side, every prime has an even exponent, except for 2, which has an odd exponent. Now we’re getting somewhere. This must also be the case on the right hand side 7, but how can we use that information? Here, I remembered that every number is coprime to its successor 8, so for any given prime $p$, its exponent in the prime factorization of $2n^2$ is equal to its exponent in the prime factorization of $m$ or to its exponent in the prime factorization of $m+1$. Therefore, every prime in the prime factorization of $m$ must have an even exponent. The same holds for $m+1$. The only exception is 2, which has an odd exponent in one of $m$ and $m+1$. This means that $m$ is twice a square and $m+1$ is a square, or vice versa! I then decided to continue my investigation by looking at each case separately.
## The First Case: $m=2k^2$
At this point, I just did a little algebra.
\begin{align*} m(m+1) &= 2k^2(2k^2 + 1) \\ &= 4k^4+2k^2\\ &= 2n^2\\ \end{align*}
\begin{align*} &n^2 = 2k^4+k^2 \end{align*}
Here was another equation to investigate. Like with the first, my initial thought was modular arithmetic. Once again, we are dealing with squares so I looked at the equation $\pmod 4$. I’ll spare you the details, but it turns out that $k$ must be congruent to 0 or 2 $\pmod 4$. In other words, $k$ is even. When I reached this point, I was out of ideas, and returned to the classic stare at the problem until inspiration hits approach. Luckily, it did. I hadn’t yet used the fact that $m+1$ was a square. This realization led to the following bit of algebra.
$$$$2k^2 + 1 = q^2\\ 2k^2 = (q-1)(q+1)$$$$
This looked promising. I tried some more mod stuff, tried thinking about what properties of $q$ I could derive from this, and considered where using the fact that 2 must divide $(q-1)$ or $(q+1)$ could get me. All of this was to no avail, however. Eventually, after some more staring, I realized that $(q-1)$ and $(q+1)$ have the same parity, and so since the left hand side is even, they must both be even as well so $q$ is odd 9. Naturally, this led to another equation, and even more algebra.
$$$$q = 2t+1\\ 2k^2 = 2t(2t+2)\\ k^2 = 2t(t+1)\\ n^2 = 2t(t+1)(2t+1)^2$$$$
At this point, I had stumbled onto a gem, although I didn’t realize it at first. Looking at that final equation, I didn’t expect more modular arithmetic to be helpful, but I had no other ideas so I tried the standard look at things $\pmod 4$ approach, and learned nothing useful. After some consideration, I returned to the idea that had gotten me this far: considering prime factorizations. The left hand side is a perfect square, so all of its prime factors have even exponents. The same must be true of the right hand side. The rightmost factor is a perfect square, so its primes obviously have even exponents. So, what remains (i.e. $2t(t+1)$ must also have primes with all even exponents. This meant that $2t(t+1)$ must be a perfect square. Symbolically,
$$2t(t+1)=s^2$$
This was exciting. If you look long enough, this equation should start to look familiar. It’s very similar to the equation this problem began with 10. In case you don’t remember, the equation I’m referring to is this one:
$$2n^2=m(m+1)$$
Very similar indeed. This gave me an idea: could solutions to the original equation lead to solutions from this equation? The answer is, luckily, yes. If $t$ is a solution to the original equation ($t(t+1)=2p^2$ is twice a perfect square), the it is also a solution to this latest equation ($2t(t+1)=4p^2=(2p)^2$ is a perfect square). This was exciting! Recall that we derived this equation for $t$ by investigation solutions to the original equation. Values for $t$ that satisfy this equation 11 can be used to generate solutions to the original equation. Since we have just shown that solutions to the original equation also solve this $t$-equation, we have shown that they also generate new solutions to the original equation! Therefore, not only is 36 not the only triangle-square number, but there are infinitely many such numbers, and once you know one of them, you know infinitely many of them! As an example, remember that $(m,n)=(8,6)$ is the solution to the original equation, so if we let $t=8$, then we can generate a new solution. The corresponding solution is $(m,n)=(288,204)$, and indeed the 288th triangle number is $204^2$.
As exciting as this news is, our work here is not done. We still do not know if this process gives all solutions or not. At this point, I thought it would. It was a really nice pattern (old solutions giving rise to new ones), and so I figured it was probably all there was to it. Still, I continued to investigate further just to make sure.
## The Second Case: $m+1=2k^2$
This case went by more quickly than the last, because it involved applying the same ideas. As always, I started with algebra.
\begin{align*} m(m+1) &= 2k^2(2k^2 - 1) \\ &= 2n^2\\ \end{align*}
\begin{align*} n^2 = k^2(2k^2-1) \end{align*}
I tried to find a restriction on the value of $k\pmod 4$ from that equation, but found none. Oh well. Modular arithmetic had not been the most useful tool during this venture, so I didn’t expect it to be very helpful here either. I then used the fact that $m$ was a perfect square to do the following.
$$$$2k^2-1=q^2\\ q=2t+1\\ 2k^2=2t+2\\ k^2=t+1\\ n^2=2(t+1)(2t+1)$$$$
Here, I returned to my good old friend prime factorizations. $n^2$, and hence $2(t+1)(2t+1)$, must only have primes with even exponents in its prime factorization. So, the exponent of 2 in the prime factorization of $(t+1)(2t+1)$ must be odd. $(2t+1)$ is odd, so 2 has an exponent of 0 in its factorization, and so the exponent of 2 must be odd in the prime factorization of $(t+1)$. Since $(t+1)$ and $(2t+1)$ are coprime, they share no prime factors. This means that the exponent of every prime (except 2) in the prime factorization of $(t+1)$ must be even. Thus $(t+1)$ is twice a square, so
$$$$t+1=2s^2\\ n^2=4s^2(4s^2-1)$$$$
This progressed nicely from here. $4s^2$ is a square, so $(4s^2-1)$ must be a square too. However, $4s^2-1=4(s^2-1)+3$, so its remainder is 3 when divided by 4 (i.e. $4s^2-1\equiv3\pmod4$). However, a perfect square will always have remainder 0 or 1 when divided by 4. Therefore, $(4s^2-1)$ cannot be a square! This means that there is no solution to the original equation where $m+1=2k^2$.
## Final Thoughts
At this point, I had found a method for generating infinitely many solutions where $m$ is twice a square, and had shown that no solutions exist when $m+1$ is twice a square. This seemed to lend evidence to the idea that all the solutions could be found by starting with a valid $m$ value (Ex. $m=1$), plugging it in for $t$ to generate a new valid $m$ value, and repeating. However, it still did not
Things were looking good, until I came across the solution $(m,n)=(49,35)$ 12. This was an issue for two reasons. One, you cannot find this solution by starting with $m=1$ and using $t$ to generate new solutions. Two, $m+1=50$ is twice a perfect square. The first issue I can deal with. All it means is that my iteration idea does not generate all solutions; it still generates valid solution, so that doesn’t bother me too much. The second issue is much more disturbing. I thought I had shown that there was no solution where $m+1$ is twice a square. This means that there must have been some flaw in my reasoning somewhere, and I need to reconsider my approach to this problem.
Once I realized I had flaws, I decided I had done enough math for one night 13, and would continue my investigation some other time. Thus, I leave you on a bit of a cliffhanger 14.
1. This post will try to show my entire approach to this problem, and not just where it worked. There will be places where my reasoning was more complicated than it needed to be. I will not point these out, because they seem obvious in hindsight.
2. duh
4. Perhaps, there would be a bound on the possible values of m or n, and so all the solutions could be found by checking every possible value for the bounded variable.
5. Equations where you are only interested in integer solutions.
6. In English, if you divide a perfect square by 4, the remainder is 0 or 1 (Ex. 9/4 = 2 R 1 and 64/4 = 16 R 0).
7. They are equal, after all.
8. n and (n+1) do not share any prime factors
9. If I were good at math, I would have realized this sooner
10. Oh yeah, this problem had a beginning. At this point, I was beginning to think it was a neverending sequence of equation manipulation with no origin and no end.
11. Values such that 2t(t+1) is a perfect square.
12. Just when I think I had done good math for once, a counterexample ruins everything.
13. I had almost filled a page, and turning to the next page in my notebook would have been too much trouble.
14. In case anyone is curious, the problem was that I forgot to square q after writing q=2t+1, so I had 2k^2=q+1 instead of 2k^2=q^2+1. Also, I don’t plan on writing another post about what I did after finding the error. |
### Factoring 2j^2+4j-15 Solution
The variable we want to find is j
We will solve for j using quadratic formula -b +/- sqrt(b^2-4ac)/(2a), graphical method and completion of squares.
${x}_{}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Where a= 2, b=4, and c=-15
Applying values to the variables of quadratic equation -b, a and c we have
${j}_{}=\frac{-4±\sqrt{{4}^{2}-4x\mathrm{2x}\mathrm{-15}}}{2x2}$
This gives
${j}_{}=\frac{-4±\sqrt{{4}^{2}-\mathrm{-120}}}{4}$
${j}_{}=\frac{-4±\sqrt{136}}{4}$
${j}_{}=\frac{-4±11.6619037897}{4}$
${\mathrm{j1}}_{}=\frac{-4+11.6619037897}{4}$
${\mathrm{j2}}_{}=\frac{-4-11.6619037897}{4}$
${j}_{1}=\frac{7.66190378969}{4}$
${j}_{1}=\frac{-15.6619037897}{4}$
The j values are
j1 = 1.91547594742 and j2 = -3.91547594742
### Factoring Quadratic equation 2j^2+4j-15 using Completion of Squares
2j^2+4j-15 =0
Step1: Divide all terms by the coefficient of j2 which is 2.
${j}^{2}+\frac{4}{2}x-\frac{15}{2}=0$
Step 2: Keep all terms containing x on one side. Move the constant to the right.
${j}^{2}+\frac{4}{2}j=\frac{15}{2}$
Step 3: Take half of the x-term coefficient and square it. Add this value to both sides.
${j}^{2}+\frac{4}{2}j+{\left(\frac{4}{4}\right)}^{2}=\frac{15}{2}+{\left(\frac{4}{4}\right)}^{2}$
Step 4: Simplify right hand sides of expression.
${j}^{2}+\frac{4}{2}j+{\left(\frac{4}{4}\right)}^{2}=\frac{136}{16}$
Step 2: Write the perfect square on the left.
${\left(j+\frac{4}{4}\right)}^{2}=\frac{136}{16}$
Step 2: Take the square root on both sides of the equation.
$j+\frac{4}{4}=±\sqrt{\frac{136}{16}}$
Step 2: solve for root j1.
${j}_{1}=-\frac{4}{4}+\frac{11.6619037897}{4}=\frac{7.66190378969}{4}$
${j}_{1}=1.91547594742$
Step 2: solve for root j2.
${j}_{2}=-\frac{4}{4}-\frac{11.6619037897}{4}=\frac{-15.6619037897}{4}$
${j}_{2}=-3.91547594742$
### Solving equation 2j^2+4j-15 using Quadratic graph
j2 + j + = 0
Solutions
how to factor polynomials?
Polynomials can be factored using this factoring calculator
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Trinomials can be solved using our quadratic solver
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Other Variants of 2j^2+4j-15 are below |
# 2.2: Dimensional Analysis
Learning Objective
• Convert from one unit to another unit of the same type.
• Understand how to use Conversion Factors
### Conversion Factors
Recall your metric table (one of two that needs to be committed to memory) while working these types of problems.
Two methods of Memorization of Metric
Using negative exponents (see above) is the traditional way to memorize the metric system. Many students who have been taught outside of the United States approach the metric system this way. For their conversion factors, students who have been taught the metric system in the United States tend to use positive exponents. Either method is fine, just pick one way and use it consistently.
Two different ways to memorize the metric systems (be sure to know all bolded conversion factors listed in Section 2.1)
• 1 gigabase = 1 x109 base
• 1 megabase = 1x106 base
• 1 kilobase = 1x10base
• 1 decibase = 1x10-1 base or 1x101 decibase = 1 base
• 1 centibase = 1x10-2 base or 1x102 centibase = 1 base
• 1 millibase = 1x10-3 base or 1x103 millibase = 1 base
• 1 microbase = 1x10-6 base or 1x106 microbase = 1 base
• 1 nanobase = 1x10-9 base or 1x109 nanobase = 1 base
• 1 picobase = 1x10-12 base or 1x1012 picobase = 1 base
Note: A base unit can be gram, liter, or meter
We have an expression,
$\dfrac{3\,ft}{1\,yd} =1 \label{1}$
Equation \ref{1} is a strange way to write 1, but it makes sense: 3 ft equal 1 yd, so the quantities in the numerator and denominator are the same quantity, just expressed with different units. The expression in Equation \ref{1} can be is called a conversion factor and it is used to formally change the unit of a quantity into another unit. (The process of converting units in such a formal fashion is sometimes called dimensional analysis or the factor label method.)
To see how this happens, let us start with the original quantity: 4 yd
Now let us multiply this quantity by 1. When you multiply anything by 1, you don’t change the value of the quantity. Rather than multiplying by just 1, let us write 1 as
$\dfrac{3 \, ft}{1 \, yd}$
$4 \, yd\times \dfrac{3 \, ft}{1 \, yd}$
The 4 yd term can be thought of as that is, it can be thought of as a fraction with 1 in the denominator. We are essentially multiplying fractions. If the same thing appears in the numerator and denominator of a fraction, they cancel. In this case, what cancels is the unit yard:
$4 \, \cancel{yd}\times \dfrac{3 \, ft}{1 \, \cancel{yd}}$
That is all that we can cancel. Now, multiply and divide all the numbers to get the final answer:
$\dfrac{4\times 3 \, ft}{1}= \dfrac{12 \, ft}{1}= 12 \, ft$
Again, we get an answer of 12 ft, just as we did originally. But in this case, we used a more formal procedure that is applicable to a variety of problems.
How many millimeters are in 14.66 m? To answer this, we need to construct a conversion factor between millimeters and meters and apply it correctly to the original quantity. One method would entail using the conversion factor of $$1\, mm = 1 \times 10^{-3}\,m$$
\begin{align} 14.66\,\cancel{m}\times \dfrac{1\,mm}{10^{-3}\,\cancel{m}} &= 14660\,mm \\[5pt] &= 1.46\times 10^{4}\,mm \end{align}
Which conversion factor do we use? The answer is based on what unit you want to get rid of in your initial quantity. The original unit of our quantity is meters, which we want to convert to millimeters. Because the original unit is assumed to be in the numerator, to get rid of it, we want the meter unit in the denominator; then they will cancel. Using the positive metric exponents, we have an alternative method to calculate the same answer as shown previously. Canceling units and performing the mathematics, we get:
\begin{align} 14.66\, \cancel{m} \times \dfrac{1000\,mm}{1\,\cancel{m}} &= 14660\,mm \\[5pt] &= 1.46\times 10^{4}\,mm \end{align}
Note how $$m$$ cancels, leaving $$mm$$, which is the unit of interest.
The ability to construct and apply proper conversion factors is a very powerful mathematical technique in chemistry. You need to master this technique if you are going to be successful in this and future science courses.
Video: Conversion Problem and Solution
Check out this fabulous light board of me modeling problem solving using this method.
Warning: my last answer is off by a factor of 10. It should be 4.8x10-2kg.
Example $$\PageIndex{1}$$
1. Convert 35.9 kL to liters.
2. Convert 555 nm to meters.
Solution
1. We will use the fact that 1 kL = 1,000 L. Of the two conversion factors that can be defined, the one that will work is 1000L/ 1kL. Applying this conversion factor, we get:
$35.9\, \cancel{kL} \times \dfrac{1000\,L}{1\, \cancel{kL}}= 35,900\,L \nonumber$
1. We will use the fact that 1 nm = 1/1,000,000,000 m, which we will rewrite as 1,000,000,000 nm = 1 m, or 109 nm = 1 m. Of the two possible conversion factors, the appropriate one has the nm unit in the denominator:
$\dfrac{1\,m}{10^{9}\,m} \nonumber$
Applying this conversion factor, we get
\begin{align*}555\,\cancel{nm} \times \dfrac{1\,m}{10^{9}\,\cancel{nm}} &= 0.000000555\,m \\[5pt] &= 5.55\times 10^{-7}\,m \end{align*}
In the final step, we expressed the answer in scientific notation.
Exercise $$\PageIndex{1}$$
1. Convert 67.08 μL to liters.
2. Convert 56.8 m to kilometers.
6.708 × 10−5 L
5.68 × 10−2 km
Exercise $$\PageIndex{2}$$
How many milliliters are in 607.8 kL?
This problem involves multiple steps that require going through the metric base of liters. Think of a pathway of a conversion before setting up these types of problems. The suggested pathway would be mL ⇒ L ⇒ kL
\begin{align*} 607.8 \,kL \times \dfrac{1000\,L}{1\,kL} &= 6.08\times 10^{5}\,L \nonumber \\[5pt] &= 6.08\times 10^5 \, L \dfrac{1000\,mL}{1\,L} \\[5pt] &= 608000000\,mL \\[5pt] &= 6.08\times 10^{8}\,mL \nonumber \end{align*}
Exercise $$\PageIndex{3}$$
Kathleen Hsu Barron (Furman University, Class of 2011) is exploring Switzerland with her husband Patrick Barron (Furman University, Class of 2011). The sign she is pointing to is a typical speed in km/hr. Convert this distance to miles/hr and feet/hr.
Again, this problem is a multi-step problem. There a mutiple ways that one could proceed to solve this problem. With the coversion factors in this course, the suggested round would be miles ⇒ kilometers ⇒ meters ⇒ centimeters ⇒ inches ⇒ feet. Another method could utilize miles ⇒ feet.
18.6 miles/hr, 9.8 x 104 ft/hr
In this course, we will not use significant figures. You will need to read the directions on your assignment, quiz, or test and round accordingly.
Applications of conversions and the Metric System
The skill of converting units can be helpful in many aspects of your life. Measuring medications can be done precisely by using milliliters instead of teaspoons (recall 1 tsp = 5.0 mL). Of course, a pharmacist would provide you with the appropriate syringe if this was needed. Sizes of syringes can vary from 1.0 to 10.0 mL for most at home applications.
Marathons are measured using the kilometer unit. If you were raised outside the United States, you would be quite familiar with how long a kilometer truly is. To appreciate this distance, most Americans need to convert this to a mileage by dividing the distance by 1.61 km.
When traveling abroad, keep in mind the majority of countries (except Myanmar and Liberia) commonly use the metric system in day to day living. Understanding how to convert Metric to English units will help you to appreciate the measurements used by a different society.
Adoption of the metric system of measurement. Countries which have officially adopted the metric system (green) and countries which have not officially adopted the metric system (US, Myanmar, Liberia)
Why has the United States not adopted the metric system as a country? Watch this video and see if you can derive some appropriate hypotheses.
### Key Takeaways
• Units can be converted to other units using the proper conversion factors.
• Conversion factors are constructed from equalities that relate two different units.
• Conversions can be a single step or multistep.
• Unit conversion is a powerful mathematical technique in chemistry that must be mastered. |
# Estimating Differences
Want to know how to estimate the differences between two numbers? So do not miss this post!
Did you know it’s possible too to utilize rounding for estimating differences? At times whenever you do subtract between two large numbers, you do not wish to find out the precise differences. You only wish to know a fast estimate. It’s possible to get an estimate by rounding each of the numbers, then discovering a difference that’s near the precise difference.
## Way to estimate Differences to the closest tenth
Here is a fast review of the rules for rounding numbers.
STEP one: If the number one puts on the right of the targeted number was lower than $$5$$, it gets rounded down ?
STEP two: If the number one position to the right of the targeted number is $$5$$ or higher, it gets rounded up ?
STEP three: Now it is possible to get the difference in-between these rounded numbers. Whenever the minuend and subtrahend in a subtraction equation are rounded, the subtraction gets easier.
Note: You merely must remember to maintain the difference near the precise answer
Tip one: Unless it says something different, constantly round figures to the maximum feasible place value.
Tip two: You should pay attention to the fact we constantly write down the word “about” before the estimated difference. That shows you the result isn’t the precise answer, merely the estimate.
### Estimating Differences – Example 1:
Estimate differences by rounding each added to the nearest ten.
$$85-34=$$___
Solution:
First, round $$85$$ to the closest ten. Check out the numeral on the right of the tens position. It is $$5$$. So, $$85$$ is rounded up to $$90$$. Do the same for $$34$$. Then $$34$$ is rounded down to $$30$$. Now subtract $$30$$ from $$90$$: $$90-30=60$$
### Estimating Differences – Example 2:
Estimate differences by rounding each added to the nearest ten.
$$53-26=$$___
Solution:
First, round $$53$$ to the closest ten. Check out the numeral on the right of the tens position. It is $$3$$ and this number is lower than $$5$$. So, $$53$$ is rounded down to $$50$$. Do the same for $$26$$. Then $$26$$ is rounded up to $$30$$. Now subtract $$30$$ from $$50$$: $$50-30=20$$
## Exercises for Estimating Differences
Estimate differences by rounding each added to the nearest ten.
1. $$\color{blue}{77-23}$$
2. $$\color{blue}{ 94-48 }$$
3. $$\color{blue}{35-12 }$$
4. $$\color{blue}{ 62-53 }$$
1. $$\color{blue}{60}$$
2. $$\color{blue}{ 40 }$$
3. $$\color{blue}{30 }$$
4. $$\color{blue}{ 10 }$$
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# Parametric Equations of a Parabola
We will learn in the simplest way how to find the parametric equations of a parabola.
The best and easiest form to represent the co-ordinates of any point on the parabola y$$^{2}$$ = 4ax is (at$$^{2}$$, 2at). Since, for all the values of ‘t’ the coordinates (at$$^{2}$$, 2at) satisfy the equation of the parabola y$$^{2}$$ = 4ax.
Together the equations x = at$$^{2}$$ and y = 2at (where t is the parameter) are called the parametric equations of the parabola y$$^{2}$$ = 4ax.
Let us discuss the parametric coordinates of a point and their parametric equations on the other standard forms of the parabola.
The following gives the parametric coordinates of a point on four standard forms of the parabola and their parametric equations.
Standard equation of the parabola y$$^{2}$$ = -4ax:
Parametric coordinates of the parabola y$$^{2}$$ = -4ax are (-at$$^{2}$$, 2at).
Parametric equations of the parabola y$$^{2}$$ = -4ax are x = -at$$^{2}$$, y = 2at.
Standard equation of the parabola x$$^{2}$$ = 4ay:
Parametric coordinates of the parabola x$$^{2}$$ = 4ay are (2at, at$$^{2}$$).
Parametric equations of the parabola x$$^{2}$$ = 4ay are x = 2at, y = at$$^{2}$$.
Standard equation of the parabola x$$^{2}$$ = -4ay:
Parametric coordinates of the parabola x$$^{2}$$ = -4ay are (2at, -at$$^{2}$$).
Parametric equations of the parabola x$$^{2}$$ = -4ay are x = 2at, y = -at$$^{2}$$.
Standard equation of the parabola (y - k)$$^{2}$$ = 4a(x - h):
The parametric equations of the parabola (y - k)$$^{2}$$ = 4a(x - h) are x = h + at$$^{2}$$ and y = k + 2at.
Solved examples to find the parametric equations of a parabola:
1. Write the parametric equations of the parabola y$$^{2}$$ = 12x.
Solution:
The given equation y$$^{2}$$ = 12x is of the form of y$$^{2}$$ = 4ax. On comparing the equation y$$^{2}$$ = 12x with the equation y$$^{2}$$ = 4ax we get, 4a = 12 ⇒ a = 3.
Therefore, the parametric equations of the given parabola are x = 3t$$^{2}$$ and y = 6t.
2. Write the parametric equations of the parabola x$$^{2}$$ = 8y.
Solution:
The given equation x$$^{2}$$ = 8y is of the form of x$$^{2}$$ = 4ay. On comparing the equation x$$^{2}$$ = 8y with the equation x$$^{2}$$ = 4ay we get, 4a = 8 ⇒ a = 2.
Therefore, the parametric equations of the given parabola are x = 4t and y = 2t$$^{2}$$.
3. Write the parametric equations of the parabola (y - 2)$$^{2}$$ = 8(x - 2).
Solution:
The given equation (y - 2)$$^{2}$$ = 8(x - 2) is of the form of (y - k)$$^{2}$$ = 4a(x - h). On comparing the equation (y - 2)$$^{2}$$ = 8(x - 2) with the equation (y - k)$$^{2}$$ = 4a(x - h) we get, 4a = 8 ⇒ a = 2 , h = 2 and k = 2.
Therefore, the parametric equations of the given parabola are x = 2t$$^{2}$$ + 2 and y = 4t + 2.
● The Parabola
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Question
A point object $\mathrm{O}$ is placed on the principal axis of a convex lens of focal length $\mathrm{f}=20\mathrm{cm}$ at a distance of $40\mathrm{cm}$ to the left of it. The diameter of the lens is $10\mathrm{cm}$. An eye is placed $60\mathrm{cm}$ to right of the lens and a distance $\mathrm{h}$ below the principal axis. The maximum value of $\mathrm{h}$ to see the image is____.
Open in App
Solution
Step 1. Given dataFocal length $\mathrm{f}=20\mathrm{cm}$Distance $=40\mathrm{cm}$We have to find the maximum value of distance $\mathrm{h}$.Step 2. Formula used.A lens is a piece of a refracting medium bounded by two surfaces, at least one of which is a curved surface.We will calculate the image distance by using the lens formula.This formula represents the relation between object distance $\mathrm{u}$, image distance $\mathrm{v}$, and focal length $\mathrm{f}$.$⇒$ $\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}$ Here, $\mathrm{f}=20\mathrm{cm}$According to the sign of convention, object distance $\mathrm{u}=-40\mathrm{cm}$ $\mathrm{v}=?$Step 3. Calculate the image distance.By using the lens formula, we get,$\frac{1}{20}=\frac{1}{\mathrm{v}}-\frac{1}{-40}$$\frac{1}{20}=\frac{1}{\mathrm{v}}+\frac{1}{40}$ $\frac{1}{\mathrm{v}}=\frac{1}{20}-\frac{1}{40}$$\frac{1}{\mathrm{v}}=\frac{1}{40}$$\mathrm{v}=40\mathrm{cm}$So, the image distance is $40\mathrm{cm}$, which means the image is formed at the center of curvature.Draw the ray diagramStep 4. Find the value of distance.According to the above figure, $∆\mathrm{CED}$ and $∆\mathrm{CAB}$ are in symmetric form.So,$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{DC}}$$\frac{5}{\mathrm{h}}=\frac{40}{20}$$\frac{5}{\mathrm{h}}=2$$\mathrm{h}=\frac{5}{2}$$\mathrm{h}=2.5\mathrm{cm}$The maximum value of $\mathrm{h}$ to see the image is $\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{cm}$.
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# Introducing Multiplication Using Arrays in Everyday Objects
Listen to this Lesson:
You might be wondering how you can teach your students to master multiplying single-digit numbers. Suppose you were asked by your student what multiplication is; you can explain that it is one of the four basic mathematical operations wherein you repeatedly combine the same values until you reach the final repetition. However, as simple as the sentence can sound, the idea may still be puzzling to your students.
It can lead to even more questions such as: What is that value being repeated? How do you combine repeated values? How many times do you repeat combining those values? When does the repetition stop?
These words, thoughts, concepts, and ideas are all abstract to a young learner. While it’s essential to learn the vocabulary associated with multiplication, a verbal explanation for an abstract concept can lead to more confusion; this is where teaching multiplication using arrays is a convenient visual tool.
In this article, we’re going to give you ideas on how to explain what an array is to your students, how it is found in everyday objects, and how it can be used for teaching multiplication. We’ll also give you a free lesson plan plus activities and tools to download and use with this lesson, so keep reading for more.
## Explaining Arrays to Kids
We see arrays in ordinary things around us, which is a great entry point into explaining what an array is. These common visual examples can help your students begin to understand how multiplication works. A dozen eggs are arranged in a tray with two rows having six eggs in each row. Imagine being in a grocery store and notice how an equal number of canned goods can be stored in the rows of shelves. Even app icons on a smartphone or tablet are in a grid arrangement. We can also use tiles on the floor to show columns and rows.
With this as an entry point, explain that arrays are simply objects structured in rows and columns. Rows are arranged horizontally, while columns are arranged vertically (it helps to remember columns like pillars). This can be demonstrated visually by drawing numbers from 1 to 3 in different orientations.
Write the numbers 1 to 3 from left to right. This shows the number of columns you can make.
There are 3 columns because you can draw 3 vertical lines. You can tell your students that counting rows start from the top to the bottom.
Write the numbers 1 to 3 from top to bottom. This shows the number of rows you can make.
There are 3 rows because you can draw 3 horizontal lines. You can tell your students that counting rows start from left to right.
Now that navigating the rows and columns of an array has been visually explained, you can give a real-life example.
For instance, a dozen eggs placed in a 2 by 6 tray can be interpreted as 2 rows of 6 eggs. Thus, 2 rows and 6 columns can be written as 2 x 6.
With this representation, we can count from left to right until 6, and we can count from top to bottom until 2. This means that there are 2 rows of 6 columns. Two rows of 6 eggs are the same as 2 x 6. Hence, when counting all the eggs in the array, there are 12 eggs all in all. We should also note that each row should have the same number of objects inside to represent an array model.
Similarly, if using a smartphone as an example, have students count how many icons are in a row and how many are in a column and have them write down the numerical representation of ____(rows) x _____ (columns).
## Arrays are Similar to Equal Groups
Let’s recall representations of objects enclosed in equal groups. Since each row contains an equal number of objects inside, a row can also be considered a grouping. Therefore, we can parallel the number of groups as the number of rows and the number of objects in each group as the number of objects in each row.
Imagine 3 containers and each container has 4 marbles. Therefore, there are 3 groups of 4 marbles.
If we arrange the marbles in rows and columns, there will be 3 rows and each row has 4 marbles in it.
Rearranging the representation of equal groups into an array model clarifies that each row contains the same number of items. This also explains the earlier questions:
What is that value being repeated? For this case, we are repeating 4 marbles 3 times in a row.
How do you combine repeated values? You combine them in rows and columns.
How many times do you repeat combining those values? There are 4 marbles in each row. There are 3 rows. Therefore, we are repeating 4 three times.
When does the repetition stop? The repetition stops when we reach the maximum number of rows. In the example above, we stop once we’ve repeated the value 4 three times.
## An Array is a Table
Another application of array models is drawing a table. No, not the table used for furniture. We are talking about a table made for data. An array is also a table because a table shows rows and columns.
Consider the game tic-tac-toe. To play this game, you first have to draw a grid that is 3 squares by 3 squares (or as we describe arrays, 3 rows of 3 columns). Because the table or grid is 3 by 3, there are 9 empty spaces wherein you can put the X and the O alternately. If all the spaces are occupied, no one wins, but we can see 3 rows of 3 items in each row.
This visualization can help students acknowledge that any drawing partitioned into a table can show an array model. To add to this knowledge, you can also easily integrate the concept of a multiplication table with your students. We can draw the multiplication table using 10 rows and 10 columns. The top leftmost corner is 1, and the bottom rightmost corner is 100.
Because you have taught your student how to read arrays, your student may understand the navigation of a multiplication table easier. For example, what is the product if you have 5 rows of 4, the student can count 5 times downward, then 4 times to the right to get 20.
To end, the total number of objects arranged in an array is the product of the multiplication expression. The numbers of rows and the numbers of objects in each row are called the factors; wherein there is a multiplicand and a multiplier. Array models can be applied to different visualizations.
Remember that arrays are all around us. Engage your students to look for different arrays in everyday objects!
## Free Lesson for Multiplication Using Arrays
When you’re ready to move on from explaining arrays, it’s time to turn to using them in a lesson to explain multiplication. One of the ways we like to teach the use of arrays is to present an introductory word problem to your students. Try something like this:
Your school is holding a science fair! The students will display their projects in their classrooms. How many displays will be in each classroom if there are four rows of desks and each row has four desks?
Feel free to tweak the word problem to something that your students can relate to or have a particular interest in. Invite them to come up with ideas of how they can solve the problem quickly.
You can demonstrate that you can count the desks as 4 + 4 + 4 + 4 = 16, but there’s an easier way by drawing an array, e.g., 4 x 4 = 16.
With the introduction and the basics of an array explained, ask your students another word problem. You can try something like this:
It’s your birthday, and you’ve baked cupcakes to give to your friends. You want to put two Hershey’s kisses on top of each cupcake, and you have 6 cupcakes. How many chocolates will you need?
## FREE Multiplication Using Arrays Worksheets and Resources
To help you get to know our Math Teacher Coach resources and learn how much time and preparation they can save you, we’ve added a bundle of free PDF resources for downloading today!
The Student Edition files are labeled SE, and the Teacher Edition files showing work and answers are labeled TE. Click the links below to download the different resources.
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# Interpolating Yourself
Suppose you have 3 cats at the beginning of the year, and 13 by the end. How many do you expect to have on July 1? In the absence of any other information, 8 is a reasonable answer. Just as July 1 is 50% of the way through the year, 8 is 50% of the way between 3 and 13.
The strategy you have just applied is linear interpolation. You are plotting a linear path between two known data points and guessing that values between the two points fall on the line. Here's the line spanning the days of the year on the x-axis and the number of cats on the y-axis:
You have recently learned that WebGL interpolates colors between texels. It does this automatically in hardware. However, there will be occasions in which you need to interpolate values in software. As a graphics developer, therefore, you should have your own linear interpolation function at the ready. Graphics developers call this function lerp, which is a contraction of linear interpolation. It is used as a verb: "You gotta lerp the bee from this flower to that one."
Here's another problem to help you develop the lerp algorithm. Suppose you are 150 centimeters tall at age 12 and 175 centimeters tall at age 17. How tall are you at age 13? 13 years is 20% of the way between 12 and 17 years. You work out your intermediate height by starting at 150 and tacking on 20% of the height difference:
$$150 + 0.2 \times (175 - 150) = 150 + 0.2 \times 25 = 155$$
This can be generalized. You must be given two coordinate pairs. Call them $$\mathrm{start}$$ and $$\mathrm{end}$$. The x-coordinates of these data points represent space or time. The y-coordinates represent size, color, position, or some other value that changes across time or space. You must also be given $$\mathrm{mid}_x$$, an x-coordinate whose y-coordinate is unknown. The job of your lerp function is to compute $$\mathrm{mid}_y$$.
The first step is to figure out the percentage of the domain that $$\mathrm{mid}_x$$ represents. This percentage is often given the name $$t$$ and is defined as follows:
$$t = \frac{\mathrm{mid}_x - \mathrm{start}_x}{\mathrm{end}_x - \mathrm{start}_x}$$
The percentage is then applied to the range as in the height problem to produce $$\mathrm{mid}_y$$:
$$\mathrm{mid}_y = \mathrm{start}_y + t \times (\mathrm{end}_y - \mathrm{start}_y)$$
That's the formula for your lerp function. You can also work it out using an argument based on similar triangles.
Sometimes you'll see the right-hand side of this formula expressed differently. Distribute $$t$$, regroup, and factor to derive this equivalent form:
\begin{aligned} \mathrm{mid}_y &= \mathrm{start}_y + t \times (\mathrm{end}_y - \mathrm{start}_y) \\ &= \mathrm{start}_y + t \times \mathrm{end}_y - t \times \mathrm{start}_y \\ &= \mathrm{start}_y - t \times \mathrm{start}_y + t \times \mathrm{end}_y \\ &= (1 - t) \times \mathrm{start}_y + t \times \mathrm{end}_y \\ \end{aligned}
Performance buffs favor this second formula because it can be compiled down to two vectorized instructions on the GPU.
Game engines usually provide a three-parameter version of lerp. Both Unity and the Unreal Engine do. You are expected to pass in $$\mathrm{start}_y$$, $$\mathrm{end}_y$$, and $$t$$.
## Bilinear Interpolation
Your lerp function only handles interpolation along one dimension and between two known data points. When the graphics card does a texture lookup, it's interpolating along two dimensions and between four known texels. Step through this breakdown of the algorithm to see how it's done:
Linearly interpolating between four points in two dimensions is called bilinear interpolation. Bilinear interpolation reduces down to three linear interpolations, two along one of the dimensions and third along the other dimension. Maybe you can guess how to interpolate between eight points in three dimensions and what you'd call it.
You're about to learn how to generate terrain from a texture. To move smoothly across the undulating but discrete terrain, you'll need a bilinear interpolation function. |
## Precalculus (6th Edition) Blitzer
The length of the side is $b=7$ and the values of the six trigonometric functions are $\sin \theta =\frac{24}{25}$ , $\cos \theta =\frac{7}{25}$ , $\tan \theta =\frac{24}{7}$ , $\csc \theta =\frac{25}{24}$ , $\sec \theta =\frac{25}{7}$ and $\cot \theta =\frac{7}{24}$.
In the right angle triangle, $a=24$ , $b$ is the side adjacent with angle $\theta$ and $c=25$. According to the Pythagoras theorem, ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$ Rearrange for $b$. \begin{align} & {{b}^{2}}={{c}^{2}}-{{a}^{2}} \\ & b=\sqrt{{{c}^{2}}-{{a}^{2}}} \\ \end{align} Substitute $25$ for $c$ and $24$ for $a$. \begin{align} & b=\sqrt{{{\left( 25 \right)}^{2}}-{{\left( 24 \right)}^{2}}} \\ & =\sqrt{625-576} \\ & =\sqrt{49} \\ & =7 \end{align} The ratio of $\sin \theta$ is $\sin \theta =\frac{a}{c}$ Substitute $25$ for $c$ and $24$ for $a$. $\sin \theta =\frac{24}{25}$ The ratio of $\cos \theta$ is $\cos \theta =\frac{b}{c}$ Substitute $25$ for $c$ and $7$ for $b$. $\cos \theta =\frac{7}{25}$ The ratio of $\tan \theta$ is $\tan \theta =\frac{a}{b}$ Substitute $24$ for $a$ and $7$ for $b$. $\tan \theta =\frac{24}{7}$ The ratio of $\csc \theta$ is $\csc \theta =\frac{c}{a}$ Substitute $25$ for $c$ and $24$ for $a$. $\csc \theta =\frac{25}{24}$ The ratio of $\sec \theta$ is $\sec \theta =\frac{c}{b}$ Substitute $25$ for $c$ and $7$ for $b$. $\sec \theta =\frac{25}{7}$ The ratio of $\cot \theta$ is $\cot \theta =\frac{b}{a}$ Substitute $24$ for $a$ and $7$ for $b$. $\cot \theta =\frac{7}{24}$ Therefore, the length of the side is $b=7$ and the values of the six trigonometric functions are, $\sin \theta =\frac{24}{25}$ , $\cos \theta =\frac{7}{25}$ , $\tan \theta =\frac{24}{7}$ , $\csc \theta =\frac{25}{24}$ , $\sec \theta =\frac{25}{7}$ and $\cot \theta =\frac{7}{24}$. |
Learn
We use number bases every day without realizing it. When we think about numbers in everyday life, we are almost always thinking of decimal numbers, or numbers in base 10, where each digit of a number can be one of ten values. People have ten fingers and ten toes (most of the time), so it seems natural to base a number system on ten. But, we are not required to use ten as a base. For example, in the movie Avatar, the Na’vi inhabitants have eight fingers and base their number system on eight because of that. Back on Earth, we see other bases use commonly in computing systems, such as binary (base 2) and hexadecimal (base 16).
When we wish to discuss non-ten bases, we use a leading notation like this: 0b1100111101 (binary for decimal 829). In common computer languages, we see, and we will use, this notation for the following bases:
• Binary (Base 2): leading 0b
• Octal (Base 8): leading 0o
• Decimal (Base 10): leading nothing (this is what we use in everyday life!)
The position of a number indicates its underlying value. The exponent starts at value zero and increases by one each time we move to the left. This statement is evident in the example.
For example, with 120, we have:
$1 \cdot 10^2 + 2 \cdot 10^1 + 0 \cdot 10^0$
Let’s revisit the pattern from right to left: ones, tens, hundreds, thousands. If we multiply it out, we go from left to right again: hundreds, tens, ones. Here are the exponents from left to right: 2, 1, 0.
In order to convert numbers in non-decimal bases to their decimal counterpart that we intuitively understand, we rely on the base number and the position of each digit. Starting from right to left, we compute the base conversion of that individual digit and add the decimal equivalent of that digit’s position.
Let’s try that with binary! Let’s look at 100 in binary, or 0b100. Remember, we’re working in base 2, so our equation will now be
$1 \cdot2^2 + 0 \cdot 2^1 + 0 \cdot 2^0$
This leaves us with 4 + 0 + 0, or 4. So 100 in binary is equal to 4 in base 10!
### Instructions
1.
Let’s say we have a number in base 10 with 5 digits, like 24601.
If we were to break this number into its components, what would be the value of the exponent for the digit 2
Assign the value to checkpoint_1 in the code editor.
2.
For any number, regardless of what base system we’re working in, what is the value of the exponent for the first digit (rightmost column)?
Assign this value to checkpoint_2 in the code editor. |
# Area of a Parallelogram
In this worksheet students will learn the formula for finding the area of a parallelogram and be able to find areas and solve simple problems.
Key stage: KS 4
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR
Curriculum topic: Geometry and Measures, Mensuration
Curriculum subtopic: Mensuration and Calculation, Area Calculations
Difficulty level:
### QUESTION 1 of 10
Parallelogram. Another 2 dimensional shape that we easily recognise, but can't always remember its name.
I bet you call it a wonky rectangle when you can't remember its proper name.
It has opposite parallel sides, which may help with remembering the name.
What examples can you think of that we see in everyday life?
Probably not so obvious to think about.
Believe it or not you can be two things at once.
For example a square has two sets of parallel sides, so is also a parallelogram but we call it a square.The same goes for a rectangle.
Basically a parallelogram is a four sided shape with their opposite sides equal in length.(the same as a square or rectangle)
Mathematicians can be awkward and make a four sided look different just for fun, so that's how we arrive at parallelogram.
There are all sorts of reasons we may need to know the area of a parallelogram
Before we can do any of this there is one thing we need to know and that is how to find the area of a parallelogram.
First of all we know that the sides of a parallelogram are not all the same length.
Two sides will have one length and the other two sides a different length.
As with a lot of shape work there is a formula that you need to learn. So here goes
Area of a parallelogram =base x height
Yes, really that is all there is to it.
You may recognise that it is the same as the formula for the area of a square and a rectangle.
Find the area of the parallelogram above.
Area = base x height
Area = 15 cm x 3 cm = 45 cm² .
As we are working with area units are always squared. e.g cm² m²
Look out. You need to work with the vertical height not the slanted height.
12 x 5 = 60 cm²
What a bargain - one formula for three shapes.
Find the area of this parallelogram.
14 cm²
36 cm²
60 cm²
64 cm²
Match these areas
## Column B
Parallelogram sides 4 cm and 3cm
30 cm²
Parallelogram sides 6 cm and 5 cm
22.26 m²
Parallelogram sides 11 cm and 3 cm
20 mm²
Parallelogram sides 8.5 cm and 7.3 cm
62.05 cm²
Parallelogram sides 4cm and 5 cm
33 cm²
Parallelogram sides 5.3 m x and 4.2 m
12 cm²
Find the area of this parallelogram.
A farmer has a parallelogram pen for his horses measuring 2.7m along one fence line and 212 cm along the other fence line. What is the area of the floor of the barn?
5724 m²
5.624 m²
5.724 m²
5.64 m²
What is the difference between a parallelogram with side lengths of 28 cm and 4 cm and a parallelogram of 37 by 5.5 cm?
Two new rugs in the shape of a parallelogram have been bought for this families living room.
One rug measures 4 m by 2 m. The other is 6m by 4m. What is the total area of the rugs?
30 m²
24 cm²
32 cm²
18 cm²
What is the area of these parallelograms?
80 cm² 18 cm² 84 cm² 19 cm² don't know Parallelogram A; Parallelogram B
This mosaic is going to have parallelogram tiles placed around it. The tiles are painted, Brown, beige and black.
The brown parallelograms have sides of 3 mm and 2 mm
The beige parallelograms have sides of 3 cm and 2.5 cm
The black parallelograms have sides of 3 cm and 1.5 cm.
4 brown, 3 beige and 2 black tiles are used. What is the total area of tiles used?
What is the area of this rectangle with sides 2x and x - 6
3x - 6
2x multiplied by x - 6
x - 6
2x² - 12x
What is the area of a parallelogram with sides 6 cm and 3x + 6.
15x
6x multiplied by 3x + 6
9x multiplied by 36x
9x multiplied by 36
• Question 1
Find the area of this parallelogram.
60 cm²
EDDIE SAYS
Did you apply the formula? Base x height When working with area of a parallelogram you are usually given both side lengths. Sometimes however there are sometimes three numbers given. Remember you need the vertical height. 12 x 5 = 60 cm²
• Question 2
Match these areas
## Column B
Parallelogram sides 4 cm and 3cm
12 cm²
Parallelogram sides 6 cm and 5 cm
30 cm²
Parallelogram sides 11 cm and 3 c...
33 cm²
Parallelogram sides 8.5 cm and 7....
62.05 cm²
Parallelogram sides 4cm and 5 cm
20 mm²
Parallelogram sides 5.3 m x and 4...
22.26 m²
EDDIE SAYS
Applying the formula is now getting easy. Make sure you read both numbers carefully 4 x 3 = 12 cm² 5 x 6 = 30 cm² 11 x 3 = 33 cm² 8 .5 x 7.3 = 62.05 cm² 4 x 5 = 20 cm² 5.2 x 4.2 = 22.26 cm² You're usually asked to round to 1 or 2 decimal places.
• Question 3
Find the area of this parallelogram.
14
EDDIE SAYS
There is nothing to it is there. 2 x 7 = 14 cm² Providing you select the correct numbers you will never go wrong.
• Question 4
A farmer has a parallelogram pen for his horses measuring 2.7m along one fence line and 212 cm along the other fence line. What is the area of the floor of the barn?
5724 m²
5.724 m²
EDDIE SAYS
Often area questions will give two different units of measurement. You will need your eagle eye for this. You can convert either way. 2.7 m = 270 cm 270 x 212 =57,240 m² or 212 cm = 2.12 m 2.7 x 2.12 =5.724 (5.73 to 3 d.p) Nay bother for you.
• Question 5
What is the difference between a parallelogram with side lengths of 28 cm and 4 cm and a parallelogram of 37 by 5.5 cm?
91.5
EDDIE SAYS
Just another example of finding area from a written question. 28 x 4 = 112 cm² 37 x 5.5 = 203.5 cm² 203.5 - 112 = 91.5 cm² The word difference in a question should lead you to subtract.
• Question 6
Two new rugs in the shape of a parallelogram have been bought for this families living room.
One rug measures 4 m by 2 m. The other is 6m by 4m. What is the total area of the rugs?
32 cm²
EDDIE SAYS
This is just a simple 2 step question. You have all the knowledge that you need. Area of rug 1 = 4 x 2 = 8 m² Area of rug 2 = 6 x 4 = 24 m² Add the two together 8 + 24 = 32 m²
• Question 7
What is the area of these parallelograms?
80 cm² 18 cm² 84 cm² 19 cm² don't know Parallelogram A; Parallelogram B
EDDIE SAYS
Did you remember the vertical height is required? Without it we can't find the area. We can only find the area of rectangle A. 10 x 8 = 80 cm²
• Question 8
This mosaic is going to have parallelogram tiles placed around it. The tiles are painted, Brown, beige and black.
The brown parallelograms have sides of 3 mm and 2 mm
The beige parallelograms have sides of 3 cm and 2.5 cm
The black parallelograms have sides of 3 cm and 1.5 cm.
4 brown, 3 beige and 2 black tiles are used. What is the total area of tiles used?
55.5
EDDIE SAYS
Find the area of each coloured parallelogram. Multiply it up by the number required for the mosaic. Add them all together. Brown 3 x 2 = 6 x 4 = 24 Beige 3 x 2.5 = 7.5 x 3 = 22.5 Black 3 x 1.5 =4.5 x 2 = 9 24 + 22.5 + 9 = 55.5 cm²
• Question 9
What is the area of this rectangle with sides 2x and x - 6
2x multiplied by x - 6
2x² - 12x
EDDIE SAYS
I know what you are thinking..eeekkk. This is easier than it looks. We know that 1 side is length 2x The other side is length x -6 We don't know what x is so all we can write is 2x multiplied by x - 6 or if you want to show of 2x times x = 2x² 2x times -6 = -12x which is 2x² -12x
• Question 10
What is the area of a parallelogram with sides 6 cm and 3x + 6.
6x multiplied by 3x + 6
9x multiplied by 36
EDDIE SAYS
Bring it on!! 6 multiplied by 3x + 6. (one side multiplied by the other) or showing off again 9x + 36 If you don't know how I got 9x + 36. Look at multiplying out brackets.
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# 8.3: Boosts and Rotations
Skills to Develop
• Explain rotations and boosts
A relative of mine fell in love. She and her boyfriend bought a house in the suburbs and had a baby. They think they’ll get married at some later point. An engineer by training, she says she doesn’t want to get hung up on the “order of operations.” For some mathematical operations, the order doesn’t matter: $$5 + 7$$ is the same as $$7 + 5$$.
# Rotations
Fgure $$\PageIndex{1}$$ shows that the order of operations does matter for rotations. Rotating around the $$x$$ axis and then $$y$$ produces a different result than $$y$$ followed by $$x$$. We say that rotations are noncommutative. This is why, in Newtonian mechanics, we don’t have an angular displacement vector $$∆θ$$; vectors are supposed to be additive, and vector addition is commutative. For small rotations, however, the discrepancy caused by choosing one order of operations rather than the other becomes small (of order $$θ^2$$), so we can define an infinitesimal displacement vector $$dθ$$, whose direction is given by the right-hand rule, and an angular velocity $$ω = dθ/dt$$.
As an example of how this works out for small rotations, let’s take the vector
$(0,0,1)$
and apply the operations shown in Figure $$\PageIndex{1}$$, but with rotations of only $$θ = 0.1$$ radians rather than $$90$$ degrees. Rotation by this angle about the $$x$$ axis is given by the transformation
$(x,y,z) → (x,y \cos θ - z \sin θ,y \sin θ + z \cos θ)$
and applying this to the original vector gives this:
$(0.00000,-0.09983,0.99500)\; \; \; (\text{after }x)$
After a further rotation by the same angle, this time about the $$y$$ axis, we have
$(0.09933,-0.09983,0.99003) \; \; \; (\text{after }x, \text{then }y)$
Starting over from the original vector in Figure $$\PageIndex{1.1}$$ and doing the operations in the opposite order gives these results:
$(0.09983,0.00000,0.99500) \; \; \; (\text{after }y)$
$(0.09983,-0.09933,0.99003) \; \; \; (\text{after }y, \text{then }x)$
The discrepancy between ($$\PageIndex{1.3}$$) and ($$\PageIndex{1.5}$$) is a rotation by very nearly $$0.005$$ radians in the $$xy$$ plane. As claimed, this is on the order of $$θ^2$$ (in fact, it’s almost exactly $$θ^2/2$$). A single example can never prove anything, but this is an example of the general rule that rotations along different axes don’t commute, and for small angles the discrepancy is a rotation in the plane defined by the two axes, with a magnitude whose maximum size is on the order of $$θ^2$$.
# Boosts
Something similar happens for boosts. In $$3 + 1$$ dimensions, we start with the vector
$(0,1,0,0)$
pointing along the $$x$$ axis. A Lorentz boost with $$v = 0.1$$ (Equation 1.4.1) in the $$x$$ direction gives
$(0.10050,1.00504,0.00000,0.00000) \; \; \; (\text{after }x)$
and a second boost, now in the $$y$$ direction, produces this:
$(0.10101,1.00504,0.01010,0.00000) \; \; \; (\text{after }x, \text{then }y)$
Starting over from ($$\PageIndex{6}$$) and doing the boosts in the opposite order, we have
$(0.00000,1.00000,0.00000,0.00000) \; \; \; (\text{after }y)$
$(0.10050,1.00504,0.00000,0.00000) \; \; \; (\text{after }y, \text{then }x)$
The discrepancy between ($$\PageIndex{8}$$) and ($$\PageIndex{10}$$) is a rotation in the $$xy$$ plane by very nearly $$0.01$$ radians. This is an example of a more general fact, which is that boosts along different axes don’t commute, and for small angles the discrepancy is a rotation in the plane defined by the two boosts, with a magnitude whose maximum size is on the order of $$v^2$$, in units of radians.
# Thomas Precession
Figure $$\PageIndex{2}$$ shows the most important physical consequence of all this. The gyroscope is sent around the perimeter of a square, with impulses provided by hammer taps at the corners. Each impulse can be modeled as a Lorentz boost, notated, e.g., $$L_x$$ for a boost in the $$x$$ direction. The series of four operations can be written as $$L_yL_xL_{-y}L_{-x}$$, using the notational convention that the first operation applied is the one on the right side of the list. If boosts were commutative, we could swap the two operations in the middle of the list, giving $$L_yL_{-y}L_xL_{-x}$$. The $$L_x$$ would undo the $$L_{-x}$$, and the $$L_y$$ would undo the $$L_{-y}$$. But boosts aren’t commutative, so the vector representing the orientation of the gyroscope is rotated in the $$xy$$ plane. This effect is called the Thomas precession, after Llewellyn Thomas (1903-1992). Thomas precession is a purely relativistic effect, since a Newtonian gyroscope does not change its axis of rotation unless subjected to a torque; if the boosts are accomplished by forces that act at the gyroscope’s center, then there is no nonrelativistic explanation for the effect.
Clearly we should see the same effect if the jerky motion in Fgure $$\PageIndex{2}$$ was replaced by uniform circular motion, and something similar should happen in any case in which a spinning object experiences an external force. In the limit of low velocities, the general expression for the angular velocity of the precession is $$Ω = a×v$$, and in the case of circular motion, $$Ω = \frac{1}{2}v^2ω$$, where $$ω$$ is the frequency of the circular motion.
If we want to see this precession effect in real life, we should look for a system in which both $$v$$ and $$a$$ are large. An atom is such a system. The Bohr model, introduced in 1913, marked the first quantitatively successful, if conceptually muddled, description of the atomic energy levels of hydrogen. Continuing to take $$c = 1$$, the over-all scale of the energies was calculated to be proportional to $$mα^2$$, where $$m$$ is the mass of the electron, and $$α$$ is the fine structure constant, defined earlier. At higher resolution, each excited energy level is found to be split into several sub-levels. The transitions among these close-lying states are in the millimeter region of the microwave spectrum. The energy scale of this fine structure is $$∼ mα^4$$. This is down by a factor of $$α^2$$ compared to the visible-light transitions, hence the name of the constant. Uhlenbeck and Goudsmit showed in 1926 that a splitting on this order of magnitude was to be expected due to the magnetic interaction between the proton and the electron’s magnetic moment, oriented along its spin. The effect they calculated, however, was too big by a factor of two.
The explanation of the mysterious factor of two had in fact been implicit in a 1916 calculation by Willem de Sitter, one of the first applications of general relativity. De Sitter treated the earth-moon system as a gyroscope, and found the precession of its axis of rotation, which was partly due to the curvature of spacetime and partly due to the type of rotation described earlier in this section. The effect on the motion of the moon was noncumulative, and was only about one meter, which was much too small to be measured at the time. In 1927, however, Thomas applied similar reasoning to the hydrogen atom, with the electron’s spin vector playing the role of gyroscope. Since the electron’s spin is $$\hbar /2$$, the energy splitting is $$\pm (\hbar /2) \Omega$$, depending on whether the electron’s spin is in the same direction as its orbital motion, or in the opposite direction. This is less than the atom’s gross energy scale $$\hbar \omega$$ by a factor of $$v^2/2$$, which is $$∼ α^2$$. The Thomas precession cancels out half of the magnetic effect, bringing theory in agreement with experiment.
Uhlenbeck later recalled: “...when I first heard about [the Thomas precession], it seemed unbelievable that a relativistic effect could give a factor of 2 instead of something of order $$v/c$$ ... Even the cognoscenti of relativity theory (Einstein included!) were quite surprised. |
# Todays Plan C1 Marks – Class Average: 76% New Lesson –Midpoint (6.4) –Division Point (6.4) –Scalar Product (6.5) Homework Questions PRACTICE.
## Presentation on theme: "Todays Plan C1 Marks – Class Average: 76% New Lesson –Midpoint (6.4) –Division Point (6.4) –Scalar Product (6.5) Homework Questions PRACTICE."— Presentation transcript:
Todays Plan C1 Marks – Class Average: 76% New Lesson –Midpoint (6.4) –Division Point (6.4) –Scalar Product (6.5) Homework Questions PRACTICE
Midpoint of a Line Segment The coordinate of M are: M B A
Properties of Midpoint of a Line Segment 1.MA + MB = 0 2.AM = MB 3.AM = ½AB 4.AB = 2AM M B A
Division point of a Line Segment Point P divides the line: –in a 3:2 ratio or 3/2 from A or –in a 2:3 ration or 2/3 from B P B A 2 3
Division point of a Line Segment We can also say: P is located at 3/5 from point A or P is located at 2/5 from point B P B A 2 3
Division point of a Line Segment The vector equality is: AP = 3/5 AB or BP = 2/5 BA P B A 2 3
Example 1 Consider A(1,4) and B(7,1), find P if AP = AB AB = (7-1,1-4) = (6,-3) AP = (x P -x A, y P -y A ) We know that AP = AB so fill in the other info… (x P - 1, y P - 4) = (6,-3)
Example 1 Consider A(1,4) and B(7,1), find P if AP = AB (x P - 1, y P - 4) = (6,-3) (x P - 1, y P - 4) = (4,-2) Look at x and y individually… x P – 1 = 4 and y P – 4 = -2 x P = 4 +1 and y P = -2 +4 x P = 5 and y P = 2 So point P is (5,2)
SCALAR PRODUCT The scalar product of two vectors u and v is a real number defined by: uv = ||u||×||v||×cosӨ u Ө v
SCALAR PRODUCT 1.When vectors u and v are orthogonal, their scalar product is ZERO 2.uu = ||u||×||u||×cos0˚ = ||u|| 2
Example 2 Consider u and v form a 30˚ angle and that ||u||=4 and ||v||=3. Find the scalar product. uv = 4 x 3 x cos30 uv = 10.39
SCALAR PRODUCT IN THE CARTESIAN PLANE Let u=(a,b) and v=(c,d), we have: uv = ac + bd If we know the angle Ө formed by vectors u and v, we have: cosӨ = uv _ ||u|| ||v||
Example 3 Let u=(4,1) and v=(2,3). Find the scalar product and the angle between the vectors uv = 4x2 + 1x3 uv = 11 What are ||u|| and ||v||? ||u||=4 2 +1 2 = 17 ||v|| = 2 2 +3 2 = 13 cosӨ = uv _ ||u|| ||v||
Example 3 Let u=(4,1) and v=(2,3). Find the scalar product and the angle between the vectors cosӨ = uv _ ||u|| ||v|| cosӨ = 11 _ = 11 = 0.74 17 x 13 221 So Ө = cos -1 (0.74) Ө = 42.3˚
PROPERTIES OF THE SCALAR PRODUCT uv = vu rusv = (rs)uv u(v+w) = uv + uw
Last homework questions? (not the quiz questions…)
HOMEWORK p. 300 #25, 26 p. 301 #27, 28, 30 p. 303 #1,2,3 p. 304 #6,7,8,9 (how do you know if two vectors are perpendicular? Scalar product is 0!) p.306-307 try them all TAKE HOME QUIZ DUE TOMORROW
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##### 6.1 Multiplying fractions and whole numbers
It’s pizza night, your friend is coming over to share it with you. She insists to split the bill based on the amount of slices that she would be eating. Your sister also wants to have at most two slices. You can eat up to 3 slices and your buddy can eat up to 4 slices. You quickly get a pad to write your shares down, 4/10, 3/10, 2/10. How much would each of you need to pay if the pizza is \$5?
Solving this problem would require you to know about fractions. The problem is quite easy to solve as long as you know the basic operations with fractions. This is why in this chapter, we will be talking all about fractions.
Fractions represent a part of a whole. Fraction is a word from Latin meaning “broken”, There two types of fractions, the proper fractions and improper fractions. Proper fractions are those that have a numerator that is smaller than the denominator; while the improper fraction, on the other hand, has a numerator that is larger than the denominator. Then, there’s also the mixed numbers which refers to the combination of both a whole number and a fraction like 5 ½, 39 4/5 and so on.
Examples of proper fractions: $\frac{2}{3}$ , $\frac{7}{{10}}$ , $\frac{{31}}{{40}}$
Examples of improper fractions: $\frac{6}{5}$, $\frac{{12}}{{11}}$, $\frac{{73}}{{45}}$
In the first part of the chapter, we will be looking at how to multiply whole numbers and fractions. In cases when we want to know how much ¼ of 20 or 5/6 of 75 is, we can be able to just multiply the two given numbers to solve the problem. Multiplying whole numbers and fractions would just involve multiplying the numerator to the whole number and simplifying the answer like this:
Example: $\frac{1}{4}\; \times 20$
Step 1: Multiply the whole number to the numerator (the number on top)
$\frac{{1 \times 20}}{4} = \;$ $\frac{{20}}{4}$
Step 2: Simplify the fraction
$\frac{{20}}{4} = \frac{5}{1} = 5$
In the second part, we will be dealing with the question of how to divide fractions with whole numbers, like 2/3 $\div$ 48 or 4/7 $\div$ 36. You simply need to turn the whole number which is your dividend into a fraction by putting 1 as its denominator. Then you get the reciprocal of the whole number turned into fraction, multiply it with the divisor which is the other fraction and then simplify the quotient if needed.
Example: $\frac{2}{3} \div 48$
Step 1: Turn the whole number into a fraction
$\frac{2}{3} \div \frac{{48}}{1}$
Step 2: Turn the second fraction upside down; and turn the division sign into multiplication sign
$\frac{2}{3} \times \frac{1}{{48}}$ $\leftarrow$ reciprocal
Step 3: Simplify the fractions
Step 4: Multiply the fractions
$\frac{1}{3} \times \frac{1}{{24}} = \;\frac{1}{{72}}$
For the third part, we will be looking at how to multiply proper fractions like ½ x ¾ or 7/10 x 5/6. The process is very much the same with how you multiply a whole number and a fraction without the part where you need to convert the whole number into a fraction.
Example: $\frac{1}{3} \times \frac{3}{5}$
Step 1: Multiply the fractions
$\frac{{1 \times 3}}{{3 \times 5}} = \;\frac{3}{{15}}$
Step 2: Simplify the fraction
$\frac{3}{{15}} = \frac{1}{5}$
The fourth part of this chapter will discuss about multiplying improper fractions and mixed numbers, a.k.a. mixed fractions. The process is similar to multiplying a proper fraction like the example given above except that the mixed numbers need to be converted into a fractional form.
Example: $\frac{4}{3} \times 2\frac{2}{5}$
Step 1: Turn the mixed numbers into improper fraction
$\frac{4}{3} \times \frac{{12}}{5}$
Step 2: Multiply the fractions
$\frac{{4\; \times \;12}}{{3\; \times \;5}} = \;\frac{{48}}{{15}}$
Step 3: Simplify the fraction
Step 4: Turn the improper fraction into mixed number (optional)
$\frac{{16}}{5} = 3\frac{1}{5}$
For the last two parts of this chapter, we will be learning how to divide fractions and mixed numbers and looking at how to use the things we learn about fractions in different math problems involving fraction operations.
Example: $1\frac{1}{5}\; \div \frac{2}{3}$
Step 1: Turn the mixed number to improper fraction
$\frac{6}{5} \div \frac{2}{3}$
Step 2: Turn the second fraction upside down; and turn the division sign into multiplication sign
$\frac{6}{5} \times \frac{3}{2}$
Step 3: Simplify the fractions
Step 4: Multiply the fractions
$\frac{{3\; \times 3}}{{5\; \times \;2}} = \frac{9}{{10}}$
We will be learning how to solve fractions operations by following the order of operations to get the correct answers. At the end of this chapter you can try to solve the pizza problem above and also try playing the free fraction games online.
### Multiplying fractions and whole numbers
We learned previously that whole numbers can be written We learned previously that whole numbers can be written as fractions with 1 as the denominator and the whole number as the numerator. To make the calculation easier, we can first make the whole numbers into fraction when we multiply whole numbers with fractions. By doing so, we turn the questions into multiplying fractions only. |
# 8th Class Mathematics Statistics Classification of Data
## Classification of Data
Category : 8th Class
### Classification of Data
In order to tabulate a large number of data, we use the frequency distribution table. Frequency distribution is of two types:
1. Discrete Frequency Distribution
2. Continuous Frequency Distribution
In discrete frequency distribution method the frequency distribution is carried out with the help of raw data using the tally marks. But in continuous frequency distribution the data is divided into small groups of class interval and corresponding frequency is found. The frequency of a data is defined as the number of times a data is repeated in the given collection of data. The small groups into which the given data is divided is known as its class interval. The difference between the upper limit and lower limit of a class interval is known as its class size.
The class mark is defined as the average of the upper limit and lower limit of a class interval. The cumulative frequency is defined as the sum of all previous frequencies of the class interval.
Class Interval Frequency 0 - 5 4 5 - 10 10 10 - 15 18 15 - 20 8 20 - 25 6
Here, 0 - 5, 5 - 10, ---are class intervals.
The difference 5 - 0 = 5 is the class size.
The average of the class interval is called class mark i.e. $\frac{0+5}{2}=\frac{5}{2}=2.5$
The cumulative frequency of class 5 - 10 is 4 + 10 = 14
There are various methods of representation of data. It can be represented in the form of bar graph. Histogram, ogive curve, frequency polygon curve or pie chart.
Frequency Distribution Table
The frequency distribution table for a certain given data can be as shown below: It consists of class interval, tally marks and frequency.
Class Interval Tally Marks Frequency
Frequency Distribution Table for an Ungrouped Data
Construct a frequency distribution table for the following data: $\text{5},\text{1},\text{ 3},\text{ 4},\text{ 2},\text{1},\text{ 3},\text{ 5},\text{ 4},\text{ 2},\text{1},\text{ 5},\text{1},\text{ 3},\text{ 2},\text{1},\text{ 5},\text{ 3},\text{ 3},\text{ 2}$
Solution:
Number Tally Marks Frequency 1 $\bcancel{||||}$ 5 2 $||||$ 4 3 $||$ 2 4 $\bcancel{||||}$ 5 5 $||||$ 4 6 $||$ 2 Total 22
The following are the marks obtained by 50 students in mathematics in their previous examination held in their school. Prepare a frequency distribution table for the data.
45, 68, 41, 87, 61, 44, 67, 30, 54, 8, 39, 60, 37, 50, 19, 6, 42, 29, 32, 61, 25, 77, 62, 98, 47, 36,15, 40, 9, 25, 34, 50, 61, 75, 51, 96, 20, 13, 18, 35, 43, 88, 25, 95, 68, 81, 29, 41, 45, 87
Solution:
To decide the length of the class interval and to take all the scores given in the problem. We find the largest value and the smallest value from the given marks scored by the various students. We can do this by merely going through all the scores. In the data given above the largest value is 98 and the smallest value is 8.
Difference = Largest value - Smallest value
$=\text{98}-\text{8 }=\text{ 9}0$
Since the difference between the largest and the smallest value is 90, so if we take class intervals of size 5 then it will be as shown : 0 - 5, 5 - 10, 10 - 15, ..., 95 100. On the other hand if we take the class interval of size 10 then it will be as shown: 0 - 10, 10 - 20, 20 - 30,...... 90 - 100. Or if we take the class interval of class size 15 then it becomes: 0 - 15, 15 - 30, ....,90 - 105. It is advisable not to over reduce the number of class intervals. If the class intervals are mentioned in the problem then we proceed as given in the problem.
For the above proceed we prepare the frequency distribution table by taking class intervals as: 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50, 50 - 60, 60 - 70, 70 - 80, 80-90 and 90-100.
Frequency Distribution Table of the Marks taken by 50 Students in a Mathematics Test
Class Intervals Tally Marks Frequency 0 - 10 $|\,|$ 2 10 - 20 $|\,|\,|\,|$ 4 20 - 30 $\bcancel{|\,|\,|\,|}\,\,|$ 6 30 - 40 $\bcancel{|\,|\,|\,|}\,\,||$ 7 40 - 50 $\bcancel{|\,|\,|\,|}\,\,||||$ 9 50 - 60 $||||$ 4 60 - 70 $\bcancel{|\,|\,|\,|}\,\,|||$ 8 70 - 80 $||$ 2 80 - 90 $\bcancel{|\,|\,|\,|}$ 5 90 - 100 3 Total 50
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# 2015 AMC 10B Problems/Problem 11
## Problem
Among the positive integers less than $100$, each of whose digits is a prime number, one is selected at random. What is the probability that the selected number is prime?
$\textbf{(A)} \dfrac{8}{99}\qquad \textbf{(B)} \dfrac{2}{5}\qquad \textbf{(C)} \dfrac{9}{20}\qquad \textbf{(D)} \dfrac{1}{2}\qquad \textbf{(E)} \dfrac{9}{16}$
## Solution 1
The one digit prime numbers are $2$, $3$, $5$, and $7$. So there are a total of $4\cdot4=16$ ways to choose a two digit number with both digits as primes and $4$ ways to choose a one digit prime, for a total of $4+16=20$ ways. Out of these $2$, $3$, $5$, $7$, $23$, $37$, $53$, and $73$ are prime. Thus the probability is $\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}$.
## Solution 2 (Listing)
Since the only primes digits are $2$, $3$, $5$, and $7$, it doesn't seem too hard to list all of the numbers out.
• 2- Prime;
• 3- Prime;
• 5- Prime;
• 7- Prime;
• 22- Composite;
• 23- Prime;
• 25- Composite;
• 27- Composite;
• 32- Composite;
• 33- Composite;
• 35- Composite;
• 37- Prime;
• 52- Composite;
• 53- Prime;
• 55- Composite;
• 57- Composite;
• 72- Composite;
• 73- Prime;
• 75- Composite;
• 77- Composite.
Counting it out, there are $20$ cases and $8$ of these are prime. So the answer is $\dfrac{8}{20}=\boxed{\textbf{(B)} \dfrac{2}{5}}$. ~JH. L
## Video Solution 1
~Education, the Study of Everything
~savannahsolver |
# Graph of a function
A graph is a visual representation of a function. If $f(x) = y$ then the point $(x,y)$ lies on the graph of $f$.
## Graphing Points
A single point is the simplest thing to graph. The graph of $(2,5)$ would be a dot 2 units to the right of $y$-axis and 5 units above the $x$-axis.
## Graphing Lines
Given two distinct points on a line, one can construct the whole line. So one way to graph a line given its equation is to just find two points on it and to draw a straight line through them.
### Problem
Graph the line $2x + 3y = 24$.
### Solution
To graph a line, it is necesasry to find two points $(x,y)$ that satisfy $2x + 3y = 24$. Letting $x=0$ gives $3y = 24\Leftrightarrow y = 8$. So $(0,8)$ is one point on the graph.
Find another point by letting $y=0$. Plugging this in and solving gives $x=12$. So $(12,0)$ is our other point.
Now plot these in the coordinate plane and draw a line through them:
The arrowheads on the ends of the line segment indicate that the line goes on infinitely in both directions.
## Graphing Polynomials
The first step in graphing a polynomial, $p(x)$, is to find the zeros of $p(x)=0$. Then a smooth curve should be drawn through the zeros accounting for multiple roots and making sure the signs match up (i.e. the graph is above the $x$-axis when the polynomial is positive and below it when the polynomial is negative). This process is best understood through examples.
### Problem
Graph the parabola $y = 2x^{2} + x - 3$.
### Solution
The quadratic equation can be written as $(2x+3)(x-1)$ making the roots $x=-\frac{3}{2}$ and $x=1$. Since the coefficient of the term with the highest power (in this case $x^2$) is $2>0$, the graph is above the $x$-axis for $(-\infty, -\frac{3}{2})$ and $(1, +\infty)$ and below the $x$-axis for $(-\frac{3}{2}, 1)$. This allows the graph to be drawn as a smooth curve curve through the zeros using this information as a guideline:
### Problem
Graph $y = x^4 - 2x^3 -7x^2 +20x -12$.
### Solution
First, we need to find the zeros of the function. Notice that if $x=1$ or $x=2$, $y=0$. Hence, the polynomial reduces to $y=(x-1)(x-2)(x^2+x-6)$. Factoring the quadratic gives $(x-1)(x-2)^2(x+3)$. So the roots are $1$ and $-3$ and a double root at $2$. The final graph looks like: |
Algebra Study Hall
## Simplifying polynomials
“Polynomials” are expressions with two or more (poly) types of numbers (nomials). An example of a polynomial expression would be:
$(6x^2+5x+3)+(x^2+3x-2)$
To simplify this expression, we need to first consider PEMDAS. Within each set of parentheses, nothing can be reduced or combined. Also, no exponents can be computed since we don’t know what x stands for—it’s already reduced as far as we take reduce it. There is no multiplication or division. There’s only addition and subtraction. So, we must now “combine like terms.” Like terms are the simply the ones “like each other.” You can ask, “How many x-squareds are there?” Well, there are 6 x-squares plus 1 more x-squared, which makes 7 x-squared. “How many x’s are there?” There are $5x$’s plus $3x$’s, to make $8x$’s. And looking at the numbers only, 3-2 is 1. So, our simplified polynomial is:
$7x^2+8x+1$
## Factoring binomials using F.O.I.L.
Factoring binomials means to multiply binomials (expressions with two types of numbers). For example, this type of problem might be:
$(x+3)(x-1)$
This type of problem is solved through the “F.O.I.L.” method. F.O.I.L. stands for “First, Outer, Inside, and Last.” Using the problem above, let’s F.O.I.L.:
• First: $x \times x$ yields $x^2$
• Outer: $x \times -1$ yields $-1x$ or just $-x$
• Inner: $3 \times x$ yields $3x$
• Last: $3 \times -1$ yields $-3$
$x^2-x+3x-3$
Lastly, we combine like terms. The only terms able to be combined are the $-x$ and the $3x$ which gives us $2x$. So, our final answer is:
$x^2+2x-3$
## Factoring trinomials
This implies doing just the opposite of what we just did. It’s like taking the expression $x^2+2x-3$ and working backwards to find the factors: $(x+3)(x-1)$
Consider this example for factoring trinomials:
$x^2-9x+20$
To find the factors for this trinomial:
• Find factors for $x^2$ : $x$ times $x$ or $(x)(x)$
• Find factors for 20: $1 \times 20$, $2 \times 10$, and $4 \times 5$
• Find the pair with a sum of -9. -4 and -5 have a sum of -9
• Complete the factors: $(x-4)(x-5)$
• Double check using F.O.I.L. Be sure the numbers are correct and the positives and negatives are correct. Using F.O.I.L., you’ll get back to $x^2-9x+20$ and know your answer is correct.
## Dividing Polynomials
Dividing polynomials can appear intimidating. Don’t get psyched out! The basic process is the same as regular old long division that you did in grade school. Consider the division problem of…
It’s a binomial (x + 1) dividing into a polynomial (the parts inside the division sign). Look at the graphic following this sentence for an explanation to solve the problem.
## Handling square roots
It’s necessary to be able to simplify, multiply and divide square roots.
Simplifying square roots:
Suppose you have an answer of $\sqrt {2}$, you’d need to simplify it on down to its most basic form. To simplify square roots, think of any perfect squares that are factors of the number inside the square root symbol. In other words, do 4, 9, 25, 36 etc. divide evenly into the number inside the symbol? If so, it needs to be simplified. In this case, it could’ve been written $\sqrt{9 \times 2}$ So, we need to “pull out” the 9. The square root of 9 is 3, so our final and most simplified answer would be $3 \sqrt{2}$
Multiplying square roots:
Multiplying square roots is simple—you multiply the numbers under the square root sign, then look to simplify. Consider the problem: $\sqrt{2} \times \sqrt{8}$. Just multiply $2 \times 8$ to get $\sqrt {16}$ The square root of 16 is 4, the answer.
Dividing square roots:
Consider the problem: $\sqrt{27} \over \sqrt{12}$
First simplify by “pulling out” the 9 in the numerator and the 4 in the denominator. The problem then becomes: $3 \sqrt{3} \over 2\sqrt{3}$ Now simplify. The two $\sqrt{3}$ factors cancel each other out, so you could just scratch them and you’re left with your answer: $3/2$
## Distance between two points in a plane
Algebra is often called “linear algebra” because the equations can be graphed on a plane, that is to say, on graph paper. This skill requires you to, when given to points on a plane, calculate the distance between those two points.
There are two methods of finding the distance between two points in a plane:
1. Graphically – you can draw the graph to visualize it.
2. Formulaically – you can use the distance formula shown below.
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
We’ll use both methods to illustrate how to compute the distance.
1. Graphically solving the problem:
Consider the example: Find the distance between point A at (-2, 6) and point B at (5,3). Refer to the graphic below to solve this problem.
2. Formulaically solving the problem:
The formula for computing the distance between two points on a graph is built on the same principles that we used in the problem above. The formula is:
$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Consider the same two points that we used up above: point A (-2,6) and point B (5,3). Placing the x and y coordinates into the formula gives us:
$d=\sqrt{(5- -2)^2+(3-6)^2}$
Which becomes…
$d=\sqrt{7^2+(-3)^2}$
Which becomes…
$d=\sqrt{49+9}$
Which becomes…
$d=\sqrt{58}$ or $d\approx 7.62$
## Ratios or proportions
Ratios or proportions are essentially the same thing. They show how numbers relate to one another. Ratios and proportions can be shown a couple of ways:
• In numbers such as 3:1. This says “three to one.”
• As fractions such as $3 \over 1$ This fraction is the same as writing 3:1. It is read as, "A ratio of 3 to 1."
• Or in words such as there are 3 teaspoons to every 1 tablespoon. It’s still “three to one.”
Often, questions on standardized tests have you solve a simple proportion problem. It may go something like this:
It took 8.5 bags of mulch to cover a playground with 500 square feet of area. How many bags of mulch would it take to cover an 800 square foot playground?
This may appear intimidating, but it’s very easy. It’s really saying 8.5 is to 500, as what is to 800? First, let “what is” be $x$ Therefore, to define the variable, $x$ = the number of bags to cover 800 square feet. Next, we put the problem in the form of a proportion—that is to say, we put it in fraction form and set them equal to one another. So, we now have:
$8.5 \over500$ = $x \over 800$
This now is read as: “8.5 is to x, as 500 is to 800.”
To solve a proportion like this, we cross-multiply and set each “line of the cross” equal to each other. So $8.5(800)=500x$ This gives us $6800=500x$ Divide both sides by 500 to get the answer, 13.6 bags.
• As a note, on a question like this, the answer will likely be “14 bags” not 13.6 because you can’t buy a “.6” bag.
page revision: 15, last edited: 12 Jul 2012 13:48
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# Find the determinant of the matrix using Gaussian elimination: \begin {bmatrix} 0&1&2 \\-1&1&3...
## Question:
Find the determinant of the matrix using Gaussian elimination: {eq}\begin {bmatrix} 0&1&2 \\-1&1&3 \\2&-2&0 \end {bmatrix} {/eq}
## Gaussian Elimination
Gaussian elimination is a technique to solve a system of linear equations by eliminating the unknowns from the equations in such a manner that the matrix of the system is an upper triangular matrix.
We can use this technique to find the determinant of a matrix, by finding an equivalent matrix which is upper triangular.
We need to know how the elementary row operations change the determinant, for example, swapping two rows multiplies the determinant by negative one, and multiplying a row by a nonzero scalar multiplies the determinant by the same scalar.
However, adding to one row a scalar multiple of another does not change the determinant.
To find the determinant of the square matrix {eq}\displaystyle \begin{align} A=\begin {bmatrix} 0&1&2 \\-1&1&3 \\2&-2&0 \end {bmatrix} \end{align} {/eq} using Gaussian elimination,
we will do row operations on the matrix and keep track of swapping rows or multiplying a row with a scalar.
We will use the notation {eq}\displaystyle R_i {/eq} for the {eq}\displaystyle i- {/eq}th row and describe the operations on the rows on top of the equivalent symbol.
{eq}\displaystyle \begin{align} det(A)=\left|\begin{array}{ccc} 0 & 1 &2\\ -1&1&3 \\ 2&-2&0 \end{array}\right| \overset{R_1 \text{ swap with }R_3}{\iff} -\left|\begin{array}{ccc} 2&-2&0\\ -1&1&3 \\ 0 & 1 &2 \end{array}\right| \overset{R_2+\frac{1}{2}\cdot R_1}{\iff} -\left|\begin{array}{ccc} 2&-2&0\\ 0&0&3 \\ 0 & 1 &2 \end{array}\right|\overset{R_2 \text{ swap with }R_3}{\iff} \left|\begin{array}{ccc} 2&-2&0\\ 0 & 1 &2\\ 0&0&3 \end{array}\right| =2\cdot 1\cdot 3=\boxed{6}. \end{align} {/eq}
We obtained an upper triangular matrix, meaning we have zeros below the diagonal terms, so the determinant is the product of the diagonal terms.
Also, when we interchanged two rows, the determinant is multiplied with negative 1 and since we interchanged twice two rows, the determinant did not change. |
# Radix Sort – Data Structures and Algorithms Tutorials
Radix Sort is a linear sorting algorithm that sorts elements by processing them digit by digit. It is an efficient sorting algorithm for integers or strings with fixed-size keys.Â
Rather than comparing elements directly, Radix Sort distributes the elements into buckets based on each digit’s value. By repeatedly sorting the elements by their significant digits, from the least significant to the most significant, Radix Sort achieves the final sorted order.
## Radix Sort Algorithm
The key idea behind Radix Sort is to exploit the concept of place value. It assumes that sorting numbers digit by digit will eventually result in a fully sorted list. Radix Sort can be performed using different variations, such as Least Significant Digit (LSD) Radix Sort or Most Significant Digit (MSD) Radix Sort.
## How does Radix Sort Algorithm work?
To perform radix sort on the array [170, 45, 75, 90, 802, 24, 2, 66], we follow these steps:
How does Radix Sort Algorithm work | Step 1
Step 1: Find the largest element in the array, which is 802. It has three digits, so we will iterate three times, once for each significant place.
Step 2: Sort the elements based on the unit place digits (X=0). We use a stable sorting technique, such as counting sort, to sort the digits at each significant place.
Sorting based on the unit place:
• Perform counting sort on the array based on the unit place digits.
• The sorted array based on the unit place is [170, 90, 802, 2, 24, 45, 75, 66].
How does Radix Sort Algorithm work | Step 2
Step 3: Sort the elements based on the tens place digits.
Sorting based on the tens place:
• Perform counting sort on the array based on the tens place digits.
• The sorted array based on the tens place is [802, 2, 24, 45, 66, 170, 75, 90].
How does Radix Sort Algorithm work | Step 3
Step 4: Sort the elements based on the hundreds place digits.
Sorting based on the hundreds place:
• Perform counting sort on the array based on the hundreds place digits.
• The sorted array based on the hundreds place is [2, 24, 45, 66, 75, 90, 170, 802].
How does Radix Sort Algorithm work | Step 4
Step 5: The array is now sorted in ascending order.
The final sorted array using radix sort is [2, 24, 45, 66, 75, 90, 170, 802].
How does Radix Sort Algorithm work | Step 5
Below is the implementation for the above illustrations:
## C++
`// C++ implementation of Radix Sort` `#include ``using` `namespace` `std;` `// A utility function to get maximum``// value in arr[]``int` `getMax(``int` `arr[], ``int` `n)``{`` ``int` `mx = arr[0];`` ``for` `(``int` `i = 1; i < n; i++)`` ``if` `(arr[i] > mx)`` ``mx = arr[i];`` ``return` `mx;``}` `// A function to do counting sort of arr[]``// according to the digit``// represented by exp.``void` `countSort(``int` `arr[], ``int` `n, ``int` `exp``)``{` ` ``// Output array`` ``int` `output[n];`` ``int` `i, count[10] = { 0 };` ` ``// Store count of occurrences`` ``// in count[]`` ``for` `(i = 0; i < n; i++)`` ``count[(arr[i] / ``exp``) % 10]++;` ` ``// Change count[i] so that count[i]`` ``// now contains actual position`` ``// of this digit in output[]`` ``for` `(i = 1; i < 10; i++)`` ``count[i] += count[i - 1];` ` ``// Build the output array`` ``for` `(i = n - 1; i >= 0; i--) {`` ``output[count[(arr[i] / ``exp``) % 10] - 1] = arr[i];`` ``count[(arr[i] / ``exp``) % 10]--;`` ``}` ` ``// Copy the output array to arr[],`` ``// so that arr[] now contains sorted`` ``// numbers according to current digit`` ``for` `(i = 0; i < n; i++)`` ``arr[i] = output[i];``}` `// The main function to that sorts arr[]``// of size n using Radix Sort``void` `radixsort(``int` `arr[], ``int` `n)``{` ` ``// Find the maximum number to`` ``// know number of digits`` ``int` `m = getMax(arr, n);` ` ``// Do counting sort for every digit.`` ``// Note that instead of passing digit`` ``// number, exp is passed. exp is 10^i`` ``// where i is current digit number`` ``for` `(``int` `exp` `= 1; m / ``exp` `> 0; ``exp` `*= 10)`` ``countSort(arr, n, ``exp``);``}` `// A utility function to print an array``void` `print(``int` `arr[], ``int` `n)``{`` ``for` `(``int` `i = 0; i < n; i++)`` ``cout << arr[i] << ``" "``;``}` `// Driver Code``int` `main()``{`` ``int` `arr[] = { 543, 986, 217, 765, 329 };`` ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` ` ``// Function Call`` ``radixsort(arr, n);`` ``print(arr, n);`` ``return` `0;``}`
## Java
`// Radix sort Java implementation` `import` `java.io.*;``import` `java.util.*;` `class` `Radix {` ` ``// A utility function to get maximum value in arr[]`` ``static` `int` `getMax(``int` `arr[], ``int` `n)`` ``{`` ``int` `mx = arr[``0``];`` ``for` `(``int` `i = ``1``; i < n; i++)`` ``if` `(arr[i] > mx)`` ``mx = arr[i];`` ``return` `mx;`` ``}` ` ``// A function to do counting sort of arr[] according to`` ``// the digit represented by exp.`` ``static` `void` `countSort(``int` `arr[], ``int` `n, ``int` `exp)`` ``{`` ``int` `output[] = ``new` `int``[n]; ``// output array`` ``int` `i;`` ``int` `count[] = ``new` `int``[``10``];`` ``Arrays.fill(count, ``0``);` ` ``// Store count of occurrences in count[]`` ``for` `(i = ``0``; i < n; i++)`` ``count[(arr[i] / exp) % ``10``]++;` ` ``// Change count[i] so that count[i] now contains`` ``// actual position of this digit in output[]`` ``for` `(i = ``1``; i < ``10``; i++)`` ``count[i] += count[i - ``1``];` ` ``// Build the output array`` ``for` `(i = n - ``1``; i >= ``0``; i--) {`` ``output[count[(arr[i] / exp) % ``10``] - ``1``] = arr[i];`` ``count[(arr[i] / exp) % ``10``]--;`` ``}` ` ``// Copy the output array to arr[], so that arr[] now`` ``// contains sorted numbers according to current`` ``// digit`` ``for` `(i = ``0``; i < n; i++)`` ``arr[i] = output[i];`` ``}` ` ``// The main function to that sorts arr[] of`` ``// size n using Radix Sort`` ``static` `void` `radixsort(``int` `arr[], ``int` `n)`` ``{`` ``// Find the maximum number to know number of digits`` ``int` `m = getMax(arr, n);` ` ``// Do counting sort for every digit. Note that`` ``// instead of passing digit number, exp is passed.`` ``// exp is 10^i where i is current digit number`` ``for` `(``int` `exp = ``1``; m / exp > ``0``; exp *= ``10``)`` ``countSort(arr, n, exp);`` ``}` ` ``// A utility function to print an array`` ``static` `void` `print(``int` `arr[], ``int` `n)`` ``{`` ``for` `(``int` `i = ``0``; i < n; i++)`` ``System.out.print(arr[i] + ``" "``);`` ``}` ` ``// Main driver method`` ``public` `static` `void` `main(String[] args)`` ``{`` ``int` `arr[] = { ``170``, ``45``, ``75``, ``90``, ``802``, ``24``, ``2``, ``66` `};`` ``int` `n = arr.length;` ` ``// Function Call`` ``radixsort(arr, n);`` ``print(arr, n);`` ``}``}`
## Python3
`# Python program for implementation of Radix Sort``# A function to do counting sort of arr[] according to``# the digit represented by exp.` `def` `countingSort(arr, exp1):` ` ``n ``=` `len``(arr)` ` ``# The output array elements that will have sorted arr`` ``output ``=` `[``0``] ``*` `(n)` ` ``# initialize count array as 0`` ``count ``=` `[``0``] ``*` `(``10``)` ` ``# Store count of occurrences in count[]`` ``for` `i ``in` `range``(``0``, n):`` ``index ``=` `arr[i] ``/``/` `exp1`` ``count[index ``%` `10``] ``+``=` `1` ` ``# Change count[i] so that count[i] now contains actual`` ``# position of this digit in output array`` ``for` `i ``in` `range``(``1``, ``10``):`` ``count[i] ``+``=` `count[i ``-` `1``]` ` ``# Build the output array`` ``i ``=` `n ``-` `1`` ``while` `i >``=` `0``:`` ``index ``=` `arr[i] ``/``/` `exp1`` ``output[count[index ``%` `10``] ``-` `1``] ``=` `arr[i]`` ``count[index ``%` `10``] ``-``=` `1`` ``i ``-``=` `1` ` ``# Copying the output array to arr[],`` ``# so that arr now contains sorted numbers`` ``i ``=` `0`` ``for` `i ``in` `range``(``0``, ``len``(arr)):`` ``arr[i] ``=` `output[i]` `# Method to do Radix Sort` `def` `radixSort(arr):` ` ``# Find the maximum number to know number of digits`` ``max1 ``=` `max``(arr)` ` ``# Do counting sort for every digit. Note that instead`` ``# of passing digit number, exp is passed. exp is 10^i`` ``# where i is current digit number`` ``exp ``=` `1`` ``while` `max1 ``/` `exp >``=` `1``:`` ``countingSort(arr, exp)`` ``exp ``*``=` `10` `# Driver code``arr ``=` `[``170``, ``45``, ``75``, ``90``, ``802``, ``24``, ``2``, ``66``]` `# Function Call``radixSort(arr)` `for` `i ``in` `range``(``len``(arr)):`` ``print``(arr[i], end``=``" "``)` `# This code is contributed by Mohit Kumra``# Edited by Patrick Gallagher`
## C#
`// C# implementation of Radix Sort``using` `System;` `class` `GFG {`` ``public` `static` `int` `getMax(``int``[] arr, ``int` `n)`` ``{`` ``int` `mx = arr[0];`` ``for` `(``int` `i = 1; i < n; i++)`` ``if` `(arr[i] > mx)`` ``mx = arr[i];`` ``return` `mx;`` ``}` ` ``// A function to do counting sort of arr[] according to`` ``// the digit represented by exp.`` ``public` `static` `void` `countSort(``int``[] arr, ``int` `n, ``int` `exp)`` ``{`` ``int``[] output = ``new` `int``[n]; ``// output array`` ``int` `i;`` ``int``[] count = ``new` `int``[10];` ` ``// initializing all elements of count to 0`` ``for` `(i = 0; i < 10; i++)`` ``count[i] = 0;` ` ``// Store count of occurrences in count[]`` ``for` `(i = 0; i < n; i++)`` ``count[(arr[i] / exp) % 10]++;` ` ``// Change count[i] so that count[i] now contains`` ``// actual`` ``// position of this digit in output[]`` ``for` `(i = 1; i < 10; i++)`` ``count[i] += count[i - 1];` ` ``// Build the output array`` ``for` `(i = n - 1; i >= 0; i--) {`` ``output[count[(arr[i] / exp) % 10] - 1] = arr[i];`` ``count[(arr[i] / exp) % 10]--;`` ``}` ` ``// Copy the output array to arr[], so that arr[] now`` ``// contains sorted numbers according to current`` ``// digit`` ``for` `(i = 0; i < n; i++)`` ``arr[i] = output[i];`` ``}` ` ``// The main function to that sorts arr[] of size n using`` ``// Radix Sort`` ``public` `static` `void` `radixsort(``int``[] arr, ``int` `n)`` ``{`` ``// Find the maximum number to know number of digits`` ``int` `m = getMax(arr, n);` ` ``// Do counting sort for every digit. Note that`` ``// instead of passing digit number, exp is passed.`` ``// exp is 10^i where i is current digit number`` ``for` `(``int` `exp = 1; m / exp > 0; exp *= 10)`` ``countSort(arr, n, exp);`` ``}` ` ``// A utility function to print an array`` ``public` `static` `void` `print(``int``[] arr, ``int` `n)`` ``{`` ``for` `(``int` `i = 0; i < n; i++)`` ``Console.Write(arr[i] + ``" "``);`` ``}` ` ``// Driver Code`` ``public` `static` `void` `Main()`` ``{`` ``int``[] arr = { 170, 45, 75, 90, 802, 24, 2, 66 };`` ``int` `n = arr.Length;` ` ``// Function Call`` ``radixsort(arr, n);`` ``print(arr, n);`` ``}` ` ``// This code is contributed by DrRoot_``}`
## Javascript
``
## PHP
`= 0; ``\$i``--) `` ``{ `` ``\$output``[``\$count``[ (``\$arr``[``\$i``] / `` ``\$exp``) % 10 ] - 1] = ``\$arr``[``\$i``]; `` ``\$count``[ (``\$arr``[``\$i``] / ``\$exp``) % 10 ]--; `` ``} ` ` ``// Copy the output array to arr[], so `` ``// that arr[] now contains sorted numbers`` ``// according to current digit `` ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++) `` ``\$arr``[``\$i``] = ``\$output``[``\$i``]; ``} ` `// The main function to that sorts arr[] ``// of size n using Radix Sort ``function` `radixsort(&``\$arr``, ``\$n``) ``{ `` ` ` ``// Find the maximum number to know`` ``// number of digits `` ``\$m` `= max(``\$arr``); ` ` ``// Do counting sort for every digit. Note `` ``// that instead of passing digit number, `` ``// exp is passed. exp is 10^i where i is `` ``// current digit number `` ``for` `(``\$exp` `= 1; ``\$m` `/ ``\$exp` `> 0; ``\$exp` `*= 10) `` ``countSort(``\$arr``, ``\$n``, ``\$exp``); ``} ` `// A utility function to print an array ``function` `PrintArray(&``\$arr``,``\$n``) ``{ `` ``for` `(``\$i` `= 0; ``\$i` `< ``\$n``; ``\$i``++) `` ``echo` `\$arr``[``\$i``] . ``" "``; ``} ` `// Driver Code ``\$arr` `= ``array``(170, 45, 75, 90, 802, 24, 2, 66); ``\$n` `= ``count``(``\$arr``); ` `// Function Call``radixsort(``\$arr``, ``\$n``); ``PrintArray(``\$arr``, ``\$n``); ` `// This code is contributed by rathbhupendra``?>`
Output
```217 329 543 765 986
```
## Complexity Analysis of Radix Sort:
Time Complexity:Â
• Radix sort is a non-comparative integer sorting algorithm that sorts data with integer keys by grouping the keys by the individual digits which share the same significant position and value. It has a time complexity of O(d * (n + b)), where d is the number of digits, n is the number of elements, and b is the base of the number system being used.
• In practical implementations, radix sort is often faster than other comparison-based sorting algorithms, such as quicksort or merge sort, for large datasets, especially when the keys have many digits. However, its time complexity grows linearly with the number of digits, and so it is not as efficient for small datasets.
Auxiliary Space:Â
• Radix sort also has a space complexity of O(n + b), where n is the number of elements and b is the base of the number system. This space complexity comes from the need to create buckets for each digit value and to copy the elements back to the original array after each digit has been sorted.
Q1. Is Radix Sort preferable to Comparison based sorting algorithms like Quick-Sort?Â
If we have log2n bits for every digit, the running time of Radix appears to be better than Quick Sort for a wide range of input numbers. The constant factors hidden in asymptotic notation are higher for Radix Sort and Quick-Sort uses hardware caches more effectively. Also, Radix sort uses counting sort as a subroutine and counting sort takes extra space to sort numbers.
Q2. What if the elements are in the range from 1 to n2?
• The lower bound for the Comparison based sorting algorithm (Merge Sort, Heap Sort, Quick-Sort .. etc) is Ω(nLogn), i.e., they cannot do better than nLogn. Counting sort is a linear time sorting algorithm that sort in O(n+k) time when elements are in the range from 1 to k.
• We can’t use counting sort because counting sort will take O(n2) which is worse than comparison-based sorting algorithms. Can we sort such an array in linear time?Â
• Radix Sort is the answer. The idea of Radix Sort is to do digit-by-digit sorting starting from the least significant digit to the most significant digit. Radix sort uses counting sort as a subroutine to sort. |
Browse Questions
# Find the equation of an ellipse that satisfies the following conditions: $a = 4$ and Foci = $(\pm3,0)$
$\begin{array}{1 1}\frac{x^2}{16}+\frac{y^2}{7}=1 \\ \frac{x^2}{16}+\frac{y^2}{7}=1 \\ \frac{x^2}{7}+\frac{y^2}{16}=1 \\ \frac{x^2}{7}-\frac{y^2}{16}=1\end{array}$
Toolbox:
• Given an ellipse as follows:
• http://clay6.com/mpaimg/Toolbar_7.png
• The equation of the ellipse is $\large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$
• $c= \sqrt {a^2 - b^2}$
• Given a and foci, we know a and c, so we can calculate b and we can substitute in the above equation and arrive at the equation.
Given Ends of Major axis $\;(\pm3,0)$, Ends of Minor Axis $\;(0, \pm 2)$, $a=4$ and $c = 3$
$c^2 = a^2 - b^2 \rightarrow 9 = 16 - b^2 \rightarrow b = \sqrt{7}$
We therefore can write the equation of the ellipse as $\;\large\frac{x^2}{4^2}$$+\large\frac{y^2}{\sqrt 7 ^2}$$=1$
$\Rightarrow$ The equation of the ellipse is $\large\frac{x^2}{16}$$+\large\frac{y^2}{7}$$=1$
edited Apr 3, 2014 |
# What is the Cartesian form of ( -9 , ( - 15pi)/2 ) ?
Jul 26, 2018
(0. -9)
#### Explanation:
We have the coordinate $\left(- 9 , \frac{- 15 \pi}{2}\right)$ in polar form.
Coordinates in polar form have the standard form (color(green)(r), color(purple)(Θ)).
To convert from polar form to Cartesian form, we use the following formulas:
• color(red)(x) = color(green)(r)coscolor(purple)(Θ)
• color(blue)(y) = color(green)(r)sincolor(purple)(Θ)
Now, let's plug stuff in. We know $\textcolor{g r e e n}{r} = - 9$ and color(purple)(Θ) = (-15pi)/2
$\textcolor{red}{x} = \left(- 9\right) \cdot \cos \left(\frac{- 15 \pi}{2}\right)$
$\textcolor{red}{x} = \left(- 9\right) \cdot \left(0\right)$
$\textcolor{red}{x} = 0$
$\textcolor{b l u e}{y} = \left(- 9\right) \cdot \sin \left(\frac{- 15 \pi}{2}\right)$
$\textcolor{b l u e}{y} = \left(- 9\right) \cdot \left(1\right)$
$\textcolor{b l u e}{y} = - 9$
We get $\textcolor{red}{x} = 0$ and $\textcolor{b l u e}{y} = - 9$, making our Cartesian coordinate $\left(0 , - 9\right) .$ |
# How can I assess the statistical probability that an individual is the one I seek?
## How do you find the probability of an individual?
Divide the number of events by the number of possible outcomes.
1. Determine a single event with a single outcome. …
2. Identify the total number of outcomes that can occur. …
3. Divide the number of events by the number of possible outcomes.
## How do you find the probability of one or another?
If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities. The probability that an event does not occur is 1 minus the probability that the event does occur.
## How do we determine if an individual is unusual?
Unusual values are values that are more than 2 standard deviations away from the µ – mean. The 68-95-99.7 rule apples only to data values that are 1,2, or 3 standard deviations from the mean. We can generalize this rule if we know precisely how many standard deviations from the mean (µ) a particular value lies.
## How do you find the probability of random selection?
P (X) = n/N; where ‘n’ is the number of the favourable outcomes and ‘N’ is the number of total possible outcomes.
## How do you find the probability of independent events?
Events A and B are independent if the equation P(A∩B) = P(A) · P(B) holds true. You can use the equation to check if events are independent; multiply the probabilities of the two events together to see if they equal the probability of them both happening together.
## How do you find the probability of a group of people?
Quote from video: Our female chart we have seven out of fourteen. If we add these two probabilities together nine. And seven we get 16 out of 14.
## How do you find the probability of either A or B?
Inclusion-Exclusion Rule: The probability of either A or B (or both) occurring is P(A U B) = P(A) + P(B) – P(AB). Conditional Probability: The probability that A occurs given that B has occurred = P(A|B). In other words, among those cases where B has occurred, P(A|B) is the proportion of cases in which event A occurs.
## How do you find the probability of both A and B?
Formula for the probability of A and B (independent events): p(A and B) = p(A) * p(B). If the probability of one event doesn’t affect the other, you have an independent event. All you do is multiply the probability of one by the probability of another.
### Confirming possible half sibling from AncestryDNA?Confirming possible half sibling from AncestryDNA?
Can a DNA test prove half-siblings? Yes, a DNA test can prove half-siblings. As a matter of fact, it’s the only accurate way to establish the biological relationship between the
### How should I show my step-mother?How should I show my step-mother?
How should I treat my step mom? Keep your calm. Focus on yourself and what you can change within yourself to better adapt to her. If your stepmom is rude
### Checking Daughters/Sons of the American Revolution (DAR/SAR) lineages?Checking Daughters/Sons of the American Revolution (DAR/SAR) lineages?
How do I find out if I am a Daughter of the American Revolution? To find out check the Descendants and Ancestor links in the GRS. If you discover a |
# Unions and Intersections When you consider all the outcomes for either of two events, A and B, you form the union of A and B. When you consider only.
## Presentation on theme: "Unions and Intersections When you consider all the outcomes for either of two events, A and B, you form the union of A and B. When you consider only."— Presentation transcript:
Unions and Intersections When you consider all the outcomes for either of two events, A and B, you form the union of A and B. When you consider only the outcomes shared by both A and B, you form the intersection of A and B. The union or intersection of two events is called the compound event.
Union A B
A B Intersection
Intersection of A and B is empty A B
Compound Events To find P(A and B) you must consider what outcomes, if any, and in the intersection of A and B. Two events are overlapping if they have one or more outcomes in common as seen in the UNION diagram. Two events are disjoint, or mutually exclusive, if they have no outcomes in common, as shown in the 3 rd diagram.
Probability of Compound Events If A and B are any two events, then the probability of A or B is: P(A or B) = P(A) + P(B) – P(A and B) If A and B are disjoint events, then the probability of A or B is: P(A or B) = P(A) + P(B)
Find probability of disjoint events A card is randomly selected from a standard deck of 52 cards. What is the probability that it is a 10 or a face card?
Find probability of disjoint events Let event A be selecting a 10 and event B be selecting a face card. A has 4 outcomes and B has 12 outcomes. Because A and B are disjoint, the probability is: P(A or B) = P(A) + P(B) =
Find probability of compound events A card is randomly selected from a standard deck of 52 cards. What is the probability that it is a face card or a spade?
Find probability of compound events Let event A be selecting a face card and event B be selecting a spade. The events are shown with the overlapping events. K Q J K Q J K Q J KQJKQJ 10 9 8 7 6 5 4 3 2 A A B
Find the probability of compound events Remember: P(A or B) = P(A) + P(B) – P(A and B) Thus the probability of drawing a spade or a face card is: P(A or B) =
Use the formula to find P(A and B) Out of 200 students in a senior class, 113 students are either varsity athletes or on the honor roll. There are 74 seniors who are varsity athletes and 51 seniors who are on the honor roll. What is the probability that a randomly selected senior is both a varsity athlete and on the honor roll?
Use a formula to find P(A and B) Let event A be selecting a senior who is a varsity athlete and event B be selecting a senior on the honor roll. From the given information you know: P(A)=P(B)=P(A or B)= Find P(A and B).
Use a formula to find P(A and B) P(A or B) = P(A) + P(B) – P(A and B) P(A and B) =
Practice A card is randomly selected from a standard deck of 52 cards. Find the probability of the given event. Selecting an ace or an eight Selecting a 10 or a diamond
Practice Answers Selecting an ace or an eight Selecting a 10 or a diamond
Complements The event A’, called the complement of event A, consists of all outcomes that are not in A. The notations A’ is read “A prime or A complement” The book uses the notation Ā for the complement and is read “A bar”. A’ = Ā
Probability of the Complement of an Event The probability of the complement of A is: P(A’) = 1 – P(A) or P(Ā) = 1 – P(A)
Find probabilities of complements When two six-sided dice are rolled, there are 36 possible outcomes as shown in the table. 123456 1234567 2345678 3456789 45678910 56789 11 6789101112
Find probabilities of complements Find the probability of the given event: The sum is not 6 The sum is less than or equal to 9 123456 1234567 2345678 3456789 45678910 56789 11 6789101112
Find probabilities of complements The sum is not 6 P(sum is not 6) = 1 – P(sum is 6) The sum is less than or equal to 9: P(sum ≤ 9) = 1 – P(sum > 9)
Use a complement in real life A restaurant gives a free fortune cookie to every guest. The restaurant claims there are 500 different messages hidden inside the fortune cookies. What is the probability that a group of 5 people receive at least 2 fortune cookies with the same message inside?
Use a complement in real life The number of ways to give messages to the 5 people is 500 5. The number of ways to give different messages to 5 people is 500 499 498 497 496. So, the probability that at least 2 of the 5 people have the same message is: P(at least 2 are the same) = 1 – P(none are the same
Practice Find P(A’) P(A) = 0.45 P(A) = ¼ P(A) = 1 P(A) = 0.03
Practice Answers P(A’) = 0.55 P(A’) = 3/4 P(A’) = 0 P(A’) = 0.97
Independent Events Two events are independent if the occurrence of one has no effect on the occurrence of the other. For instance, if a coin is tossed twice, the outcome of the first toss (heads or tails) has no effect on the outcome of the second toss.
Probability of Independent Events If A and B are independent events, then the probability that both A and B occur is: P(A and B) = P(A) P(B) More generally, the probability that n independent events occur is the product of the n probabilities of the individual events.
Probability of Independent Events For a fundraiser, a class sells 150 raffle tickets for a mall gift certificate and 200 raffle tickets for a booklet of movie passes. You buy 5 raffle tickets for each prize. What is the probability that you win both prizes?
Probability of Independent Events Let events A and B be getting the winning ticket for the gift certificate and movie passes, respectively. The events are independent. So, the probability is: P(A and B) = P(A) P(B) =
Find the probability of 3 independent events In a BMX meet, each heat consists of 8 competitors who are randomly assigned lanes from 1 to 8. What is the probability that a racer will draw lane 8 in the 3 heats in which the racer participates?
Find the probability of 3 independent events Let events A, B, and C be drawing lane 8 in the first, second, and third heats, respectively. The 3 events are independent. So, the probability is: P(A and B and C) = P(A) P(B) P(C)
Use a complement to find a probability While you are riding to school, your portable CD player randomly plays 4 different songs from a CD with 16 songs on it. What is the probability that you will hear your favorite song on the CD at least once during the week (5 days)?
Use a complement to find a probability For one day, the probability of not hearing you favorite song is: P(not hearing song) =
Use a complement to find a probability Hearing or not hearing your favorite song on Monday, on Tuesday, and so on are independent events. So, the probability of hearing the song at least once is: P(hearing song) = 1 – [P(not hearing song)] 5 =
Dependent Events Two events are A and B are dependent events if the occurrence of one affects the occurrence of the other. The probability that B will occur given that A has occurred is called the conditional probability of B given A and is written as P(B|A).
Probability of Dependent Events If A and B are dependent events, then the probability that both A and B occur is: P(A and B) = P(A) P (B|A)
Find a conditional probability The table shows the numbers of tropical cyclones that formed during the hurricane seasons from 1988 to 2004. Use the table on the next slide to estimate: The probability that a future tropical cyclone is a hurricane The probability that a future tropical cyclone in the Northern Hemisphere is a hurricane.
Find a conditional probability Type of Tropical Cyclone Northern Hemisphere Southern Hemisphere Tropical depression19918 Tropical storm398200 Hurricane545215 P(hurricane) = # of hurricanes Total # of cyclones P(hurricane | Northern Hemisphere) = # of hurricanes in Northern Hemisphere Total # of cyclones in Northern Hemisphere
Comparing independent and dependent events You randomly select two cards from a standard deck of 52 cards. What is the probability that the first card is not a heart and the second card is a heart? Find the probability first WITH REPLACEMENT, then WITHOUT REPLACEMENT.
Comparing independent and dependent events Let A be “the first card is not a heart” and B be “the second card is a heart”. If you replace the first card before selecting the second card, then A and B are independent events. So, the probability is: P(A and B) = P(A) P(B)
Comparing independent and dependent events If you do not replace the first card before selecting the second card, the A and B are dependent events. So, the probability is: P(A and B) = P(A) P (B|A)
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### How many consecutive odd integers beginning with 5 will sum to 480?
How many consecutive odd integers beginning with 5 will sum to 480?
Solution :
Consecutive integers starting from 5 are 5, 7, 9, ............
5 + 7 + 9 + 11 + ............. = 480
Sn = 480
(n/2)[2a + (n - 1)d] = 480
a = 5, d = 7 - 5 = 2
(n/2)[2(5) + (n - 1)(2)] = 480
(n/2)[10 + 2n - 2] = 480
(n/2)[2n + 8] = 480
2n2 + 8n = 960
Dividing by 2, we get
n2 + 4n - 480 = 0
(n + 24) (n - 20) = 0
n = -24 or n = 20
Hence by finding the sum of 20 terms we will get 480.
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# The fundamental theorem of arithmetic
## The Fundamental Theorem of Arithmetic
### Euclidian division
Euclidian division is the theory for any pair of natural numbers, we can divide one by the other and have a remainder less than the divisor. Formally: $$\forall a \in \mathbb{N} ,\forall b \in \mathbb{N}^+ ,\exists q \in \mathbb{N},\exists r\in \mathbb{N} [(a=bq+r)\land (0\le r < b)]$$
Where $$\mathbb{N}^+$$ refers to natural numbers excluding $$0$$.
That is, every natural number $$a$$ is a multiple $$q$$ of any other natural number $$b$$, plus another natural number $$r$$ less than the other natural number $$b$$.
These are unique. For each jump in $$q$$, $$r$$ falls by $$b$$. As the range of $$r$$ is $$b$$ there is only one solution.
$$17=2.8+1$$
$$9=3.3+0$$
### Bezout’s identity
For any two non-zero natural numbers $$a$$ and $$b$$ we can select natural numbers $$x$$ and $$y$$ such that
$$ax+by=c$$
The value of $$c$$ is always a multiple of the greatest common denominator of $$a$$ and $$b$$.
In addition, there exist $$x$$ and $$y$$ such that $$c$$ is the greatest common denominator itself. This is the smallest positve value of c..
Let’s take two numbers of the form $$ax+by$$:
$$d=as+bt$$
$$n=ax+by$$
Where $$n>d$$. And $$d$$ is the smallest non-zero natural number form.
We know from Euclidian division above that for any numbers $$i$$ and $$j$$ there is the form $$i=jq+r$$.
So there are values for $$q$$ and $$r$$ for $$n=dq+r$$.
If $$r$$ is always zero that means that all values of $$ax+by$$ are multiples of the smallest value.
$$n=dq+r$$ so $$r=n-dq$$.
$$r=ax+by-(as+bt)q$$
$$r=a(x-sq)+b(y-tq)$$
This is also of the form $$ax+by$$. Recall that $$r$$ is the remainder for the division of $$d$$ and $$n$$, and that $$d=ax+by$$ is the smallest positive value.
$$r$$ cannot be above or equal to $$d$$ due to the rules of euclidian division and so it must be $$0$$.
As a result we know that all solutions to $$ax+by$$ are multiples of the smallest value.
As every possible $$ax+by$$ is a multiple of $$d$$, $$d$$ must be a common divisor to both numbers. This is because $$a.0+b.1$$ and $$a.1+b.0$$ are also solutions, and $$d$$ is their divisor.
So we know that the smallest positive solution is a common mutliple of both numbers.
We now need to show that that $$d$$ is the largest common denominator. Consider a common denominator $$c$$.
$$a=pc$$
$$b=qc$$
And as before:
$$d=ax+by$$
So:
$$d=pcx+qcy$$
$$d=c(px+qy)$$
So $$d\ge c$$
### Euclid’s lemma
#### Statement
If a prime number $$p$$ divides product $$a.b$$ then $$p$$ must divide at least of one of $$a$$ or $$b$$.
#### Proof
From Bezout’s identity we know that:
$$d=px+by$$
Where $$p$$ and $$b$$ are natural numbers and $$d$$ is their greatest common denominator.
Let’s choose a prime number for $$p$$. There are no common divisors, other than one. As a result there are exist values for $$x$$ and $$y$$ such that:
$$1=px+by$$
Now, we are trying to prove that if $$p$$ divides $$a.b$$ then $$p$$ must divide at least one of $$a$$ and $$b$$, so let’s multiply this by $$a$$.
$$a=pax+aby$$
We know that $$p$$ divides $$pax$$, and $$p$$ divides $$ab$$ by definition. As a result $$p$$ can divide $$a$$.
### Fundamental Theorem of Arithmetic
#### Statement
Each natural number is a prime or unique product of primes.
#### Proof: existance of each number as a product of primes
If $$n$$ is prime, no more is needed.
If $$n$$ is not prime, then $$n=ab$$, $$a,b\in \mathbb{N}$$.
If $$a$$ and $$b$$ are prime, this is complete. Otherwise we can iterate to find:
$$n=\prod_{i=1} p_i$$
#### Proof: this product of primes is unique
Consider two different series of primes for the same number:
$$s=\prod_{i=1}^n p_i = \prod_{i=1}^m q_i$$
We need to show that $$n=m$$ and $${p}={q}$$.
We know that $$p_i$$ divides $$s$$. We also know that through Euclid’s lemma that if a prime number divides a non-prime number, then it must also divide one of its components. As a result $$p_i$$ must divide one of $${q}$$.
But as all of $${q}$$ are prime then $$p_i$$=$$q_j$$.
We can repeat this process to to show that $${p}={q}$$ and therefore $$n=m$$.
### Existence of an infinite number of prime numbers
#### Existence of an infinite number of prime numbers
If there are a finite number of primes, we can call the set of primes $$P$$.
We identify a new natural number $$a$$ by taking the product of existing primes and adding $$1$$.
$$a=1+\prod_{p\in P} p$$
From the fundamental theorem of arithmetic we know all numbers are primes or the products of primes.
If $$a$$ is not a prime then it can be divided by one of the existing primes to form number $$n$$:
$$\dfrac{\prod^n p_i +1}{p_j}=n$$
$$\dfrac{p_j \prod^n_{i\ne j} p_i +1}{p_j}=n$$
$$\prod^n_{i\ne j} p_i +\dfrac{1}{p_j}=n$$
As this is not a whole number, $$n$$ must prime.
We can do this process for any finite number of primes, so there are an infinite number. |
# Trigonometric Addition and Subtraction formulae
Trigonometric Addition ( Sum ) and Subtraction ( Difference ) formula:
The formulae
which are popularly
known as addition ( sum )
and subtraction( difference )
formulae are as follows:
Sine of sum of angles:
$\sin (A+B) = \sin A . \cos B + \cos A . \sin B$
Cosine of sum of angles:
$\cos (A+B) = \cos A . \cos B - \sin A . \sin B$
Sine of difference of angles:
$\sin (A-B) = \sin A . \cos B - \cos A . \sin B$
Cosine of difference of angles:
$\cos (A-B) = \cos A . \cos B + \sin A . \sin B$
And similarly the sum and difference of angle formula of Tangent are:
$\tan (A+B) = \dfrac{\tan A + \tan B}{1 - \tan A . \tan B}$
and,
$\tan (A-B) = \dfrac{\tan A - \tan B}{1 + \tan A . \tan B}$
Proof of Trigonometric Sum and Difference Formulae:
Now let us prove the identities or formulae listed above.
In the approach to prove these identities we should first prove the identity of $\cos (A - B)$ and rest of the identities can be derived from the identity:
Let $P(x_1 , y_1)$ and $Q(x_2 , y_2)$ be two points , different from the origin , on the terminal arms of two angles A and B such that $A > B$ placed in the standard position as shown in the figure below:
cosine difference formula
Now,
If: $OP = r_1$ and $OQ = r_2$ then,
$x_1 = r_1 . \cos A$ , $y_1 = r_1 . \sin A$
and $x_2 = r_2 . \cos B$ , $y_2 = r_2 . \sin B$
Hence , Using the Distance formula:
$PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$
or , $PQ^2 = (x^2_1 + y^2_1) + (x^2_2 + y^2_2) - 2(x_1 x_2 + y_1 y_2$
or, $PQ^2 = r^2_1 + r^2_2 - 2 r_1 r_2 (\cos A . \cos B + \sin A \sin B)$ ————– Expression 1 for PQ^2
Now , Let us rotate the coordinate system so that OQ coincides with the positive x-axis. This is equivalent to place the angle (A-B) in the standard position with OQ along the positive x-axis. As shown in figure below:
cosine difference formula
Then , the new coordinates of $P(x^| , y^|)$ and $Q(x^{||} , y^{||})$ are given by:
$x^| = r_1 \cos (A-B)$
, $y^| = r_1 \sin (A-B)$
and :
$x^{||} = r_2$ and $y^{||} = 0$
Hence now:
$PQ^2 = (x^{||} - x^|)^2 + (y^{||} - y^|)^2$
Or , $PQ^2 = x^{|2} + y^{|2} + x^{||2} - 2 x^| x^{||}$
Or , $PQ^2 = r^2_1 + r^2_2 - 2 r_1 r_2 . \cos (A - B)$ ————– Expression 2 for PQ^2
Now comparing the two expression for $PQ^2$ , we have:
$2 r_1 r_2 . \cos (A - B) = 2 r_1 r_2 (\cos A . \cos B + \sin A \sin B)$
Hence: $\cos (A - B) = \cos A . \cos B + \sin A . \sin B$
This is the subtraction or difference formula for cosine , true for arbitrary angles A and B.
Now we can derive other sum and difference formula using this formula, before deriving those formulae let us consider following special cases:
Special Cases:
Note: All the numbers and variables eg: A , B , 90 are in Degrees , but we can also apply this in other angle measurement system by replacing the value of numbers used with corresponding value.
Case 1> If A=o then , we get:
$\cos(-B) = \cos B$
Case 2> If A=90 then, we get:
$\cos(90 - B) = \sin B$
Case 3> If B is replaced by 90-B in Case 2 then we get:
$\cos B = \sin (90 - B)$
Case 4> If B is replaced by -90 and A is replaced by -A then , we get:
$- \sin (-A) = \sin A$ and $\sin (-A) = - \sin A$
Now let us Derive other general formulas using the special cases above and cosine difference of angle formula:
Formula for $\cos (A+B)$
Replacing B by -B in $\cos (A - B) = \cos A . \cos B + \sin A . \sin B$ we get:
$\cos (A + B) = \cos A . \cos (-B) + \sin A . \sin (-B)$
Now using case 2 and 5 we get:
$\cos (A + B) = \cos A . \cos B - \sin A . \sin B$
Formula for $\sin (A+B)$
Replacing A by 90-A to get:
$\cos (90-A-B) = \cos (90-A) . \cos B + \sin (90 - A) . \sin B$
Then , using Case 3 and 4 we get:
$\sin (A+B) =\sin A \cos B + \cos A \sin B$
Formula for $\sin (A-B)$
Replacing B by -B in above formula or $\sin (A+B) =\sin A \cos B + \cos A \sin B$ and the uisng case 2 and 5 we get:
$\sin (A-B) =\sin A \cos B - \cos A \sin B$
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# Work Rate Problems with Solutions
A set of problems related to work and rate of work is presented with detailed solutions.
Problem 1: It takes 1.5 hours for Tim to mow the lawn. Linda can mow the same lawn in 2 hours. How long will it take John and Linda, work together, to mow the lawn? Solution to Problem 1: We first calculate the rate of work of John and Linda John: 1 / 1.5 and Linda 1 / 2 Let t be the time for John and Linda to mow the Lawn. The work done by John alone is given by t * (1 / 1.5) The work done by Linda alone is given by t * (1 / 2) When the two work together, their work will be added. Hence t * (1 / 1.5) + t * (1 / 2) = 1 Multiply all terms by 6 6 (t * (1 / 1.5) + t * (1 / 2) ) = 6 and simplify 4 t + 3 t = 6 Solve for t t = 6 / 7 hours = 51.5 minutes. Problem 2: It takes 6 hours for pump A, used alone, to fill a tank of water. Pump B used alone takes 8 hours to fill the same tank. We want to use three pumps: A, B and another pump C to fill the tank in 2 hours. What should be the rate of pump C? How long would it take pump C, used alone, to fill the tank? Solution to Problem 2: The rates of pumps A and B can be calculated as follows: A: 1 / 6 and B: 1 / 8 Let R be the rate of pump C. When working together for 2 hours, we have 2 ( 1 / 6 + 1 / 8 + R ) = 1 Solve for R R = 1 / 4.8 , rate of pump C. Let t be the time it takes pump C, used alone, to fill the tank. Hence t * (1 / 4.8) = 1 Solve for t t = 4.8 hours , the time it takes pump C to fill the tank. Problem 3: A tank can be filled by pipe A in 5 hours and by pipe B in 8 hours, each pump working on its own. When the tank is full and a drainage hole is open, the water is drained in 20 hours. If initially the tank was empty and someone started the two pumps together but left the drainage hole open, how long does it take for the tank to be filled? Solution to Problem 3: Let's first find the rates of the pumps and the drainage hole pump A: 1 / 5 , pump B: 1 / 8 , drainage hole: 1 / 20 Let t be the time for the pumps to fill the tank. The pumps ,add water into the tank however the drainage hole drains water out of the tank, hence t ( 1 / 5 + 1 / 8 - 1 / 20) = 1 Solve for t t = 3.6 hours. Problem 4: A swimming pool can be filled by pipe A in 3 hours and by pipe B in 6 hours, each pump working on its own. At 9 am pump A is started. At what time will the swimming pool be filled if pump B is started at 10 am? Solution to Problem 4: the rates of the two pumps are pump A: 1 / 3 , pump B: 1 / 6 Working together, If pump A works for t hours then pump B works t - 1 hours since it started 1 hour late. Hence t * (1 / 3) + (t - 1) * (1 / 6) = 1 Solve for t t = 7 / 3 hours = 2.3 hours = 2 hours 20 minutes. The swimming pool will be filled at 9 + 2:20 = 11:20 More math problems with detailed solutions in this site.
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Updated: 2 April 2013 |
Did you know 37 is the 12th prime number? Hence, it has actually just 2 determinants, 1 and also the number itself (37). In this mini-leskid let us learn to calculate the square root of 37 utilizing the long division and approximation approaches and also to expush the square root of 37 in the most basic radical form.
You are watching: What is the square root of 37
Square Root of 37: 37 = 6.082Square of 37: 372 = 1369
1 What Is the Square Root of 37? 2 Is Square Root of 37 Rational or Irrational? 3 How to Find the Square Root of 37? 4 Important Notes on Square Root of 37 5 Challenging Questions 6 FAQs on Square Root of 37
## What Is the Square Root of 37?
6.082 × 6.082 = 37 and also - 6.082 × - 6.082 = 3737 = ± 6.082The square root of 37 in its easiest radical form = 37
## How to Find the Square Root of 37?
The square root of 37 or any number can be calculated in many ways. To mention a few: Prime factorization strategy, approximation method and also the lengthy division approach.
### Square Root of 37 by Approximation method
√37 lies between √36 and also √49 Clearly on, √36 lies closer to 6, as we recognize 6 × 6 = 36. Use the average method to determine the approximate worth of √37.
The square root of 37 lies between the square root of 36 and also the square root of 49. Hence, 36 Divide 37 by 7. 37 ÷ 7 = 5.28Find the average between 5.28 and 7. (5.28 + 7)/ 2 = 12.28 /2 = 6.04 Thus, 37 ≈ 6.04
### Square Root of 37 by the Long Division Method
The long division technique helps us to find the even more exact worth of the square root of any type of number. Let"s see exactly how to find the square root of 37 by the long division strategy. Here are the desirable actions to be adhered to.
Write 37 as 37. 00 00 00.Take 37 as a pair. Find a number × number such that the product is much less than or equal to 37.We determine that 6 × 6 = 36. subtract this from 37. Get the remainder as 1 and also carry down the first pair of zeros. 1 00 is our brand-new divisor.Place the decimal point after 6 in the quotient. Multiply the quotient by 2 and have 12x as the brand-new divisor.Find a number in the location of x such that 12x × x offers 100 or less than that. We find no such number. So 120 × 0 is 0. Subtract it from 100 and also acquire the remainder as 10. Bring dvery own the next pair of zeros. 1 00 00 is our brand-new divisor.Multiply the quotient by 2 and also have actually 120x as the new divisor. Find a number in the location of x such that 120x × x gives 10 00 00 or less than that. We uncover 1208 × 8 is 94 64. Subtract it from 10 00 00 and also get the remainder as 3 36.Repeat the procedure until we approximate to 3 decimal locations.Thus √37 = 6.082 to the nearest thousandths.
Explore Square roots utilizing illustrations and also interenergetic examples
Important Notes
The square root of 37 is √37 = 37 ½ = ± 6.082. The square root of 37 lies between the 2 perfect squares 36 and also 49. √37 is irrational.
See more: Suppliers Are Those Organizations Or Individuals Who Provide Procurement Services.
Challenging Questions
Find the smallest integer that hregarding be added to 37 and subtracted from 37 to make it a perfect square. Also find the square root of those perfect squares.Find the square root of √37. |
# Ordering Fractions
4 teachers like this lesson
Print Lesson
## Objective
Students will use common denominators and equivalent fractions to order fractions with unlike denominators.
#### Big Idea
Given a set of 3 or more fractions the student will be able to place them in order from least to greatest.
## Anticipatory
8 minutes
My students have been learning how to compare fractions with like and unlike denominators. To assess students' knowledge of fractions so far, I write two fractions with the same denominator on the board (3/12 and 6/12) I ask students to tell me how the fractions are alike. (Students determine that both fractions have the same denominator.) Then, I write a 9/24 which is equivalent to 3/12 and 6/12. I ask students how are 9/24 like the other two fractions. (Students are able to say that the fractions are equivalent.)
I may probe a bit more to see if they are ready to order fractions. For instance, I ask the following questions: What did you notice about the fractions? Explain? Can someone think of a way we can organize fractions by their size? What other math concept can we use to arrange fractions?
Student Talk:
Students see that we can use greater than and less than to arrange numbers in order from least to greatest.
We will be using the following Mathematical Practices in this lesson:
MP.2. Reason abstractly and quantitatively.
MP.4. Model with mathematics.
MP.7. Look for and make use of structure.
MP.8. Look for and express regularity in repeated reasoning.
## Modeling
10 minutes
I invite students to the carpet. Then I say, we have learned how to find equivalent fractions, and today you will learn how to use equivalent fractions to compare and order fractions.
Teacher and Student Talk:
How can you determine if two fractions are equivalent? (They are different names for the same amount.) I take note of students who are unable to respond to the given question. After that, I model how to determine if two fractions are equivalent.
I say:
The best way to think about equivalent fractions is that they are fractions that have the same overall value. Equivalent fractions represent the same part of a whole. For example, if we cut a pie exactly down the middle, into two equally sized pieces, one piece is the same as one half of the pie.
And if another pie (the same size) is cut into 4 equal pieces, then two pieces of that pie represent the same amount of pie that 1/2 did. So we can say that 1/2 is equivalent (or equal) to 2/4. I draw an illustration on the board to point out that it is the same size, however, the pieces are different.
Pose a Question:
Joe skated 2/3 mile, Carl skated 3/5 mile, and Pete skated ½ miles. Write these distances in order from least to greatest.
To model, I use large fraction strips to create models of the given fractions. Then I place the fraction strips side by side to show students a visual picture of the given order.
Then I write the fractions in order from least to greatest. I ask students to explain how they would solve this given problem using other solution methods. It is my goal to only model what students are expected to do at this point.
Struggling Students:
I ask them to arrange the fraction strips in order by size. (small to large)
## Guided Practice
20 minutes
In a whole group discussion I ask students to explain the given question.
How would you write 3/10, 5/10, and 4/10 in order from least to greatest?
I want to see if students will illustrate fraction strips to solve their problem, or will they be able to solve it on their own. I also want to check their understanding of ordering fractions. This helps me to adjust the complexity of the lesson.
After that, I pair students with a partner. I want them to have time ordering fractions from least to greatest. I give them a set of fraction strips, and I write several fractions on the board. I ask students to arrange the listed fractions in order from least to greatest. I notice some students use fraction strips to determine the size, and then write the fraction on their paper in the correct order. Some students ask is there another way to order fractions. I encourage them to explore new ways, however, the concept I am to help them understand is that using visual models is beneficial when ordering fractions. As students continue to work, I circle the room to check for understanding.
Probing Questions:
What did you notice?
Can you explain?
Did your partner find another way to order fractions? Explain?
Students Response:
Without the fraction strips it was kind of hard to determine which fraction was larger/ smaller.
## Independent/Closing
20 minutes
After that, students are asked to return to their assigned seats. I remind students to list the fractions in order from least to greatest, not greatest to least, since this was one of the misconceptions I noticed while students were working in pairs. Then, I give each student their Work Assignment, and remind them to draw strips to assist them if they need help with understanding.
As students are working, I circle the room to monitor students' thinking. I take notes of any student who seem to be having difficulty understanding how to order fractions from least to greatest. I will use the notes to help clear up misunderstandings in a smaller group setting. |
# Height and Distance Questions Answers
• #### 1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30 degree and 45 degree respectively. If the lighthouse is 100 m high, the distance between the two ships is:
1. 276 metre
2. 273 metre
3. 270 metre
4. 263 metre
Explanation:
Let AB be the lighthouse and C and D be the positions of the ships.
\begin{aligned}
\text{AB = 100m}, \angle{ACB}=30^{\circ}, \\
\frac{AB}{AC} = tan&30{\circ} = \frac{1}{\sqrt{3}} \\
=> AC = AB*\sqrt{3} = 100\sqrt{3}m\\
\frac{AB}{AD} = tan&45^{\circ} = 1 \\
=> AB = AD = 100m\\
= (100\sqrt{3}+100)&m \\
= 100(\sqrt{3}+1)&m\\
= 100*2.73&m
= 273m
\end{aligned}
• #### 2. The angle of elevation of the sun, when the length of the shadow of a tree is \begin{aligned}\sqrt{3}\end{aligned} times the height of tree, is :
1. 30 degree
2. 45 degree
3. 60 degree
4. 9 degree
Explanation:
Let AB be the tree and AC be its shadow, So as per question, shadow of a tree is \begin{aligned}\sqrt{3}\end{aligned} times the height of tree. Let h be the height of the tree, then
\begin{aligned}
\frac{AB}{AC} = tan\theta \\
=> \frac{h}{\sqrt{3}h} = tan\theta \\
=> tan\theta = \frac{1}{\sqrt{3}} \\
=> \theta = 30^{\circ}
\end{aligned}
• #### 3. From a point C on a level ground, the angle of elevation of the top of a tower is 30 degree. If the tower is 100 meter high, find the distance from point C to the foot of the tower.
1. 170 meter
2. 172 meter
3. 173 meter
4. 167 meter
Explanation:
Let AB be the tower.
\begin{aligned}
then \angle{ACB}= 30^{\circ} \\
AB = 100&meter \\
\frac{AB}{AC} = tan&30^{\circ} \\
=>\frac{100}{AC} = \frac{1}{\sqrt{3}} \\
=> AC = \sqrt{3}*100 \\
=> AC = 1.73*100 \\
=> AC = 173 meter
\end{aligned}
Please always remember the value of square root 3 is 1.73, and value of square root 2 is 1.41.
This will be very helpful while solving height and distance questions and saving your time.
• #### 4. A man standing at a point C is watching the top of a tower, which makes an angle of elevation of 30 degree. The man walks some distance towards the tower to watch its top and the angle of elevation become 60 degree. What is the distance between the base of the tower and point C.
1. \begin{aligned} 4\sqrt{3} meter \end{aligned}
2. \begin{aligned} 2\sqrt{3} meter \end{aligned}
3. \begin{aligned} \sqrt{3} meter \end{aligned}
Explanation:
Please refer to the image. One of the AB, AD and CD must be given. So data is inadequate.
• #### 5. A man is watching form the top of the tower a boat speeding away from the tower. The boat makes the angle of depression of 45 degree with the man's eye when at a distance of 60 metres from the tower. After 5 seconds the angle of depression becomes 30 degree. What is the approximate speed of the boat, assuming that it is running in still water ?
1. 22 Km/Hr
2. 28 Km/Hr
3. 32 Km/Hr
4. 36 Km/Hr
Explanation:
Let AB be the tower and C and D be the positions of the boat.
\begin{aligned}
AC = 60&m \\
Let& AB = h \\
=>\frac{AB}{AC} = tan&45^{\circ} = 1 \\
=> AB = AC = 60&m \\
and, \frac{AB}{AD} = tan&30^{\circ} = \frac{1}{\sqrt{3}} \\
=> AD = AB*\sqrt{3} = 60\sqrt{3}&m \\
=> CD = AD-AC = 60\sqrt{3}-60\\
=> CD = 60(\sqrt{3}-1) m \\
\text{We Know Speed = }\frac{Distance}{Time} \\
=> Speed = \left[\frac{60(\sqrt{3}-1)}{5} \right]m/sec
= 12*0.73&m/sec \\
=> Speed = 12*0.73*\frac{18}{5} Km/Hr \\
=> Speed = 31.5 & Km/hr, \\
\text{Which is approx 32 Km/Hr}
\end{aligned}
• #### 6. The Top of a 15 metre high tower makes an angle of elevation of 60 degree with the bottom of an electric pole and angle of elevation of 30 degree with the top of pole. Find the height of the electric pole.
1. 7 metre
2. 8 metre
3. 9 metre
4. 10 metre
Explanation:
Let AB be the tower and CD be the electric pole.
\begin{aligned}
\angle{ACB}=60^{\circ}, \angle{EDB}=30^{\circ}, \\
AB = 15&m \\
Let & CD = h, \text{ then}, \\
BE = AB-AE = AB - AE = 15-h\\
\frac{AB}{AC} = tan 60^{\circ} = \sqrt{3} \\
=> AC = \frac{AB}{\sqrt{3}} \\
=> AC = \frac{15}{\sqrt{3}} \\
and, & \frac{BE}{DE} = tan 30^{\circ} = \frac{1}{\sqrt{3}} \\
=> DE = (BE*\sqrt{3}) \\
= \sqrt{3}(15-h) \\
Now, & AC = DE \\
=> \frac{15}{\sqrt{3}} = \sqrt{3}(15-h) \\
=> 3h = 45-15 \\
=> h = \frac{30}{3} = 10 m
\end{aligned}
• #### 7. A toy leaves the earth at a point A and rises vertically at uniform speed. After two minutes of vertical rise boy finds the angular elevation of the balloon as 60°.If the point at which boy is standing is 150 m away from point A, what is the speed of the toy ?
1. .98 meter/second
2. 1.08 meter/second
3. 1.16 meter/second
4. 2.08 meter/second
5. 2.16 meter/second
Explanation:
Let boy is standing at C position and A is that position from where toy left eartg and B is the position of the toy after 2 minutes.
\begin{aligned}
\text{ Given that CA = 150 m } \\
\text{Angle is }\angle{ 60 ^{\circ} } \\
tan 60 ^{\circ} = \frac{BA}{CA} \\
BA = 150 \sqrt{3}
\end{aligned}
So distance travelled by toy is,
\begin{aligned}
150 \sqrt{3} \\
\text { Total time taken is = 2 min} \\
= \text {2 * 60 = 120 seconds } \\
Speed = \frac{Distance}{Time} \\
= \frac{150\sqrt{3}}{120} = 1.25\sqrt{3}\\
= \text {1.25 * 1.73 = 2.16 mtr/sec}
\end{aligned}
• #### Shankar Krish 6 years ago
Thanks very helpful. I thought prepared aptitude for exam properly but now only found I just answered only 3.
• #### Gautam jayasurya 7 years ago
Thank you. Very useful explanations. |
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Important Questions Maths Class 8 Chapter 4: Practical Geometry
Important questions for class 8 Maths Chapter 4 Practical Geometry are made available here for students to make them prepared for final exams 2020. These questions are available with solutions, provided by our subject experts to help them understand the concept well. Since the problems here are given based on CBSE(NCERT) latest syllabus, thus it will be easy for students to score good marks.
In this chapter, students of 8th standard will learn to construct the quadrilaterals based on certain given conditions such as: when the dimensions are given for its sides and diagonals or for angles. Let us see some of the important problems with their answers to have a proper revision here. Also, get Maths important questions for class 8 for all the chapters at BYJU’S.
## Important Questions With Solutions For Class 8th Maths Chapter 4 (Practical Geometry)
Q.1: Construct a quadrilateral ABCD where AB = 4.5 cm BC = 5.5 cm CD = 4 cm AD = 6 cm AC = 7 cm.
Solution:
Q.2: Construct a quadrilateral JUMP where JU = 3.5 cm UM = 4 cm MP = 5 cm PJ = 4.5 cm PU = 6.5 cm.
Solution:
Q.3: Construct a Rhombus BEND with BN = 5.6 cm DE = 6.5 cm.
Solution:
As we know, the diagonals of a rhombus bisect each other at a right angle.
Let us consider that these diagonals intersect each other at a point O in rhombus.
Hence, EO = OD = 3.25 cm
Here is a rough sketch of the rhombus BEND:
Step 1: Draw a line segment BN of 5.6 cm and also draw its perpendicular bisector. It will intersect the line segment BN at point O.
Step 2: Taking O as the centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at point D and E.
Step 3: Join points D and E to points B and N.
BEND is the required Rhombus.
Q.4: Construct a Quadrilateral PLAN with PL = 4 cm LA = 6.5 cm ∠P = 90° ∠A = 110° ∠N = 85°.
Solution:
The sum of four angles of a quadrilateral is 360°.
∠P + ∠L + ∠A + ∠N = 360°
90° + ∠L + 110° + 85° = 360°
285° + ∠L = 360°
∠L = 360° − 285° = 75°
Q.5: Construct a Parallelogram HEAR where HE = 5 cm EA = 6 cm ∠R = 85° .
Solution:
Q.6: Construct a Quadrilateral DEAR where DE = 4 cm EA = 5 cm AR = 4.5 cm ∠E = 60°.
Solution:
Step 1: Draw a line segment DE of 4 cm and an angle of 60º at point E. As A is 5 cm away from E, cut the line segment EA 5 cm from point E on this ray.
Step 2: Again draw an angle of 90º at point A. As R is 4.5 cm away from A, cut a line segment RA of 4.5 cm from A on this ray.
Step 3: Join D to R.
Also check:
### Class 8 Chapter 4 -Practical Geometry Extra Questions
1. Construct a quadrilateral PQRS where PQ = 4.4 cm, QR = 4 cm, RS = 6.4 cm, SP = 3.8 cm and PR = 6.6 cm.
2. Construct a quadrilateral PQRS where PQ = 5.4 cm, QR = 2.5 cm, RS = 4 cm, SP =6.5 cm and PR = 5cm.
3. Construct a quadrilateral PQRS where PQ = 4 cm, QR = 5.6 cm, RS = 4.5 cm, SP = 5 cm AND PR = 6.5cm.
4. Construct a quadrilateral PQRS where PQ = 3.6 cm, QR = 5.5 cm, RS = 5 cm ∠B = 125° and ∠C= 80°.
5. Construct a quadrilateral PQRS where PQ = 6 cm, QR = 4 cm, RS = 4 cm, ∠Q = 95° and ∠R= 90°.
6. Construct a quadrilateral PQRS where PQ = 4 cm, QR = 3 cm, ∠P = 75° ∠Q = 80° and ∠R= 120°
7. Construct a quadrilateral PQRS where PQ = 4 cm, QR = 5 cm, ∠P = 50°,∠Q = 110° and ∠R= 70°.
8. Construct a quadrilateral PQRS where PQ = 6 cm, QR = 6 cm, RS = SP = 4.5 cm ∠Q = 120°
9. Construct a quadrilateral PQRS where PQ = 7.5 cm, QR = 6 cm, RS = 6 cm, SP =5 cm and PR = 10 cm.
10. Construct a quadrilateral PQRS where PQ = 5 cm, QR = 5.5 cm, RS = 2.5 cm, SP = 7.1 cm and PR = 8cm.
11. Construct the following Quadrilateral ABCD
AB = 4.5 cm
BC = 5.5 cm
CD = 4 cm |
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Math can be a challenging subject for many students. But there is help available in the form of Slope problem solver. Integral equations are a powerful tool for solving mathematical problems. However, they can be difficult to solve. In general, an integral equation is an equation that involves an integral. The most common type of integral equation is a differential equation. A differential equation is an equation that involves a derivative. For example, the equation y'=y^2 is a differential equation. To solve a differential equation, you first need to find the integrating factor. The integrating factor is a function that multiplies the derivatives in the equation. It allows you to rewrite the equation as an equivalent first-order differential equation. Once you have found the integrating factor, you can use it to rewrite the original equation as an equivalent first-order differential equation. You can then solve the new equation using standard methods. In general, solving an integral equation requires significant mathematical knowledge and skill. However, with practice, it is possible to master this technique and use it to solve complex problems.
To solve a factorial, simply multiply the given number by every number below it until you reach one. So, to solve 5!, you would multiply 5 by 4, then 3, then 2, and then 1. The answer would be 120. It is important to start with the given number and work your way down, rather than starting with one and working your way up. This is because the factorial operation is not commutative - that is, 5! is not the same as 1 x 2 x 3 x 4 x 5. When solving factorials, always start with the given number and work your way down to one.
A radical is a square root or any other root. The number underneath the radical sign is called the radicand. In order to solve a radical, you must find the number that when multiplied by itself produces the radicand. This is called the principal square root and it is always positive. For example, the square root of 16 is 4 because 4 times 4 equals 16. The symbol for square root is . To find other roots, you use division. For example, the third root of 64 is 4 because 4 times 4 times 4 equals 64. The symbol for the third root is . Sometimes, you will see radicals that cannot be simplified further. These are called irrational numbers and they cannot be expressed as a whole number or a fraction. An example of an irrational number is . Although radicals can seem daunting at first, with a little practice, they can be easily solved!
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# Circumference of a Circle
In Mathematics, the circumference of any shape defines the path or the boundary that surrounds the shape. In other words, the circumference is also called the perimeter, which helps to identify the length of the outline of any shape. As we know, the perimeter and area of circle are the two important parameters of a circle. In this article, we will discuss the “Circumference of a circle” or “Perimeter of circle” with its definition, formula, methods to find the circle’s circumference with many solved examples.
## Circle’s Circumference
Circumference of the circle or perimeter of the circle is the measurement of the boundary of the circle. Whereas the area of circle defines the region occupied by it. If we open a circle and make a straight line out of it, then its length is the circumference. It is usually measured in units, such as cm or unit m.
When we use the formula to calculate the circumference of the circle, then the radius of the circle is taken into account. Hence, we need to know the value of the radius or the diameter to evaluate the perimeter of the circle.
## Circumference of a Circle Formula
The Circumference (or) perimeter of circle = 2πR
where,
R is the radius of the circle
π is the mathematical constant with an approximate (up to two decimal points) value of 3.14
Again,
Pi (π) is a special mathematical constant; it is the ratio of circumference to diameter of any circle.
where C = π D
C is the circumference of the circle
D is the diameter of the circle
For example: If the radius of the circle is 4cm then find its circumference.
Circumference = 2πr
= 2 x 3.14 x 4
= 25.12 cm
Also, check:
### Area of a Circle Formula
Area of any circle is the region enclosed by the circle itself or the space covered by the circle. The formula to find the area of the circle is;
A = πr2
Where r is the radius of the circle, this formula is applicable to all the circles with different radii.
## Perimeter of Semi-Circle
The semi-circle is formed when we divide the circle into two equal parts. Therefore, the perimeter of the semi-circle also becomes half.
Hence, Perimeter = πr +2r
## Area of Semi-Circle
Area of the semi-circle is the region occupied by a semi-circle in a 2D plane. The area of the semi-circle is equal to half of the area of a circle, whose radii are equal.
Therefore, Area = πr2/2
Thus, we can define three different formulas to find the perimeter of circle (i.e. circumference of a circle).
Formula 1: When the radius of a circle is known.
Circumference of a circle = 2πr
Formula 2: When the diameter of a circle is known.
Circumference = πd
Formula 3: When the area of a circle is known, we can write the formula to find the perimeter of the circle as:
C = √(4πA)
Here,
C = Circumference of the circle
A = Area of the circle
Summary
Circumference of Circle 2πr Area of circle πr2 Perimeter of semi-circle πr + 2r Area of semi-circle πr2/2
The distance from the centre to the outer line of the circle is called a radius. It is the most important quantity of the circle based on which formulas for the area and circumference of the circle are derived. Twice the radius of a circle is called the diameter of the circle. The diameter cuts the circle into two equal parts, which is called a semi-circle.
## What is the Circumference of Circle?
The meaning of circumference is the distance around a circle or any curved geometrical shape. It is the one-dimensional linear measurement of the boundary across any two-dimensional circular surface. It follows the same principle behind finding the perimeter of any polygon, which is why calculating the circumference of a circle is also known as the perimeter of a circle.
A circle is defined as a shape with all the points are equidistant from a point at the centre. The circle depicted below has its centre lies at point A.
The value of pi is approximately 3.1415926535897… and we use a Greek letter π (pronounced as Pi) to describe this number. The value π is a non-terminating value.
For circle A (as given below), the circumference and the diameter will be-
In other words, the distance surrounding a circle is known as the circumference of the circle. The diameter is the distance across a circle through the centre, and it touches the two points of the circle perimeter. π shows the ratio of the perimeter of a circle to the diameter. Therefore, when you divide the circumference by the diameter for any circle, you obtain a value close enough to π. This relationship can be explained by the formula mentioned below.
C/d = π
Where C indicates circumference and d indicates diameter. A different way to put up this formula is C = π × d. This formula is mostly used when the diameter is mentioned, and the perimeter of a circle needs to be calculated.
Circumference to Diameter
We know that the diameter of a circle is twice the radius. The proportion between the circumference of a circle and its diameter is equal to the value of Pi(π). Hence, we say that this proportion is the definition of the constant π.
(i.e) C= 2πr
C= πd (As, d = 2r)
If we divide both sides by the diameter of the circle, we will get the value that is approximately close to the value of π.
Thus, C/d = π.
## How to Find Circumference?
Method 1: Since it is a curved surface, we can’t physically measure the length of a circle using a scale or ruler. But this can be done for polygons like squares, triangles and rectangles. Instead, we can measure the circumference of a circle using a thread. Trace the path of the circle using the thread and mark the points on the thread. This length can be measured using a normal ruler.
Method 2: An accurate way of knowing the circumference of a circle is to calculate it. For this, the radius of the circle has to be known. The radius of a circle is the distance from the centre of the circle and any point on the circle itself. The figure below shows a circle with radius R and centre O. The diameter is twice the radius of the circle.
## Solved Examples on Perimeter of Circle
Example 1:
What is the circumference of the circle with diameter 4 cm?
Solution:
Since the diameter is known to us, we can calculate the radius of the circle,
Therefore, Circumference of the Circle = 2 x 3.14 x 2 = 12.56 cm.
Example 2:
Find the radius of the circle having C = 50 cm.
Solution:
Circumference = 50 cm
As per formula, C = 2 π r
This implies, 50 = 2 π r
50/2 = 2 π r/2
25 = π r
or r = 25/π
Therefore, the radius of the circle is 25/π cm.
Example 3:
Find the perimeter of circle whose radius is 3 cm?
Solution:
We know that the circumference or the perimeter of a circle is 2πr units.
Now, substitute the radius value in the formula, we get
C = (2)(22/7)(3) cm
C = 18.857 cm
Therefore, the circumference of circle is 18.857 cm.
Example 4:
Calculate the perimeter of circle in terms of π, whose diameter is 10m.
Solution:
Given: Diameter = 10m.
Hence, radius = diameter/2 = 10/2 = 5 m.
We know that, perimeter of circle = 2πr units
C = 2π(5) = 10π m.
Therefore, the perimeter of circle in terms of π, whose diameter 10 cm is 10π m.
### Practice Questions
1. Calculate the perimeter of circle whose diameter is 8 cm.
2. What will be the diameter of a circle if it’s C = 10 cm?
3. If C = 12 cm, what will be its radius?
4. What is the circumference of a 16-inch circle?
5. What is the circumference of a 6 mm circle?
#### Watch The Below Video to Learn The Basics of Circles
To learn all concepts in Math in a more engaging way, register at BYJU’S. Also, watch interesting videos on various Maths topics by downloading BYJU’S– The Learning App from Google Play Store or the app store.
### What is the Circumference of a Circle?
The circumference of a circle is defined as the linear distance around it. In other words, if a circle is opened to form a straight line, then the length of that line will be the circle’s circumference.
### How to Calculate the Circumference of a Circle?
To calculate the circumference of a circle, multiply the diameter of the circle with π (pi). The circumference can also be calculated by multiplying 2×radius with pi (π=3.14).
### How to Calculate Diameter from Circumference?
The formula for circumference = diameter × π
Or, diameter = circumference/π
So, the diameter of the circle in terms of circumference will be equal to the ratio of the circumference of the circle and pi.
### What is the Circumference of a Circle with Radius 24 inches?
Circumference = 2×π×r
C = 2×3.14×24
C = 150.72 inches |
The notion of cryptography stems from the word”Crypto” as well as the contemporary idea of cryptography is all about privacy, security and the capacity to transact securely.
Most people would concur that all these things fall under the notion of cryptography, although there are additional uses for cryptography.
Very well, you can use numbers and mathematics to help in the understanding paper writing of cryptography and this I will show you how exactly is mathematics. You will find various means of employing the mathematics and statistics involved with cryptography, also you may use it if you would like to provide a key, or when you would like to greatly help in the comprehension of a key. When you want to help provide the security for your data and defend it and you also might also use it.
First let’s look at how is math. Let’s think about a key. In cryptography we use a top secret key decrypt it and then to encrypt information. There is A key composed of several factors, the quantity of keys that will have to set up a piece of information.
It’s possible to figure out just how many keys that there are and how many my sources that there are demanded, therefore the primary element is more than the second issue. It’s likely to calculate the square root of two, and then to do the rest by dividing by 2, thus multiplying the last variable and then multiplying by the variety of keys required to decrypt the info. That’s the way the mathematics.
Then there was certainly just another way of applying will be math and numbers to help in the understanding of just the way exactly is mathematics used in cryptography. There is A factor needed to assist in the calculation of a secret. You definitely ought to choose the square foot of the number of factors necessary to multiply the secret, to find the number of keys required to decrypt the 43, In the event you require a key to be multiplied with some factors.
Utilizing the amounts to calculate a secret may possibly be inadequate, because you’ll find different approaches to determine it. For instance, you may calculate the square root of two, divide the range of keys necessary to decrypt the info from two, and then multiply by the number of keys necessary to encrypt the data, then divide by twoand multiply the result https://knowledge.wharton.upenn.edu/article/student-loan-debt-crisis/ by the variety of keys required to authenticate the information. That’s how is mathematics.
You may use exactly precisely the exact identical way is math and numbers to find just how is math used in cryptography. You are able to use it in order to come across the aspects that are needed to multiply the people key, to get the variety of keys required to decrypt the info. |
# A Quartic Equation
### Problem
Solve $(x+1)^{4}+(x+5)^{4}=82.$
### The way you've been taught
Being handed down an equation with integer coefficients of degree greater than 1, there is always a hope that the equation has integer solutions. If it does, they can be found via Viète's formulas, assisted by some guessing, division of polynomials, and good luck.
A quartic - fourth degree polynomial - with roots $\alpha,\beta,\gamma,\delta$ equals $a(x-\alpha)(x-\beta)(x-\gamma)(x-\delta),$ for some $a\ne 0.\;$ It follows that its free (of $x)$ coefficient equals $a\alpha\beta\gamma\delta,$ so that if it's an integer then the roots may be found among its divisors.
By direct verification,
$P(x)=(x+1)^{4}+(x+5)^{4}-82 = 2x^{4}+24x^{3}+156x^{2}+504x+544=0.$
$544=2^{5}\cdot 17.\;$ Since all coefficients of the equation are positive, it can't have positive roots.
Start checking:
\begin{align} P(-1)&=0^{4}+4^{4}=256\ne 82,\\ P(-2)&=(-1)^{4}+3^{4}=1+81=82\\ P(-4)&=(-3)^{4}+1^{4}=81+1=82\\ P(-8)&=(-7)^{4}+(-3)^{4}\gt 82. \end{align}
It is clear that, for other possible integer values of $x,$ $-16,-32,$ and the multiples of $17$ the polynomial exceeds $82.\;$ So there are two integer solutions $x=-2$ and $x=-4.\;$ A quadratic polynomial with these roots equals $Q(x)=x^{2}+6x+8,$ implying $P(X)=Q(x)R(x),$ where $R(x)$ is another quadratic polynomial, which can be found by the "long division algorithm" to be $Q(x)=2x^{2}+12x+68=2(x^{2}+6x+34).\;$ We may now safely apply the quadratic formula to solving $x^{2}+6x+34=0:$
$x_{1,2}=-3\pm\sqrt{9-34}=-3\pm 5i.$
where $i^{2}=-1.\;$ The equation thus has four roots: $-2,-4,-3\pm 5i.$
### Clever substitution 1
Set $a=x+1$ and $b=x+5.\;$ So defined $a$ and $b$ satisfy a system of equations:
$a^{4}+b^{4}=82\\ b - a = 4.$
From the second equation, $a^{2}+b^{2}=(a-b)^{2}+2ab=2(8 + ab).\;$ In the same vein,
\begin{align} a^{4}+b^{4} &= (a^{2}+b^{2})^{2}-2(ab)^{2}\\ &=4(8+ab)^{2}-2(ab)^{2}\\ &=256+64ab+2(ab)^{2}, \end{align}
which leads to the equation $(ab)^{2}+32(ab)+87=0.\;$ With the quadratic formula, $(ab)_{1,2}=-16\pm\sqrt{256-87}=-16\pm 13;$ so that there are two possible values for the product $ab=(x+1)(x+5):$ $-3$ and $-29.\;$ The quadratic equation $(x+1)(x+5)=-3$ has two roots $-2$ and $-4.\;$ The second equation $(x+1)(x+5)=-29$ gives the complex pair $-3\pm 5i.$
### Clever substitution 2
Let $y=x+3.\;$ Then $x+1=y-2$ and $x+5=y+2.\;$ In terms of $y$ the equation becomes $(y-2)^{4}+(y+2)^{4}=82.\;$ From the binomial theorem, the odd powers of $y$ cancel out, leaving $y^{4}+24y^{2}-25=0.\;$ This is an example of biquadratic. Setting, say, $z = y^2$ reduces it to a quadratic equation in $z,$ which has two solutions $1$ and $-25.\;$ $y^{2}=1$ gives $y_{1,2}=\pm 1,$ so that $x+2=\pm 1.\;$ $y^{2}=25$ leads to two complex values, as above.
### Acknowledgment
The problem has been posted at the Short Mathematical Idea facebook group. The first clever solution is by Kunihiko Chikaya, the second by Regragui El Khammal. |
# NCERT solutions for class 10 Maths chapter 1 Real Numbers (Exercise – 1.1)
## Ex 1.1 Question 1.
Use Euclid’s Division Algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 × 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to obtain
135 = 90 × 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
90 = 2 × 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 × 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.
(iii) 867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255 × 3 + 102
Since remainder 102 ≠ 0, we apply the division lemma to 255 and 102 to obtain
255 = 102 × 2 + 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
102 = 51 × 2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51, Therefore, HCF of 867 and 255 is 51.
## Ex 1.1 Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solution:
Let a be any positive integer and b = 6.
Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 × 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer 6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 is an integer Clearly,
6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5 .
## Ex 1.1 Question 3.
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid’s algorithm to find the HCF.
616 = 32 × 19 + 8
32 = 8 × 4 + 0
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.
## Ex 1.1 Question 4.
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Solution:
Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2 Or,
a² = (3q)² or (3q + 1)² or (3q + 2)²
= (3q)² or 9q² + 6q + 1 or 9q² + 12q + 4
= 3 × (3q²) or 3 × (3q² + 2q) + 1 or 3 × (3q² + 4q + 1) + 1
= 3k1 or 3k2 + 1 or 3k3 + 1
Where k1, k2, and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
## Ex 1.1 Question 5.
Use Euclid’s Division Lemma to show that the cube of any positive integer is either of the form 9m, 9m + 1 or 9m + 8.
Solution:
Let a be any positive integer and b = 3 a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
a = 3q or 3q + 1 or 3q + 2
Therefore, every number can be represented as these three forms.
There are three cases.
Case 1: When a = 3q, a³ = (3q)³ = 27q³ = 9(3q³ )= 9m
Where m is an integer such that m = 3q3
Case 2: When a = 3q + 1, a³ = (3q +1)³ a³ = 27q³ + 27q² + 9q + 1 a³ = 9(3q³ + 3q² + q) + 1 a³ = 9m + 1
Where m is an integer such that m = (3q3 + 3q2 + q)
Case 3: When a = 3q + 2, a³ = (3q +2)³ a³ = 27q³ + 54q² + 36q + 8 a³ = 9(3q³ + 6q² + 4q) + 8 a³ = 9m + 8
Where m is an integer such that m = (3q³ + 6q² + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
### UKPSC Forest Guard Exam 09 April 2023 – Answer Key
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# How do you simplify the expression sqrt3(2+3sqrt6)?
May 13, 2017
See a solution process below:
#### Explanation:
First, expand the parentheses by multiplying each term within the parentheses by the factor outside the parentheses:
$\textcolor{red}{\sqrt{3}} \left(2 + 3 \sqrt{6}\right) \implies$
$\left(\textcolor{red}{\sqrt{3}} \cdot 2\right) + \left(\textcolor{red}{\sqrt{3}} \cdot 3 \sqrt{6}\right) \implies$
$2 \sqrt{3} + 3 \sqrt{3} \sqrt{6}$
We can now use this rule for radicals to simplify the term on the right:
$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$
$2 \sqrt{3} + 3 \sqrt{3} \sqrt{6} \implies$
$2 \sqrt{3} + 3 \sqrt{3 \cdot 6} \implies$
$2 \sqrt{3} + 3 \sqrt{18}$
We can use this same rule in reverse to further simplify the term on the right:
$2 \sqrt{3} + 3 \sqrt{18} \implies$
$2 \sqrt{3} + 3 \sqrt{9 \cdot 2} \implies$
$2 \sqrt{3} + 3 \sqrt{9} \sqrt{2} \implies$
$2 \sqrt{3} + \left(3 \cdot 3\right) \sqrt{2} \implies$
$2 \sqrt{3} + 9 \sqrt{2}$ |
Before your child can begin to practice division problems, they first need to understand the concept of division. Explain it to them by relating the idea of division to the idea of sharing. Help them conceptualize it by explaining how a number of items can be shared equally between groups, and give them examples.
Subsequently, What is division simple words?
The division is a method of distributing a group of things into equal parts. It is one of the four basic operations of arithmetic, which gives a fair result of sharing. … If 3 groups of 4 make 12 in multiplication; 12 divided into 3 equal groups give 4 in each group in division.
Also, How do you explain division to Year 2?
How do you explain division to a child?
Before your child can begin to practice division problems, they first need to understand the concept of division. Explain it to them by relating the idea of division to the idea of sharing. Help them conceptualize it by explaining how a number of items can be shared equally between groups, and give them examples.
Last Review : 10 days ago.
What are some Division words?
The terms used in division are dividend, divisor, quotient and remainder.
How do you explain division in words?
The number which is divided is called the dividend. The number which divides is called the divisor. The number which is the result of the division is called the quotient. If there is any number left over, it is called the remainder.
How do you explain division?
Division is breaking a number up into an equal number of parts. Example: 20 divided by 4 = ? If you take 20 things and put them into four equal sized groups, there will be 5 things in each group.
How do you divide in simple method?
– Setup the division problem (84/7).
– Divide 8 by 7 to get 1. …
– Multiply 1 and 7 to get 7. …
– Subtract 7 from 8 to get 1.
– Carry down the 4.
– Divide 14 by 7 to get 2. …
– Multiply 2 by 7 to get 14.
– Subtract 14 from 14 to get 0.
How do you explain division to a 5 year old?
Before your child can begin to practice division problems, they first need to understand the concept of division. Explain it to them by relating the idea of division to the idea of sharing. Help them conceptualize it by explaining how a number of items can be shared equally between groups, and give them examples.
How do you divide step by step?
– Step 1: D for Divide. How many times will 5 go into 65? …
– Step 2: M for Multiply. You multiply your answer from step 1 and your divisor: 1 x 5 = 5. …
– Step 3: S for Subtract. Next you subtract. …
– Step 4: B for Bring down. …
– Step 1: D for Divide. …
– Step 2: M for Multiply. …
– Step 3: S for Subtract.
How do you introduce a division?
When you start teaching division to your child you should introduce division as being a sharing operation where objects are shared (or divided) into a number of groups of equal number. Once you have built an understanding of the concept of division you can try using these division worksheets.
How do you solve division step by step?
– Divide the tens column dividend by the divisor.
– Multiply the divisor by the quotient in the tens place column.
– Subtract the product from the divisor.
– Bring down the dividend in the ones column and repeat.
What are the steps to teach division?
– Divide the tens column dividend by the divisor.
– Multiply the divisor by the quotient in the tens place column.
– Subtract the product from the divisor.
– Bring down the dividend in the ones column and repeat.
What are the 5 steps of division?
It follows the same steps as that of long division, namely, – divide, multiply, subtract, bring down and repeat or find the remainder. Here’s an example of long division with decimals. 123454321 when divided by 11111 gives a quotient of 11111 and remainder 0.
What is the meaning of divide in math?
more … To split into equal parts or groups. It is “fair sharing”. Example: there are 12 chocolates, and 3 friends want to share them, how do they divide the chocolates? Answer: They should get 4 each.
How do you introduce division to a child?
Since your child has a basic understanding of multiplication and most of their times tables memorized, this is a good way to introduce division. The number 24, for example, can be made by multiplying 3 and 8. Show your child that 24 can be broken into 3 groups of 8 by reversing their multiplication facts. |
NCERT Class 10 Maths Surface Areas and Volumes
# NCERT Class 10 Maths Surface Areas and Volumes
The chapter 13 begins with the introduction of solids and solids made up of combinations of two or more basic solids which we encounter in day-to-day life. Then the formula to determine the total surface area of a combination of solids is introduced. An exercise follows this concept which has basic problems that can be solved using the formula discussed in the previous section. Next, the concept of the volume of a combination of solids is introduced and the formula to calculate the same is presented. Then we have the concept of conversion of solid from one shape to another and an exercise follows it. The concept of the frustum of a right circular cone is introduced and the formula to determine its volume, curved surface area and total surface area is discussed.
## Chapter 13 Ex.13.1 Question 1
$$2$$ cubes each of volume $$64 \;\rm{cm}^3$$ are joined end to end. Find the surface area of the resulting cuboid.
### Solution
What is known?
Two cubes each of volume $$64 \;\rm{cm}^3$$ are joined end to end.
What is unknown?
Surface area of the resulting cuboid when two cubes are joined end to end.
Reasoning:
We will find the length of the edge of each cube by using the formula for volume of a cube $$=a^3$$ , where length of the edge is $$a.$$
As the cubes are joined end to end, they will appear as follows
Using the formula for Surface area of a cuboid \begin{align} = 2\left( {lb + bh + lh} \right)\end{align} ,
where $$l,b\,\,{\rm{ and }}\,\,h$$ are length, breadth and height respectively. We’ll be able to get the answer.
Steps:
Let the length of the edge of each cube is $$a$$
Therefore, volume of the cube $$=a^3$$
volume of the cube,
\begin{align} {a^3} &= 64 \rm {c{m^3}}\\{a^3} &= 64 \rm {c{m^3}}\\a &= \sqrt[3]{{64 \rm c{m^3}}}\\a &= \sqrt[3]{{{{\left( {4 \rm {cm}} \right)}^3}}}\\a &= 4 \rm {cm}\end{align}
Therefore,
Length of the resulting cuboid,$$l = a = 4\rm{cm}$$
Breadth of the resulting cuboid,$$b = a = 4\rm{cm}$$
Height of the resulting cuboid,$$h = 2a = 2 \times 4\rm{cm} = 8\rm{cm}$$
Surface area of the resulting cuboid
\begin{align} &= 2\left( {lb + bh + lh} \right) \\&= 2 ( \rm 4{cm}\! \times\! 4{cm}\! +\! 4{cm}\! \times\! 8 \rm {cm} \!+ \!4{cm} \!\times\! 8{cm} ) \\&= 2 (16 \rm {c{m^2}} + 32{c{m^2}} + 32 \rm {c{m^2}}) \\\ &= 2 \times 80 \rm {c{m^2}}\\&= 160{\rm c{m^2}}\end{align}
## Chapter 13 Ex.13.1 Question 2
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is $$14\;\rm{cm}$$ and the total height of the vessel is $$13 \;\rm{cm}$$. Find the inner surface area of the vessel.
### Solution
What is known?
The diameter of the hemisphere is $$14\rm{ cm}$$ and total height of the vessel is $$13\rm {cm.}$$
What is unknown?
The inner surface area of the vessel.
Reasoning:
Create a figure of the vessel according to the given description
From the figure it’s clear that the inner surface area of the vessel includes the CSA of the hemisphere and the cylinder.
Inner surface area of the vessel $$=$$ CSA of the hemisphere $$+$$ CSA of the cylinder
We will find the area of the vessel by using formulae;
CSA of the hemisphere $$= 2\pi {r^2}$$
where $$r$$ is the radius of the hemisphere
CSA of the cylinder $$= 2\pi rh$$
where $$r$$ and $$h$$ are the radius and height of the cylinder respectively.
Height of the cylinder $$=$$ Total height of the vessel $$–$$ height of the hemisphere
Steps:
Diameter of the hemisphere,$$d = 14 \rm cm$$
Radius of the hemisphere, \begin{align}r = \frac{{14 \rm cm}}{2} = 7\rm{cm}\end{align}
Height of the hemisphere $$=$$ radius of the hemisphere, $$r = 7 \rm cm$$
Radius of the cylinder,$$r = 7 \rm cm$$
Height of the cylinder $$=$$ Total height of the vessel $$–$$ height of the hemisphere
$h = 13 \rm cm - 7cm = 6cm$
Inner surface area of the vessel $$=$$ CSA of the hemisphere $$+$$ CSA of the cylinder
\begin{align}&= 2\pi {r^2} + 2\pi rh\\&= 2\pi r\left( {r + h} \right)\\&= 2 \times \frac{{22}}{7} \times 7\rm{cm} \left( {7{cm} + 6{cm}} \right)\\&= 2 \times 22 \times 13 \rm {c{m^2}}\\&= 572 \rm {c{m^2}}\end{align}
## Chapter 13 Ex.13.1 Question 3
A toy is in the form of a cone of radius $$3.5\; \rm{cm}$$ mounted on a hemisphere of same radius. The total height of the toy is $$15.5 \;\rm{cm}$$. Find the total surface area of the toy.
### Solution
What is known?
The toy is in the form of a cone of radius $$=3.5\, \rm{cm}$$ mounted on a hemisphere with the same radius. The total height of the toy is $$15.5 \,\rm{cm}$$.
What is unknown?
The total surface area of the toy.
Reasoning:
We can create the figure of the toy as per given information
From the figure it’s clear that total surface area of the toy includes CSA of the cone and hemisphere.
Total surface area of the toy $$=$$ CSA of the hemisphere $$+$$ CSA of the cone
We will find the total area of the toy by using formulae;
CSA of the hemisphere $$= 2\pi {r^2}$$
where $$r$$ is the radius of the hemisphere
CSA of the cone $$= \pi rl$$
where $$r$$ and $$l$$ are the radius and slant height of the cone respectively.
Slant height of the cone, $$l = \sqrt {{r^2} + {h^2}}$$
Height of the cone, $$h =$$ total height of the toy $$–$$ height of the hemisphere
Steps:
Radius of the hemisphere, $$r = 3.5 \rm cm$$
Height of the hemisphere = radius of the hemisphere, $$r = 3.5 \rm cm$$
Radius of the cone, $$r = 3.5 \rm cm$$
Height of the cone $$=$$ Total height of the toy $$–$$ height of the hemisphere
$h = 15.5 \rm cm - 3.5cm = 12cm$
Slant height of the cone, $$l = \sqrt {{r^2} + {h^2}}$$
\begin{align} l &= \sqrt {{r^2} + {h^2}} \\l &= \sqrt {{{\left( {3.5 \rm {cm}} \right)}^2} + {{\left( {12{cm}} \right)}^2}} \\l &= \sqrt {12.25 \rm {c{m^2}} + 144{c{m^2}}} \\l &= \sqrt {156.25 \rm {c{m^2}}} \\l &= 12.5 \rm {cm}\end{align}
Total surface area of the toy $$=$$ CSA of the hemisphere $$+$$ CSA of the cone
\begin{align}&= 2\pi {r^2} + \pi rl\\&= \pi r\left( {2r + l} \right)\\&= \frac{{22}}{7} \!\times \!3.5 \rm {cm} \!\times\! ( 2 \!\times\! 3.5 \rm {cm}\!+\! 12.5{cm}) \\&= \frac{{22}}{7} \!\times\! \frac{7}{2} \rm {cm} \!\times \! \left( {7 \rm {cm} \!+ \!12.5{cm}} \right) \\&= 11 \rm {cm} \times 19.5{cm}\\&= 214.5 \rm {c{m^2}}\end{align}
## Chapter 13 Ex.13.1 Question 4
A cubical block of side $$7 \,\rm{cm}$$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
### Solution
What is known?
A cubical block of side $$7 \,\rm{cm}$$ is surmounted by a hemisphere.
What is unknown?
The greatest diameter the hemisphere can have and the surface area of the solid.
Reasoning:
We can create the figure of the solid as per given information
From the figure its clear that the greatest diameter the hemisphere can have is equal to the cube’s edge. otherwise a part of hemisphere’s rim (circumference of its circular base) will lie outside the top part of the cube.
As it’s clear from the top view of the figure that the top part of the cube is partially covered by hemispherical part.
Total surface area of the solid $$=$$ Surface Area of the cubical part $$-$$ Area of base of hemispherical part $$+$$ CSA of the hemispherical part
We will find the total area of the solid by using formulae;
CSA of the hemisphere $$= 2\pi {r^2}$$
Area of the base of the hemisphere $$= \pi {r^2}$$
where $$r$$ is the radius of the hemisphere
Surface area of the cube $$= 6{l^2}$$
where $$l$$ is the length of the edge of the cube.
Steps:
Length of the edge of the cube, $$l = 7 \rm cm$$
From the figure it’s clear that the greatest diameter the hemisphere can have is equal to the cube’s edge
Diameter of the hemisphere, $$d = l = 7 \rm cm$$
Radius of the hemisphere, \begin{align}r = \frac{d}{2} = \frac{7}{2} \rm cm\end{align}
Total surface area of the solid $$=$$ Surface area of the cubical part $$–$$ Area of the base of the
hemispherical part $$+$$ CSA of the hemispherical part
\begin{align}&= 6{l^2} - \pi {r^2} + 2\pi {r^2}\\&= 6{l^2} + \pi {r^2}\\&= \!6 \!\times \! {\left( {7\rm {cm}} \right)^2} \!+ \! \frac{{22}}{7} \times \! {\left( {\frac{7}{2} \rm {cm}} \right)^2}\\&= 6 \times 49{ \rm c{m^2}} + \frac{{22}}{7} \times \frac{{49}}{4}{ \rm c{m^2}}\\&= 294{ \rm c{m^2}} + 38.5{ \rm c{m^2}}\\&= 332.5{ \rm c{m^2}}\end{align}
## Chapter 13 Ex.13.1 Question 5
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $$l$$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
### Solution
What is known?
Diameter $$l$$ of the hemisphere is equal to the edge of the cube.
What is unknown?
The surface area of the remaining solid.
Reasoning:
We can create the figure of the solid as per given information
From the figure it’s clear that the surface area of the remaining solid includes TSA of the cube, CSA of the hemisphere and excludes base of the hemisphere.
Surface area of the remaining solid $$=$$ TSA of the cubical part$$+$$ CSA of the hemisphericalpart $$–$$ Area of the base of the hemispherical part
We will find the remaining area of the solid by using formulae;
TSA of the cube $$= 6{l^2}$$
where $$l$$ is the length of the edge of the cube
CSA of the hemisphere $$= 2\pi {r^2}$$
Area of the base of the hemisphere $$= \pi {r^2}$$
where $$r$$ is the radius of the hemisphere
Steps:
Diameter of the hemisphere $$=$$ Length of the edge of the cube $$= l$$
Radius of the hemisphere, $$r = \frac{l}{2}$$
Surface area of the remaining solid $$=$$ TSA of the cubical part $$+$$ CSA of the hemisphericalpart $$–$$ Area of the base of the hemispherical part
\begin{align}&= 6{l^2} + 2\pi {r^2} - \pi {r^2}\\&= 6{l^2} + \pi {r^2}\\&= 6{l^2} + \pi {\left( {\frac{l}{2}} \right)^2}\\&= 6{l^2} + \frac{{\pi {l^2}}}{4}\\&= \frac{1}{4}{l^2}\left( {\pi + 24} \right)\end{align}
## Chapter 13 Ex.13.1 Question 6
A medicine capsule is in the shape of cylinder with two hemispheres stuck to each of its ends (see Fig. 13.10). The length of the entire capsule is $$14 \,\rm{mm}$$ and the diameter of the capsule is $$5 \,\rm{mm}$$. Find its surface area.
### Solution
What is known?
A medicine capsule is in the shape of a cylinder with two hemispheres stuck to its ends. The length of the entire capsule $$=14 \,\rm{mm}$$, diameter of the capsule $$=5 \,\rm{mm}$$.
What is the unknown?
The surface area of the capsule.
Reasoning:
Since the capsule is in shape of a cylinder with $$2$$ hemispheres stuck to its ends,
Diameter of the capsule $$=$$ diameter of its cylindrical part $$=$$ diameter of its hemispherical part.
From the figure, it’s clear that the capsule has the curved surface of two hemispheres and the curved surface of a cylinder.
Surface area of the capsule $$= 2 \times$$ CSA of hemispherical part $$+$$ CSA of cylindrical part
We will find the surface area of the capsule by using formulae;
CSA of the hemisphere $$= 2\pi {r^2}$$
where $$r$$ is the radius of the hemisphere
CSA of the cylinder $$= 2\pi rh$$
where $$r$$ and $$h$$ are radius and height of the cylinder respectively.
Length of the cylindrical part $$=$$ Length of the capsule $$- 2 \times$$ radius of the hemispherical part
Steps:
Diameter of the capsule, $$d = 5 \rm mm$$
Radius of the hemisphere, \begin{align}r = \frac{d}{2} = \frac{5}{2} \rm mm\end{align}
Radius of the cylinder, \begin{align}r = \frac{5}{2} \rm mm\end{align}
Length of the cylinder = Length of the capsule$$- 2 \times$$ radius of the hemisphere
$h = 14 \rm mm - 2 \times \frac{5}{2}mm = 9mm$
Surface area of the capsule$$- 2 \times$$ CSA of hemispherical part $$+$$ CSA of cylindrical part
\begin{align}& = 2 \times 2\pi {r^2} + 2\pi rh\\ &= 2\pi r\left( {2r + h} \right)\\ &= \begin{bmatrix} 2 \times \frac{{22}}{7} \times \frac{5}{2} \rm {mm} \times \\ \left( {2 \times \frac{5}{2}\rm {mm} + 9\rm {mm}} \right) \end{bmatrix} \\&= \frac{{110}}{7} \rm{mm} \times 14{mm}\\&= 220 \rm {m{m^2}}\end{align}
## Chapter 13 Ex.13.1 Question 7
A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $$2.1\,\rm{ m}$$ and $$4 \,\rm{m}$$ respectively, and the slant height of the top is $$2.8 \,\rm{m}$$, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs $$500 \;\rm{per}\, \rm{m}^2$$. (Note that the base of the tent will not be covered with canvas.)
### Solution
What is known?
The height and diameter of the cylindrical part are $$2.1 \rm m$$ and $$4 \rm m$$ respectively, and the slant height of the conical top is $$2.8 \rm m$$. The rate of the canvas of the tent is $$₹ 500$$ per $$\rm m^2$$.
What is unknown?
Area and cost of the canvas used for making the tent
Reasoning:
We can create the figure of the tent as per given information
From the figure it’s clear that, diameter of the cylindrical part is equal to the diameter of the conical part as the cylindrical part is surmounted by the conical part.
Since its given that the base of the tent will not be covered by the canvas then the base of the cylindrical part of the tent is not included to get the area of the canvas required.
Visually, the surface of the tent includes the curved surface of the cylindrical part and curved surface of the conical part only.
Area of the canvas used for making the tent is the surface area of the tent.
Area of the canvas used $$=$$ CSA of the cylindrical part $$+$$ CSA of the conical part
We will find the area of the canvas by using formulae;
CSA of the cone $$= \pi rl$$
where $$r$$ and $$l$$ are radius and slant height of the cone respectively.
CSA of the cylinder $$= 2\pi rh$$
where $$r$$ and $$h$$ are radius and height of the cylinder respectively.
Cost of the canvas of the tent $$=$$ Area of the canvas $$\times$$ Rate of the canvas
Steps:
Given
Height of the cylindrical part $$h$$ $$= 2.1 \,\rm{m}$$
Diameter of the cylindrical part, $$d= 4 \,\rm{m}$$
Radius of the cylindrical, \begin{align}r = \frac{d}{2} = \frac{{4 \rm m}}{2} = 2 \rm m\end{align}
Radius of the cone,$$r = 2 \rm m$$
Slant height of the cone, $$l = 2.8 \rm m$$
Area of the canvas used $$=$$ CSA of the cylindrical part $$+$$ CSA of the conical part
\begin{align}&= 2\pi rh + \pi rl\\&= \pi r\left( {2h + l} \right)\\&= \frac{{22}}{7}\! \times\! 2 \rm m \!\times\! ( 2 \!\times\! 2.1 \rm m \!+\! 2.8 \rm m) \\&= \frac{{44}}{7} \rm m \times 7m\\&= 44 \rm {m^2}\end{align}
Cost of the canvas of the tent $$=$$ Area of the canvas $$\times$$ Rate of the canvas
\begin{align}&= 44 \rm {m^2} \times ₹\,500/{m^2}\\&= ₹ 22000\end{align}
## Chapter 13 Ex.13.1 Question 8
From a solid cylinder whose height is $$2.4 \,\rm{cm}$$ and diameter $$1.4\,\rm{ cm}$$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $$\rm{cm}^2$$.
### Solution
What is known?
A solid cylinder of height $$2.4 \,\rm{cm}$$ and diameter $$1.4\,\rm{ cm}$$ from which a conical cavity of the same height and same diameter is hollowed out.
What is unknown?
Total surface area of the remaining solid.
Reasoning:
We can create the figure of the solid as per given information
Since the conical cavity of the same diameter has been hollowed out, its clear that one of the bases of the cylinder is not included in total surface area of solid.
TSA of the remaining solid $$=$$ CSA of the cylindrical part $$+$$ CSA of conical part $$+$$ Area of a cylinderical base
We will find the area of the remaining solid by using formulae;
CSA of the cylinder $$= 2\pi rh$$
Area of the base of the cylinder $$= \pi {r^2}$$
where $$r$$ and $$h$$ are radius and height of the cylinder respectively.
CSA of the cone $$= \pi rl$$
Slant height of the cone, $$l = \sqrt {{r^2} + {h^2}}$$
where $$r, h$$ and $$l$$ are radius, height and slant height of the cone respectively.
Steps:
Height of the cylinder $$=$$ Height of the cone $$= h = 2.4 \rm cm$$
Diameter of the cylinder $$=$$ diameter of the cone $$= d = 1.4 \rm cm$$
Radius of the cylinder $$=$$ radius of the cone \begin{align} = r = \frac{{1.4 \rm cm}}{2} = 0.7 \rm cm\end{align}
Slant height of the cone,$$l = \sqrt {{r^2} + {h^2}}$$
\begin{align}l &= \sqrt {{{\left( {0.7\rm cm} \right)}^2} + {{\left( {2.4cm} \right)}^2}} \\ &= \sqrt {0.49 \rm c{m^2} + 5.76c{m^2}} \\&= \sqrt {6.25 \rm c{m^2}} \\&= 2.5 \rm cm\end{align}
TSA of the remaining solid $$=$$ CSA of the cylindrical part $$+$$ CSA of conical part $$+$$Area of the cylindrical base
\begin{align}&= 2\pi rh + \pi rl + \pi {r^2}\\& = \pi r\left( {2h + l + r} \right)\\&= \!\!\frac{{22}}{7}\! \!\times\! 0.7 \rm cm \!\times\!\! ( 2 \!\times\! 2.4 \rm cm \!+ \! 2.5 \rm cm \!+\! 0.7cm ) \\&= 2.2 \rm cm \times 8cm\\&= 17.6 \rm c{m^2}\end{align}
Hence, the total surface area of the remaining solid to the nearest $$\rm cm^2$$ is $$18 \, \rm cm^2.$$
## Chapter 13 Ex.13.1 Question 9
A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11. If the height of the cylinder is $$10 \rm{cm}$$, and its base is of radius $$3.5 \rm{cm,}$$find the total surface area of the article.
### Solution
What is known?
The height of the cylinder is $$10 \rm{cm}$$, and radius of its base is $$3.5 \rm{cm}$$
What is unknown?
Total surface area of the article.
Reasoning:
From the figure it’s clear that radius of the hemispheres scooped out is same as the radius of base of the cylinder since both the hemispheres are of equal radius.
So, total surface area of the article only includes the CSA of both the hemispheres and the cylinder.
TSA of the article $$= 2 \times$$ CSA of the hemispherical part $$+$$ CSA of the cylindrical part
We will find the TSA of the article by using formulae;
CSA of the hemisphere $$= 2\pi {r^2}$$
where r is the radius of the hemisphere
CSA of the cylinder $$= 2\pi rh$$
where $$r$$ and $$h$$ are radius and height of the cylinder respectively.
Steps:
Height of the cylinder $$= h = 10 \rm cm$$
Radius of the cylinder $$=$$ radius of the hemisphere $$= r = 3.5 \rm cm$$
TSA of the article $$=$$ $$2 \times$$ CSA of the hemispherical part $$+$$ CSA of the cylindrical part
\begin{align}&= 2 \times 2\pi {r^2} + 2\pi rh\\&= 2\pi r\left( {2r + h} \right)\\&= 2\! \times \!\!\frac{{22}}{7}\!\! \times \!3.5 \rm cm \!\times\! ( 2 \!\times\! 3.5 \rm cm\!+ \!10 \rm cm )\\&= 22 \rm cm \!\times \!17 \rm cm\\&= 374 \rm c{m^2}\end{align}
Surface Area and Volumes | NCERT Solutions
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# PROBABILITY, PERMUTATION & COMBINATION SET 1
Q.1 A bag contains four black and five red balls, if three balls are picked at random one after another WITH replacement, what is the chance that they’re all black?
A) 64/729
B) 64/730
C) 63/729
D) 63/730
Ans. A
With replacement means you pick up a ball note down its color and then put it back in the bag again. So Total Number of balls remain same for each event. And Hence probability of picking a black ball (4/9) remains the same in every case.
1st Pick up:4/9;
2nd Pick up: 4/9 ;because we put the ball back in the bag, So probability is same as “1st Pick”.
3rd Pick up: 4/9 ;because we put the ball back in the bag.
So final probability
=1st x 2nd x 3rd
=4/9 x 4/9 x 4/9
=Cube of 4/9
=64/729
Q.2 If you have a spinning wheel with 3 green sectors, 1 blue sector and 1 red sector, what is the probability of getting a green sector? What is the probability of getting a non blue sector?
A) Green = 3/5 ; Non-blue = 4/5
B) Green = 2/5 ; Non-blue = 3/5
C) Green = 4/5 ; Non-blue = 3/5
D) Green = 3/5 ; Non-blue = 2/5
Ans. A
Total number of events = 5
Q.3 A bag has 4 red balls and 2 yellow balls. (The balls are identical in all
respects other than colour). A ball is drawn from the bag without looking into the bag. What is probability of getting a red ball?
A) 2/3
B) 2/4
C) 2/5
D) 2/6
Ans. A
There are in all (4 + 2 =) 6 outcomes of the event. Getting a red ball
consists of 4 outcomes.
Therefore, the probability of getting a red ball is
4/6 = 2/3
Q.4 There are 100 students in a particular class. 60% students play cricket, 30% student play football and 10% students play both the games. What is the number of students who play neither cricket nor football?
A) 25
B) 20
C) 18
D) 15
Ans. B
100- (50+20+10)
= 100-80 = 20
Q.5 In a group of persons, 70% of the persons are male and 30% of the persons are
married. If two-sevenths of the males are married, what fraction of the females is
single?
A) 2/7
B) 1/3
C) 3/7
D) 2/3
Ans. D
Let total no. of persons =100
Therefore total no. of males= 70
Total no. of females= 30
Given that, no. of unmarried persons =30
So, Number of married males=
2/7 * 70=20
Therefore, No. of married females
=30-20 =10
Therefore, No. of unmarried females
=30-10 = 20
Therefore, Required fraction of single females=
20/30 = 2/3 |
# Video: Finding the Area of a Triangle Using Trigonometry
Finding the Area of a Triangle Using Trigonometry
14:12
### Video Transcript
In this video, we’ll see how we can find the area of a triangle using trigonometry. To do this, we’ll need to know the lengths of two sides and their included angle. But before we do that, let’s consider a rectangle. We know that a rectangle has four 90-degree angles. It has a length and a width. And the area is equal to the length times the width. If we draw a diagonal across this rectangle, we’ve divided it in half. And each of these triangles have an area of one-half the length times the width.
Let’s look more closely at the right triangle. When we’re dealing with the triangle, that width becomes a height. And instead of calling it the length, we call it the base. And so, we say that the area of a triangle is one-half the height times the base. The key to using this formula is that the height is perpendicular to the base. If we’re not given a perpendicular height, we can’t use this formula.
Let’s say we’re dealing with a triangle that looks like this. We could call this side the base. But the height of this triangle is the perpendicular distance from the base to the vertex opposite that base. This distance is the height. It’s still true that the area of this triangle is one-half times the height times the base. That formula is true because every triangle is half of a rectangle.
If we double this triangle and move the pieces around, we have a rectangle. This proves that the formula area equals one-half times height times base will find the area of any triangle as long as you have the perpendicular height. But this video is going to help us learn how to solve the area of triangles when we aren’t given the height. In these cases, we need trigonometry to come to the rescue.
Okay, let’s go back. Area equals one-half the height times the base. And we know that the height is a perpendicular intersection of the base. Now, in this triangle, our base is 𝑎. So, let’s just leave that there. We want to use some trigonometry to help us find the height. The height of this triangle actually makes the larger triangle divided into two smaller triangles. And the two smaller triangles that make up the larger triangle are both right-angled triangles.
If we call the point where the height intersects the base point 𝑃, we could say triangle 𝐴𝑃𝐶 is a right triangle. Let’s just move that information up a bit. We can also say that the line segment 𝐴𝑃 is the height of the larger triangle 𝐴𝐵𝐶. We can say that side 𝑏 is the hypotenuse of the smaller triangle, 𝐴𝑃𝐶. And now, let’s consider angle 𝐶. Opposite angle 𝐶 is the line segment 𝐴𝑃 or what we’ve labelled as ℎ.
And we’ve already said that side 𝑏 is the hypotenuse. If we know an opposite and a hypotenuse in a right triangle, what trig value should we be thinking of? We can choose from the sine, cosine, or tangent relationships. But the sine relationship is the one that is the opposite over the hypotenuse. This means we can say that sin of angle 𝜃 equals ℎ over 𝑏. This means that the height over the side length 𝑏 will be equal to the sin of the angle 𝜃.
We’re trying to solve for the height. If we multiply both sides of this equation by the side length 𝑏, then we can say that the height is equal to the sin of 𝜃 times 𝑏. And in case of this triangle, we know that 𝜃 is the sin of angle 𝐶. We can now substitute in for the height, sin of 𝐶 times 𝑏. Another way to find the area of a triangle is one-half times sin 𝐶 times 𝑏 times 𝑎. We can rearrange it to its more common form, the area of a triangle is found by one-half times 𝑎 times 𝑏 times sin of 𝐶.
Before we move on, let’s just walk through this one more time. For our triangle, 𝐴𝐵𝐶, we can find its area if we know the lengths of two sides and the value of its included angle. If we knew side length 𝑎 and 𝑏 and the value of angle 𝐴, we could not use this rule. Nor could we use this rule if we knew the value of side length 𝑎 and side length 𝑏 and the angle 𝐵. We must know the included angle, the angle between the two side lengths we know. To use this formula, we must have two side lengths and an included angle. It does not matter which two sides, as long as the angle we’re given is included between those two sides. Now, let’s move on to some examples.
Which of the following is a formula that can be used to find the area of a triangle? A) one-half 𝑎𝑏 cos 𝐶, B) one-half 𝑎𝑏 sin 𝐶, C) one-third 𝑎𝑏 sin 𝐶, D) one-fourth 𝑎𝑏 cos 𝐶, or E) one-fourth 𝑎𝑏 sin 𝐶.
If we sketch a triangle and label it 𝐴, 𝐵, and 𝐶, the side length opposite vertex 𝐴 is usually labelled with a lower case 𝑎. The side length opposite vertex 𝐵 is labelled with a lower case 𝑏. And we label lower case 𝑐 the side length opposite vertex 𝐶. We have to remember that a triangle is half of a rectangle. And so, it’s unlikely that options C through E would be the answer.
We noticed that options A and B are dealing with the angle at vertex 𝐶, that’s this angle, and the lengths 𝑎 and 𝑏. At this point, we recognize that we have two sides and an included angle. And we know that the height of this triangle will be equal to 𝑏 times the sin of 𝐶. To use trigonometry to solve for the area of a triangle, we take one-half times 𝑎 times 𝑏 times sin of 𝐶, which is option B here.
For this example, we need to apply the formula we’ve been talking about.
In the given figure, work out the area of the triangle to two decimal places.
In this figure, we were given two side lengths and an included angle. Since we have this information, we can use the formula 𝐴 equals one-half times 𝑎𝑏 times sin of 𝐶 to find the area, where 𝑎 and 𝑏 are side lengths and 𝐶 is an included angle. The area of this triangle is found by one-half times 10 times seven times sin of 136 degrees. If you plug this into your calculator, you get 24.3130429 continuing. If your calculator did not give you this answer, you should check and make sure that your calculator is set to degree mode and not to radians.
To get our final answer, though, we want it rounded to two decimal places. There’s a one in the second decimal place. To the right of that, in the third decimal place, the thousandths place, is a three. This means we should round down to 24.31. We weren’t given any units, so we can just say that the area of this triangle is 24.31 units squared.
This example might look just as simple as the last one on the surface, but it’s going to take a few more steps to find the area.
The figure shows a triangular field with sides 670 meters, 510 meters, and 330 meters. Find the area of the field giving the answer to the nearest square meter.
At this point, we’re not given the angle measure of any of the angles inside this triangle. That means we can’t check for a perpendicular distance that could be the height. Before we do anything here, we’re going to have to find at least one of the angles. Because we know all three sides, there is one rule we can use here. We can use an application of the cosine rule, which tells us 𝑐 squared is equal to 𝑎 squared plus 𝑏 squared minus two 𝑎𝑏 times cos of 𝐶.
Now, we actually want to find the measure of angle 𝐶. We don’t know what that is. But we can rearrange this rule so that the cos of 𝐶 is the term that we’re looking for. We can subtract 𝑎 squared and 𝑏 squared from both sides of the equation. Then, we have the statement 𝑐 squared minus 𝑎 squared minus 𝑏 squared equals negative two 𝑎𝑏 cos of 𝐶. And if we divide both sides by negative two 𝑎𝑏, we can say that cos of 𝐶 is equal to 𝑐 squared minus 𝑎 squared minus 𝑏 squared over negative two 𝑎𝑏.
Our goal is to find 𝐶. And the 𝑏 side is equal to 510 meters. And the 𝑎 side is equal to 670 meters. And our third side is 𝑐. 𝑐 squared is 330 squared minus 670 squared minus 510 squared over negative two times 670 times 510. All of this will equal the cos of angle 𝐶. If you enter all of that into your calculator, you will get 0.878109453 continuing is equal to the cos of angle 𝐶.
We need to be careful here. Angle 𝐶 is not equal to 0.878109453 continuing degrees. The cos of angle 𝐶 is equal to this decimal value. To find angle 𝐶, we need to take the inverse cos of 0.878109453 continuing. On most calculators, you can hit cos inverse of the answer, of the previous answer. Angle 𝐶 is equal to 28.58485793 continuing degrees. If your calculator did not give you this answer, you should check and make sure that you’re calculating in degrees and not in radians.
Now that we have a value for one of the angles, we have the value of two side lengths, and an included angle. Which means we can use the area is equal to one-half times 𝑎 times 𝑏 times sin of 𝐶, where 𝑎 and 𝑏 are side lengths and 𝐶 is an included angle. To find the area of this triangle, we want to say the area is equal to one-half times 670 times 510 times the sin of the answer already in your calculator. Why would we do this? This gives us the most accurate answer before we round.
It’s calculating the sin of 28.58485793 continuing. When we do that, we get 81744.85833 continuing. And this is the point where we want to round to the nearest square meter. We wanna round to the nearest whole number. We have a four in the ones place, and the digit to the right of that is an eight, telling us we need to round up. That four rounds up to a five, and everything to the left of the five stays the same. So, we have 81745 square meters. The area of this playing field is 81745 meters squared when we round to the nearest square meter.
Let’s do a quick recap of the key points of finding the area of a triangle using trigonometry. The area of a triangle can be found by multiplying one-half times side length 𝑎 times side length 𝑏 times the sin of the included angle 𝐶. So, we say that this formula requires us to have two sides and an included angle. |
Factors of 100 are the repertoire of both positive and an unfavorable numbers which have the right to be evenly divided into 100. The word hundred was developed in 1920 by nine-year-old Milton Sirotta (1911-1981), nephew that Edward Kasner, a U.S. Mathematician. Learning about the factors of 100 is beneficial in learning progressed Maths concepts. In this lesson, we will calculate the factors of 100, its element factors, its components in pairs, and also we will finish by fixing some examples for far better understanding.
You are watching: What is the prime factorization of 100
Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, and 100Factors that -100: -1, -2, -4, -5, -10, -20, -25, -50 and -100Prime administrate of 100: 100 = 22 × 52
1 What are determinants of 100? 2 How to Calculate determinants of 100? 3 Factors the 100 by element Factorization 4 Factors that 100 in Pairs 5 Important Notes 6 FAQs on determinants of 100
## What are determinants of 100?
The determinants of 100 are all the integers 100 have the right to be split into. The number 100 is an also composite number.
As that is even, it will have actually 2 together its factor. To recognize why the is composite, let"s recall the definition of a composite number. A number having actually a complete count of components in overfill of two is identified as a composite number. ~ above the other hand, a number such as 17 is a element number because it has only 2 determinants i.e. 1 and 17.
Now appropriately the components of 100 room 1, 2, 4, 5, 10, 20, 25, 50, and also 100.
## How to calculation the determinants of 100?
Let"s start calculating the components of 100, beginning with the smallest totality number, i.e., 1. Divide 100 v this number. Is the remainder 0? yes! So, we will get:
100 ÷ 1 = 100100 × 1 = 100
The next entirety number is 2. Currently divide 100 through this number:
100 ÷ 2 = 502 × 50 = 100
Proceeding in a comparable manner, we gain other numbers 100 deserve to be split by. They have the right to be written as:
1 × 100 = 1002 × 50 = 1004 × 25 = 1005 × 20 = 10010 × 10 = 100
Explore components using illustrations and also interactive examples:
## Factors the 100 by prime Factorization
Prime factorization means expressing a composite number as the product the its element factors.
Step 1: To obtain the element factorization of 100, we divide it by its the smallest prime factor, 2 favor 100 ÷ 2 = 50.Step 2: Now, 50 is separated by its smallest prime factor, and the quotient is obtained.Step 3: This process goes ~ above till we get the quotient together 1.The prime factorization of 100 in the form of a element tree is presented below.
The above factorization is the tree diagram depiction of components of 100. Therefore, factors that 100 = 2 × 2 × 5 × 5
Q: Now the we have done the prime factorization of our number, we can multiply them and also get the other factors. Can you shot and uncover out if every the determinants are extended or not?
A: And as you could have already guessed, for prime numbers, there are no other factors.
## Factors that 100 in Pairs
The pair, consisting of numbers that offer 100 once multiplied, is well-known as the variable pair the 100. The complying with are the determinants of 100 in pairs:
The product type of 100 Pair factor 1 × 100 = 100 (1, 100) 2 × 50 = 100 (2, 50) 4 × 25 = 100 (4, 25) 5 × 20 = 100 (5, 20) 10 × 10 = 100 (10, 10) 20 × 5 = 100 (20, 5) 25 × 4 = 100 (25, 4) 50 × 2 = 100 (50, 2) 100 × 1 = 100 (100, 1)
Observe in the table above, ~ 10 × 10, the factors start repeating. So, the is sufficient to find factors until (10,10). If we consider an adverse integers, then both the numbers in the pair factors will be negative i.e. - ve (×) - ve = + ve.
So, we can have an adverse factor bag of 100 as (-1,-100), (-2,-50), (-4,-25), (-5,-20), and (-10,-10).
Important Notes:
The number which us multiply to get 100 room the components of 100.
See more: How Long Is 7 Inches Compared To An Object S? 9 Common Things That Are 7 Inches Long
Factors of 100 space written together 1, 2, 4, 5, 10, 20, 25, 50, and 100.Factor pairs are the pairs of two numbers that, as soon as multiplied, provide the initial number. The pair aspect of 100 space (1,100), (2,50), (4,25), (5,20), and (10,10). |
# 14 in Roman Numerals
The number 14 in Roman numerals is XIV. As we all know, Roman numerals are a technique of representing numbers using the Roman alphabet. In this article, we’ll look at how to write the number 14 in Roman numerals with numerous examples and solutions.
14 in Roman Numerals = XIV
## How to Write 14 in Roman Numerals?
The following method can be used to write the number 14 in Roman numeral:
• Break the number 14 into its simplest form first.
• As a result, 14 = 10 + 5 – 1
• Write the appropriate Roman numerals and add/subtract them.
• Thus,14 = 10 + 5 – 1 becomes XIV = X + V – I.
• As a result, 14 is written as XIV in Roman numerals.
### Roman Numerals Related to the Number 14
Some of the Roman numerals related to the number 14 are as follows:
Roman Numeral for 14 = XIV
Roman Numeral for 114 = 100 + 14 = CXIV
Roman Numeral for 214 = 100 + 100 + 14 = CCXIV
Roman Numeral for 314 = 100 + 100 + 100 + 14 = CCCXIV
Roman Numeral for 414 = 400 + 14 = CDXIV
Roman Numeral for 514 = 500 + 14 = DXIV
Roman Numeral for 614 = 500 + 100 + 14 = DCXIV
Roman Numeral for 714 = 500 + 100 + 100 +14 = DCCXIV
Roman Numeral for 814 = 500 + 100 + 100 + 100 + 14 = DCCCXIV
Roman Numeral for 914 = 900 + 14 = CMXIV
## Video Lesson on Roman Numerals
### 14 in Roman Numerals – Solved Examples
Example 1:
What is the value of 20 – 6? Express the value in Roman numerals.
Solution:
The value of 20 – 6 is 14.
Thus, 14 in Roman numerals is XIV.
Example 2:
Find the sum of L and XIV. Express the sum in Roman numerals.
Solution:
We know that,
L = 50
XIV = 14
So, L + XIV = 50 + 14 = 64.
Hence, 64 in Roman numerals is LXIV.
Visit BYJU’S – The Learning App and download the app to learn all Maths concepts easily by watching more learning videos.
## Frequently Asked Questions on 14 in Roman Numerals
Q1
### What is 14 in Roman numerals?
14 in Roman numerals is XIV.
Q2
### Express the value of 10 + 4 in Roman numerals.
The value of 10 + 4 is 14. So, X + IV = XIV. Hence, 14 in Roman numerals is XIV.
Q3
### What should be added to 6 to get the Roman numerals, XIV? Express the value in words.
8 should be added to 6 to get the Roman numerals XIV, which is 14. Hence, 8 in Roman numerals is VIII.
Test your Knowledge on 14 in Roman Numerals |
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# How do you solve $6 + ( - 6)$?
Last updated date: 08th Aug 2024
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Hint: Here we will take the given expression and open the bracket and will solve accordingly. Since there is a positive sign outside the bracket therefore there will be no change in the sign of the terms inside the bracket. Apply this concept and solve accordingly for the resultant required value.
Complete step-by-step solution:
Take the given expression: $6 + ( - 6)$
Open the bracket given in the above expression, remember when there is a positive sign outside the bracket then the sign of the terms inside the bracket does not change. Positive terms will remain positive and the negative term will remain negative when brackets are opened and hence, there is no change of the sign of the terms.
$6 + ( - 6) = 6 - 6$
Like terms with the same value and opposite sign cancel each other.
$6 + ( - 6) = 0$
This is the required solution.
Additional information: Be careful about the sign while doing simplification among the like terms.
i) Addition of two positive terms gives the positive term
ii) Addition of one negative and positive term, you have to do subtraction and give signs of bigger numbers, whether positive or negative.
iii) Addition of two negative numbers gives a negative number but in actual you have to add both the numbers and give a negative sign to the resultant answer.
Note: Since there was a positive sign outside the bracket there was no change in the sign of the terms inside the bracket but when there is a negative sign outside the bracket then there will be a change of sign for all the terms inside the bracket while opening the bracket. Positive terms become negative and the negative term becomes positive. |
Mathematics » Math Models and Geometry I » Use Properties of Rectangles, Triangles, and Trapezoids
# Using the Properties of Triangles
## Using the Properties of Triangles
We now know how to find the area of a rectangle. We can use this fact to help us visualize the formula for the area of a triangle. In the rectangle from the previous lesson, we’ve labeled the length $$b$$ and the width $$h,$$ so it’s area is $$bh.$$
The area of a rectangle is the base, $$b,$$ times the height, $$h.$$
We can divide this rectangle into two congruent triangles (see the figure below). Triangles that are congruent have identical side lengths and angles, and so their areas are equal. The area of each triangle is one-half the area of the rectangle, or $$\frac{1}{2}bh.$$ This example helps us see why the formula for the area of a triangle is $$A=\frac{1}{2}bh.$$
A rectangle can be divided into two triangles of equal area. The area of each triangle is one-half the area of the rectangle.
The formula for the area of a triangle is $$A=\frac{1}{2}bh,$$ where $$b$$ is the base and $$h$$ is the height.
To find the area of the triangle, you need to know its base and height. The base is the length of one side of the triangle, usually the side at the bottom. The height is the length of the line that connects the base to the opposite vertex, and makes a $$\text{90°}$$ angle with the base. The figure below shows three triangles with the base and height of each marked.
The height $$h$$ of a triangle is the length of a line segment that connects the the base to the opposite vertex and makes a $$\text{90°}$$ angle with the base.
### Definition: Triangle Properties
For any triangle $$\text{Δ}ABC,$$ the sum of the measures of the angles is $$\text{180°}.$$
$$m\text{∠}A+m\text{∠}B+m\text{∠}C=\text{180°}$$
The perimeter of a triangle is the sum of the lengths of the sides.
$$P=a+b+c$$
The area of a triangle is one-half the base, $$b,$$ times the height, $$h.$$
$$A=\frac{1}{2}\phantom{\rule{0.1em}{0ex}}bh$$
## Example
Find the area of a triangle whose base is $$11$$ inches and whose height is $$8$$ inches.
### Solution
Step 1. Read the problem. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the area of the triangle Step 3. Name. Choose a variable to represent it. let A = area of the triangle Step 4.Translate. Write the appropriate formula. Substitute. Step 5. Solve the equation. Step 6. Check: Step 7. Answer the question. The area is 44 square inches.
## Example
The perimeter of a triangular garden is $$24$$ feet. The lengths of two sides are $$4$$ feet and $$9$$ feet. How long is the third side?
### Solution
Step 1. Read the problem. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. length of the third side of a triangle Step 3. Name. Choose a variable to represent it. Let c = the third side Step 4.Translate. Write the appropriate formula. Substitute in the given information. Step 5. Solve the equation. Step 6. Check: Step 7. Answer the question. The third side is 11 feet long.
## Example
The area of a triangular church window is $$90$$ square meters. The base of the window is $$15$$ meters. What is the window’s height?
### Solution
Step 1. Read the problem. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. height of a triangle Step 3. Name. Choose a variable to represent it. Let h = the height Step 4.Translate. Write the appropriate formula. Substitute in the given information. Step 5. Solve the equation. Step 6. Check: Step 7. Answer the question. The height of the triangle is 12 meters.
## Isosceles and Equilateral Triangles
Besides the right triangle, some other triangles have special names. A triangle with two sides of equal length is called an isosceles triangle. A triangle that has three sides of equal length is called an equilateral triangle. The figure below shows both types of triangles.
In an isosceles triangle, two sides have the same length, and the third side is the base. In an equilateral triangle, all three sides have the same length.
### Definition: Isosceles and Equilateral Triangles
An isosceles triangle has two sides the same length.
An equilateral triangle has three sides of equal length.
## Example
The perimeter of an equilateral triangle is $$93$$ inches. Find the length of each side.
### Solution
Step 1. Read the problem. Draw the figure and label it with the given information. Perimeter = 93 in. Step 2. Identify what you are looking for. length of the sides of an equilateral triangle Step 3. Name. Choose a variable to represent it. Let s = length of each side Step 4.Translate. Write the appropriate formula. Substitute. Step 5. Solve the equation. Step 6. Check: Step 7. Answer the question. Each side is 31 inches.
## Example
Arianna has $$156$$ inches of beading to use as trim around a scarf. The scarf will be an isosceles triangle with a base of
$$60$$ inches. How long can she make the two equal sides?
### Solution
Step 1. Read the problem. Draw the figure and label it with the given information. P = 156 in. Step 2. Identify what you are looking for. the lengths of the two equal sides Step 3. Name. Choose a variable to represent it. Let s = the length of each side Step 4.Translate. Write the appropriate formula. Substitute in the given information. Step 5. Solve the equation. Step 6. Check: Step 7. Answer the question. Arianna can make each of the two equal sides 48 inches long.
### Optional Video: Area of a Triangle with Fractions
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Is 4 a prime number? What are the divisors of 4?
## Parity of 4
4 is an even number, because it is evenly divisible by 2: 4 / 2 = 2.
Find out more:
## Is 4 a perfect square number?
A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 4 is 2.
Therefore, the square root of 4 is an integer, and as a consequence 4 is a perfect square.
As a consequence, 2 is the square root of 4.
## What is the square number of 4?
The square of a number (here 4) is the result of the product of this number (4) by itself (i.e., 4 × 4); the square of 4 is sometimes called "raising 4 to the power 2", or "4 squared".
The square of 4 is 16 because 4 × 4 = 42 = 16.
As a consequence, 4 is the square root of 16.
## Number of digits of 4
4 is a single-digit number, because it is strictly less than 10; 4 is in fact itself a digit.
## What are the multiples of 4?
The multiples of 4 are all integers evenly divisible by 4, that is all numbers such that the remainder of the division by 4 is zero. There are infinitely many multiples of 4. The smallest multiples of 4 are:
• 0: indeed, 0 is divisible by any natural number, and it is thus a multiple of 4 too, since 0 × 4 = 0
• 4: indeed, 4 is a multiple of itself, since 4 is evenly divisible by 4 (we have 4 / 4 = 1, so the remainder of this division is indeed zero)
• 8: indeed, 8 = 4 × 2
• 12: indeed, 12 = 4 × 3
• 16: indeed, 16 = 4 × 4
• 20: indeed, 20 = 4 × 5
• etc.
## How to determine whether an integer is a prime number?
To determine the primality of a number, several algorithms can be used. The most naive technique is to test all divisors strictly smaller to the number of which we want to determine the primality (here 4). First, we can eliminate all even numbers greater than 2 (and hence 4, 6, 8…). Then, we can stop this check when we reach the square root of the number of which we want to determine the primality (here the square root is 2). Historically, the sieve of Eratosthenes (dating from the Greek mathematics) implements this technique in a relatively efficient manner.
More modern techniques include the sieve of Atkin, probabilistic algorithms, and the cyclotomic AKS test.
## Numbers near 4
• Preceding numbers: …2, 3
• Following numbers: 5, 6
### Nearest numbers from 4
• Preceding prime number: 3
• Following prime number: 5
Find out whether some integer is a prime number |
# Applied Mathematics ms
Applied math is the part of math. That is mostly of the college level students are using the applied mathematics problems. Applied mathematics is very helpful for students for learning about math. The following topics are covered in applied mathematics; they are functions of a complex variable, calculus, ordinary differential equations, partial differential equations and the calculus of variations. Here we will discuss about the applied mathematics ms.
## Example problems for applied mathematics ms:
Applied mathematics ms – Example: 1
Solve: g(s) = f(s) + \lambda \int_0^{2\pi} \sin(s) \cos(t) g(t) dt
Solution:
g(s) = f(s) + \lambda \sin(s) \int_0^{2\pi} \cos(t) g(t) dt
Let c = \int_0^{2\pi} \cos(t) g(t) dt.
Write a new equation:
c = \int_0^{2\pi} \cos(t) $f(t) + \lambda \sin(t) c$ dt
c = \int_0^{2\pi} \cos(t) f(t) dt + \lambda c \int_0^{2\pi} \cos(t) \sin(t) dt
The last integrand is odd and the limits of integration are a multiple of its period, so the integral equals 0.
c = \int_0^{2\pi} \cos(t) f(t) dt
From
g(s) = f(s) + \lambda \sin(s) c
g(s) = f(s) + \lambda \sin(s) \int_0^{2\pi} \cos(t) f(t) dt
Applied mathematics ms – Example: 2
Evaluate \int_{C} F\cdotdr where F=zi+xj+yk and C is the arc of the curve r=\cos t i+\sin t j+tk from t=0 to t=\pi
Solution:
The given curve is r=\cos t i+\sin t j+tk
Hence its parametric equations are x=\cos t,y=\sin t,z=t --(1)
\int_{C}F\cdot,dr=\int_{C}(zi+xj+yk)\cdot(dxi+dyj+dzk)=\int_{C}(zdx+xdy+ydz)
=\int_{0}^{2\pi}[z\frac{dx}{dt}+x\frac{dy}{dt}+y\frac{dz}{dt}]dt
=\int_{0}^{2\pi}[t(-\sin t)+\cos t (\cos t)+\sin t]dt
=-\int_{0}^{2\pi}t\sin tdt+\frac{1}{2}\int_{0}^{2\pi}(1+\cos 2t)dt+\int_{0}^{2\pi}\sin tdt
=-[t(-\cos t)_{0}^{2\pi}+(\sin t)_{0}^{2\pi}]+\frac{1}{2}(2\pi)+[-\cos 2\pi+\cos 0]=2\pi+\pi=3\pi
Applied mathematics ms – Example: 3
Prove that \frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}=2\sec\theta
Proof:
LHS = \ frac{1+\sin\theta}{\cos\theta}+\frac{\cos\theta}{1+\sin\theta}
Simplifying,we get
\frac((1+\sin\theta)^2+cos^2 \theta) (\cos\theta(1+\sin\theta))
\frac{1+\sin^2 \theta+2\sin\theta+\cos^2 \theta}{\cos\theta(1+\sin\theta)}
\frac{1+1+2\sin\theta}{\cos\theta(1+\sin\theta)}
\frac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)}
\frac{2}{\cos\theta}=2\sec\theta=RHS
## Practice problems for applied mathematics ms:
1. Solve: u(x) = \int_0^x e^{x-y} u(y) dy, u(0)=0
Answer: u(x) = 0
2. Evaluate \int_{C} F\cdotdr where F=yzi+zxj+xyk and C is the portion of the curve r=a\cos t i+b\sin t j+ctk from t=0 to t=\frac{\pi}{2}
Answer: 0
3. Prove that \frac{\tan A}{\sec A-1}+\frac{\tan A}{\sec A+1}=2\csc A |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Solving Rational Equations using Cross-Multiplication
## Solve equations that are fractions on both sides
Estimated10 minsto complete
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Practice Solving Rational Equations using Cross-Multiplication
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Solving Rational Equations using Cross-Multiplication
A scale model of a racecar is in the ratio of 1:x to the real racecar. The length of the model is \begin{align*}2x-21\end{align*} units, and the length of the real racecar is \begin{align*}x^2\end{align*} units. What is the value of x?
### Guidance
A rational equation is an equation where there are rational expressions on both sides of the equal sign. One way to solve rational equations is to use cross-multiplication. Here is an example of a proportion that we can solve using cross-multiplication.
If you need more of a review of cross-multiplication, see the Proportion Properties concept. Otherwise, we will start solving rational equations using cross-multiplication.
#### Example A
Solve \begin{align*}\frac{x}{2x-3}=\frac{3x}{x+11}\end{align*}.
Solution: Use cross-multiplication to solve the problem. You can use the example above as a guideline.
Check your answers. It is possible to get extraneous solutions with rational expressions.
#### Example B
Solve \begin{align*}\frac{x+1}{4}=\frac{3}{x-3}\end{align*}.
Solution: Cross-multiply and solve.
\begin{align*}\frac{5+1}{4}=\frac{3}{5-3} \rightarrow \frac{6}{4}=\frac{3}{2}\end{align*} and
\begin{align*}\frac{-3+1}{4}=\frac{3}{-3-3} \rightarrow \frac{-2}{4}=\frac{3}{-6} \end{align*}
#### Example C
Solve \begin{align*}\frac{x^2}{2x-5}=\frac{x+8}{2}\end{align*}.
Solution: Cross-multiply.
Check the answer: \begin{align*}\frac{\left(\frac{40}{11}\right)^2}{\frac{80}{11}-5}=\frac{\frac{40}{11}+8}{2} \rightarrow \frac{1600}{121} \div \frac{25}{11}=\frac{128}{11} \div 2 \rightarrow \frac{64}{11}=\frac{128}{22}\end{align*}
Intro Problem Revisit We need to set up a rational equation and solve for x.
\begin{align*}\frac{1}{x} = \frac{2x-21}{x^2}\end{align*}
Now cross-multiply.
However, x is a ratio so it must be greater than 0. Therefore x equals 21 and the model is in the ratio 1:21 to the real racecar.
### Guided Practice
Solve the following rational equations.
1. \begin{align*}\frac{-x}{x-1}=\frac{x-8}{3}\end{align*}
2. \begin{align*}\frac{x^2-1}{x+2}=\frac{2x-1}{2}\end{align*}
3. \begin{align*}\frac{9-x}{x^2}=\frac{4}{3x}\end{align*}
1.
2.
3.
\begin{align*}x = 0\end{align*} is not actually a solution because it is a vertical asymptote for each rational expression, if graphed. Because zero is not part of the domain, it cannot be a solution, and is extraneous.
### Problem Set
1. Is \begin{align*}x=-2\end{align*} a solution to \begin{align*}\frac{x-1}{x-4}=\frac{x^2-1}{x+4}\end{align*}?
Solve the following rational equations.
1. \begin{align*}\frac{2x}{x+3}=\frac{8}{x}\end{align*}
2. \begin{align*}\frac{4}{x+1}=\frac{x+2}{3}\end{align*}
3. \begin{align*}\frac{x^2}{x+2}=\frac{x+3}{2}\end{align*}
4. \begin{align*}\frac{3x}{2x-1}=\frac{2x+1}{x}\end{align*}
5. \begin{align*}\frac{x+2}{x-3}=\frac{x}{3x-2}\end{align*}
6. \begin{align*}\frac{x+3}{-3}=\frac{2x+6}{x-3}\end{align*}
7. \begin{align*}\frac{2x+5}{x-1}=\frac{2}{x-4}\end{align*}
8. \begin{align*}\frac{6x-1}{4x^2}=\frac{3}{2x+5}\end{align*}
9. \begin{align*}\frac{5x^2+1}{10}=\frac{x^3-8}{2x}\end{align*}
10. \begin{align*}\frac{x^2-4}{x+4}=\frac{2x-1}{3}\end{align*}
Determine the values of a that make each statement true. If there no values, write none.
1. \begin{align*}\frac{1}{x-a}=\frac{x}{x+a}\end{align*}, such that there is no solution.
2. \begin{align*}\frac{1}{x-a}=\frac{x}{x-a}\end{align*}, such that there is no solution.
3. \begin{align*}\frac{x-a}{x}=\frac{1}{x+a}\end{align*}, such that there is one solution.
4. \begin{align*}\frac{1}{x+a}=\frac{x}{x-a}\end{align*}, such that there are two integer solutions.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 9.14.
### Vocabulary Language: English
Rational Equation
Rational Equation
A rational equation is an equation that contains a rational expression. |
# How many grams of glucose are needed to prepare 400 mL of a?
Glucose solutions are commonly used in medical settings as a source of calories and water for patients who are unable to eat or drink normally. Preparing the right concentration of glucose solution is important for delivering the correct dose of glucose to the patient. This article will provide a step-by-step explanation of how to calculate the grams of glucose needed to prepare 400 mL of a 5% glucose solution.
To prepare 400 mL of a 5% glucose solution, 20 grams of glucose are needed.
## Step-by-Step Calculation
Follow these steps to calculate the grams of glucose needed:
1. Identify the desired concentration of the glucose solution as 5%
2. Identify the desired final volume of glucose solution as 400 mL
3. Convert the percentage concentration to a decimal: 5% = 0.05
4. Use the formula: grams of solute = (concentration in decimal form) x (final volume in mL)
5. Plug in the numbers:
• Concentration = 0.05
• Final volume = 400 mL
6. Calculate:
• Grams of glucose = (0.05) x (400 mL)
• Grams of glucose = 20 grams
Therefore, to prepare 400 mL of a 5% glucose solution, 20 grams of glucose are required.
## Explanation of Calculations
Let’s go through an explanation of each step in the calculation:
### 1. Identify the concentration and final volume
We are told that the desired concentration of the glucose solution should be 5%. This means 5 grams of glucose per 100 mL of solution.
We are also told the desired final volume is 400 mL.
### 2. Convert percentage to decimal
To use the formula, we need to convert the percentage concentration to a decimal. 5% equals 0.05 when converted to a decimal.
### 3. Use the formula
The formula for calculating grams of solute is:
Grams of solute = (Concentration in decimal form) x (Final volume in mL)
### 4. Plug in the numbers
We can plug our given values into the formula:
• Concentration = 0.05
• Final volume = 400 mL
### 5. Calculate
Performing the calculation gives us:
Grams of glucose = (0.05) x (400 mL) = 20 grams
Therefore, 20 grams of glucose are required to prepare the 5% glucose solution with a final volume of 400 mL.
## Why the Volume and Concentration Matter
It’s important to accurately calculate the required grams of glucose based on the desired concentration and final volume. Here is why the volume and concentration are important:
### Delivering the correct dose
The concentration and volume determine the total dose of glucose delivered to the patient. A 5% glucose solution contains 5 grams of glucose per 100 mL. If the concentration is too low, the patient may not receive adequate glucose. If it’s too high, the patient may receive excess glucose.
Excessively high or low glucose concentrations can have adverse effects on the patient. Side effects of excessive glucose include high blood sugar, dehydration, and hyperosmolar coma. Low concentrations may lead to hypoglycemia and electrolyte imbalances.
### Achieving appropriate fluid balance
The volume of the solution determines the amount of fluid the patient receives. This affects their fluid and electrolyte balance. An appropriate volume should be prepared to avoid under or over hydrating the patient.
## Tips for Preparing Glucose Solutions
Here are some useful tips for preparing glucose solutions accurately:
### Use an accurate scale
Measure the glucose powder carefully using a calibrated scale to ensure the correct mass is obtained.
### Use volumetric glassware
Prepare the solution in volumetric flasks or graduated cylinders to accurately measure volumes.
### Label clearly
Label the solution container clearly with the concentration, volume, and expiration date.
### Check concentrations
Use glucose test strips to verify the final concentration, especially for very dilute or concentrated solutions.
### Store properly
Store solutions according to manufacturer’s instructions to maintain stability.
## Clinical Uses of Glucose Solutions
Here are some common clinical uses of intravenous glucose solutions:
### Fluid replacement
Glucose solutions can help replace fluid volume losses from vomiting, diarrhea, wounds, burns, or surgery.
### Calorie supplementation
Glucose solutions provide a source of calories when patients cannot meet needs through diet alone.
### Low blood sugar
Glucose solutions can help increase blood sugar levels in hypoglycemic patients.
### Electrolyte imbalance
Glucose solutions may be used to correct electrolyte imbalances like low potassium or low phosphorus.
High glucose doses help maximize glycogen stores before surgeries, burns, and other stresses.
### Drug delivery
Glucose solutions can serve as a diluent and vehicle for intravenous medications.
## Examples of Glucose Solution Preparation
Let’s look at a few examples of calculating grams of glucose for different volumes and concentrations:
### Example 1
How many grams of glucose are needed to prepare 500 mL of a 10% solution?
Steps:
1. Convert 10% to decimal: 10% = 0.10
2. Volume is 500 mL
3. Use formula: Grams glucose = (0.10) x (500 mL) = 50 grams
### Example 2
How many grams of glucose are needed to prepare 1 liter of a 15% solution?
Steps:
1. 15% = 0.15
2. 1 liter = 1000 mL
3. Grams glucose = (0.15) x (1000 mL) = 150 grams
### Example 3
How many grams of glucose are required to make 250 mL of a 2.5% solution?
Steps:
1. 2.5% = 0.025
2. Volume is 250 mL
3. Grams glucose = (0.025) x (250 mL) = 6.25 grams
As shown in the examples above, the grams of glucose required depends on the desired percentage concentration and the final volume of solution.
## Considerations for Specific Patient Populations
Special considerations may be necessary when calculating glucose solution doses for certain patient populations:
### Pediatric patients
– Require lower volumes and infusion rates compared to adults.
– Doses are calculated based on weight or body surface area.
### Renal impairment
– May require decreased concentration and volume to avoid fluid overload.
– Glucose may accumulate due to impaired excretion.
### Diabetes
– Need careful glucose monitoring to avoid hyperglycemia.
– Insulin therapy should be coordinated with glucose administration.
### Hepatic impairment
– Impaired glucose metabolism may require lower glucose doses.
– Monitor for signs of hepatic encephalopathy which can be exacerbated by glucose.
### Pregnancy
– Gestational diabetes requires tight glucose control.
– Fetal effects from maternal hyperglycemia may occur.
## Safety Precautions for Glucose Solutions
Take the following safety precautions when working with glucose solutions:
• Double check glucose calculations to prevent errors.
• Use aseptic technique to avoid bacterial contamination of solutions.
• Don’t use damaged or expired glucose containers.
• Administer with an infusion pump to control flow rate.
• Monitor patient’s glucose levels and watch for signs of excess or insufficient glucose.
• Follow hospital protocols for safe handling and preparation of intravenous solutions.
## Conclusion
Determining the correct amount of glucose when preparing intravenous solutions requires careful calculation based on the desired concentration and final volume. A simple formula allows calculation of the grams of glucose needed, but clinical judgement is also required to adapt doses for individual patients. Strict adherence to safety protocols helps avoid errors and prevent patient harm when working with glucose solutions. |
I found the secret, the key to the vault
Q. We will have 600 people at a conference. How many possible two-person pairs does that allow?
A. In order to solve this problem, let’s solve some easier problems first.
Let’s have all 600 people line up in a row. How many ways are there to line up 600 people? Well, first we have to choose the first person. There are 600 choices for that first person, that is, any of those 600 could go first. Then, who goes second? We have 599 people left. So, to figure out how many ways we can choose the first and second people in a line of 600 people, we calculate $600 \times 599$.
Now if we continue this logic through all 600 people, choosing the first, the second, the third and so on, we have $600 \times 599 \times 598 \times ... \times 2 \times 1$ ways to line up 600 people in a row.
In other words, there are:
$n!$
ways to line up n people in a row.
But that’s not the right answer to our original problem. Let’s try to get that answer closer to the original answer we wanted. Let’s say we divided each of those $n!$ rows of people evenly into pairs, taking them each two by two in order of the row. In that case, it wouldn’t matter if the first pair contained Alice or Bob, or Bob and Alice. Within a single pair, we don’t care what the ordering of the people in that pair is.
For 600 people, there are $\frac{600}{2}$ pairs or 300 pairs. So, the number of ways those pairs might have swapped the first for the second is:
$2^{(n/2)}$
or
$2^{300}$
Therefore, the number of ways to order 600 people, ignoring the ways to merely swap a pair of people, is:
$\frac{n!}{2^{(n/2)}}$
We’ve only got one more step to get an answer. Notice that while we’ve taken care of the case where a single pair of people are swapped, we haven’t taken care of the case where a pair of people is swapped with another pair of people. In other words, we don’t care whether it’s Alice and Bob followed by Carol and Dan, or if it’s Carol and Dan followed by Alice and Bob. So, we need to also divide by the number of ways to order 300 pairs of people. As we know from above, there are $300!$ ways to order 300 things. To ignore swapping pairs of n things, we need to divide by:
$(\frac{n}{2})!$
So let’s take a look at our final formula, which takes into account the ways we can choose n people, ignoring a swapped pair of two people, and also ignoring swapped pairs of people:
$\frac{n!}{2^{(n/2)}(\frac{n}{2})!}$
Let’s plug in $600$ for $n$:
$\frac{600!}{2^{(600/2)}(\frac{600}{2})!}$
You can calculate this answer yourself if you use this online calculator and type in this formula:
factorial(600)/((2^300)*factorial(300)) |
Question 204503
# 1
{{{f(x+h)=(x+h)^2+3}}} Replace each "x" with "x+h"
{{{f(x+h)=x^2+2xh+h^2+3}}} FOIL (ie expand)
--------------------------------
{{{(f(x+h)-f(x))/h}}} Move onto the given difference quotient.
{{{(x^2+2xh+h^2+3-(x^2+3))/h}}} Plug in {{{f(x+h)=x^2+2xh+h^2+3}}} and {{{f(x)=x^2+3}}}
{{{(x^2+2xh+h^2+3-x^2-3)/h}}} Distribute
{{{(2xh+h^2)/h}}} Combine like terms.
{{{(h(2x+h))/h}}} Factor out the GCF "h" from the numerator.
{{{(cross(h)(2x+h))/cross(h)}}} Cancel out the common terms.
{{{2x+h}}} Simplify
So {{{(f(x+h)-f(x))/h=2x+h}}} when {{{f(x)=x^2+3}}}
=====================================================================================
# 2
Take note that when {{{x=-3}}} it is less than zero. Because {{{f(x)=-x}}} when {{{x<0}}}, this means that we simply plug in {{{x=-3}}} to get: {{{f(-3)=-(-3)=3}}}.
So {{{f(-3)=3}}}
Also, when {{{x=2}}} it is greater than zero. Since {{{f(x)=x^3}}} when {{{x>=0}}}, we just plug in {{{x=2}}} to get: {{{f(2)=(2)^3=8}}}.
So {{{f(2)=8}}}
================================
# 3
{{{x+2y=3}}} Add 3 to both sides.
{{{2y=3-x}}} Subtract {{{x}}} from both sides.
{{{2y=-x+3}}} Rearrange the terms.
{{{y=(-x+3)/2}}} Divide both sides by {{{2}}} to isolate y.
{{{y=-(1/2)x+3/2}}} Break up the fraction and simplify.
We can see that the equation {{{y=-(1/2)x+3/2}}} has a slope {{{m=-1/2}}} and a y-intercept {{{b=3/2}}}.
Now to find the slope of the perpendicular line, simply flip the slope {{{m=-1/2}}} to get {{{m=-2/1}}}. Now change the sign to get {{{m=2}}}. So the perpendicular slope is {{{m=2}}}.
Now let's use the point slope formula to find the equation of the perpendicular line by plugging in the slope {{{m=-1/2}}} and the coordinates of the given point *[Tex \LARGE \left$$6,-4\right$$].
{{{y--4=2(x-6)}}} Plug in {{{m=2}}}, {{{x[1]=6}}}, and {{{y[1]=-4}}}
{{{y+4=2(x-6)}}} Rewrite {{{y--4}}} as {{{y+4}}}
{{{y+4=2x+2(-6)}}} Distribute
{{{y+4=2x-12}}} Multiply
{{{y=2x-12-4}}} Subtract 4 from both sides.
{{{y=2x-16}}} Combine like terms.
So the equation of the line perpendicular to {{{x+2y-3=0}}} that goes through the point *[Tex \LARGE \left$$6,-4\right$$] is {{{y=2x-16}}}.
Here's a graph to visually verify our answer:
{{{drawing(500, 500, -10, 10, -10, 10,
graph(500, 500, -10, 10, -10, 10,-(1/2)x+3/2,2x-16)
circle(6,-4,0.08),
circle(6,-4,0.10),
circle(6,-4,0.12))}}}
Graph of the original equation {{{x+2y-3=0}}} (red) and the perpendicular line {{{y=2x-16}}} (green) through the point *[Tex \LARGE \left$$6,-4\right$$].
=================================
# 5
{{{(x^2-8x)+(y^2-6y)=0}}} Group like terms.
{{{(x^2-8x+highlight(16))+(y^2-6y)=0+highlight(16)}}} Take half of the x-coefficient -8 to get -4. Square -4 to get 16. Add this value to both sides.
{{{(x^2-8x+16)+(y^2-6y+highlight(9))=0+16+highlight(9)}}} Take half of the y-coefficient -6 to get -3. Square -3 to get 9. Add this value to both sides.
{{{(x-4)^2+(y-3)^2=25}}} Combine like terms.
{{{(x-4)^2+(y-3)^2=5^2}}} Rewrite 25 as {{{5^2}}}
Now the equation is in the form {{{(x-h)^2+(y-k)^2=r^2}}} (which is a circle) where (h,k) is the center and "r" is the radius
In this case, {{{h=4}}}, {{{k=3}}}, and {{{r=5}}}
So the center is (4,3) and the radius is 5 units.
Here's the graph:
{{{drawing(500, 500, -10, 10, -10, 10,
grid(1),
graph(500, 500, -10, 10, -10, 10,0),
circle(4,3,5)
)}}} |
# An introduction to differentiation: What is it and how to solve it?
In mathematics, several branches are used to solve various kinds of problems. These branches are algebra, geometry, calculus, set theory, etc. Calculus is one of the main branches of mathematics that deal with the study of differentiation and integration.
The subtypes of calculus are very essential in finding the solution to complex problems. In this post, we are going to describe differentiation along with calculations.
## What is differentiation?
In calculus, the process of finding the derivative of a function w.r.t its independent variable and considering the dependent variables as a constant is said to be the differentiation. In geometry, it is widely used to find the slope of the tangent line.
It is a wide concept that is very helpful in finding complex calculus problems. There are two ways in differentiation for solving the derivative of the function, one is by the first principle method and the other is by using the rules of differentiation.
### Formula of differentiation
The formula for finding the differential of the function by using limits is:
d/dx (f(x)) = limh→0 [f(x + h) – f(h)] / h
## Laws of differentiation
There are various laws of differentiation that are helpful in solving complex calculus problems. Here are a few laws of differentiation.
### 1. Constant Law
The constant law of differentiation states that the derivative of any constant function gives zero as an output. Hence the derivative of any constant coefficient and dependent variables always give zero.
d/dx [k] = 0
where k is any constant.
### 2. Sum law
In differentiation, sum law is frequently used in complex calculus problems. according to this law, the notation of differential must be applied to each function separately. Whenever 2, 3, or more functions are given with a plus sign among them then the notation is applied separately.
d/dx [f(x) + p(x) + q(x)] = d/dx [f(x)] + d/dx [p(x)] + d/dx [q(x)]
### 3. Difference law
In differentiation, difference law is frequently used in complex calculus problems. According to this law, the notation of differential must be applied to each function separately. Whenever 2, 3, or more functions are given with a minus sign among them then the notation is applied separately.
d/dx [f(x) – p(x) – q(x)] = d/dx [f(x)] – d/dx [p(x)] – d/dx [q(x)]
### 4. Product law
The product law is widely used in differentiation to calculate the derivative of two or more functions with a multiply sign among them. According to this law, the differential is applied to the first function while the second function remains the same.
After that, the differential function is applied to the second function while the function remains the same with a plus sign between them.
d/dx [f(x) * p(x)] = p(x) d/dx [f(x)] + f(x) d/dx [p(x)]
### 5. Quotient law
The quotient law is widely used in differentiation to calculate the derivative of two or more functions with a division sign among them. According to this law, the differential is applied to the first function while the second function remains the same.
After that, the differential function is applied to the second function while the function remains the same with a minus sign between them. And divide it by the square of the second function.
d/dx [f(x) / p(x)] = 1/[p(x)]2 [p(x) d/dx [f(x)] – f(x) d/dx [p(x)]]
### 6. Power Law
In differentiation, the power law is used with the functions present with exponents. According to this law, the exponent must multiply with the function and the power must be reduced by 1.
d/dx [f(x)]m = m[f(x)]m-1 d/dx [f(x)]
## How do calculate differentiation problems?
The problems of differentiation can be solved easily with the help of its laws. Let us take a few examples to learn how to calculate differentiation problems.
Example 1
Evaluate the differential of the given function w.r.t “x”.
p(x) = 2x4 + 5x6 – 12x2 + 3x + 12t
Solution
Step-1: First of all, apply the notation of differentiation to the given function.
d/dt p(x) = d/dt [2x4 + 5x6 – 12x2 + 3x + 12t]
Step-2: Use the sum and difference laws of differentiation to apply the notation separately to each function.
d/dt [2x4 + 5x6 – 12x2 + 3x + 12t] = d/dt [2x4] + d/dt [5x6] – d/dt [12x2] + d/dt [3x] + d/dt [12t]
Step-3: Now use the constant function law to take the constant coefficients outside the notation.
d/dt [2x4 + 5x6 – 12x2 + 3x + 12t] = 2d/dt [x4] + 5d/dt [x6] –`12d/dt [x2] + 3d/dt [x] + d/dt [12t]
Step-4: Use the power rule to find the differential of the above function.
= 2 [4 x4-1] + 5 [6 x6-1] –`12 [2 x2-1] + 3 [x1-1] + [0]
= 2 [4 x3] + 5 [6 x5] –`12 [2 x1] + 3 [x0] + [0]
= 2 [4 x3] + 5 [6 x5] –`12 [2 x] + 3 [1] + [0]
= 8 [x3] + 30 [x5] –`24 [x] + 3 [1]
= 8x3 + 30x5 –`24x + 3
A differentiation calculator can also be used to solve the above problem to ease up the calculations.
Example 2
Evaluate the differential of the given function w.r.t “u”.
p(u) = 5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3
Solution
Step-1: First of all, apply the notation of differentiation to the given function.
p(u) = 5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3
d/du p(u) = d/du [5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3]
Step-2: Use the sum and difference laws of differentiation to apply the notation separately to each function.
d/du [5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3] = d/du [5u2] + d/du [12cos(u)] – d/du [15u12] + d/du [5sin(u)] + d/du [2u3]
Step-3: Now use the constant function law to take the constant coefficients outside the notation.
d/du [5u2 + 12cos(u) – 15u12 + 5sin(u) + 2u3] = 5d/du [u2] + 12d/du [cos(u)] – 15d/du [u12] + 5d/du [sin(u)] + 2d/du [u3]
Step-4: Use the power rule to find the differential of the above function.
= 5 [2 u2-1] + 12 [-sin(u)] – 15 [12 u12-1] + 5 [cos(u)] + 2 [3 u3-1]
= 5 [2 u1] + 12 [-sin(u)] – 15 [12 u11] + 5 [cos(u)] + 2 [3 u2]
= 5 [2 u] + 12 [-sin(u)] – 15 [12 u11] + 5 [cos(u)] + 2 [3 u2]
= 10 [u] – 12 [sin(u)] – 180 [u11] + 5 [cos(u)] + 6 [u2]
= 10u – 12sin(u) – 180u11 + 5cos(u) + 6u2
## Wrap up
In this post, we have discussed all the basics of differentiation and methods for solving it either by using an online calculator or manually. Now you can complete assignments and prepare for your exams without any difficulty. |
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## Exercise 3.3 Page: 50
1. Given a parallelogram ABCD. Complete each statement along with the definition or property used.
(i) AD = …… (ii) ∠DCB = ……
(iii) OC = …… (iv) m ∠DAB + m ∠CDA = ……
Solution:
(i) AD = BC (Opposite sides of a parallelogram are equal)
(ii) ∠DCB = ∠DAB (Opposite angles of a parallelogram are equal) (iii) OC = OA (Diagonals of a parallelogram are equal)
(iv) m ∠DAB + m ∠CDA = 180°
2. Consider the following parallelograms. Find the values of the unknown x, y, z
Solution:
(i)
4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution:
ABCD is a figure of quadrilateral that is not a parallelogram but has exactly two opposite
angles that is ∠B = ∠D of equal measure. It is not a parallelogram because ∠A ≠ ∠C.
5. The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Solution:
Let the measures of two adjacent angles ∠A and ∠B be 3x and 2x respectively in
parallelogram ABCD.
∠A + ∠B = 180°
⇒ 3x + 2x = 180°
⇒ 5x = 180°
⇒ x = 36°
We know that opposite sides of a parallelogram are equal.
∠A = ∠C = 3x = 3 × 36° = 108°
∠B = ∠D = 2x = 2 × 36° = 72°
6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution:
Let ABCD be a parallelogram.
Sum of adjacent angles of a parallelogram = 180°
∠A + ∠B = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
also, 90° + ∠B = 180°
⇒ ∠B = 180° – 90° = 90°
∠A = ∠C = 90°
∠B = ∠D = 90
°
7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
Solution:
y = 40° (alternate interior angle)
∠P = 70° (alternate interior angle)
∠P = ∠H = 70° (opposite angles of a parallelogram)
z = ∠H – 40°= 70° – 40° = 30°
∠H + x = 180°
⇒ 70° + x = 180°
⇒ x = 180° – 70° = 110°
8. The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
Solution:
(i) SG = NU and SN = GU (opposite sides of a parallelogram are equal) 3x = 18
x = 18/3
⇒ x =6
3y – 1 = 26 an
d,
⇒ 3y = 26 + 1
⇒ y = 27/3=9
x = 6 and y = 9
(ii) 20 = y + 7 and 16 = x + y (diagonals of a parallelogram bisect each other) y + 7 = 20
⇒ y = 20 – 7 = 13 and,
x + y = 16
⇒ x + 13 = 16
⇒ x = 16 – 13 = 3
x = 3 and y = 13
9. In the above figure both RISK and CLUE are parallelograms. Find the value of x.
Solution:
∠K + ∠R = 180° (adjacent angles of a parallelogram are supplementary)
⇒ 120° + ∠R = 180°
⇒ ∠R = 180° – 120° = 60°
also, ∠R = ∠SIL (corresponding angles)
⇒ ∠SIL = 60°
also, ∠ECR = ∠L = 70° (corresponding angles) x + 60° + 70° = 180° (angle sum of a triangle)
⇒ x + 130° = 180°
⇒ x = 180° – 130° = 50°
10. Explain how this figure is a trapezium. Which of its two sides are parallel? (Fig 3.32)
Solution:
When a transversal line intersects two lines in such a way that the sum of the adjacent angles on the same side of transversal is 180° then the lines are parallel to each other. Here, ∠M + ∠L = 100° + 80° = 180°
Thus, MN || LK
As the quadrilateral KLMN has one pair of parallel line therefore it is a trapezium. MN and LK are parallel lines.
Solution:
m∠C + m∠B = 180° (angles on the same side of transversal)
⇒ m∠C + 120° = 180°
⇒ m∠C = 180°- 120° = 60°
12. Find the measure of ∠P and ∠S if SP || RQ ? in Fig 3.34. (If you find m∠R, is there more than one method to find m∠P?)
Solution:
∠P + ∠Q = 180° (angles on the same side of transversal)
⇒ ∠P + 130° = 180°
⇒ ∠P = 180° – 130° = 50°
also, ∠R + ∠S = 180° (angles on the same side of transversal)
⇒ 90° + ∠S = 180°
⇒ ∠S = 180° – 90° = 90°
Thus, ∠P = 50° and ∠S = 90°
Yes, there are more than one method to find m∠P.
PQRS is a quadrilateral. Sum of measures of all angles is 360°.
Since, we know the measurement of ∠Q, ∠R and ∠S.
∠Q = 130°, ∠R = 90° and ∠S = 90°
∠P + 130° + 90° + 90° = 360°
⇒ ∠P + 310° = 360°
⇒ ∠P = 360° – 310° = 50° |
# Division with Whole Numbers and Decimals
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## Transcription
1 Grade 5 Mathematics, Quarter 2, Unit 2.1 Division with Whole Numbers and Decimals Overview Number of Instructional Days: 15 (1 day = minutes) Content to be Learned Divide multidigit whole numbers using a variety of strategies based on place value, properties of operations, and the relationship between multiplication and division. Explain calculations using rectangular arrays, area models, and equations. Divide decimals to the hundredths, using concrete models or drawings and strategies based on place value and properties of operations. Converting units of measurement within a given measurement system. Use conversions to solve single-step and multistep real-world problems. Essential Questions How can the methods and strategies you used to learn whole number division help you to divide decimals? How can you use what you know about place value to explain the relationship between expressions such as 14 2, , 14 20, 1.4 2? How can you model the quotient of _ _? What happens to your quotient as your dividend gets larger? What happens to your quotient as your divisor gets larger? Mathematical Practices to Be Integrated Attend to precision. Calculate accurately and efficiently. Give carefully formulated explanations as to why a procedure works. Look for and express regularity in repeated reasoning. See repeated calculations and look for generalizations and shortcuts. Continually evaluate reasonableness of answer. How is division with decimals similar to dividing with whole numbers? How is it different? What is your strategy for converting from a larger unit to a smaller unit (gallons to quarts)? From a smaller unit to a larger unit (quarts to gallons)? How do you know when a conversion is needed to solve a problem? 13
2 Grade 5 Mathematics, Quarter 2 Unit 2.1 Division With Whole Numbers and Decimals (15 days) Written Curriculum Common Core State Standards for Mathematical Content Number and Operations in Base Ten 5.NBT Perform operations with multi-digit whole numbers and with decimals to hundredths. 5.NBT.6 5.NBT.7 Find whole-number quotients of whole numbers with up to four-digit dividends and two-digit divisors, using strategies based on place value, the properties of operations, and/or the relationship between multiplication and division. Illustrate and explain the calculation by using equations, rectangular arrays, and/or area models. Add, subtract, multiply, and divide decimals to hundredths, using concrete models or drawings and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Measurement and Data 5.MD Convert like measurement units within a given measurement system. 5.MD.1 Convert among different-sized standard measurement units within a given measurement system (e.g., convert 5 cm to 0.05 m), and use these conversions in solving multi-step, real world problems. Common Core Standards for Mathematical Practice 6 Attend to precision. Mathematically proficient students try to communicate precisely to others. They try to use clear definitions in discussion with others and in their own reasoning. They state the meaning of the symbols they choose, including using the equal sign consistently and appropriately. They are careful about specifying units of measure, and labeling axes to clarify the correspondence with quantities in a problem. They calculate accurately and efficiently, express numerical answers with a degree of precision appropriate for the problem context. In the elementary grades, students give carefully formulated explanations to each other. By the time they reach high school they have learned to examine claims and make explicit use of definitions. 8 Look for and express regularity in repeated reasoning. Mathematically proficient students notice if calculations are repeated, and look both for general methods and for shortcuts. Upper elementary students might notice when dividing 25 by 11 that they are repeating the same calculations over and over again, and conclude they have a repeating decimal. By paying attention to the calculation of slope as they repeatedly check whether points are on the line through (1, 2) with slope 3, middle school students might abstract the equation (y 2)/(x 1) = 3. Noticing the regularity in the way terms cancel when expanding (x 1)(x + 1), (x 1)(x 2 + x + 1), and (x 1)(x 3 + x 2 + x + 1) might lead them to the general formula for the sum of a geometric series. As they work to solve a problem, mathematically proficient students maintain oversight of the process, while attending to the details. They continually evaluate the reasonableness of their intermediate results. 14
4 Grade 5 Mathematics, Quarter 2 Unit 2.1 Division With Whole Numbers and Decimals (15 days) 16
8 Grade 5 Mathematics, Quarter 2 Unit 2.2 Addition and Subtraction of Fractions with Unlike Denominators (15 days) Additionally, Curriculum documents sometimes give the sense that the ultimate goal of teaching fractions is that the students will be able to carry out the four operations with them. We believe that students need to be given time to understand what fractions are about (rather than move too quickly to computation) (p. 374). 20
9 Grade 5 Mathematics, Quarter 2, Unit 2.3 Understanding Fractions as Division Overview Number of Instructional Days: 10 (1 day = minutes) Content to be Learned Understand that a fraction is the division of the numerator by the denominator (3/4 is 3 divided by 4). Solve word problems involving division of whole numbers leading to answers in the form of fractions or mixed numbers. Solve word problems using visual fraction models or equations to represent the problem. Mathematical Practices to Be Integrated Construct viable arguments and critique the reasoning of others. Use objects, drawings, or diagrams to construct arguments. Ask questions to clarify an argument. Show/state justification. Prove two ways (i.e., using concrete referents and algorithms). Essential Questions How is fraction notation related to division? How can you use a visual model, such as a fraction bar, to represent division? How would you represent a fraction as division? 21
11 Grade 5 Mathematics, Quarter 2 Unit 2.3 Understanding Fractions as Division (10 days) calculations by using equations, rectangular arrays, and/or area models, and interpreted remainders within the context of the problem. Current Learning Students understand a fraction as a way to represent the division of two quantities. They are expected to demonstrate their understanding using concrete materials and models, and to clearly explain their thinking. They read 3/5 as three fifths and, after extensive experiences working through sharing problems, they understand that 3/5 can also be interpreted as 3 divided by 5 resulting in fractional parts. This connects back to the idea that division is about decomposing/sharing out. It is essential that students have multiple, varied opportunities to work with this concept. Focus should not be on the algorithm, but rather on building solid understanding through visual representation. Finally, word problems should represent students real life experiences. Future Learning In sixth grade, students will extend their understanding of multiplication and division to divide fractions by fractions. They will use models and will write and solve story problems involving fractions divided by fractions. They will also use this reasoning to solve ratio problems. In seventh grade, students will convert rational numbers to decimals using long division. Additional Findings According to Principles and Standards for School Mathematics, as students encounter a new meaning of a fraction as a quotient of two whole numbers they can see another way to arrive at this equivalence (p. 150). Furthermore, students develop understanding of fractions as parts of unit wholes, as parts of a collection, as locations on number lines, and as division of whole numbers (p. 148). 23
12 Grade 5 Mathematics, Quarter 2 Unit 2.3 Understanding Fractions as Division (10 days) 24
### Understanding Place Value of Whole Numbers and Decimals Including Rounding
Grade 5 Mathematics, Quarter 1, Unit 1.1 Understanding Place Value of Whole Numbers and Decimals Including Rounding Overview Number of instructional days: 14 (1 day = 45 60 minutes) Content to be learned
### Decimals in the Number System
Grade 5 Mathematics, Quarter 1, Unit 1.1 Decimals in the Number System Overview Number of Instructional Days: 15 (1 day = 45 minutes) Content to Be Learned Recognize place value relationships. In a multidigit
### Integer Operations. Overview. Grade 7 Mathematics, Quarter 1, Unit 1.1. Number of Instructional Days: 15 (1 day = 45 minutes) Essential Questions
Grade 7 Mathematics, Quarter 1, Unit 1.1 Integer Operations Overview Number of Instructional Days: 15 (1 day = 45 minutes) Content to Be Learned Describe situations in which opposites combine to make zero.
### Understanding Place Value
Grade 5 Mathematics, Quarter 1, Unit 1.1 Understanding Place Value Overview Number of instructional days: 7 (1 day = 45 minutes) Content to be learned Explain that a digit represents a different value
### CCSS: Mathematics. Operations & Algebraic Thinking. CCSS: Grade 5. 5.OA.A. Write and interpret numerical expressions.
CCSS: Mathematics Operations & Algebraic Thinking CCSS: Grade 5 5.OA.A. Write and interpret numerical expressions. 5.OA.A.1. Use parentheses, brackets, or braces in numerical expressions, and evaluate
### Common Core State Standards. Standards for Mathematical Practices Progression through Grade Levels
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In this chapter, you will revise work you have done on squares, cubes, square roots and cube roots. You will learn about laws of exponents that will enable you to do calculations using numbers written in exponential form.
Very large numbers are written in scientific notation. Scientific notation is a convenient way of writing very large numbers as a product of a number between 1 and 10 and a power of 10.
## Revision
### Exponential notation
1. Calculate.
1. $$2 \times 2 \times 2$$
2. $$2 \times 2 \times 2 \times 2 \times 2 \times 2$$
3. $$3 \times 3 \times 3$$
4. $$3 \times 3 \times 3 \times 3 \times 3 \times 3$$
Instead of writing $$3 \times 3 \times 3 \times 3 \times 3 \times 3$$ we can write $$3^6$$.
We read this as "3 to the power of 6". The number 3 is the base, and 6 is the exponent.
When we write $$3 \times 3 \times 3 \times 3 \times 3 \times 3$$ as $$3^6$$, we are using exponential notation.
1. Write each of the following in exponential form:
1. $$2 \times 2 \times 2$$
2. $$2 \times 2 \times 2 \times 2 \times 2 \times 2$$
3. $$3 \times 3 \times 3$$
4. $$3 \times 3 \times 3 \times 3 \times 3 \times 3$$
2. Calculate.
1. $$5^2$$
2. $$2^5$$
3. $$10^2$$
4. $$15^2$$
5. $$3^4$$
6. $$4^3$$
7. $$2^3$$
8. $$3^2$$
### Squares
To square a number is to multiply it by itself. The square of 8 is 64 because $$8 \times 8$$ equals 64.
We write $$8 \times 8$$ as $$8^2$$ in exponential form.
We read $$8^2$$ as eight squared.
1. Complete the table.
Number Square the number Exponential form Square (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 (f) 6 (g) 7 (h) 8 $$8 \times 8$$ $$8^2$$ 64 (i) 9 (j) 10 (k) 11 (l) 12
2. Calculate the following:
1. $$3^2 \times 4^2$$
2. $$2^2 \times 3^2$$
3. $$2^2 \times 5^2$$
4. $$2^2 \times 4^2$$
3. Complete the following statements to make them true:
1. $$3^2 \times 4^2 = \text{______}^2$$
2. $$2^2 \times 3^2 = \text{______}^2$$
3. $$2^2 \times 5^2 = \text{______}^2$$
4. $$2^2 \times 4^2 = \text{______}^2$$
### Cubes
To cube a number is to multiply it by itself and then by itself again. The cube of 3 is 27 because 3 \times 3 \times 3 equals 27.
We write $$3 \times 3 \times 3$$ as $$3^3$$ in exponential form.
We read $$3^3$$ as three cubed.
1. Complete the table.
Number Cube the number Exponential form Cube (a) 1 (b) 2 (c) 3 $$3 \times 3 \times 3$$ $$3^3$$ 27 (d) 4 (e) 5 (f) 6 (g) 7 (h) 8 (i) 9 (j) 10
2. Calculate the following:
1. $$2^3 \times 3^3$$
2. $$2^3 \times 5^3$$
3. $$2^3 \times 4^3$$
4. $$1^3 \times 9^3$$
3. Which of the following statements are true? If a statement is false, rewrite it as a true statement.
1. $$2^3 \times 3^3 = 6^3$$
2. $$2^3 \times 5^3 = 7^3$$
3. $$2^3 \times 4^3 = 8^3$$
4. $$1^3 \times 9^3 = 10^3$$
### Square and cube roots
To find the square root of a number we ask the question: Which number was multiplied by itself to get a square?
The square root of 16 is 4 because $$4 \times 4 = 16$$.
The question: Which number was multiplied by itself to get 16? is written mathematically as $$\sqrt{16}$$.
The answer to this question is written as $$\sqrt{16} = 4$$.
1. Complete the table.
Number Square of the number Square root of the square of the number Reason (a) 1 (b) 2 (c) 3 (d) 4 16 4 $$4 \times 4 = 16$$ (e) 5 (f) 6 (g) 7 (h) 8 (i) 9 (j) 10 (k) 11 (l) 12
1. $$\sqrt{144}$$
2. $$\sqrt{100}$$
3. $$\sqrt{81}$$
4. $$\sqrt{64}$$
To find the cube root of a number we ask the question: Which number was multiplied by itself and again by itself to get a cube?
The cube root of 64 is 4 because $$4 \times 4 \times 4 = 64$$.
The question: Which number was multiplied by itself and again by itself (or cubed) to get 64? is written mathematically as $$\sqrt[3]{64}$$.
The answer to this question is written as $$\sqrt[3]{64} = 4$$.
1. Complete the table.
Number Cube of the number Cube root of the cube of the number Reason (a) 1 (b) 2 (c) 3 (d) 4 64 4 $$4 \times 4 \times 4 = 64$$ (e) 5 (f) 6 (g) 7 (h) 8 (i) 9 (j) 10
1. $$\sqrt[3]{216}$$
2. $$\sqrt[3]{8}$$
3. $$\sqrt[3]{125}$$
4. $$\sqrt[3]{27}$$
5. $$\sqrt[3]{64}$$
6. $$\sqrt[3]{1000}$$
## Working with integers
### Representing integers in exponential form
1. Calculate the following, without using a calculator:
1. $$-2 \times -2 \times -2$$
2. $$-2 \times -2 \times -2 \times -2$$
3. $$-5 \times -5$$
4. $$-5 \times -5 \times -5$$
5. $$-1 \times -1 \times -1 \times -1$$
6. $$-1 \times -1 \times -1$$
2. Calculate the following:
1. $$-2^2$$
2. $$(-2)^2$$
3. $$(-5)^2$$
4. $$-5^3$$
1. Are your answers to question 2(a) and (b) different or the same as those of the calculator?
2. If your answers are different to those of the calculator, try to explain how the calculator did the calculations differently from you.
The calculator "understands" $${\bf-5^2}$$ and $${\bf(-5)^2}$$ as two different numbers.
I understands $${\bf-5^2}$$ as $${\bf-5 \times 5 = -25}$$ and $${\bf(-5)^2}$$ as $${\bf-5 \times -5 = 25}$$
1. Write the following in exponential form:
1. $$-2 \times -2 \times -2$$
2. $$-2 \times -2 \times -2 \times -2$$
3. $$-5 \times -5$$
4. $$-5 \times -5 \times -5$$
5. $$-1 \times -1 \times -1 \times -1$$
6. $$-1 \times -1 \times -1$$
2. Calculate the following:
1. $$(-3)^2$$
2. $$(-3)^3$$
3. $$(-2)^4$$
4. $$(-2)^6$$
5. $$(-2)^5$$
6. $$(-3)^4$$
3. Say whether the sign of the answer is negative or positive. Explain why.
1. $$(-3)^6$$
2. $$(-5)^{11}$$
3. $$(-4)^{20}$$
4. $$(-7)^5$$
4. Say whether the following statements are true or false. If a statement is false rewrite it as a correct statement.
1. $$(-3)^2 = -9$$
2. $$-3^2 = 9$$
3. $$(-5^2) = 5^2$$
4. $$(-1)^3 = -1^3$$
5. $$(-6)^3 = -18$$
6. $$(-2)^6 = 2^6$$
## Laws of exponents
### Product of powers
1. A product of 2s is given below. Describe it using exponential notation, that is, write it as a power of 2.
$$2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$$
2. Express each of the following as a product of the powers of 2, as indicated by the brackets.
1. $$(2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2)$$
2. $$(2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2)$$
3. $$(2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2) \times (2 \times 2)$$
4. $$(2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2) \times (2 \times 2 \times 2)$$
5. $$(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2) \times (2 \times 2)$$
3. Complete the following statements so that they are true. You may want to refer to your answers to question 2 (a) to (e) to help you.
1. $$2^3 \times \text{______} = 2^{12}$$
2. $$2^5 \times \text{______} \times 2^2 = 2^{12}$$
3. $$2^2 \times 2^2 \times 2^2 \times 2^2 \times 2^2 \times 2^2 = \text{______}$$
4. $$2^8 \times \text{______} = 2^{12}$$
5. $$2^3 \times 2^3 \times 2^3 \times \text{______} = 2^{12}$$
6. $$2^6 \times \text{______} = 2^{12}$$
7. $$2^2 \times 2^{10} \times = \text{______}$$
Suppose we are asked to simplify: $$3^2 \times 3^4$$.
\begin{align} \text{The solution is: }3^3 \times 3^4 & = 9 \times 81 \\ &= 729 \\ &= 3^6 \end{align}
The base (3) is a repeated factor. The exponents (2 and 4) tell us the number of times each factor is repeated.
We can explain this solution in the following manner:
\begin{align} 3^2 \times 3^4 = \underbrace{3 \times 3}_\text{2 factors} \times \underbrace{3 \times 3 \times 3 \times 3}_\text{4 factors} = \underbrace{3 \times 3 \times 3 \times 3 \times 3 \times 3}_\text{6 factors} = 3^6 \end{align}
1. Complete the table.
Product of powers Repeated factor Total number of times the factor is repeated Simplified form (a) $$2^7 \times 2^3$$ (b) $$5^2 \times 5^4$$ (c) $$4^1 \times 4^5$$ (d) $$6^3 \times 6^2$$ (e) $$2^8 \times 2^2$$ (f) $$5^3 \times 5^3$$ (g) $$4^2 \times 4^4$$ (h) $$2^1 \times 2^9$$
When you multiply two or more powers that have the same base, the answer has the same base, but its exponent is equal to the sum of the exponents of the numbers you are multiplying.
We can express this symbolically as $$a^m \times a^n = a^{m+n}$$, where m and n are natural numbers and a is not zero.
1. What is wrong with these statements? Correct each one.
1. $$2^3 \times 2^4 = 2^{12}$$
2. $$10 \times 10^2 \times 10^3 \times = 10^{1 \times 2 \times 3} = 10^6$$
3. $$3^2 \times 3^2 = 3^6$$
4. $$5^3 \times 5^2 = 15 \times 10$$
2. Express each of the following numbers as a single power of 10.
Example: 1 000 000 as a power of 10 is $$10^6$$.
1. $$100$$
2. $$10 00$$
3. $$100 000$$
4. $$10^2 \times 10^3 \times 10^4$$
5. $$100 \times 1000 \times 10000$$
6. $$1000000000$$
3. Write each of the following products in exponential form:
1. $$x \times x \times x \times x \times x \times x \times x \times x \times x$$
2. $$(x \times x) \times (x \times x \times x) \times (x \times x \times x \times x)$$
3. $$(x \times x \times x \times x) \times (x \times x) \times (x \times x) \times x$$
4. $$(x \times x \times x \times x \times x \times x) \times (x \times x \times x)$$
5. $$(x \times x \times x) \times (y \times y \times y)$$
6. $$(a \times a) \times (b \times b)$$
4. Complete the table.
Product of powers Repeated factor Total number of times the factor is repeated Simplified form (a) $$x^7 \times x^3$$ (b) $$x^2 \times x^4$$ (c) $$x^1 \times x^5$$ (d) $$x^3 \times x^2$$ (e) $$x^8 \times x^2$$ (f) $$x^3 \times x^3$$ (g) $$x^1 \times x^9$$
### Raising a power to a power
1. Complete the table of powers of 2.
x 1 2 3 4 5 6 7 8 9 10 11 $$2^x$$ 2 4 $$2^1$$ $$2^2$$ $$2^3$$
x 12 13 14 15 16 17 18 $$2^x$$
2. Complete the table of powers of 3.
x 1 2 3 4 5 6 7 8 9 $$3^x$$ $$3^1$$ $$3^2$$ $$3^3$$
$$x$$ 10 11 12 13 14 $$3^x$$
3. Complete the table. You can read the values from the tables you made in questions 1 and 2.
Product of powers Repeated factor Power of powernotation Total number of repetitions Simplified form Value $$2^4 \times 2^4 \times 2^4$$ 2 $$(2^4)^3$$ 12 $$2^{12}$$ 4 096 $$3^2 \times 3^2 \times 3^2 \times 3^2$$ $$2^3 \times 2^3 \times 2^3 \times 2^3 \times 2^3$$ $$3^4 \times 3^4 \times 3^4$$ $$2^6 \times 2^6 \times 2^6$$
4. Use your table of powers of 2 to find the answers for the following:
1. $$2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = \text{______} = \text{______}$$
2. $$(2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) = \text{______} = \text{______}$$
3. $$16^3 = \text{______} = \text{_______} = \text{______}$$
5. Use your table of powers of 2 to find the answers for the following:
1. Is $$16^3 = 2^{12}$$?
2. Is $$2^4 \times 2^4 \times 2^4 = 2^{12}$$?
3. Is $$2^4 \times 2^3 = 2^{12}$$?
4. Is $$(2^4)^3 = 2^4 \times 2^4 \times 2^4$$?
5. Is $$(2^4)^3 = 2^{12}$$?
6. Is $$(2^4)^3 = 2^{4+3}$$?
7. Is $$(2^4)^3 = 2^{4 \times 3}$$?
8. Is $$(2^2)^5 = 2^{2+5}$$?
1. Express $$8^5$$ as a power of 2. It may help to first express 8 as a power of 2.
2. Can $$(2^3) \times (2^3) \times (2^3) \times (2^3) \times (2^3)$$ be expressed as $$(2^3)^5$$?
3. Is $$(2^3)^5 = 2^{3+5}$$ or is $$(2^3)^5 = 2^{3 \times 5}$$?
1. Express $$4^3$$ as a power of 2.
2. Calculate $$2^2 \times 2^2 \times 2^2$$ and express your answer as a single power of 2.
3. Can $$(2^2) \times (2^2) \times (2^2)$$ be expressed as $$(2^2)^3$$?
4. Is $$(2^2)^3 = 2^{2+3}$$ or is $$(2^2)^3 = 2^{2 \times 3}$$?
6. Simplify the following.
Example: $$(10^2)^2 = 10^2 \times 10^2 = 10^{2+2}=10^4 = 10000$$
1. $$(3^2)^2$$
2. $$(4^3)^2$$
3. $$(2^4)^2$$
4. $$(9^2)^2$$
5. $$(3^3)^3$$
6. $$(4^3)^3$$
7. $$(5^4)^3$$
8. $$(9^2)^3$$
$$(a^m)^n = a^{m \times n}$$, where m and n are natural numbers and a is not equal to zero.
1. Simplify.
1. $$(5^4)^{10}$$
2. $$(10^4)^5$$
3. $$(6^4)^4$$
2. Write $$5^{12}$$ as a power of powers of 5 in two different ways.
To simplify $$(x^2)^5$$ we can write it out as a product of powers or we can use a shortcut.
\begin{align} (x^2)^5 &= (x^2) \times (x^2) \times (x^2) \times (x^2) \times (x^2) \\ \ &= \underbrace{x \times x} \times \underbrace{x \times x} \times \underbrace{x \times x} \times \underbrace{x \times x} \times \underbrace{x \times x} = x^{10} \\ & 2 \times 5 \text{ factors} = 10 \text{ factors} \end{align}
1. Complete the table.
Expression Write as a product of the powers and then simplify Use the rule $${\bf(a^m)^n}$$ to simplify (a) $$(a^4)^5$$ $a^4 \times a^4 \times a^4 \times a^4 \times a^4 \\ a^{4 + 4 +4 + 4 + 4} = a^{20}$ $$(a^4)^5 = a^{4 \times 5} =a ^{20}$$ (b) $$(b^{10})^5$$ (c) $$(x^7)^3$$ (d) $s^6 \times s^6 \times s^6 \times s^6 \\ = s^{6+6+6+6} \\ = s^{24}$ (e) $$y^{3 \times 7} = y^{21}$$
### Power of a product
1. Complete the table. You may use your calculator when you are not sure of a value.
$$x$$ 1 2 3 4 5 (a) $$2^x$$ $$2^1 = 2$$ (b) $$3^x$$ $$3^2 = 9$$ (c) $$6^x$$ $$6^3 = 216$$
2. Use the table in question 1 to answer the questions below. Are these statements true or false? If a statement is false rewrite it as a correct statement.
1. $$6^2 = 2^2 \times 3^2$$
2. $$6^3 = 2^3 \times 3^3$$
3. $$6^5 = 2^5 \times 3^5$$
4. $$6^8 = 2^4 \times 3^4$$
3. Complete the table.
Expression The bases of the expression are factors of . . . Equivalent expression (a) $$2^6 \times 2^5$$ 10 $$10^6$$ (b) $$3^2 \times 4^2$$ (c) $$4^2 \times 2^2$$ (d) $$56^5$$ (e) $$30^3$$ (f) $$3^5 \times x^5$$ $$3x$$ $$(3x)^5$$ (g) $$7^2 \times z^2$$ (h) $$4^3 \times y^3$$ (i) $$(2m)^6$$ (j) $$(2m)^3$$ (k) $$2^{10} \times y^{10}$$ $$(2y)^{10}$$
$$12^2$$can be written in terms of its factors as $$(2 \times 6)^2$$ or as $$(3 \times4)^2$$
We already know that $$12^2$$ = 144.
What this tells us is that both $$(2 \times 6)^2$$ and $$(3 \times4)^2$$ also equal 144.
We write 12:
\begin{align} 12^2 &= (2 \times 6)^2 \\ &= 2^2 \times 6^2 \\ &= 4 \times 36 \\ &= 144 \end{align}
or
\begin{align} 12^2 &= (3 \times 4)^2 \\ &= 3^2 \times 4^2 \\ &= 9 \times 16 \\ &= 144 \end{align}
A product raised to a power is the product of the factors each raised to the same power.
Using symbols, we write $$(a\times b)^m = a^m \times b^m$$, where m is a natural number and a and b are not equal to zero
1. Write each of the following expressions as an expression with one base:
Example: $$3^{10} \times 2^{10} = (3 \times 2)^{10} = 6^{10}$$
1. $$3^2 \times 5^2$$
2. $$5^3 \times 2^3$$
3. $$7^4 \times 4^4$$
4. $$2^3 \times 6^3$$
5. $$4^4 \times 2^4$$
6. $$5^2 \times 7^2$$
2. Write the following as a product of powers:
Example:$$(3x)^3 = 3^3 \times x^3 = 27x^3$$
1. $$6^3$$
2. $$15^2$$
3. $$21^4$$
4. $$6^5$$
5. $$18^2$$
6. $$(st)^7$$
7. $$(ab)^3$$
8. $$(2x)^2$$
9. $$(3y)^5$$
10. $$(3c)^2$$
11. $$(gh)^4$$
12. $$(4x)^3$$
3. Simplify the following expressions:
Example: $$3^2 \times m^2 = 9 \times m^2 = 9m^2$$
1. $$3^5 \times b^5$$
2. $$2^6 \times y^6$$
3. $$x^2 \times y^2$$
4. $$10^4 \times x^4$$
5. $$3^3 \times x^3$$
6. $$5^2 \times t^2$$
7. $$6^3 \times m^7$$
8. $$12^2 \times a^2$$
9. $$n^3 \times p^9$$
### A quotient of powers
Consider the following table:
$$x$$ 1 2 3 4 5 6 $$2^x$$ 2 4 8 16 32 64 $$3^x$$ 3 9 27 81 243 729 $$5^x$$ 5 25 125 625 3 125 15 625
Answer questions 1 to 4 by referring to the table when you need to.
1. Give the value of each of the following:
1. $$3^4$$
2. $$2^5$$
3. $$5^6$$
1. Calculate $$3^6 \div 3^3$$ (Read the values of $$3^6$$ and $$3^3$$ from the table and then divide. You may use a calculator where necessary.)
To calculate $$4^{5-3}$$ we first do the calculation in the exponent, that is, we subtract 3 from 5. Then we can calculate $$4^2$$ as $$4 \times 4 = 16$$.
2. Calculate $$3^{6-3}$$
3. Is $$3^6 \div 3^3$$ equal to $$3^3$$? Explain.
1. Calculate the value of $$2^{6-2}$$
2. Calculate the value of $$2^6 \div 2^2$$
3. Calculate the value of $$2^{6\div 2}$$
4. Read from the table the value of $$2^3$$
5. Read from the table the value of $$2^4$$
6. Which of the statements below is true? Give an explanation for your answer.
A. $$2^6 \div 2^2 = 2^{6-2} = 2^4$$
B. $$2^6 \div 2^2 = 2^{6 \div 2} = 2^3$$
2. Say which of the statements below are true and which are false. If a statement is false rewrite it as a correct statement.
1. $$5^6 \div 5^4 = 5^{6 \div 4}$$
2. $$3^{4-1} = 3^4 \div 3$$
3. $$5^6 \div 5 = 5^{6 - 1}$$
4. $$2^5 \div 2^3 = 2^2$$
$$a^m \div a^n = a^{m-n}$$
where m and n are natural numbers and m is a number greater than n and a is not zero.
1. Simplify the following. Do not use a calculator.
Example: $$3^{17} \div 3^{12} = 3^{17-12} = 3^5 = 243$$
1. $$2^{12} \div 2^{10}$$
2. $$6^{17} \div 6^{14}$$
3. $$10^{20} \div 10^{14}$$
4. $$5^{11} \div 5^{8}$$
2. Simplify:
1. $$x^{12} \div x^{10}$$
2. $$y^{17} \div y^{14}$$
3. $$t^{20} \div t^{14}$$
4. $$n^{11} \div n^{8}$$
### The power of zero
1. Simplify the following:
1. $$2^{12} \div 2^{12}$$
2. $$6^{17} \div 6^{17}$$
3. $$6^{14} \div 6^{14}$$
4. $$2^{10} \div 2^{10}$$
We define $$a^0 = 1$$
Any number raised to the power of zero is always equal to 1.
2. Simplify the following:
1. $$100^0$$
2. $$x^0$$
3. $$(100x)^0$$
4. $$(5x^3)^0$$
## Calculations
### Mixed operations
Simplify the following:
1. $$3^3 + \sqrt[3]{-27} \times 2$$
2. $$5 \times (2+3)^2 + (-1)^0$$
3. $$3^2 \times 2^3 + 5 \times \sqrt{100}$$
4. $$\frac{ \sqrt[3]{1000}}{\sqrt{100}} + (4-1)^2$$
5. $$\sqrt{16} \times \sqrt{16} + \sqrt[3]{216} + 3^2 \times 10$$
6. $$4^3 \div 2^3 + \sqrt{144}$$
## Squares, cubes and roots of rational numbers
### Squaring a fraction
Squaring or cubing a fraction or a decimal fraction is no different from squaring or cubing an integer.
1. Complete the table.
Fraction Square the fraction Value of the square of the fraction (a) $$\frac{1}{2}$$ $$\frac{1}{2} \times \frac{1}{2}$$ $$\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$ (b) $$\frac{2}{3}$$ (c) $$\frac{3}{4}$$ (d) $$\frac{2}{5}$$ (e) $$\frac{3}{5}$$ (f) $$\frac{2}{6}$$ (g) $$\frac{3}{7}$$ (h) $$\frac{11}{12}$$
2. Calculate the following:
1. $$(\frac{3}{2})^2$$
2. $$(\frac{4}{5})^2$$
3. $$(\frac{7}{8})^2$$
1. Use the fact that 0,6 can be written$$\frac{6}{10}$$ to calculate $$(0,6)^2$$.
2. Use the fact that 0,8 can be written as $$\frac{8}{10}$$ to calculate $$(0,8)^2$$.
### Finding the square root of a fraction
1. Complete the table.
Fraction Writing the fraction as a product of factors Square root (a) $$\frac{81}{121}$$ (b) $$\frac{64}{81}$$ (c) $$\frac{49}{169}$$ (d) $$\frac{100}{225}$$
2. Determine the following:
1. $$\sqrt{\frac{25}{16}}$$
2. $$\sqrt{\frac{81}{144}}$$
3. $$\sqrt{\frac{400}{900}}$$
4. $$\sqrt{\frac{36}{81}}$$
1. Use the fact that 0,01 can be written as $$\frac{1}{100}$$ to calculate $$\sqrt{0.01}$$.
2. Use the fact that 0,49 can be written as$$\frac{49}{100}$$ to calculate $$\sqrt{0.49}$$.
3. Calculate the following
1. $$\sqrt{0.09}$$
2. $$\sqrt{0.64}$$
3. $$\sqrt{1.44}$$
### Cubing a fraction
One half cubed is equal to one eighth.
We write this as $$(\frac{1}{2})^3 = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}$$
1. Calculate the following:
1. $$(\frac{2}{3})^3$$
2. $$(\frac{5}{10})^3$$
3. $$(\frac{5}{6})^3$$
4. $$(\frac{4}{5})^3$$
1. Use the fact that 0,6 can be written$$\frac{6}{10}$$ to calculate $$(0,6)^3$$.
2. Use the fact that 0,8 can be written as $$\frac{8}{10}$$ to calculate $$(0,8)^3$$.
3. Use the fact that 0,7 can be written as $$\frac{7}{10}$$ to calculate $$(0,7)^3$$.
## Scientific notation
### Very large numbers
1. Express each of the following as a single number. Do not use a calculator.
Example: $$7,56 \times 100$$ can be written as 756.
1. $$3,45 \times 100$$
2. $$3,45 \times 10$$
3. $$3,45 \times 1 000$$
4. $$2,34 \times 10^2$$
5. $$2,34 \times 10$$
6. $$2,34 \times 10^3$$
7. $$10^4 \times 10^2$$
8. $$10^0 \times 10^6$$
9. $$3,4 \times 10^5$$
We can write 136 000 000 as $$1,36 \times 10^8$$.
$$1,36 \times 10^8$$ is called the scientific notation for 136 000 000.
In scientific notation, a number is expressed in tow parts: a number between 1 and 10 multiples by a power of 10. The exponent must always be an integer.
1. Write the following numbers in scientific notation:
1. 367 000 000
2. 21 900 000
3. 600 000 000 000
4. 178
2. Write each of thefollowing numbers in the ordinary way.
For example: $$3,4 \times 10^5$$ written in the ordinary way is 340 000.
1. $$1,24 \times 10^8$$
2. $$9,2074 \times 10^4$$
3. $$1,04 \times 10^6$$
4. $$2,05 \times 10^3$$
3. The age of the universe is 15 000 000 000 years. Express the age of the universe in scientific notation.
4. The average distance from the Earth to the Sun is 149 600 000 km. Express this distance in scientific notation.
Because it is easier to multiply powers of ten without a calculator, scientific notation makes it possible to do calculations in your head.
5. Explain why the number $$24 \times 10^3$$ is not in scientific notation.
6. Calculate the following. Do not use a calculator.
Example: $$3 000 000 \times 90 000 000 = 3 \times 10^6 \times 9 \times 10^7 = 3 \times 9 \times 10^{6 + 7} \\ = 27 \times 10^{13} = 270 000 000 000 000$$
1. $$13 000 \times 150 000$$
2. $$200 \times 6 000 000$$
3. $$120 000 \times 120 000 000$$
4. $$2,5 \times 40 000 000$$
7. Use > or < to compare these numbers:
1. $$1,3 \times 10^9$$ ☐ $$2,4 \times 10^7$$
2. $$6,9 \times 10^2$$ ☐ $$4,5 \times 10^3$$
3. $$7,3 \times 10^4$$ ☐ $$7,3 \times 10^2$$
4. $$3,9 \times 10^6$$ ☐ $$3,7 \times 10^7$$
1. Calculate:
1. $$11^2$$
2. $$3^2 \times 4^2$$
3. $$6^3$$
4. $$\sqrt{121}$$
5. $$(-3)^2$$
6. $$\sqrt[3]{125}$$
2. Simplify
1. $$3^4 \times m^6$$
2. $$b^2 \times n^6$$
3. $$y^{12} \div y^5$$
4. $$(10^2)^3$$
5. $$(2w^2)^3$$
6. $$(3d^5)(2d)^3$$
3. Calculate
1. $$(\frac{2}{5})^2$$
2. $$\sqrt{\frac{9}{25}}$$
3. $$(6^4y^2)^0$$
4. $$(0.7)^2$$
4. Simplify
1. $$(2^2 + 4)^2 + \frac{6^2}{3^3}$$
2. $$\sqrt[3]{-125} -5 \times 3^2$$
5. Write $$3 \times 10^9$$ in the ordinary way.
6. The first birds appeared on Earth about 208 000 000 years ago. Write this number in scientific notation. |
# What is 10 to the 4th Power?
Mathematics is an exciting field consisting of a wide range of concepts and principles; one such concept is exponential notation. Exponential notation is a fundamental concept that involves raising a base number to a specific power (or exponent); these are potent concepts used to express vast and tiny numbers more appropriately; one such example is discussed in this article, i.e., 10 to the 4th Power. So, in this article, we will learn the concept of exponential notation, discuss the significance of the 10 to the 4thPower, and learn about various ways of expressing this value.
## Defining Exponential Notation
The term exponential notation, also known as scientific notation, allows us to express numbers in a standard and organized way. It is mainly used when we deal with vast and small numbers because, with this expression, we can easily represent huge and tiny numbers.
### Parts of Scientific Notation
There are two parts to this notation:
1. The Base Number
2. The Exponent (or power) Number
The base is 10 in our case, which shows we are dealing with the Power of 10. The exponent or Power is 4, which shows how many times the base number (10) is multiplied by itself; this implies that 10 to the Power of 4 equals 10 multiplied by itself 4 times. 10 x 10 x 10 x 10 = 104
### Significant
10 to the 4th Power has been critical in various fields, especially scientific research, technology, and everyday life. It represents the value of ten thousand, equal to 1, followed by four zeros, i.e., 10,000. In scientific notations, this value is often used when we measure large quantities, such as counting the number of stars in the universe, the distance between planets and celestial objects, and the world population. It helps scientists and researchers to express these huge quantities more easily and makes it convenient for them to comprehend and analyze that data.
## Expressing 10 to the 4th Power
There are several ways to express or represent 10 to the Power of 4, depending on the field it is used in or personal preference of use.
1. Numerical Representation: Another way to express 10 to the Power of 4 is by writing the number in its expanded form, 10,000. This format breaks down the number into its digits and highlights its value to understand how huge this number is.
2. Scientific Notation: The most simple and straight representation is 104, where 10 is base, and 4 is exponent or Power; this format is mainly recognized and used in mathematical and scientific books.
3. Long-form: The long-form representation of 10 to the Power of 4 is "ten thousand," we use this format in our daily language, allowing us to get the magnitude of the number more easily.
## Applications of 10 to the 4th Power
Now that we understand the calculation let's explore some practical applications where the 10 to the 4th Power finds relevance:
1. Computing Large Numbers: In fields such as finance, physics, and engineering, large numerical calculations are usually involved. Exponential notation is valuable in simplifying these calculations, as it allows for compact representation and reduces the chances of error.
2. Scientific Notation: 10 to the 4th Power is a crucial component of scientific notation. This notation expresses very large or minimal numbers in a more manageable format.
3. Digital Technology: In computers and digital technology, powers of 10 play a significant role. Computers store and process information using binary code based on a two-digit system (0 and 1). Powers of 2, closely related to powers of 10, are used to represent the size of the memory card, data transfer rates, and speed of the processors.
4. Scale and Magnitude: Powers of 10 are often employed to describe the vastness or minuteness of objects or phenomena. For example, astronomers use the concept of a light-year, which is the distance light travels in one year, approximately equal to 9.461 × 1012 Similarly, the atomic radius of an atom is in the range of 0.1 to 0.5 nanometers (10-10 meters).
5. Population Growth: Exponential growth is frequently observed in natural phenomena, including population growth. The concept of 10 to the 4th Power can help us understand how populations can increase rapidly over time; for instance, if a population of 10,000 individuals grows at a rate of 10% per year, it would reach approximately 14,641 after four years.
## Conclusion
Understanding and expressing the 10 to the 4th Power is crucial in comprehending the wideness of numbers or numerical values used in scientific research, technology, and our daily lives. We can express this large number in various forms, but all these forms are used to convey and comprehend such large numerical values. |
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# Converting Fractions Study Guide (page 2)
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## Comparing Fractions
Which fraction is larger, or ? Don't be fooled into thinking that is larger just because it has the larger bottom number. There are several ways to compare two fractions, and they can be best explained by example.
Use your intuition: "pizza" fractions. Visualize the fractions in terms of two pizzas, one cut into 8 slices and the other cut into 5 slices. The pizza that's cut into 5 slices has larger slices. If you eat 3 of them, you're eating more pizza than if you eat 3 slices from the other pizza. Thus, is larger than . Compare the fractions to known fractions like . Both and are close to . However, is more than , while is less than . Therefore, is larger than . Comparing fractions to is actually quite simple. The fraction is a little less than , which is the same as ; in a similar fashion, is a little more than , which is the same as . ( may sound like a strange fraction, but you can easily see that it's the same as by considering a pizza cut into 5 slices. If you were to eat half the pizza, you'd eat slices.) Change both fractions to decimals. Remember the fraction definition at the beginning of this lesson? A fraction means divide: Divide the top number by the bottom number. Changing to decimals is simply the application of this definition. Because 0.6 is greater than 0.375, the corresponding fractions have the same relationship: is greater than . Raise both fractions to higher terms. If both fractions have the same denominator, then you can compare their top numbers. Because 24 is greater than 15, the corresponding fractions have the same relationship: is greater than . Shortcut: cross multiply. "Cross multiply" the top number of one fraction with the bottom number of the other fraction, and write the result over the top number. Repeat the process using the other set of top and bottom numbers. Since 24 is greater than 15, the fraction under it, , is greater than .
#### Tip
It's time to take a look at your pocket change again! Only this time, you need less than a dollar. So if you found extra change in your pocket, now is the time to be generous and give it away. After you gather a pile of change that adds up to less than a dollar, write the amount of change you have in the form of a fraction. Then reduce the fraction to its lowest terms.
You can do the same thing with time intervals that are less than an hour. How long till you have to leave for work, go to lunch, or begin your next activity for the day? Express the time as a fraction, and then reduce to lowest terms.
## Converting Fractions Sample Questions
1. Reduce to lowest terms.
2. Raise to 16ths.
### Solutions
#### Question 1
Divide by 3:
#### Question 2
1 Divide the old bottom number (8) into the new one (16): 2 Multiply the quotient (2) by the old top number (3): 2 ×3=6 3 Write the product (6) over the new bottom number (16): 4 Check: Reduce the new fraction to make sure you get back the original.
Find practice problems and solutions for these concepts at Converting Fractions Practice Questions.
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### Related Questions
#### Q:
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# Pascal’s Triangle – Definition, Pattern & Examples
The binomial theorem is a method of expanding an expression that has been raised to any finite power. Using a binomial theorem, any algebraic expression $(a + b)$ with non-negative power can be expanded into a sum of the form $(a+b)^n = ^n \text{C}_0 a^nb^0 + ^n \text{C}_1 a^{n – 1}b^1 + ^n \text{C}_2 a^{n – 2}b^2 + … + ^n \text{C}_{n – 1} a^{1}b^{n – 1} + ^n \text{C}_n a^0b^n$. In this expression, each term with a numeric value called a coefficient can be obtained using Pascal’s Triangle.
Let’s understand what is Pascal’s Triangle and how to create Pascal’s Triangle.
## What is Pascal’s Triangle?
Pascal’s Triangle is a kind of number pattern. Pascal’s Triangle is the triangular arrangement of numbers that gives the coefficients in the expansion of any binomial expression. The numbers are so arranged that they reflect a triangle.
A Pascal’s triangle is an arrangement of numbers in a triangular array such that the numbers at the end of each row are $1$ and the remaining numbers are the sum of the nearest two numbers in the above row. This concept is used widely in probability, combinatorics, and algebra. Pascal’s triangle is used to find the likelihood of the outcome of the toss of a coin, coefficients of binomial expansions in probability, etc.
## How to Create a Pascal’s Triangle?
Pascal’s triangle is a special triangle that is named after Blaise Pascal. In this triangle, we start with a number $1$ at the top, then $1$s at both sides of the triangle until the end. The middle numbers are so filled that each number is the sum of the two numbers just above it.
The number of elements in the $n^\text{th}$ row is equal to $(n + 1)$ elements. Pascal’s triangle can be constructed by writing $1$ as the first and the last element of a row and the other elements of the row are obtained from the sum of the two consecutive elements of the previous row. Pascal’s triangle can be constructed easily by just adding the pair of successive numbers in the preceding lines and writing them in the new line.
The above figure shows Pascal’s Triangle up to the $6^{th}$ row.
You can see that the number at the top is $1$, from where the creation of Pascal’s triangle starts. This is called the $0^\text{th}$ row. Next, $1$s is added on both sides which form the $1^\text{st}$ row.
The coefficients of the binomial expansion start from the $2^\text{nd}$, which contain the numbers, $1$, $2$, and $1$. The number $2$ is obtained by adding the two $1$s in the previous row and after that $1$s are added on both sides.
Similarly, for the $3^\text{rd}$ row, the two $3$s in the middle are obtained by adding $1$, $2$ and $2$, $1$ of $2^\text{nd}$ row.
Moving on to the $4^\text{th}$ row, $4 = 1 + 3$, $6 = 3 + 3$, $4 = 3 + 1$, and then again two $1$s are added on both sides.
## Pascal’s Triangle Formula
The formula to fill the number in the nth column and mth row of Pascal’s triangle we use Pascal’s triangle formula. The formula requires the knowledge of the elements in the $(n-1)^\text{th}$ row, and $(n-1)^\text{th}$ and $n^\text{th}$ columns.
The elements of the $n^\text{th}$ row of Pascal’s triangle are given by, $^n\text{C}_0$, $^n\text{C}_1$, $^n\text{C}_2$, …, $^n\text{C}_n$.
The formula for Pascal’s triangle is $^n \text{C}_m = ^{n-1} \text{C}_{m-1} + ^{n-1} \text{C}_{m}$
where
$^n \text{C}_m$ represents the $(m+1)^\text{th}$ element in the $n^\text{th}$ row
$n$ is a non-negative integer, and $0 \le m \le n$
### Examples of Pascal’s Triangle Formula
Example 1: Find the value of $^4 \text{C}_2$ using Pascal’s Triangle.
To get the value of $^4 \text{C}_2$ using Pascal’s Triangle, find the $3^\text{rd}$ element in the $4^\text{th}$ row. From the triangle, we see that the value is $6$.
Therefore, $^4 \text{C}_2 = 6$.
Example 2: Find the value of $^6 \text{C}_4$ using Pascal’s Triangle.
To get the value of $^6 \text{C}_4$ using Pascal’s Triangle, find the $5^\text{rd}$ element in the $6^\text{th}$ row. From the triangle, we see that the value is $15$.
Therefore, $^6 \text{C}_4 = 15$.
## Pascal’s Triangle in Binomial Expansion
Pascal’s triangle can also be used to find the coefficient of the terms in the binomial expansion $(x+y)n = a_0 x^n y^0+ a_1 x^{n-1} y^1 + a_2 x^{n-2} y^2 + … + a_n x^0 y^n$. Pascal’s triangle is a handy tool to quickly verify if the binomial expansion of the given polynomial is done correctly or not.
### Examples of Pascal’s Triangle in Binomial Expansion
Example 1: Expand $(x+y)^2$ using Pascal’s triangle.
Since the power in $(x+y)^2$ is $2$, therefore, we will look into the $2^\text{nd}$ row of Pascal’s triangle. The numbers in the second row of Pascal’s triangle are $1$, $2$, and $1$. These numbers will be the coefficients of the terms in the expansion of $(x+y)^2$.
Therefore, $(x+y)^2 = 1 \times x^2 \times y^0+ 2 \times x^1 \times y^1 + 1 \times x^0 \times y^2 = x^2 + 2xy + y^2$.
Example 2: Expand $(x+y)^5$ using Pascal’s triangle.
Since the power in $(x+y)^5$ is $5$, therefore, we will look into the $5^\text{th}$ row of Pascal’s triangle. The numbers in the second row of Pascal’s triangle are $1$, $5$, $10$, $10$, $5$, and $1$. These numbers will be the coefficients of the terms in the expansion of $(x+y)^5$.
Therefore, $(x+y)^5 = 1 \times x^5 \times y^0 + 5 \times x^4 \times y^1 + 10 \times x^3 \times y^2 + 10 \times x^2 \times y^3 + 5 \times x^1 \times y^4 + 1 \times x^0 \times y^5$
$= x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5$.
## Pascal’s Triangle in Probability
Pascal’s triangle can be used in various places in the field of mathematics. Pascal’s triangle is used in probability and can be used to find the number of combinations, etc. It gives us the number of combinations of heads or tails that are possible from the number of tosses.
For example, if we toss a coin two times, we get $\text{HH}$ $1$ time, $\text{HT}$ or $\text{TH}$ $2$ times, and $\text{TT}$ $1$ time, which is the exact match of the elements in the second row of Pascal’s triangle.
The following table shows the results in the various number of tosses.
## Fibonacci Series in Pascal’s Triangle
By adding the numbers in the diagonals of Pascal’s triangle, we obtain the Fibonacci sequence.
There are various ways to show the Fibonacci numbers on Pascal’s triangle. R. Knott was able to find the Fibonacci appearing as sums of “rows” in Pascal’s triangle. He moved all the rows over by one place and here the sums of the columns would represent the Fibonacci numbers.
## Properties of Pascal’s Triangle
The following are some of the important properties of Pascal’s triangle.
• Each number is the sum of the two numbers above it
• The outside numbers are all $1$
• The triangle is symmetric
• The first diagonal shows the counting numbers
• The sums of the rows give the powers of $2$
• Each row gives the digits of the powers of $11$
• Each entry is an appropriate “combinatorics number” and is the “binomial coefficient”
• The Fibonacci numbers are there along diagonals
## Practice Problems
1. Find the following numbers in Pascal’s triangle
• $2^\text{nd}$ row, $1^\text{st}$ column
• $4^\text{th}$ row, $3^\text{rd}$ column
• $6^\text{th}$ row, $4^\text{th}$ column
• $10^\text{th}$ row, $8^\text{th}$ column
2. Expand the following using Pascal’s triangle
• $(x + y)^4$
• $(x + y)^7$
• $(2x – 3y)^6$
## FAQs
### What is Pascal’s Triangle?
Pascal’s triangle is a special triangle that is named after Blaise Pascal. In this triangle, we start with a number $1$ at the top, then $1$s at both sides of the triangle until the end. The middle numbers are so filled that each number is the sum of the two numbers just above it.
### What is Pascal’s Triangle rule?
In this triangle, we start with a number $1$ at the top, then $1$s at both sides of the triangle until the end. The middle numbers are so filled that each number is the sum of the two numbers just above it.
### Why is Pascal’s triangle important?
The importance of Pascal’s Triangle lies in the fact that it is widely used in probability, combinatorics, and algebra.
### Why is it called Pascal’s triangle?
Pascal’s triangle is named after the famous 17th-century French mathematician Blaise Pascal because of his work on many triangle properties.
### How do you use Pascal’s triangle in a binomial expansion?
Pascal’s triangle can also be used to find the coefficient of the terms in the binomial expansion $(x+y)n = a_0 x^n y^0+ a_1 x^{n-1} y^1 + a_2 x^{n-2} y^2 + … + a_n x^0 y^n$. The numbers in each row of Pascal’s triangle correspond to coefficients in the expansion of binomial expression.
### What is Pascal’s triangle formula?
The formula for Pascal’s triangle is $^n \text{C}_m = ^{n-1} \text{C}_{m-1} + ^{n-1} \text{C}_{m}$
where
$^n \text{C}_m$ represents the $(m+1)^\text{th}$ element in the $n^\text{th}$ row
$n$ is a non-negative integer, and $0 \le m \le n$
### What is the first element in each row of Pascal’s triangle?
The first, as well as the last element of each row of Pascal’s triangle, is $1$.
## Conclusion
Pascal’s triangle is a special triangle that is named after Blaise Pascal. In this triangle, we start with a number $1$ at the top, then $1$s at both sides of the triangle until the end. The middle numbers are so filled that each number is the sum of the two numbers just above it. Pascal’s Triangle is used widely in probability, combinatorics, and algebra. |
# Show that the middle term in the expansion
Question:
Show that the middle term in the expansion of $\left(x-\frac{1}{x}\right)^{2 n}$ is
$\frac{1 \times 3 \times 5 \times \ldots \times(2 n-1)}{n !} \times(-2)^{n}$
Solution:
Given, expression is $\left(x-\frac{1}{x}\right)^{2 n}$
Since the index is $2 n$, which is even. So, there is only one middle term, i.e.,
$\left(\frac{2 n}{2}+1\right)$ th term $=(n+1)$ th term
$T_{n+1}={ }^{2 n} C_{n}(x)^{2 n-n}\left(-\frac{1}{x}\right)^{n}={ }^{2 n} C_{n}(-1)^{n}=(-1)^{n} \frac{(2 n !)}{n ! \cdot n !}$
$=(-1)^{n} \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \ldots(2 n-1) \cdot(2 n)}{n ! \cdot n !}=(-1)^{n} \frac{[1 \cdot 3 \cdot 5 \ldots(2 n-1)] \cdot[2 \cdot 4 \cdot 6 \ldots(2 n)]}{(1 \cdot 2 \cdot 3 \ldots n) \cdot n !}$
$=(-1)^{n} \frac{[1 \cdot 3 \cdot 5 \ldots(2 n-1)] \cdot 2^{n}[1 \cdot 2 \cdot 3 \ldots n]}{(1 \cdot 2 \cdot 3 \ldots n) \cdot n !}=(-2)^{n} \frac{[1 \cdot 3 \cdot 5 \ldots(2 n-1)] \cdot 2^{n}}{n !}$ |
# How do you translate the following statement "Seven times a number, added to the sum of the number and six" into an algebraic expression?
Mar 18, 2016
$7 x + \left(x + 6\right)$
#### Explanation:
Let us assume that the number is $x$
Now the statement:
"$\textcolor{g r e e n}{\text{Seven times a number}} ,$ $\textcolor{red}{\text{added to}}$ the $\textcolor{b l u e}{\text{sum of the number}}$$\textcolor{b l u e}{\text{ and six}}$"
Seven times a number $\to$$7$ multiplied by the number $x \to 7 x$
added to $\to +$ the number in the part following of statement
sum of the number and six$\to x + 6$
All three steps written together
$7 \times x$ $+$ $\left(x + 6\right)$ , gives us the required algebraic expression
$7 x + \left(x + 6\right)$ |
# 36 Divided by 4
36 Divided by 4 = 9. If you typed “36/4” into a calculator, you would get “9.” The division is a numerical term used to down the 36 into equivalent 4 pieces. Along these lines, we can express that in the division cycle, 36 divided by four makes 36 divided into nine pieces concerning 4. The long advance division is a direct and simple interaction for the division after number cruncher.
## Portions of Division Process
How about we see the main thing we believe should do is make sense of the terms with the objective that you know what each piece of the division is:
The chief number, 36, is known as the profit.
The next number, four, is known as the divisor.
What we’ll do here is independent every movement of the long division process for 36 disengaged by four and figure out all of them, so you see definitively the specific thing going on.
### 36 Divided by 4 Long Haul Division
To understand Long haul division, we have to go through many processes. Here we will discuss them one by one.
1. Process 1 of division
The underlying advance is to set up our division issue with the divisor on the left side and the profit on the right side, like what we have under:
36÷4
2. Process 2 of division
We can figure out that the divisor (4) goes into the primary digit of the dividend (4) nine times (s). As of now, we realize that we can put nine at the top:
36÷4
As 4*9=36
We write nine at the top of the division in the answer section
3. Process 3 of division
Accepting we copy the divisor by the result in the past development (4 x 9 = 36), we can now add that reaction under the profit:
As 4*9=36
Now write 36 below 36 and subtract them.
4. Process 4 of division
Then, we will remove the result from the past development from the second digit of the profit (36 - 36= 0) and form that reaction underneath:
After subtracting 36-36 we got 0 remainder with answer 9
So 36÷4=9
In any case, what is the reaction from 36 detached by 4’s perspective?
Expecting you made it this far into the educational activity; nicely done! There are no more digits to drop down from the profit, which implies we have completed the long division issue.
Your reaction is the top number; any extra piece will be the base number. In this way, for 36 isolated by 4, the last course of action is 9. The extra part implies the remaining portion is 0.
Summary: The number 36 is the numerator or profit, and the number 4 is the denominator or divisor. The remaining part of 36 and 4, the extension of 36 and 4, and the little piece of 36 and 4 all mean (almost) the same thing: The common way to write the number 36 divided by four is 36/4.
## Additional Calculations of 36 Divided by 4
By and by, you’ve taken the long division method for managing 36 divided by 4; coming up next are two or three substitute ways you could do the calculation:
Using a little PC, expecting that you created in 36 parcelled by 4, you’d get 9. You could in like manner impart 36/4 as a mixed part: 9 0/4
Expecting you look at the mixed part 9 0/4, you’ll see that the numerator is identical to the remainder of), (the denominator is our one-of-a-kind divisor (4), and the whole number is our last reaction (9).
### 1. What is 36 Divided by 4?
We provide you with the result of division 36 by four right away:
• 36 isolated by 4 = 9
The result of 36/4 is a number, which is number that may be formed without decimal places.
• 36 detached by 4 in decimal = 9
• 36 disengaged by 4 to some extent = 36/4
• 36 separated by 4 in rate = 900%
Note that you could use our top-tier mini-computer above to get the remainder of any two numbers or decimals, including 36 and 4.
Repented, if any, are demonstrated in ().
The change is done normally once the nominator, for instance, 36, and the denominator, 4, have been installed. Avoid pressing the button, aside from assuming that you want to start. Look at it now with a similar division by 4.
### 2. Remainder and Quotient of 36 Divided by 4
Here we outfit you with the result of the division with an extra part, generally called Euclidean division, recollecting the terms more or less:
The remainder and rest of the 36 isolated by 4 = 9 Remainder 0. The remainder (number division) of 36/4 reciprocals 9; the remainder of (“over”) is 0. 36 is the profit, and 4 is the divisor.
To sum up, 36/4 = 9. It is a whole number with no fragmentary part. As division with an extra piece, the delayed consequence of 36 ÷ 4 = 9 with leftover portion 0.
### 3. Division in Mathematical Terms
The division is one of the major arithmetic errands in math in which a greater number is separated into additional unobtrusive social events having a comparative number of things. If 36 pupils are sorted into groups of 4, hard and fast groups will be formed.
The division movement provides a nuanced framework for dealing with such challenges. Here we need to seclude 36 by 4. The result will be 36 ÷ 4 = 9. Like this, there will be nine social events for four students each. You can affirm this value by copying 9 and 4, giving you the principal number, 36.
### 4. Division Definition
The division is the course of inauspicious subtraction. It is something contrary to the multiplication action. It is portrayed as the exhibit of forming identical social occasions.
While isolating numbers, we separate a greater number into other unassuming numbers so much that the multiplication of those more humble numbers will be identical to the greater number taken. For example, 36 ÷ 4 = 9. It can be formed as a multiplication reality as nine × 4 = 36.
### 5. Division Symbol
A picture of a little straight line with a dot above and below the line shows the division. Two major division pictures address the division of two numbers. They are ÷ and/.
For example, 36 ÷ 4 = 9, and 36/4 = 9.
Summary: Using a little PC, expecting you created in 36 parcelled by 4, you’d get 9. You could, in like manner, impart 36/4 as a mixed part. 9 0/4. Use our top-tier mini-computer above to get the remainder of any two numbers or decimals, including 36 and 4. The division is the course of inauspicious subtraction. It is portrayed as the exhibit of forming identical social occasions. A picture of a little straight line with a dot above and below the line shows the division. It can be formed as a multiplication reality as 9 × 4 = 36.
## Parts of Division
Parts of division mean the name of the terms connected with the division association. There are four bits of the division, which are profit, divisor, remainder, and remaining part. Permit us to look at an outline of the division given underneath and handle the ramifications of these four bits of the division.
Segments of division: Dividend, Divisor, Quotient, Remainder
Here, when we divide 36 by 4, we get the potential gains of a divisor, profit, remainder, and remaining part. Look at the table underneath to get a handle on the significance of these terms.
Terms Descriptions Values
Dividend The number divided 36
Divisor The number of that areas to be formulated, or the number by which we split the dividend 4
Quotient The answer got after the process of division 9
Remainder The remaining part of the dividend that isn’t a piece of part of the division 0
### What is Division Algorithm?
The division estimation is a condition that shapes an association between every one of the four bits of the division. In every division truth, the dividend equals the divisor, quotient, and remainder total.
Therefore, the general condition of division is:
Dividend = (Divisor × Quotient) + Remainder.
This is known as division estimation.
The criterion above confirms the gains of the quotient and additional piece after division. We can substitute the potential gains of the quotient, extra piece, and divisor in the above condition and check whether or not the result is identical to the dividend.
Expecting we get the dividend, it suggests we have done the method for division precisely. If not, it suggests there is a screw-up in our assessments that we need to alter. Permit us to take one model and check whether it satisfies the above division estimation.
In 36 divided by four models, 36 divided by four will give us nine as the quotient and 0 like the rest.
Dividend = (Divisor × Quotient) + Remainder
36= (9 × 4) + 0
36 = 36 + 0
36 = 36
### Characteristics of Division in Maths
By and by letting us look at a part of the properties of division movement that will help you figure out this action by a long shot predominant. Recorded under are several properties of division:
• Division by 1: Any number parcelled by 1 in the genuine number. By the day’s end, if divisor = 1, dividend = quotient.
• Division by 0: The value of a number isolated by 0 isn’t described; for instance, n/0 = not portrayed, where n is any number.
• Division without assistance from any other individual: If we segment a number without any other person, we will consistently find one as the arrangement. With everything taken into account, in the occasion, that dividend = divisor, quotient = 1.
• Division of 0 by any number: 0 apportioned by any number, by and large, achieves 0. A couple of models are 0 ÷ 15 = 0, 0 ÷ 18 = 0, 0 ÷ 5757 = 0, etc.
• Division by 10: If we segment a number by 10, the digit at the spot will continually be the other extra digits on the left will be the quotient. For example, 900 ÷ 10 = 90 R 0.
• Division by 100: If we segment a number by 100, the number outlined from the spot and the tens place digits will consistently be the other abundance digits on the left will be the quotient. For example, 9000 ÷ 100 = 90 R 0.
Keep In Mind: There are four division bits: profit, divisor, remainder, and remaining part. The general condition of division is: Dividend = (Divisor × Quotient) + Remainder. It is known as division estimation. Division by 0: The value of a number isolated by 0 isn’t described; for instance, n/0 = not portrayed, where n is any number. Division by 10: If we segment a number by 10, then the digit at the spot will continually be the other extra digits on the left will be the quotient.
Here are some questions that are asked frequently about divisions which are as follows.
1. How is division carried out in Mathematics?
In maths, we have four major arithmetic errands, i.e., addition, division, multiplication, and subtraction. Among these four errands, the division is one of the huge undertakings we use in our everyday activities. It is the technique associated with separating a huge social affair into comparable, more unobtrusive get-togethers.
2. What are the Two Basic Types of Division process?
The division is separated into two areas, i.e., partitive and quotative models. Partitive is used while disengaging a number into a known number of spaces.
For example, expecting we segment four into two spaces, we can sort out the number of things in each open. Quotative division is used while parcelling a number into openings of a conscious sum. For example, when we parcel four into openings of 2, we can conclude the number of spaces that can be made.
3. What are the three essentials of division?
The three essential bits of division are dividend, quotient, and divisor. In addition, when the divisor is anything but a variable of the dividend, we get a non-zero spare part which is the fourth piece of the division.
4. What is Long Division Method?
The long division method is the unique system used to handle division issues. In this association, the divisor is made outer the division picture, while the dividend is set inside. The quotient is created over the overbar on top of the dividend.
5. What are the processes of division?
The resources to separate are recorded underneath:
• Process 1: Take the most important number from the dividend. Check if this number is more or less important than the divisor.
• Process 2: Then segment it by the divisor and create the reaction on top.
• Process 3: Take the answer away from the digit and bring it down.
• Process 4: Again, repeat a comparative cycle.
6. what happens when the divisor is larger than the dividend?
For this circumstance of division, we can keep adding zeros aside from the dividend until it becomes reasonable to isolate further. In addition, we can segment the quotient by comparative powers of 10 for the last reaction once we finish the division precisely.
7. How to Divide Decimals in Mathematics?
Isolating decimals is additionally essential, as straightforward as secluding other numbers. You should copy the decimal with powers of ten until you get a number. Then you can finish the commonplace division process. When you find your old plan, try to divide it up with the powers of 10 that you used to separate it before.
8. How to Use Division Calculator?
A division number cruncher is a device used to deal with division and gives quickly in seconds. Endeavour this division number on the advanced mini-computer for dealing with issues considering division and track down your answers in seconds just by a singular snap.
9. What are the basic rules of Multiplication and Division of Integers?
The norms for the multiplication and division of the whole numbers are given under:
• Positive ÷/× positive = Positive
• Negative ÷/× negative = Positive
• Negative ÷/× positive = Negative
• Positive ÷/× negative = Negative
10. What is the Division Symbol used in mathematics?
There are two pictures of division which are: ÷ and/. ÷ picture is drawn by putting two little spots on the top and lower part of a little even line. Likewise,/sign is used generally with divisions, extents, and rates.
11. When division is Undefined by zero?
Division by zero is unclear because one can’t separate any number by nothing. It is because when any number is expanded to nothing, the reaction is 0. As of now, I believe it’s the inverse. 1/0 will have infinite worth. We can not quantify this value in science. In this way, the division of any number by zero is undefined.
12. How do you explain division?
It is possible to divide something into equal portions using this procedure. Among the four fundamental arithmetic operations, it yields a just distribution of resources. The division is the opposite of the multiplication operation.
13. What are the four ways to divide?
When you divide, you need to know four important terms. They are the dividend, the divisor, the quotient, and the remainder.
14. What is an example of division?
Mathematical division involves slicing a number into equal pieces and counting how many such pieces are. To divide 15 by 3, for instance, one must create three groups of five.
15. What are the two different types of division?
Partitive division and quotation division are the two types of division. The partitive division is splitting a number into a set number of groups. In quotative division, a number is split into a certain amount.
## Conclusion:
Using a PC, expecting you created in 36 parcelled by 4, you’d get 9. The division is one of the major arithmetic errands in math. It is when a greater number is separated into additional unobtrusive social events. Two major division pictures address the division of two numbers, ÷ and/. For example, 36 ÷ 4 = 9, and 36/4 = 9. Parts of division are the terms connected with the division association. There are four bits of the division, which are profit, divisor, remainder, and remaining part. The general condition of division is: Dividend = (Divisor × Quotient) + Remainder. The division is the technique associated with separating a huge social affair into comparable, more unobtrusive get-togethers. There are four division bits: profit, divisor, remainder, and remaining part. Partitive division and quotation division are the two types of division. In quotative division, a number is split into a certain amount - 15 by 3, for instance.
## Related Articled:
Optimized by Mohammad Waqar on 22/08/22 |
# Octave reduction
Octave reduction is the process of replacing an interval by the unique equivalent interval between the unison and the octave. In practice, this is done by adding or subtracting octaves from the starting interval as necessary.
## Practical methods
An easy way to find the reduced form of an interval is to use a specialized calculator such as xen-calc. This is especially useful when working with very complex ratios.
There are also several methods that can be followed. The choice of an appropriate method depends on the interval size measure being used: linear measures (e.g. frequency ratios), or logarithmic measures (e.g. scale steps or cents).
### Linear measures
Stacking intervals expressed as ratios corresponds to multiplying those ratios. For instance, going up an octave means multiplying by 2, while going down an octave means dividing by 2.
1. If the starting interval is greater or equal to the unison (1) and less than the octave (2), it is already in reduced form.
2. If the starting interval is less than 1, multiply it by 2. Repeat until the resulting interval is greater than 1.
3. If the starting interval is greater than 2, divide it by 2. Repeat until the resulting interval is less than 2.
Examples:
• 3/4 is less than 1, so multiply by 2 to get 3/2.
• 7/2 is greater than 2, so divide by 2 to get 7/4.
• 4/1 is greater than 2, so divide by 2 to get 2/1, which is equal to 2, so divide by 2 to get 1/1.
• Adding 4 just perfect fifths (3/2) corresponds to (3/2)4, thus 81/16 (or 5.0625), which is greater than 2 octaves (22 = 4), but less than 3 octaves (23 = 8), so divide by 2 twice to get 81/64.
• Subtracting a just perfect fourth (4/3) from a classic minor third 6/5 corresponds to 6/5 divided by 4/3, thus 9/10 (or 0.9). This interval is less than a unison (20 = 1) but greater than one octave down (2-1 = 1/2), so multiply by 2 once to get 9/5.
### Logarithmic measures
Stacking intervals expressed with logarithmic measures corresponds to adding those measures. For instance, when working in cents, going up an octave means adding 1200 ¢, while going down an octave means subtracting 1200 ¢.
1. Find the logarithmic measure of the octave in the same unit as the one used for your starting interval; e.g. 1200 ¢, 19 steps of 19edo, 1900 , etc.
2. If the starting interval is positive and less than the octave (e.g. 1200 ¢), it is already in reduced form.
3. If the starting interval is negative, add the octave. Repeat until the result is positive.
4. If the starting interval is greater than the octave, subtract the octave. Repeat until the result is less than the octave.
Examples:
• 1442¢ is greater than 1200 ¢, so subtract 1200 ¢ to get 242 ¢.
• In 31edo, the octave is 31 steps and the patent val of the fifth harmonic is 72 (steps). This interval is greater than the octave, so subtract 31 to get 41, so subtract 31 again to get 10.
## General formulas
### Linear measures
For a starting interval $r$ expressed as a ratio, the reduced form $\text{red}(r)$ of that interval can be found using this formula: $\text{red}(r) = r \cdot 2^{-\left\lfloor{\log_2 r}\right\rfloor}$.
Example:
• Octave-reducing 4900/243 can be done by using the formula with $r = 4900/243$:
\begin{align}\text{red}(4900/243) &= 4900/243 \cdot 2^{-\left\lfloor{\log_2 4900/243}\right\rfloor} \\ &= 4900/243 \cdot 2^{-\left\lfloor{4.33375\ldots}\right\rfloor} \\ &= 4900/243 \cdot 2^{-4} \\ &= 4900/243 \cdot 1/16 \\ &= 1225/972\end{align}
### Logarithmic measures
For a starting interval $l$ and octave $e$ expressed in the same units, the reduced form $\text{red}(l, e)$ of that interval can be found using this formula: $\text{red}(l, e) = r \bmod e$, where $\bmod$ is the modulo operation.
Example:
• Octave-reducing 412 steps of 97edo can be done by using the formula with $r = 412$, $e = 97$:
\begin{align}\text{red}(412, 97) &= 412 \bmod 97 \\ &= 24\end{align}
This formula can also be written without the modulo operation: $\text{red}(l, e) = r - e\left\lfloor{l/e}\right\rfloor$.
Example:
• Octave-reducing 412 steps of 97edo again:
\begin{align}\text{red}(412, 97) &= 412 - 97\left\lfloor{412/97}\right\rfloor \\ &= 412 - 97\left\lfloor{4.24742\ldots}\right\rfloor \\ &= 412 - 97 \cdot 4 \\ &= 412 - 388 \\ &= 24\end{align}
## Generalization
### Other equaves
Octave reduction is mainly used in the context of octave-equivalent tunings (eg. 12edo), where equivalent notes are separated by octaves. However, this operation can be generalized to any periodic tuning by replacing the octave by the interval of equivalence or equave of that tuning.
For example, tritave reduction is the analog of octave reduction in a tritave-equivalent tuning (eg. Bohlen-Pierce), where the equave is the tritave. Therefore, a tritave-reduced interval is always obtained through transposition by tritaves, and the reduced interval lies between the unison (1/1) and the tritave (3/1).
The general formula for an interval $r$ and an equave $e$ is as follows: $\text{red}(r, e) = r \cdot e^{-\left\lfloor{\log_e r}\right\rfloor}$. Note that $e$ is a variable and not Euler's number.
Examples:
• Consider a tritave-equivalent tuning; 7/9 is less than 3, so multiply by 3 to get 7/3.
• Consider a just perfect fifth-equivalent tuning; $\text{red}(81/64, 3/2) = 81/64 \cdot (3/2)^{-\left\lfloor{\log_{3/2} 81/64}\right\rfloor} = 1$.
• In the equal-tempered Bohlen-Pierce tuning, a tritave can be expressed as 1300 hekts and a BP fifth down as -500 hekts. This interval is less than the unison, so add 1300 hekts to get 800 hekts.
### Balanced reduction
Balanced reduction is an alternate operation where the values are equally distributed around the unison instead of being situated between the unison and the octave (or the equave).
Examples:
• Balanced octave-reduction with ratios will lead to values greater than or equal to $\frac{1}{\sqrt{2}}$, but less than $\sqrt{2}$.
• Balanced octave-reduction with cents will lead to values greater than or equal to -600 ¢, but less than 600 ¢.
• Balanced tritave-reduction with ratios will lead to values greater than or equal to $\frac{1}{\sqrt{3}}$, but less than $\sqrt{3}$.
Here are some formulas for balanced reduction:
• Balanced octave-reduction of an interval $r$ expressed as a ratio: $\text{reb}(r)=\frac{1}{\sqrt{2}} \text{red}(\sqrt{2} \cdot \text{red}(r))$[1].
• Balanced reduction of an interval $r$ and an equave $e$ expressed as ratios: $\text{reb}(r, e)=\frac{1}{\sqrt{e}} \text{red}(\sqrt{e} \cdot \text{red}(r, e), e)$.
• Balanced reduction of an interval $l$ and an equave $e$ expressed as logarithmic measures in the same units: $\text{reb}(l, e)= \text{red}(\text{red}(l, e) + e/2, e) - e/2$. |
# 4.1 Algebraic expressions
Page 2 / 3
## Sample set b
Identify the factors in each term.
$9{a}^{2}-6a-12$ contains three terms. Some of the factors in each term are
$\begin{array}{ll}\text{first}\text{\hspace{0.17em}}\text{term:}\hfill & 9\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}{a}^{2},\text{\hspace{0.17em}}\text{or},\text{\hspace{0.17em}}9\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}a\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}a\hfill \\ \text{second}\text{\hspace{0.17em}}\text{term:}\hfill & -6\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}a\hfill \\ \text{third}\text{\hspace{0.17em}}\text{term:}\hfill & -12\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}1,\text{\hspace{0.17em}}\text{or},\text{\hspace{0.17em}}12\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}-1\hfill \end{array}$
$14{x}^{5}y+{\left(a+3\right)}^{2}$ contains two terms. Some of the factors of these terms are
$\begin{array}{ll}\text{first}\text{\hspace{0.17em}}\text{term:}\hfill & 14,\text{\hspace{0.17em}}{x}^{5},\text{\hspace{0.17em}}y\hfill \\ \text{second}\text{\hspace{0.17em}}\text{term:}\hfill & \left(a+3\right)\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\left(a+3\right)\hfill \end{array}$
## Practice set b
In the expression $8{x}^{2}-5x+6$ , list the factors of the
first term:
second term:
third term:
8, $x$ , $x$ ; $-5$ , $x$ ; 6 and 1 or 3 and 2
In the expression $10+2\left(b+6\right){\left(b-18\right)}^{2}$ , list the factors of the
first term:
second term:
10 and 1 or 5 and 2; 2, $b+6$ , $b-18$ , $b-18$
## Common factors
Sometimes, when we observe an expression carefully, we will notice that some particular factor appears in every term. When we observe this, we say we are observing common factors . We use the phrase common factors since the particular factor we observe is common to all the terms in the expression. The factor appears in each and every term in the expression.
## Sample set c
Name the common factors in each expression.
$5{x}^{3}-7{x}^{3}+14{x}^{3}$ .
The factor ${x}^{3}$ appears in each and every term. The expression ${x}^{3}$ is a common factor.
$4{x}^{2}+7x$ .
The factor $x$ appears in each term. The term $4{x}^{2}$ is actually $4xx$ . Thus, $x$ is a common factor.
$12x{y}^{2}-9xy+15$ .
The only factor common to all three terms is the number 3. (Notice that $12=3\cdot 4,\text{\hspace{0.17em}}9=3\cdot 3,\text{\hspace{0.17em}}15=3\cdot 5$ .)
$3\left(x+5\right)-8\left(x+5\right)$ .
The factor $\left(x+5\right)$ appears in each term. So, $\left(x+5\right)$ is a common factor.
$45{x}^{3}{\left(x-7\right)}^{2}+15{x}^{2}\left(x-7\right)-20{x}^{2}{\left(x-7\right)}^{5}$ .
The number 5, the ${x}^{2}$ , and the $\left(x-7\right)$ appear in each term. Also, $5{x}^{2}\left(x-7\right)$ is a factor (since each of the individual quantities is joined by a multiplication sign). Thus, $5{x}^{2}\left(x-7\right)$ is a common factor.
$10{x}^{2}+9x-4$ .
There is no factor that appears in each and every term. Hence, there are no common factors in this expression.
## Practice set c
List, if any appear, the common factors in the following expressions.
${x}^{2}+5{x}^{2}-9{x}^{2}$
${x}^{2}$
$4{x}^{2}-8{x}^{3}+16{x}^{4}-24{x}^{5}$
$4{x}^{2}$
$4{\left(a+1\right)}^{3}+10\left(a+1\right)$
$2\left(a+1\right)$
$9ab\left(a-8\right)-15a{\left(a-8\right)}^{2}$
$3a\left(a-8\right)$
$14{a}^{2}{b}^{2}c\left(c-7\right)\left(2c+5\right)+28c\left(2c+5\right)$
$14c\left(2c+5\right)$
$6\left({x}^{2}-{y}^{2}\right)+19x\left({x}^{2}+{y}^{2}\right)$
no common factor
## Coefficient
In algebra, as we now know, a letter is often used to represent some quantity. Suppose we represent some quantity by the letter $x$ . The notation $5x$ means $x+x+x+x+x$ . We can now see that we have five of these quantities. In the expression $5x$ , the number 5 is called the numerical coefficient of the quantity $x$ . Often, the numerical coefficient is just called the coefficient. The coefficient of a quantity records how many of that quantity there are.
## Sample set d
$12x$ means there are $12x\text{'}\text{s}$ .
$4ab$ means there are four $ab\text{'}\text{s}$ .
$10\left(x-3\right)$ means there are ten $\left(x-3\right)\text{'}\text{s}$ .
$1y$ means there is one $y$ . We usually write just $y$ rather than $1y$ since it is clear just by looking that there is only one $y$ .
$7{a}^{3}$ means there are seven ${a}^{3\text{'}}\text{s}$ .
$5ax$ means there are five $ax\text{'}\text{s}$ . It could also mean there are $5ax\text{'}\text{s}$ . This example shows us that it is important for us to be very clear as to which quantity we are working with. When we see the expression $5ax$ we must ask ourselves "Are we working with the quantity $ax$ or the quantity $x$ ?".
$6{x}^{2}{y}^{9}$ means there are six ${x}^{2}{y}^{9\text{'}}\text{s}$ . It could also mean there are $6{x}^{2}{y}^{9\text{'}}\text{s}$ . It could even mean there are $6{y}^{9}{x}^{2\text{'}}\text{s}$ .
explain and give four Example hyperbolic function
The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
Pawel
2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
Pawel
ok
Ifeanyi
on number 2 question How did you got 2x +2
Ifeanyi
combine like terms. x + x + 2 is same as 2x + 2
Pawel
Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
Pawel
how do I set up the problem?
what is a solution set?
Harshika
find the subring of gaussian integers?
Rofiqul
hello, I am happy to help!
Abdullahi
hi mam
Mark
find the value of 2x=32
divide by 2 on each side of the equal sign to solve for x
corri
X=16
Michael
Want to review on complex number 1.What are complex number 2.How to solve complex number problems.
Beyan
yes i wantt to review
Mark
use the y -intercept and slope to sketch the graph of the equation y=6x
how do we prove the quadratic formular
Darius
hello, if you have a question about Algebra 2. I may be able to help. I am an Algebra 2 Teacher
thank you help me with how to prove the quadratic equation
Seidu
may God blessed u for that. Please I want u to help me in sets.
Opoku
what is math number
4
Trista
x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
can you teacch how to solve that🙏
Mark
Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
Brenna
(61/11,41/11,−4/11)
Brenna
x=61/11 y=41/11 z=−4/11 x=61/11 y=41/11 z=-4/11
Brenna
Need help solving this problem (2/7)^-2
x+2y-z=7
Sidiki
what is the coefficient of -4×
-1
Shedrak
the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Jeannette has $5 and$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
Please keep in mind that it's not allowed to promote any social groups (whatsapp, facebook, etc...), exchange phone numbers, email addresses or ask for personal information on QuizOver's platform. |
# Lesson 8
### Lesson Narrative
By now students have encountered a variety of situations that can be modeled with quadratic functions. They are also familiar with some features of the expressions, tables, and graphs that represent such functions. This lesson transitions students from reasoning concretely and contextually about quadratic functions to reasoning about their representations in ways that are more abstract and formal (MP2).
In earlier grades, students reasoned about multiplication by thinking of the product as the area of a rectangle where the two factors being multiplied are the side lengths of the rectangle. In this lesson, students use this familiar reasoning to expand expressions such as $$(x+4)(x+7)$$, where $$x+4$$ and $$x+7$$ are side lengths of a rectangle with each side length is decomposed into $$x$$ and a number. They use the structure in the diagrams to help them write equivalent expressions in expanded form, for example, $$x^2 +11x + 28$$ (MP7). Students recognize that finding the sum of the partial areas in the rectangle is the same as applying the distributive property to multiply out the terms in each factor.
After this lesson, students move to think more abstractly about such diagrams. Rather than reasoning in terms of area, they use the diagrams to organize and account for all terms when applying the distributive property.
The terms “standard form” and “factored form” are not yet used and will be introduced in an upcoming lesson, after students have had some experience working with the expressions.
### Learning Goals
Teacher Facing
• Use area diagrams to reason about the product of two sums and to write equivalent expressions.
• Use the distributive property to write equivalent quadratic expressions.
### Student Facing
• Let’s use diagrams to help us rewrite quadratic expressions.
### Student Facing
• I can rewrite quadratic expressions in different forms by using an area diagram or the distributive property.
Building On |
# Factors of 71: Prime Factorization, Methods, and Examples
The factors of 71 are the integers that divide 71 evenly without any residue. Since it can be noted that 71 is an odd number as well as a prime number. Therefore factorization of 71 is very easy as it can only be factored as the product of 1 and 71 itself.
The factors of the given number can be positive and negative provided that the product of any of those two is always the factored number.
### Factors of 71
Here are the factors of number 71.
Factors of 71: 1, and 71
### Negative Factors of 71
The negative factors of 71 are similar to its positive factors, just with a negative sign.
Negative Factors of 71: -1 and -71
### Prime Factorization of 71
The prime factorization of 71 is the way of expressing its prime factors in the form of its product.
Prime Factorization: 1 x 71
In this article, we will learn about the factors of 71 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 71?
The factors of number 71 are 1, and 71. Both of these numbers are the factors as they do not leave any remainder when divided by 71.
The factors of 71 are classified as prime numbers as 71 itself is a prime number. The prime factors of the number 71 can be determined using the technique of prime factorization.
## How To Find the Factors of 71?
You can find the factors of 71 by using the rules of divisibility. The rule of divisibility states that any number when divided by any other natural number then it is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 71, create a list containing the numbers that are exactly divisible by 71 with zero remainders. One important thing to note is that 1 and numbers themselves are the factors of 71 as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 71 are determined as follows:
$\dfrac{71}{1} = 71$
$\dfrac{71}{71} = 1$
Therefore, 1, and 71 are the factors of 71.
### Total Number of Factors of 71
For 71 there are 2 positive factors as found above and 2 negative factors. So in total, there are 4 factors of 71
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure the total number of factors of 71 is given as:
Prime Factorization of 71 is 1 x 71.
The exponent of both 1 and 71 is 1.
Adding 1 to each and multiplying them together results in 4.
Therefore, the total number of factors of 71 is 4.
### Important Notes
Here are some important points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor which is the smallest prime factor.
## Factors of 71 by Prime Factorization
The number 71 is a prime number. Prime factorization is a useful technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 71 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 71, start dividing by its smallest prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 71 can be expressed as:
Prime Factorization of 71 = 1 x 71
## Factors of 71 in Pairs
The factor pairs are the duplet of numbers that when multiplied together result in the factorized number. Depending upon the total number of factors of the given numbers, factor pairs can be more than one.
71 is a prime number that has only two factors therefore there can be only a 1-factor pair of 71.
For 71, the factor pair can be found as:
1 x 71 = 71
The possible factor pair of 71 is (1, 71).
Both of these numbers in pairs, when multiplied, give 71 as the product.
The negative factor pair of 71 is given as:
-1 x -71 = 71
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign due to which the resulting product is the original positive number. Therefore, -1, and -71 are called negative factors of 71.
The list of all the factors of number 71 including positive as well as negative numbers is given below.
Factor List of 71: 1, -1, 71, and -71
## Factors of 71 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 71 are there?
### Solution
The total number of Factors of 71 is 4. Positive factors are 1 and 71.
Negative factors are -1 and -71.
### Example 2
Find the factors of 71 using prime factorization.
### Solution
The prime factorization of 71 is given as:
71 $\div$ 1 = 71
So the prime factorization of 71 can be written as:
1 x 71 = 71
### Example 3
What is the sum of factors of 71?
### Solution
The sum of the factors of 71 is 1 + 71 = 72.
Therefore the sum of the factors of 71 is 72 which is an even number. |
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## Course: 8th grade > Unit 3
• Linear graphs word problems
• Modeling with tables, equations, and graphs
• Linear graphs word problem: cats
• Linear equations word problems: volcano
• Linear equations word problems: earnings
• Modeling with linear equations: snow
• Linear equations word problems: graphs
## Linear equations word problems
• Linear function example: spending money
• Linear models word problems
• Fitting a line to data
• an integer, like 6
• a simplified proper fraction, like 3 / 5
• a simplified improper fraction, like 7 / 4
• a mixed number, like 1 3 / 4
• an exact decimal, like 0.75
• a multiple of pi, like 12 pi or 2 / 3 pi
## Word Problems on Linear Equations
Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.
There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.
Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.
## Step-by-step application of linear equations to solve practical word problems:
1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.
Solution: Then the other number = x + 9 Let the number be x. Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9) ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides) ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.
2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers? Solution: Let the common ratio be x. Let the common ratio be x. Their difference = 48 According to the question, 7x - 3x = 48 ⇒ 4x = 48 ⇒ x = 48/4 ⇒ x = 12 Therefore, 7x = 7 × 12 = 84 3x = 3 × 12 = 36 Therefore, the two numbers are 84 and 36.
3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle. Solution: Let the breadth of the rectangle be x, Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72 ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.
4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages.
Solution: Let Ron’s present age be x. Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4. According to the question; Ron will be twice as old as Aaron. Therefore, x + 4 = 2(x - 5 + 4) ⇒ x + 4 = 2(x - 1) ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.
5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts. Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x ⇒ x = 30/2 ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40 The two parts are 15 and 25.
More solved examples with detailed explanation on the word problems on linear equations.
6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages. Solution: Let Robert’s age be x years. Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question, 4x + 5 = 3(x + 5) ⇒ 4x + 5 = 3x + 15 ⇒ 4x - 3x = 15 - 5 ⇒ x = 10 ⇒ 4x = 4 × 10 = 40 Robert’s present age is 10 years and that of his father’s age = 40 years.
7. The sum of two consecutive multiples of 5 is 55. Find these multiples. Solution: Let the first multiple of 5 be x. Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2 ⇒ x = 25 Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.
8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles. Solution: Let the angle be x. Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51 Therefore, the two complementary angles are 39° and 51°
9. The cost of two tables and three chairs is \$705. If the table costs \$40 more than the chair, find the cost of the table and the chair. Solution: The table cost \$ 40 more than the chair. Let us assume the cost of the chair to be x. Then the cost of the table = \$ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x) Total cost of 2 tables and 3 chairs = \$705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is \$125 and that of each table is \$165.
10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number? Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2 According to the question, 3/5 ᵗʰ of the number is 4 more than 1/2 of the number. ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.
Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.
● Equations
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Worksheet on Word Problems on Linear Equation
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## Linear Equations
• An equation which has only one variable is called Linear Equation
## Solving equations which have linear equations on one side and numbers on the other side
EXAMPLE 1: Solve 3x + 15 = 1
• We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport 15 to the right side by changing its sign i.e., -15. => 3x + 15 = 1 => 3x = 1-15
• We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport -12 to the other side by changing its sign i.e., +12.
## Solving equations having variables having the variable on both sides
EXAMPLE 1: Solve .
• Simplify the given equation. => =>
• Bring the variable terms on the left side of the equation and the other numerical terms on the right side of the equation. => =>
• Now, to find the value of ‘x’ we need to divide both sides of the equation by 6 to maintain equality.
## Application of Linear Equations
Linear equations are used to find the value of an unknown quantity. Have a look at the following examples:
EXAMPLE 1: The sum of the digits of a 2 digit number 13. The numbers obtained by interchanging the digits is 14 more than the given number. Find the number.
SOLUTION: Let the digit at units place be x and the number at tens place be y.
=> y + x = 13 [sum is 13 given] => y = 13 – x [Transposing x to the other side by changing its sign] Thus, the formed number is= [Since x is at ones place and y=13-x is at tens place] After interchanging the digits, the number is= [Now x is at tens place & y=13-x is at one's place] The interchanged number is greater than the original number by 14. [Given]
New number Old number Difference
• Simplify => => => =>
• Transpose the variable term ‘x’ on the left side of the equation and other numerical terms on the right side of the equation by changing their sign. => 18x – 117 = 14 => 18x = 131
EXAMPLE 2: The distance between town A and town B is 123 km. Two buses begin their journey from these towns and move directly toward each other. From town A, the bus is moving at a speed of 45 km per hour and from town B, the bus is moving at 67 km per hour. Assuming the buses start at the same time, find how far is their meeting point from town A. SOLUTION: Let the buses meet after t hours.
We know that distance= speed X time Distance covered by bus 1 = 45 X t Distance covered by bus 2 = 67 X t Therefore, 45t + 67t = 123
The distance travelled by bus 1 from city A to the meeting point= speed of bus 1 X time taken by it to reach the meeting point. =45 X 1.098= 49.41 km Thus, the distance of reaching point from town A is 49.41km. [ANS]
## Equations reducible to linear form
• Now, simplify => 6x + 12 + 12x + 15 = 14x + 16 => 18x + 27 = 14x +16
• Transpose the variable term ‘x’ to the left side and the numerical terms on the right side of the equation by changing their sign. => 18x – 14x = 16 – 27 => 4x = 11
## Practice these questions
Q3) Three numbers are in the ratio 1:2:3. If the sum of the largest and the smallest equals the second and 45. Find the numbers.
Q4) Find the number whose 1/6 th part decreased 7 equals its 8/9 th part diminished by 1.
Q5) The difference between two numbers is 23. And the quotient obtained by dividing the larger number by the smaller one is 4. Find the numbers.
Q6) A man cycles to the office from his house at a speed of 5km per hour and reaches 6 minutes late. If he cycles at a speed of 7km/hr, he reaches 8 minutes early. What is the distance between the office and his house? Q7) Suraj is now half as old as his father. 20 years ago, Suraj’s father was six times Suraj’s age. What are their ages now? Q8) The perimeter of an isosceles triangle is 91cm. If the length of each equal side is 2cm more than the length of its base. Find the lengths of the sides of the triangle. Q9) The age of a boy in months is equal to the age of his grandfather in years. If the difference between their ages is 66 years, find their ages.
• The basic principle used in solving any linear equation is that any operation performed on one side of the equation must also be performed on the other side of the equation.
• Any term in an equation can be transposed from one side to other side by changing its sign.
• In cross multiplication, we multiply the numerator of LHS by the denominator of RHS and the denominator of LHS by the numerator of RHS and the resultant expression are equal to each other.
• Practical problems are based on the relations between some known and unknown quantities. We convert such problems into equations and then solve them.
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## CLASS-8 WORD PROBLEM OF LINEAR EQUATION
Word Problem Of Linear Equations -
Many word problems can be solved easily with the help of linear equations. To solve a word problem this way, take the following steps.
Step.1) Read the problem carefully to analyze the fact given
Step.2) Denote the unknown quantity by ‘x’ or by any other variable.
Step.3) Express all other quantities mentioned in the problem in terms of the variable.
Step.4) Frame an equation by using the conditions of the problems.
Step.5) Solve the equation to find the value of the variable or the unknown, If the conditions of the given problem are satisfied by the value of the unknown, the solution is correct.
Example.1) Fours less than five times a number is 10 more than thrice the number, find the number.
Ans.) Let the number is = ‘x’
Then four less than five times a number = 5x - 4
And 10 more than thrice the number is = 3x + 10
From the given condition, the required equation would be –
5x - 4 = 3x + 10
Or, 5x – 3x = 10 + 4
Or, 2x = 14
Or, x = 7
The desired number is 7 ( Ans. )
Example.2) The difference between the squares of two consecutive numbers is 121, find out the numbers.
Ans.) As per the given condition, let the number is ‘x’
So, the two consecutive number is x and x + 1
As per the given condition, we can find the desired equation is –
(x + 1)² - x² = 121
Or, x² + 2x + 1 - x² = 121
Or, 2x = 121 – 1 = 120
Or, x = 60
So, the required numbers are 60, 61 ( Ans. )
Example.3) Two numbers add up to 50. One-fourth of the larger number is 20 more than one-sixth of the smaller number, find the desired numbers.
Ans.) Let the larger number be x, then the smaller number be = 50 – x
x
One-fourth of the larger number is ---------
4
50 - x
20 more than one-sixth of the smaller number = ------------ + 20 6
From the given condition, the required equation would be –
x 50 - x
--------- = ----------- + 20
4 6
x 50 – x + 120
Or, --------- = ------------------
4 6
By, cross multiplication, we find –
Or, 6x = 4 (170 – x)
Or, 6x = 680 – 4x
Or, 10x = 680
Or, x = 68
So, the required number is 68 ( Ans. )
Example.4) If the same number be added to the numbers 5, 15, 20, and 25 then the resultant numbers are in proportion. Find the numbers
Ans.) Let the number be ‘x’
Then, as per the given condition proportion number would be 5 + x, 15 + x, 20 + x, and 25 + x
5 + x 20 + x
The required equation would be ------------ = -------------
15 + x 25 + x
By, cross multiplication we find –
(5 + x) (25 + x) = (20 + x) (15 + x)
Or, 125 + 25x + 5x + x² = 300 + 15x + 20x + x²
Or, 30x + 125 = 300 + 35x
Or, - 300 + 125 = 35x – 30x
Or, 5x = - 175
Or, x = - 35 ( Ans. )
Example.5) If a number is subtracted from the numerator of the fraction 5/6 and thrice that number is added to the denominator, the fraction becomes 1/5. Find the number.
Ans.) Let the number be ‘x’
5 – x 1
So, as per the given condition ------------ = ---------
6 + 3x 5
Via cross multiplication -
Or, 5 (5 – x) = 6 + 3x
Or, 25 – 5x = 6 + 3x
Or, - 5x – 3x = - 25 + 6
Or, - 8x = - 19
Or, x = 19 / 8 ( Ans. )
Example.6) The sum of the digits of a two-digit numbers is 10. If 5 is subtracted from the number formed interchanging the digits, the result is triple the original number. Find the original number.
Ans.) Let the digit in the units place be = ‘x’
As per the given condition, the sum of two-digit is = 10
Then the digit in the tens place = 10 – x
So, the number is = 10 (10 – x) + x
Now, as per the given condition if the number formed by the interchanging the digits = 10x + (10 - x)
From the question, as per the given condition 10x + (10 - x) – 5 = 3 {10 (10 – x) + x}
So, 10x + (10 - x) – 5 = 3 {10 (10 – x) + x}
Or, 10x + 10 – x – 5 = 30 (10 – x) + 3x
Or, 9x + 5 = 300 – 30x + 3x
Or, 9x + 27x = 300 – 5
Or, 36x = 295
Or, x = 295 / 36 = 8.19 = 8 ( Ans. )
So, the digit in the units place = 295 / 36 = 8.19 = 8 (Approximately in round figure)
Digit in the tens place = 10 – 8 = 2
Hence, the original number = 10 (10 – x) + x = 10 (10 – 8) + 8 = 10 X 2 + 8 = 28 ( Ans. )
Example.7) In a two-digit number, the digit in the tens place exceeds the digit in the units place by 5. If 3 more than six times the sum of the digits is subtracted from the number, the digits are reserved. Find the original number.
Ans.) let the digit in the unit place be = ‘x’
Then the digit in the tens place = x + 5
So, the number = 10(x + 5) + x
And, the sum of digits = x + (x + 5) = 2x + 5
After reversing the digit, the new number = 10x + (x + 5)
As per the given condition, we can get the required equation is –
{10(x + 5) + x} – { 3 + 6(2x + 5)} = 10x + (x + 5)
Or, 10x + 50 + x - 3 - 12x - 30 = 10x + x + 5
Or, 17 - x = 11x + 5
Or, 11x + x = 17 - 5
Or, 12x = 12
Or, x = 1
So, the original number is 2x + 5 = 2.1 + 5 = 7 ( Ans. )
Example.8) The present age of a man is twice that of his son. 5 years hence their ages will be in the ratio 3 : 2. Find the son’s present age
Ans.) Let the age of the son is ‘x’, so as per the given condition the age of the person would be 2x
After 5 years the age of the son and his father’s age will be (x + 5) and (2x + 5) respectively
So, as per the given condition –
2x + 5 3
--------------- = ----------
x + 5 2
by, cross multiplication –
2 (2x + 5) = 3 (x + 5)
Or, 4x + 10 = 3x + 15
Or, 4x – 3x = 15 – 10
Or, x = 5
The present age of the son is 5 years ( Ans. )
Example.9) A car covers the distance between two cities in 6 hours. A van covers the same distance in 5.5 hours by traveling 4 km/h faster than the car. What is the distance between the two cities? find the speeds of the car and the van
Ans.) Let the distance between two cities is ‘x’ km
As we know, the formula for the speed of any moving object is (here the speed of car)
Distance x
Speed = -------------- = ----------- km/h
Time 6
x
Speed of car = ---------- km/h
6
x 10x 2x
Speed of van is = -------- km/h = --------- km/h = --------- km/h
5.5 55 11
Now, as per the given condition, the desired equation would be –
2x x
----------- = ---------- + 4
11 6
2x (x + 24)
Or, ----------- = ---------------
11 6
Or, 12x = 11 (x + 24)
Or, 12x – 11x = 24 X 11
Or, x = 264
x 264
Now, the speed of the car is = --------- = ---------- = 44 km/h
6 6
2x 2 x 264
And, the speed of Van is = -------- = ----------
11 11
= 2 X 24 = 48 km/h ( Ans. )
Example.10) The length of a rectangular field is 15 m more than its width. If the length is decreased by 10 meters and the width is decreased by 20 m, the area decreased by 250 m². Find the length and the width of the field.
Ans.) Let the width of the field is = ‘x’ m
Then the length as per given condition is = (x + 15) m
As per formula, the area of field is = x (x + 15) m²
If the length of the field is decreased by 10 m then the length would be = (x + 15) – 10 = (x + 5) m
The width of the field when it has been decreased by 20 m then the desired width would be = (x – 20) m
The new area of same field would be = (x + 5) (x – 20) m²
As per the given condition, the new area will be decreased by 250 m²
So, the equation would be –
x (x + 15) – {(x + 5) (x – 20)} = 250
or, x² + 15x – (x² + 5x – 20x – 100) = 250
or, x² + 15x – x² + 15x + 100 = 250
or, 30x = 250 – 100 = 150
or, x = 150 / 30 = 5
so, the width of the field is 5m
and the length of the field is = (x + 15) m = (5 + 15) m = 20 m ( Ans. )
## Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths - Word problems that involve linear equations | 8th Maths : Chapter 3 : Algebra
Chapter: 8th maths : chapter 3 : algebra.
Word problems that involve linear equations
The challenging part of solving word problems is translating the statements into equations.
Collect as many such problems and attempt to solve them.
Example 3.33
The sum of two numbers is 36 and one number exceeds another by 8. Find the numbers.
Let the smaller number be x and the greater number be x +8
Given: the sum of two numbers = 36
x + ( x +8) = 36
2 x +8 = 36
2 x = 36 – 8
x = 28/2 = 14
(i) The smaller number, x =14
(ii) The greater number, x +8=14+8 = 22
Example 3.34
A bus is carrying 56 passengers with some people having ₹ 8 tickets and the remaining having ₹ 10 tickets. If the total money received from these passengers is ₹ 500, find the number of passengers with each type of tickets.
Let the number of passengers having ₹ 8 tickets be y . Then, the number of passengers with ₹ 10 tickets is (56−y).
Total money received from the passengers = ₹ 500
That is, y × 8 + (56 - y ) × 10 = 500
8y +560 −10 y = 500
8y−10 y = 500 – 560
− 2 y = −60
Hence, the number of passengers having,
(i) ₹ 8 tickets =30
(ii) ₹ 10 tickets =56−30 =26
Example 3.35
The length of a rectangular field exceeds its breadth by 9 metres. If the perimeter of the field is 154 m , find the length and breadth of the field.
Let the breadth of the field be ‘ x ’ metres; then its length (x+9) metres.
Perimeter of the P = 2(length + breadth) = 2( x + 9 + x )= 2(2 x + 9)
Given that, 2(2 x + 9) = 154.
4 x + 18 = 154
4 x =154−18
(i) Thus, breadth of the rectangular field = 34 m
(ii) length of the rectangular field = x +9 = 34+9 = 43 m
Example 3.36
There is a wooden piece of length 2 m . A carpenter wants to cut it into two pieces such that the first piece is 40 cm smaller than twice the other piece. What is the length of the smaller piece ?
Let us assume that the length of the first piece is x cm .
Then the length of the second piece is (200 cm – x cm ) i.e., (200 − x ) cm.
According to the given statement (change m to cm ),
First piece = 40 less than twice the second piece.
x = 2× (200 − x ) – 40
x = 400 − 2 x – 40
x + 2 x = 360
(i) Thus the length of the first piece is 120 cm
(ii) The length of second piece is 200 cm − 120 cm = 80 cm , which happens to be the smaller.
Suppose we take the second piece to be x and the first piece to be (200 − x), how will the steps vary? Will the answer be different?
Let 2 nd piece be ‘ x ’ & 1 st piece is 200 − x
Given that 1st piece is 40 cm smaller than hence the other piece
∴ 200 − x = 2 × x − 40
200 − x = 2 x – 40
∴ 200 + 40 = 2 x + x
∴ x = 240 / 3 = 80
∴ 1 st piece = 200 – x = 200 – 80 = 120 cm
2 nd piece = x = 80 cm
Example 3.37
Mother is five times as old as her daughter. After 2 years, the mother will be four times as old as her daughter. What are their present ages?
Given condition: After two years, Mother’s age = 4 times of Daughter's age
5 x + 2 = 4 ( x + 2)
5 x + 2 = 4 x + 8
5 x − 4 x = 8 - 2
Hence daughter’s present age = 6 years;
and mother’s present age = 5 x = 5 × 6 = 30 years
Example 3.38
The denominator of a fraction is 3 more than its numerator. If 2 is added to the numerator and 9 is added to the denominator, the fraction becomes 5/6. Find the original fraction.
Let the original fraction be x/y .
Given that y = x + 3. (Denominator = Numerator + 3).
Therefore, the fraction can be written as x /( x + 3).
As per the given condition,
[( x + 2) ] / [( x + 3) + 9] = 5/ 6
By cross multiplication, 6( x +2) = 5 ( x +3+9)
6 x +12 = 5( x +12)
6 x +12 = 5 x +60
6 x - 5 x =60−12
x / x + 3 = 48 / 48+3 = 48/51
Example 3.39
The sum of the digits of a two-digit number is 8. If 18 is added to the value of the number, its digits get reversed. Find the number.
Let the two digit number be xy (i.e., ten’s digit is x , ones digit is y )
Its value can be expressed as 10 x +y.
Given, x + y = 8 which gives y = 8 – x
Therefore its value is 10 x +y
= 10 x + 8 – x
The new number is yx with value is 10 y + x
= 10(8 − x ) + x
Given, when 18 is added to the given number ( xy ) gives new number ( yx )
(9 x + 8) + 18 = 80 – 9 x
This simplifies to 9 x + 9 x = 80 –8–18
x = 3 ⇒ y = 8 – 3 = 5
The two digit number is xy = 10 x+y ⇒ 10(3)+5 = 30+5 = 35
Example 3.40
From home, Rajan rides on his motorbike at 35 km/hr and reaches his office 5 minutes late. If he had ridden at 50 km/hr, he would have reached his office 4 minutes earlier. How far is his office from his home?
Let the distance be ‘ x ’ km. (Recall that, time = Distance / Speed )
Time taken to cover ‘ x ’ km at 35 km/hr: T 1 = x/ 35 hr
Time taken to cover ‘ x ’ km at 50 km/hr: T 2 = x /50 hr
Speed 1 = 35 km/hr
Speed 2 = 50 km/hr
According to the problem, the difference between two timings
= 4+5 =9 minutes
= 9/60 hour (changing minutes to hour)
Given, T 1 – T 2 = 9/60
The distance to his office x = 17(1/2) km.
Related Topics
## Extra questions: CBSE Class 8 | Mathematics - Linear equations in one variable.
Extra questions and detailed answers from the CBSE 8th-grade Mathematics chapter on Linear equations in One Variable.
May 6, 2023
Summary of cbse class 8 mathematics - linear equations in one variable.
CBSE Class 8 Mathematics - Linear Equations in One Variable includes the following topics:
• Introduction to linear equations
• Solving equations having the variable on both sides
• Some applications
• Solving equations reducible to the linear form
• Equations of the form ax + b = cx + d and their solutions
• Word problems leading to equations
• Direct and inverse variations
• Solving equations with the help of models
• Solving algebraic equations of more than one variable and their word problems
## Key Notes for CBSE Class 8 Mathematics - Linear Equations in One Variable
Key notes from the chapter on Linear Equations in One Variable in CBSE Class 8 Mathematics.
• Linear equations of one variable are equations that can be written in the form of ax+b=0, where a and b are some constants, and x is the variable.
• Linear equations can be solved by simplifying and rearranging them using inverse operations.
• You can solve equations of one variable on both sides by adding or subtracting a constant from both sides.
• Equations can also be reduced to a linear form by simplifying using cross-multiplication or factorization.
• It is essential to understand the real-life applications of linear equations while solving word problems.
• You should be able to identify the direct and inverse variations for better understanding of the concept.
• The use of models and diagrams helps to solve and explain equations in a more clear and concise way.
• Solving algebraic equations with more than one variable is also essential as they help in solving real-life problems.
## Practice Questions for CBSE Class 8 Mathematics - Linear Equations in One Variable
1. Solve the equation: x + 7 = 12
Answer: x + 7 - 7 = 12 - 7 (Subtract 7 from both sides); x = 5
2. Solve the equation: 2x = 10
Answer: 2x/2 = 10/2 (Divide by 2 on both sides); x = 5
3. If 7y - 5 = 4y + 3, find the value of y.
Answer: 7y - 5 - 4y = 3 (Subtract 4y and add 5 on both sides) 3y = 8 y = 8/3 (Divide by 3 on both sides)
4. Find the value of x in the equation: 3x - 2 = 1
Answer: 3x - 2 + 2 = 1 + 2 (Add 2 on both sides) 3x = 3 (Simplify) x = 1 (Divide by 3 on both sides)
5. A train takes 6 hours to travel a distance of 360 km. Find the speed of the train.
Answer: speed = distance ÷ time = 360 ÷ 6 = 60 kmph
6. A boat covers a certain distance downstream in 2 hours. It covers the same distance upstream in 3 hours. Find the speed of the boat in still water and the speed of the stream.
Answer: Let the speed of boat in still water be x and the speed of the stream be y. Then, distance = speed x time 2(x + y) = 3(x - y) (Downstream speed = Upstream speed) 2x + 2y = 3x - 3y y = (x/5) (Simplify) Substituting the value of y in the first equation, we get 10x = 4x + 20 x = 5 kmph (Speed of boat in still water) y = x/5 = 1 kmph (Speed of the stream)
7. Find the value of m in the equation: (18 - m)/3 = 5
Answer: (18 - m)/3 = 5 (Cross multiply) 18 - m = 15 * 3 m = 18 - 45 m = -27
8. The difference between two numbers is 10 and their sum is 40. Find the two numbers.
Answer: Let the two numbers be x and y. Then, x - y = 10 (Given: The difference between the two numbers is 10) x + y = 40 (Given: Their sum is 40) Adding both equations, 2x = 50 x = 25 Substituting the value of x in any of the equations, y = 15
9. Find the value of x in the equation: (3/5)x - 2 = 1
Answer: (3/5)x - 2 + 2 = 1 + 2 (Add 2 on both sides) (3/5)x = 3 (Simplify) x = 5
10. A boat travels 45 km upstream and 55 km downstream in 7 hours. If the speed of the stream is 5 kmph, find the speed of the boat in still water.
Answer: Let the speed of the boat in still water be x. Then, speed upstream = (x - 5) kmph and speed downstream = (x + 5) kmph Time taken for upstream journey + time taken for downstream journey = 7 hours (45/(x-5)) + (55/(x+5)) = 7 45(x+5) + 55(x-5) = 7(x^2 - 25) 45x + 225 + 55x - 275 = 7x^2 - 175 7x^2 - 100x - 225 = 0 (7x + 15)(x - 15) = 0 x = 15 kmph
11. The length of a rectangular plot is 20 meters more than its breadth. If the perimeter of the plot is 160 meters, find the length and breadth of the plot.
Answer: Let the breadth of the plot be x. Then, length of the plot = (x + 20) Perimeter = 2 (Length + Breadth) 160 = 2 (x + x + 20) x = 30 Length = x + 20 = 50 m and Breadth = x = 30 m
12. Divya bought some chocolates for Rs 240. If the price of each chocolate is Rs 4 less, then she could have bought 8 more chocolates for the same amount. Find the number of chocolates she bought originally.
Answer: Let the original price of each chocolate be x. Then, the number of chocolates bought = 240/x According to the question, 240/(x-4) = 240/x + 8 240x + 8x x-4 = 240 248x - 960 = x * 240 248x - 240x = 960 8x = 960 x = 120 Number of chocolates bought originally = 240/120 = 2
13. 2/3 of a number subtracted from 15 gives 9. Find the number.
Answer: Let the number be x. 15 - (2/3)x = 9 (2/3)x = 15 - 9 (2/3)x = 6 x = 9
14. A father is four times as old as his son. In 12 years, he will be twice as old as his son. Find their present ages.
Answer: Let the present age of the son be x and the present age of the father be y. Then, y = 4x (Given: The father is four times as old as his son) y + 12 = 2(x + 12) (Given: In 12 years, the father will be twice as old as his son) Substituting the value of y in the second equation, 4x + 12 = 2x + 24 2x = 12 x = 6 y = 4x = 4 * 6 =24 Present age of the son = 6 years and Present age of the father = 24 years
15. A train takes 10 seconds to pass a pole and takes 8 seconds to cross a bridge of length 150 m. Find the length of the train.
Answer: Let the length of the train be x. Then, speed of the train = x/10 m/s Total time taken to cross the bridge = 8 seconds Distance travelled by train in 8 seconds = length of train + length of bridge x + 150 = (x/10) * 8 80x + 1200 = 8x 72x = 1200 x = 16.67 m Length of the train = 16.67 m (approx.)
## Higher difficulty questions for Linear Equations in One Variable
1. A man invested Rs 30,000 in a scheme for two years at a simple interest rate of 12% per annum. How much more interest would he have earned if the interest rate was compounded annually?
Answer: Simple interest for two years = (30,000 * 12 * 2)/100 = Rs 7,200 Amount after two years = 30,000 + 7,200 = Rs 37,200 Compound interest for two years at 12% per annum = 37,200 - 30,000 = Rs 7,200 Therefore, he wouldn't have earned any more interest if the interest was compounded annually.
2. In how many ways can the letters of the word 'ZENITH' be arranged?
Answer: Total number of letters = 6 Number of ways to arrange 6 letters = 6! But, the letter 'Z' and 'E' appear twice. Therefore, the total number of ways = 6!/(2! * 2!) = 180
3. A shopkeeper marks his goods at a price 20% above the cost price. He sells the goods at a discount of 5%. Find his profit percentage.
Answer: Let the cost price be Rs 100 Marked price = 120 Selling price = 120 - (5/100 * 120) = Rs 114 Profit = 114 - 100 = 14 Profit percentage = (Profit/Cost Price) * 100 = (14/100) * 100 = 14%
4. Find the value of x in the equation: 2^(3x+2) = 64
Answer: 2^(3x+2) = 64 2^(3x+2) = 2^6 3x + 2 = 6 3x = 4 x = 4/3
5. If (x+y+z)^2 = x^2 + y^2 + z^2, prove that x + y + z = 0.
Answer: (x+y+z)^2 = x^2 + y^2 + z^2 x^2 + y^2 + z^2 + 2xy + 2xz + 2yz = x^2 + y^2 + z^2 2xy + 2xz + 2yz = 0 2(x+y+z)z = 0 x+y+z = 0
6. In a triangle ABC, AD is a median and BE is an altitude. If the length of AD is 62.5 cm and the length of BE is 24 cm, find the length of AB
Answer: Let the length of AB be x cm. In triangle ABD, using Pythagoras' theorem, (AD)^2 = (BD)^2 + (AB/2)^2 (62.5)^2 = (BD)^2 + (x/2)^2 (1) In triangle BEC, BE = (BC * EC)/ AB 24 = (BC * BD)/ AB (2) From (1) and (2), we get (62.5)^2 = (24 * BD)^2 + (x/2)^2 * (24)^2 (62.5)^2 = 576 * BD^2 + 144 * (x/2)^2 15625 = 36 * BD^2 + 9x^2 4375 = BD^2 + x^2/4 4375 = ((AB/2)^2 + (BC)^2) + x^2/4 (Using Pythagoras' theorem) 4375 = ((x/2)^2 + (BD)^2) + x^2/4 (From equation (1)) 4375 = (x^2/4) + (BD^2) + (x^2/4) 4375 = (x^2/2) + (BD^2)
Substituting the value of BD^2 from equation (2), 4375 = (x^2/2) + (24 * AB)^2/AB^2 4375 = (x^2/2) + 576 x^2/2 = 3799 x^2 = 7598 x = sqrt(7598) Therefore, the length of AB = sqrt(7598) cm (approx. 87.19 cm)
7. The altitude of an equilateral triangle is 7 cm. Find the perimeter of the triangle.
Answer: Let the length of one side of the equilateral triangle be x cm. Using Pythagoras' theorem, we get, x^2 = (7^2) + ((x/2)^2) x^2 = 49 + (x^2/4) 3x^2/4 = 49 x^2 = 196/3 x = sqrt(196/3) cm Therefore, the perimeter of the equilateral triangle = 3x = 3sqrt(196/3) cm (approx. 33.94 cm)
8. If (a+b+c)(1/a + 1/b + 1/c) = 10, find the value of (a/b + b/c + c/a).
Answer: Using the identity (a+b+c)^2 = (a^2 + b^2 + c^2) + 2(ab + bc + ac) (a/b + b/c + c/a) = (a^2/bc + b^2/ac + c^2/ab) (a^2/bc + b^2/ac + c^2/ab) + 2 = (a^2/bc + b^2/ac + c^2/ab) + (2a/b + 2b/c + 2c/a) from given equation, 10 = a/b + b/a + b/c + c/b + c/a + a/c 10 = a^2/bc + b^2/ac + c^2/ab + 2(a/b + b/c + c/a) Therefore, (a/b + b/c + c/a) = (10 - a^2/bc - b^2/ac - c^2/ab)/2 9. A shopkeeper sold an item for Rs. 450 and incurred a loss of 10%. What would have been the selling price if he had incurred a profit of 10% on it?
Answer: Let the cost price of the item be Rs x Selling price = Rs 450 Loss = 10% Cost price = selling price/(1-loss percentage) = 450/(1-10/100) = Rs 500 Profit = 10% Selling price = cost price + profit = 500 + 10% of 500 = Rs 550 Therefore, the selling price would have been Rs 550 if a profit of 10% would have been incurred.
10. Find the value of y if 8y^3 + 36y^2 + 54y + 27 = 0.
Answer: Given expression = 8y^3 + 36y^2 + 54y + 27 = (2y + 3)^3 Therefore, (2y + 3)^3 = 0 2y + 3 = 0 y = -3/2 Therefore, the value of y is -3/2.
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• RD Sharma Solutions
• Chapter 9 Linear Equation In One Variable
## RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equation in One Variable
Mathematics is one of the scoring subjects where students secure maximum marks in the exam. When it comes to preparing for the annual exam, it is the toughest time when most students struggle to solve problems. So, here at BYJU’S, our expert faculty have formulated RD Sharma Class 8 Maths Solutions , which help students prepare for their exams effortlessly. All the solutions are well designed, keeping in mind the latest CBSE syllabus and exam pattern. Also, students can learn easy tricks and shortcut methods by practising these solutions on a regular basis. The PDFs of this chapter are available here, and students can download them for free from the links provided below.
Chapter 9 – Linear Equation in One Variable contains four exercises, and the RD Sharma Solutions available on this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts covered in this chapter.
• Linear equation and its definitions.
• A solution of a linear equation.
• Solving equations having variable terms on one side and number(s) on the other side.
• Transposition method for solving linear equations in one variable.
• Cross-multiplication method for solving equations.
• Applications of linear equations to practical problems.
• RD Sharma Solutions for Class 8 Maths Chapter 1 Rational Numbers
• RD Sharma Solutions for Class 8 Maths Chapter 2 Powers
• RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots
• RD Sharma Solutions for Class 8 Maths Chapter 4 Cubes and Cube Roots
• RD Sharma Solutions for Class 8 Maths Chapter 5 Playing with Numbers
• RD Sharma Solutions for Class 8 Maths Chapter 6 Algebraic Expressions and Identities
• RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization
• RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions
• RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equations in One Variable
• RD Sharma Solutions for Class 8 Maths Chapter 10 Direct and Inverse Variations
• RD Sharma Solutions for Class 8 Maths Chapter 11 Time and Work
• RD Sharma Solutions for Class 8 Maths Chapter 12 Percentage
• RD Sharma Solutions for Class 8 Maths Chapter 13 Profit, Loss, Discount and Value Added Tax (VAT)
• RD Sharma Solutions for Class 8 Maths Chapter 14 Compound Interest
• RD Sharma Solutions for Class 8 Maths Chapter 15 Understanding Shapes – I (Polygons)
• RD Sharma Solutions for Class 8 Maths Chapter 16 Understanding Shapes – II (Quadrilaterals)
• RD Sharma Solutions for Class 8 Maths Chapter 17 Understanding Shapes – II (Special Types of Quadrilaterals)
• RD Sharma Solutions for Class 8 Maths Chapter 18 Practical Geometry (Constructions)
• RD Sharma Solutions for Class 8 Maths Chapter 19 Visualising Shapes
• RD Sharma Solutions for Class 8 Maths Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)
• RD Sharma Solutions for Class 8 Maths Chapter 21 Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)
• RD Sharma Solutions for Class 8 Maths Chapter 22 Mensuration – III (Surface Area and Volume of a Right Circular Cylinder)
• RD Sharma Solutions for Class 8 Maths Chapter 23 Data Handling – I (Classification and Tabulation of Data)
• RD Sharma Solutions for Class 8 Maths Chapter 24 Data Handling – II (Graphical Representation of Data as Histograms)
• RD Sharma Solutions for Class 8 Maths Chapter 25 Data Handling – III (Pictorial Representation of Data as Pie Charts)
• RD Sharma Solutions for Class 8 Maths Chapter 26 Data Handling – IV (Probability)
• RD Sharma Solutions for Class 8 Maths Chapter 27 Introduction to Graphs
• Exercise 9.1 Chapter 9 Linear Equations in One Variable
• Exercise 9.2 Chapter 9 Linear Equations in One Variable
• Exercise 9.3 Chapter 9 Linear Equations in One Variable
• Exercise 9.4 Chapter 9 Linear Equations in One Variable
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## Access Answers to Maths RD Sharma Solutions for Class 8 Chapter 9 Linear Equation in One Variable
Exercise 9.1 page no: 9.5.
Solve each of the following equations and also verify your solution:
1. 9 ¼ = y – 1 1/3
9 ¼ = y – 1 1/3
37/4 = y – 4/3
Upon solving, we get,
y = 37/4 + 4/3
By taking LCM for 4 and 3, we get 12
y = (37×3)/12 + (4×4)/12
= 111/12 + 16/12
= (111 + 16)/12
∴ y = 127/12
Verification
RHS = y – 1 1/3
= 127/12 – 4/3
= (127 – 16)/12
= 9 ¼
2. 5x/3 + 2/5 = 1
5x/3 + 2/5 = 1
5x/3 = 1 – 2/5 (by taking LCM)
By using cross-multiplication, we get,
5x = (3×3)/5
x = 9/(5×5)
∴ x = 9/25
LHS = 5x/3 + 2/5
= 5/3 × 9/25 + 2/5
= 3/5 + 2/5
= (3 + 2)/5
3. x/2 + x/3 + x/4 = 13
x/2 + x/3 + x/4 = 13
let us take LCM for 2, 3 and 4, which is 12
(x×6)/12 + (x×4)/12 + (x×3)/12 = 13
6x/12 + 4x/12 + 3x/12 = 13
(6x+4x+3x)/12 = 13
13x/12 = 13
13x = 12×13
∴ x = 12
LHS = x/2 + x/3 + x/4
= 12/2 + 12/3 + 12/4
= 6 + 4 + 3
4. x/2 + x/8 = 1/8
x/2 + x/8 = 1/8
let us take LCM for 2 and 8, which is 8
(x×4)/8 + (x×1)/8 = 1/8
4x/8 + x/8 = 1/8
∴ x = 1/5
LHS = x/2 + x/8
= (1/5)/2 + (1/5)/8
= 1/10 + 1/40
= (4 + 1)/40
5. 2x/3 – 3x/8 = 7/12
2x/3 – 3x/8 = 7/12
By taking LCM for 3 and 8, we get 24
(2x×8)/24 – (3x×3)/24 = 7/12
16x/24 – 9x/24 = 7/12
(16x-9x)/24 = 7/12
7x/24 = 7/12
7x×12 = 7×24
x = (7×24)/(7×12)
∴ x = 2
LHS = 2x/3 – 3x/8
= 2(2)/3 – 3(2)/8
= 4/3 – 6/8
= 4/3 – 3/4
= (16 – 9)/ 12
6. (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0
(x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1) = 0
Upon expansion, we get,
x 2 + 5x + 6 + x 2 – 5x +6 – 2x 2 – 2x =0
-2x + 12 = 0
By dividing the equation using -2, we get,
x – 6 = 0
∴ x = 6
LHS = (x + 2) (x + 3) + (x – 3) (x – 2) – 2x(x + 1)
= (6 + 2) (6 + 3) + (6 – 3) (6 – 2) – 2(6) (6 + 1)
= (8) (9) + (3) (4) – 12(7)
= 72 + 12 – 84
= 84 – 84
7. x/2 – 4/5 + x/5 + 3x/10 = 1/5
x/2 – 4/5 + x/5 + 3x/10 = 1/5
x/2 + x/5 + 3x/10 = 1/5 + 4/5
by taking LCM for 2, 5 and 10, which is 10
(x×5)/10 + (x×2)/10 + (3x×1)/10 = 5/5
5x/10 + 2x/10 + 3x/10 = 1
(5x+2x+3x)/10 = 1
∴ x = 1
LHS = x/2 – 4/5 + x/5 + 3x/10
= ½ – 4/5 + 1/5 + 3(1)/10
= (5 – 8 + 2 + 3)/10
= (10 – 8)/10
8. 7/x + 35 = 1/10
7/x + 35 = 1/10
7/x = 1/10 – 35
= ((1×1) – (35×10))/10
= (1 – 350)/10
7/x = -349/10
x = -70/349
∴ x = -70/349
LHS = 7/x + 35
= 7/(-70/349) + 35
= (-7 × 349)/70 + 35
= -349/10 + 35
= (-349 + 350)/ 10
9. (2x-1)/3 – (6x-2)/5 = 1/3
(2x-1)/3 – (6x-2)/5 = 1/3
By taking LCM for 3 and 5, which is 15
((2x-1)×5)/15 – ((6x-2)×3)/15 = 1/3
(10x – 5)/15 – (18x – 6)/15 = 1/3
(10x – 5 – 18x + 6)/15 = 1/3
(-8x + 1)/15 = 1/3
(-8x + 1)3 = 15
-24x + 3 = 15
-24x = 15 – 3
∴ x = -1/2
LHS = (2x – 1)/3 – (6x – 2)/5
= [2(-1/2) – 1]/3 – [6(-1/2) – 2]/5
= (- 1 – 1)/3 – (-3 – 2)/5
= – 2/3 – (-5/5)
= (-2 + 3)/3
10. 13(y – 4) – 3(y – 9) – 5(y + 4) = 0
13(y – 4) – 3(y – 9) – 5(y + 4) = 0
13y – 52 – 3y + 27 – 5y – 20 = 0
13y – 3y – 5y = 52 – 27 + 20
∴ y = 9
LHS = 13(y – 4) – 3 (y – 9) – 5 (y + 4)
= 13 (9 – 4) – 3 (9 – 9) – 5 (9 + 4)
= 13 (5) – 3 (0) – 5 (13)
= 65 – 0 – 65
11. 2/3(x – 5) – 1/4(x – 2) = 9/2
2/3(x – 5) – 1/4(x – 2) = 9/2
2x/3 – 10/3 – x/4 + 2/4 = 9/2
2x/3 – 10/3 – x/4 + 1/2 = 9/2
2x/3 – x/4 = 9/2 + 10/3 – 1/2
By taking LCM for (3 and 4 is 12) (2 and 3 is 6)
(2x×4)/12 – (x×3)/12 = (9×3)/6 + (10×2)/6 – (1×3)/6
8x/12 – 3x/12 = 27/6 + 20/6 – 3/6
(8x-3x)/12 = (27+20-3)6
5x/12 = 44/6
5x×6 = 44×12
LHS = 2/3 (x – 5) – ¼ (x – 2)
= 2/3 [(88/5) – 5] – ¼ [(88/5) – 2]
= 2/3 [(88 – 25)/5] – ¼ [(88 – 10)/5]
= 2/3 × 63/5 – ¼ × 78/5
= 42/5 – 39/10
= (84 – 39)/10
## EXERCISE 9.2 PAGE NO: 9.11
Solve each of the following equations and also check your results in each case:
1. (2x+5)/3 = 3x – 10
(2x+5)/3 = 3x – 10
Let us simplify,
(2x+5)/3 – 3x = – 10
By taking LCM
(2x + 5 – 9x)/3 = -10
(-7x + 5)/3 = -10
-7x + 5 = -30
-7x = -30 – 5
Let us verify the given equation now,
By substituting the value of ‘x’, we get,
(2×5 + 5)/3 = 3(5) – 10
(10+5)/3 = 15-10
Hence, the given equation is verified
2. (a-8)/3 = (a-3)/2
(a-8)/3 = (a-3)/2
(a-8)2 = (a-3)3
2a – 16 = 3a – 9
2a – 3a = -9 + 16
By substituting the value of ‘a’ we get,
(-7 – 8)/3 = (-7 – 3)/2
-15/3 = -10/2
3. (7y + 2)/5 = (6y – 5)/11
(7y + 2)/5 = (6y – 5)/11
(7y + 2)11 = (6y – 5)5
77y + 22 = 30y – 25
77y – 30y = -25 – 22
By substituting the value of ‘y’, we get,
(7(-1) + 2)/5 = (6(-1) – 5)/11
(-7 + 2)/5 = (-6 – 5)/11
-5/5 = -11/11
4. x – 2x + 2 – 16/3x + 5 = 3 – 7/2x
x – 2x + 2 – 16/3x + 5 = 3 – 7/2x
Let us rearrange the equation
x – 2x – 16x/3 + 7x/2 = 3 – 2 – 5
By taking LCM for 2 and 3, which is 6
(6x – 12x – 32x + 21x)/6 = -4
-17x/6 = -4
By cross-multiplying
-17x = -4×6
x = -24/-17
24/17 – 2(24/17) + 2 – (16/3)(24/17) + 5 = 3 – (7/2)(24/17)
24/17 – 48/17 + 2 – 384/51 + 5 = 3 – 168/34
By taking 51 and 17 as the LCM we get,
(72 – 144 + 102 – 384 + 255)/51 = (102 – 168)/34
-99/51 = -66/34
-33/17 = -33/17
5. 1/2x + 7x – 6 = 7x + 1/4
1/2x + 7x – 6 = 7x + 1/4
1/2x + 7x – 7x = 1/4 + 6 (by taking LCM)
1/2x = (1+ 24)/4
1/2x = 25/4
4x = 25 × 2
(1/2) (25/2) + 7(25/2) – 6 = 7(25/2) + 1/4
25/4 + 175/2 – 6 = 175/2 + 1/4
By taking LCM for 4 and 2 is 4
(25 + 350 – 24)/4 = (350+1)/4
351/4 = 351/4
6. 3/4x + 4x = 7/8 + 6x – 6
3/4x + 4x = 7/8 + 6x – 6
3/4x + 4x – 6x = 7/8 – 6
By taking 4 and 8 as LCM
(3x + 16x – 24x)/4 = (7 – 48)/8
-5x/4 = -41/8
-5x(8) = -41(4)
-40x = -164
x = -164/-40
(3/4)(41/10) + 4(41/10) = 7/8 + 6(41/10) – 6
123/40 + 164/10 = 7/8 + 246/10 – 6
(123 + 656)/40 = (70 + 1968 – 480)/80
779/40 = 1558/80
779/40 = 779/40
7. 7x/2 – 5x/2 = 20x/3 + 10
7x/2 – 5x/2 = 20x/3 + 10
7x/2 – 5x/2 – 20x/3 = 10
By taking LCM for 2 and 3 is 6
(21x – 15x – 40x)/6 = 10
-34x/6 = 10
(7-/2)(-30/17) – (5/2)(-30/17) = (20/3)(-30/17) + 10
-210/34 +150/34 = -600/51 + 10
-30/17 = (-600+510)/51
-30/17 = -30/17
8. (6x+1)/2 + 1 = (7x-3)/3
(6x+1)/2 + 1 = (7x-3)/3
(6x + 1 + 2)/2 = (7x – 3)/3
(6x + 3)3 = (7x – 3)2
18x + 9 = 14x – 6
18x – 14x = -6 – 9
(6(-15/4) + 1)/2 + 1 = (7(-15/4) – 3)/3
(3(-15/2) + 1)/2 + 1 = (-105/4 -3)/3
(-45/2 + 1)/2 + 1 = (-117/4)/3
(-43/4) + 1 = -117/12
(-43+4)/4 = -39/4
-39/4 = -39/4
9. (3a-2)/3 + (2a+3)/2 = a + 7/6
(3a-2)/3 + (2a+3)/2 = a + 7/6
(3a-2)/3 + (2a+3)/2 – a = 7/6
((3a-2)2 + (2a+3)3 – 6a)/6 = 7/6
(6a – 4 + 6a + 9 – 6a)/6 = 7/6
(6a + 5)/6 = 7/6
By substituting the value of ‘a’, we get,
(3(1/3)-2)/3 + (2(1/3) + 3)/2 = 1/3 + 7/6
(1-2)/3 + (2/3 + 3)/2 = (2+7)/6
-1/3 + (11/3)/2 = 9/6
-1/3 + 11/6 = 3/2
(-2+11)/6 = 3/2
10. x – (x-1)/2 = 1 – (x-2)/3
x – (x-1)/2 = 1 – (x-2)/3
x – (x-1)/2 + (x-2)/3 = 1
(6x – (x-1)3 + (x-2)2)/6 = 1
(6x – 3x + 3 + 2x – 4)/6 = 1
(5x – 1)/6 = 1
5x – 1 = 6
7/5 – (7/5 – 1)/2 = 1 – (7/5 – 2)/3
7/5 – (2/5)/2 = 1 – (-3/5)/3
7/5 – 2/10 = 1 + 3/15
(14 – 2)/10 = (15+3)/15
12/10 = 18/15
11. 3x/4 – (x-1)/2 = (x-2)/3
3x/4 – (x-1)/2 = (x-2)/3
3x/4 – (x-1)/2 – (x-2)/3 = 0
By taking LCM for 4, 2 and 3, which is 12
(9x – (x-1)6 – (x-2)4)/12 = 0
(9x – 6x + 6 – 4x + 8)/12 = 0
(-x + 14)/12 = 0
-x + 14 = 0
3(14)/4 – (14-1)/2 = (14-2)/3
42/4 – 13/2 = 12/3
(42 – 26)/4 = 4
12. 5x/3 – (x-1)/4 = (x-3)/5
5x/3 – (x-1)/4 = (x-3)/5
5x/3 – (x-1)/4 – (x-3)/5 = 0
By taking LCM for 3, 4 and 5, which is 60
((5x×20) – (x-1)15 – (x-3)12)/60 = 0
(100x – 15x + 15 -12x + 36)/60 = 0
(73x + 51)/60 = 0
73x + 51 = 0
(20x – (x-1)3)/12 = (-51/73 – 3)/5
(20x – 3x + 3)/12 = (-270/73)/5
(17x + 3)/12 = -270/365
(17(-51/73) + 3)/12 = -54/73
(-867/73 + 3)/12 = -54/73
((-867 + 219)/73)/12 = -54/73
(-648)/876 = -54/73
-54/73 = -54/73
13. (3x+1)/16 + (2x-3)/7 = (x+3)/8 + (3x-1)/14
(3x+1)/16 + (2x-3)/7 = (x+3)/8 + (3x-1)/14
(3x+1)/16 + (2x-3)/7 – (x+3)/8 – (3x-1)/14 = 0
By taking LCM for 16, 7, 8 and 14, which is 112
((3x+1)7 + (2x-3)16 – (x+3)14 – (3x-1)8)/112 = 0
(21x + 7 + 32x – 48 – 14x – 42 – 24x + 8)/112 = 0
(21x + 32x – 14x – 24x + 7 – 48 – 42 + 8)/112 = 0
(15x – 75)/112 = 0
15x – 75 = 0
(3(5)+1)/16 + (2(5)-3)/7 = (5+3)/8 + (3(5)-1)/14
(15+1)/16 + (10-3)/7 = 8/8 + (15-1)/14
16/16 + 7/7 = 8/8 + 14/14
1 + 1 = 1 + 1
14. (1-2x)/7 – (2-3x)/8 = 3/2 + x/4
(1-2x)/7 – (2-3x)/8 = 3/2 + x/4
(1-2x)/7 – (2-3x)/8 – x/4 = 3/2
By taking LCM for 7, 8 and 4, which is 56
((1-2x)8 – (2-3x)7 – 14x)/56 = 3/2
(8 – 16x – 14 + 21x – 14x)/56 = 3/2
(-9x – 6)/56 = 3/2
2(-9x-6) = 3(56)
-18x – 12 = 168
-18x = 168+12
x = 180/-18
(1-2(-10))/7 – (2-3(-10))/8 = 3/2 + (-10)/4
(1+20)/7 – (2+30)/8 = 3/2 – 5/2
21/7 – 32/8 = 3/2 – 5/2
3 – 4 = -2/2
15. (9x+7)/2 – (x – (x-2)/7) = 36
(9x+7)/2 – (x – (x-2)/7) = 36
Let us simplify the given equation into a simple form
(9x+7)/2 – (7x-x+2)/7 = 36
(9x+7)/2 – (6x+2)/7 = 36
By taking LCM for 2 and 7 is 14
(7(9x+7) – 2(6x+2))/14 = 36
(63x+49 – 12x – 4)/14 = 36
(51x + 45)/14 = 36
51x + 45 = 36(14)
51x + 45 = 504
51x = 504-45
(9(9)+7)/2 – (6(9)+2)/7 = 36
(81+7)/2 – (54+2)/7 = 36
88/2 – 56/7 = 36
44 – 8 = 36
16. 0.18(5x – 4) = 0.5x + 0.8
0.18(5x – 4) = 0.5x + 0.8
0.18(5x – 4) – 0.5x = 0.8
0.90x – 0.72 – 0.5x = 0.8
0.90x – 0.5x = 0.8 + 0.72
0.40x = 1.52
x = 1.52/0.40
0.18(5(3.8)-4) = 0.5(3.8) + 0.8
0.18(19-4) = 1.9 + 0.8
17. 2/3x – 3/2x = 1/12
2/3x – 3/2x = 1/12
By taking LCM for 3x and 2x, which is 6x
((2×2) – (3×3))/6x = 1/12
(4-9)/6x = 1/12
-5/6x = 1/12
2/3(-10) – 3/2(-10) = 1/12
-2/30 + 3/20 = 1/12
((-2×2) + (3×3))/60 = 1/12
(-4+9)/60 = 1/12
5/60 = 1/12
1/12 = 1/12
18. 4x/9 + 1/3 + 13x/108 = (8x+19)/18
4x/9 + 1/3 + 13x/108 = (8x+19)/18
4x/9 + 13x/108 – (8x+19)/18 = -1/3
By taking LCM for 9, 108 and 18, which is 108
((4x×12) + 13x×1 – (8x+19)6)/108 = -1/3
(48x + 13x – 48x – 114)/108 = -1/3
(13x – 114)/108 = -1/3
(13x – 114)3 = -108
39x – 342 = -108
39x = -108 + 342
4(6)/9 + 1/3 + 13(6)/108 = (8(6)+19)/18
24/9 + 1/3 + 78/108 = 67/18
8/3 + 1/3 + 13/18 = 67/18
((8×6) + (1×6) + (13×1))/18 = 67/18
(48 + 6 + 13)/18 = 67/18
67/18 = 67/18
19. (45-2x)/15 – (4x+10)/5 = (15-14x)/9
(45-2x)/15 – (4x+10)/5 = (15-14x)/9
By rearranging
(45-2x)/15 – (4x+10)/5 – (15-14x)/9 = 0
By taking LCM for 15, 5 and 9, which is 45
((45-2x)3 – (4x+10)9 – (15-14x)5)/45 = 0
(135 – 6x – 36x – 90 – 75 + 70x)/45 = 0
(28x – 30)/45 = 0
28x – 30 = 0
(45-2(15/14))/15 – (4(15/14) + 10)/5 = (15 – 14(15/14))/9
(45- 15/7)/15 – (30/7 + 10)/5 = (15-15)/9
300/105 – 100/35 = 0
(300-300)/105 = 0
20. 5(7x+5)/3 – 23/3 = 13 – (4x-2)/3
5(7x+5)/3 – 23/3 = 13 – (4x-2)/3
(35x + 25)/3 + (4x – 2)/3 = 13 + 23/3
(35x + 25 + 4x – 2)/3 = (39+23)/3
(39x + 23)/3 = 62/3
(39x + 23)3 = 62(3)
39x + 23 = 62
39x = 62 – 23
(35x + 25)/3 – 23/3 = 13 – (4x-2)/3
(35+25)/3 – 23/3 = 13 – (4-2)/3
60/3 – 23/3 = 13 – 2/3
(60-23)/3 = (39-2)/3
37/3 = 37/3
21. (7x-1)/4 – 1/3(2x – (1-x)/2) = 10/3
(7x-1)/4 – 1/3(2x – (1-x)/2) = 10/3
Upon expansion
(7x-1)/4 – (4x-1+x)/6 = 10/3
(7x-1)/4 – (5x-1)/6 = 10/3
By taking LCM for 4 and 6, we get 24
((7x-1)6 – (5x-1)4)/24 = 10/3
(42x – 6 – 20x + 4)/24 = 10/3
(22x – 2)/24 = 10/3
22x – 2 = 10(8)
22x – 2 = 80
(7(41/11)-1)/4 – (5(41/11)-1)/6 = 10/3
(287/11 – 1)/4 – (205/11 – 1)/6 = 10/3
(287-11)/44 – (205-11)/66 = 10/3
276/44 – 194/66 = 10/3
69/11 – 97/33 = 10/3
((69×3) – (97×1))/33 = 10/3
(207 – 97)/33 = 10/3
110/33 = 10/3
10/3 = 10/3
22. 0.5(x-0.4)/0.35 – 0.6(x-2.71)/0.42 = x + 6.1
0.5(x-0.4)/0.35 – 0.6(x-2.71)/0.42 = x + 6.1
Let us simplify
(0.5/0.35)(x – 0.4) – (0.6/0.42)(x – 2.71) = x + 6.1
(x – 0.4)/0.7 – (x – 2.71)/0.7 = x + 6.1
(x – 0.4 – x + 2.71)/0.7 = x + 6.1
-0.4 + 2.71 = 0.7(x + 6.1)
0.7x = 2.71 – 0.4 – 4.27
x = -1.96/0.7
0.5(-2.8 – 0.4)/0.35 – 0.6(-2.8 – 2.71)/0.42 = -2.8 + 6.1
-1.6/0.35 + 3.306/0.42 = 3.3
-4.571 + 7.871 = 3.3
23. 6.5x + (19.5x – 32.5)/2 = 6.5x + 13 + (13x – 26)/2
6.5x + (19.5x – 32.5)/2 = 6.5x + 13 + (13x – 26)/2
6.5x + (19.5x – 32.5)/2 – 6.5x – (13x – 26)/2 = 13
(19.5x – 32.5)/2 – (13x – 26)/2 = 13
(19.5x – 32.5 – 13x + 26)/2 = 13
(6.5x – 6.5)/2 = 13
6.5x – 6.5 = 13×2
6.5x – 6.5 = 26
6.5x = 26+6.5
6.5x = 32.5
x = 32.5/6.5
6.5(5) + (19.5(5) – 32.5)/2 = 6.5(5) + 13 + (13(5) – 26)/2
32.5 + (97.5 – 32.5)/2 = 32.5 + 13 + (65 – 26)/2
32.5 + 65/2 = 45.5 + 39/2
(65 + 65)/2 = (91+39)/2
130/2 = 130/2
24. (3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
(3x – 8) (3x + 2) – (4x – 11) (2x + 1) = (x – 3) (x + 7)
9x 2 + 6x – 24x – 16 – 8x 2 – 4x + 22x + 11 = x 2 + 7x – 3x – 21
9x 2 + 6x – 24x – 16 – 8x 2 – 4x + 22x + 11 – x 2 – 7x + 3x + 21 = 0
9x 2 – 8x 2 – x 2 + 6x – 24x – 4x + 22x – 7x + 3x – 16 + 21 + 11 = 0
-4x + 16 = 0
(3(4) – 8) (3(4) + 2) – (4(4) – 11) (2(4) + 1) = (4 – 3) (4 + 7)
(12-8) (12+2) – (16-11) (8+1) = 1(11)
4 (14) – 5(9) = 11
56 – 45 = 11
25. [(2x+3) + (x+5)] 2 + [(2x+3) – (x+5)] 2 = 10x 2 + 92
Let us simplify the given equation
By using the formula (a+b) 2
9x 2 + 48x + 64 + x 2 – 4x + 4 = 10x 2 + 92
9x 2 – 10x 2 + x 2 + 48x – 4x = 92 – 64 – 4
(106/11) 2 + (-16/11) 2 = (360 + 11132)/121
11236/121 + 256/121 = 11492/121
11492/121 = 11492/121
## EXERCISE 9.3 PAGE NO: 9.17
1. (2x-3) / (3x+2) = -2/3
(2x-3) / (3x+2) = -2/3
Let us perform cross-multiplication we get,
3(2x – 3) = -2(3x + 2)
6x – 9 = -6x – 4
When rearranged,
6x + 6x = 9 – 4
Now let us verify the given equation,
(2(5/12) – 3) / (3(5/12) + 2) = -2/3
((5/6)-3) / ((5/4) + 2) = -2/3
((5-18)/6) / ((5+8)/4) = -2/3
(-13/6) / (13/4) = -2/3
(-13/6) × (4/13) = -2/3
-4/6 = -2/3
-2/3 = -2/3
2. (2-y) / (y+7) = 3/5
(2-y) / (y+7) = 3/5
Let us perform cross-multiplication, we get,
5(2-y) = 3(y+7)
10 – 5y = 3y + 21
10 – 21 = 3y + 5y
8y = – 11
(2 – (-11/8)) / ((-11/8) + 7) = 3/5
((16+11)/8) / ((-11+56)/8) = 3/5
(27/8) / (45/8) = 3/5
(27/8) × (8/45) = 3/5
27/45 = 3/5
3. (5x – 7) / (3x) = 2
(5x – 7) / (3x) = 2
5x – 7 = 2(3x)
5x – 7 = 6x
5x – 6x = 7
(5(-7) – 7) / (3(-7)) = 2
(-35 – 7) / -21 = 2
-42/-21 = 2
4. (3x+5) / (2x + 7) = 4
(3x+5) / (2x + 7) = 4
3x + 5 = 4(2x+7)
3x + 5 = 8x + 28
3x – 8x = 28 – 5
(3(-23/5) + 5) / (2(-23/5) + 7) = 4
(-69/5 + 5) / (-46/5 + 7) = 4
(-69+25)/5 / (-46+35)/5 = 4
-44/5 / -11/5 = 4
-44/5 × 5/-11 = 4
5. (2y + 5) / (y + 4) = 1
(2y + 5) / (y + 4) = 1
2y + 5 = y + 4
2y – y = 4 – 5
(2(-1) + 5) / (-1 + 4) = 1
(-2+5) / 3 = 1
6. (2x + 1) / (3x – 2) = 5/9
(2x + 1) / (3x – 2) = 5/9
9(2x + 1) = 5(3x – 2)
18x + 9 = 15x – 10
18x – 15x = -10 – 9
(2(-19/3) + 1) / (3(-19/3) – 2) = 5/9
(-38/3 + 1) / (-57/3 – 2) = 5/9
(-38 + 3)/3 / (-57 – 6)/3 = 5/9
-35/3 / -63/3 = 5/9
-35/3 × 3/-63 = 5/9
-35/-63 = 5/9
7. (1 – 9y) / (19 – 3y) = 5/8
(1 – 9y) / (19 – 3y) = 5/8
8(1- 9y) = 5(19-3y)
8 – 72y = 95 – 15y
8 – 95 = 72y – 15y
(1 – 9(-29/19)) / (19 – 3(-29/19)) = 5/8
(19+261)/19 / (361+87)/19 = 5/8
280/19 × 19/448 = 5/8
280/ 448 = 5/8
8. 2x / (3x + 1) = 1
2x / (3x + 1) = 1
2x = 1(3x + 1)
2x = 3x + 1
2x – 3x = 1
2(-1) / (3(-1) + 1) = 1
-2 /(-3+1) = 1
9. y – (7 – 8y)/9y – (3 + 4y) = 2/3
y – (7 – 8y)/9y – (3 + 4y) = 2/3
(y – 7 + 8y) / (9y – 3 – 4y) = 2/3
(-7 + 9y) / (5y – 3) = 2/3
3(-7 + 9y) = 2(5y – 3)
-21 + 27y = 10y – 6
27y – 10y = 21 – 6
15/17 – (7-8(15/17))/ 9(15/17) – (3 + 4(15/17)) = 2/3
15/17 – (7 – 120/17) / 135/17 – (3 + 60/17) = 2/3
15/17 – ((119-120)/17) / 135/17 – ((51+60)/17) = 2/3
15/17 – (-1/17) / 135/17 – (111/17) = 2/23
((15 + 1)/17) / ((135-111)/17) = 2/3
16/17 / 24/17 = 2/3
16/24 = 2/3
10. 6/ 2x – (3 – 4x) = 2/3
6/ 2x – (3 – 4x) = 2/3
6/(2x – 3 + 4x) = 2/3
6/(6x – 3) = 2/3
3(6) = 2(6x – 3)
18 = 12x – 6
12x = 18 + 6
6/ (6(2) – 3) = 2/3
6/(12-3) = 2/3
11. 2/3x – 3/2x = 1/12
4-9/6x = 1/12
By cross-multiplying, we get,
12(-5) = 1 (6x)
2/-30 – 3/-20 = 1/12
-4+6/60 = 1/12
12. (3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)
(3x + 5)/ (4x + 2) = (3x + 4)/(4x + 7)
(3x + 5)/ (4x + 2) – (3x + 4)/(4x + 7) = 0
By taking LCM as (4x + 2) (4x + 7)
((3x + 5) (4x + 7) – (3x + 4) (4x + 2)) / (4x + 2) (4x + 7) = 0
(3x + 5) (4x + 7) – (3x + 4) (4x + 2) = 0
12x 2 + 21x + 20x + 35 – 12x 2 – 6x – 16x – 8 = 0
19x + 35 – 8 = 0
(3(-27/19) +5) / (4(-27/19) + 2) = (3(-27/19) + 4) / (4(-27/19) + 7)
(-81/19 + 5) / (-108/19 + 2) = (-81/19 + 4) / (-108/19 + 7)
((-81+95)/19) / ((-108+38)/19) = ((-81+76)/19) / ((-108+133)/19)
14/19 / -70/19 = -5/19 / 25/19
-14/70 = -5/25
-1/5 = -1/5
13. (7x – 2) / (5x – 1) = (7x +3)/(5x + 4)
(7x – 2) / (5x – 1) = (7x +3)/(5x + 4)
(7x – 2) / (5x – 1) – (7x +3)/(5x + 4) = 0
By taking LCM as (5x – 1) (5x + 4)
((7x-2) (5x+4) – (7x+3)(5x-1)) / (5x – 1) (5x + 4) = 0
(7x-2) (5x+4) – (7x+3)(5x-1) = 0
Upon simplification,
35x 2 + 28x – 10x – 8 – 35x 2 + 7x – 15x + 3 = 0
10x – 5 = 0
(7(1/2) – 2) / (5(1/2) – 1) = (7(1/2) + 3) /(5(1/2) + 4)
(7/2 – 2) / (5/2 – 1) = (7/2 + 3) / (5/2 + 4)
((7-4)/2) / ((5-2)/2) = ((7+6)/2) / ((5+8)/2)
(3/2) / (3/2) = (13/2) / (13/2)
14. ((x+1)/(x+2)) 2 = (x+2) / (x + 4)
((x+1)/(x+2)) 2 = (x+2) / (x + 4)
(x+1) 2 / (x+2) 2 – (x+2) / (x + 4) = 0
By taking LCM as (x+2) 2 (x+4)
((x+1) 2 (x+4) – (x+2) (x+2) 2 ) / (x+2) 2 (x+4) = 0
(x+1) 2 (x+4) – (x+2) (x+2) 2 = 0
Let us expand the equation
(x 2 + 2x + 1) (x + 4) – (x + 2) (x 2 + 4x + 4) = 0
x 3 + 2x 2 + x + 4x 2 + 8x + 4 – (x 3 + 4x 2 + 4x + 2x 2 + 8x + 8) = 0
x 3 + 2x 2 + x + 4x 2 + 8x + 4 – x 3 – 4x 2 – 4x – 2x 2 – 8x – 8 = 0
-3x – 4 = 0
(x+1) 2 / (x+2) 2 = (x+2) / (x + 4)
(-4/3 + 1) 2 / (-4/3 + 2) 2 = (-4/3 + 2) / (-4/3 + 4)
((-4+3)/3) 2 / ((-4+6)/3) 2 = ((-4+6)/3) / ((-4+12)/3)
(-1/3) 2 / (2/3) 2 = (2/3) / (8/3)
1/9 / 4/9 = 2/3 / 8/3
15. ((x+1)/(x-4)) 2 = (x+8)/(x-2)
((x+1)/(x-4)) 2 = (x+8)/(x-2)
(x+1) 2 / (x-4) 2 – (x+8) / (x-2) = 0
By taking LCM as (x-4) 2 (x-2)
((x+1) 2 (x-2) – (x+8) (x-4) 2 ) / (x-4) 2 (x-2) = 0
(x+1) 2 (x-2) – (x+8) (x-4) 2 = 0
(x 2 + 2x + 1) (x-2) – ((x+8) (x 2 – 8x + 16)) = 0
x 3 + 2x 2 + x – 2x 2 – 4x – 2 – (x 3 – 8x 2 + 16x + 8x 2 – 64x + 128) = 0
x 3 + 2x 2 + x – 2x 2 – 4x – 2 – x 3 + 8x 2 – 16x – 8x 2 + 64x – 128 = 0
45x – 130 = 0
(x+1) 2 / (x-4) 2 = (x+8) / (x-2)
(26/9 + 1) 2 / (26/9 – 4) 2 = (26/9 + 8) / (26/9 – 2)
((26+9)/9) 2 / ((26-36)/9) 2 = ((26+72)/9) / ((26-18)/9)
(35/9) 2 / (-10/9) 2 = (98/9) / (8/9)
(35/-10) 2 = (98/8)
(7/2) 2 = 49/4
49/4 = 49/4
16. (9x-7)/(3x+5) = (3x-4)/(x+6)
(9x-7)/(3x+5) = (3x-4)/(x+6)
(9x-7)/(3x+5) – (3x-4)/(x+6) = 0
By taking LCM as (3x+5) (x+6)
((9x-7) (x+6) – (3x-4) (3x+5)) / (3x+5) (x+6) = 0
(9x-7) (x+6) – (3x-4) (3x+5) = 0
9x 2 + 54x – 7x – 42 – (9x 2 + 15x – 12x – 20) = 0
44x – 22 = 0
(9(1/2) – 7) / (3(1/2) + 5) = (3(1/2) – 4) / ((1/2) + 6)
(9/2 – 7) / (3/2 + 5) = (3/2 – 4) / (1/2 + 6)
((9-14)/2) / ((3+10)/2) = ((3-8)/2) / ((1+12)/2)
-5/2 / 13/2 = -5/2 / 13/2
-5/13 = -5/13
17. (x+2)/(x+5) = x/(x+6)
(x+2)/(x+5) = x/(x+6)
(x+2)/(x+5) – x/(x+6) = 0
By taking LCM as (x+5) (x+6)
((x+2) (x+6) – x(x+5)) / (x+5) (x+6) = 0
(x+2) (x+6) – x(x+5) = 0
Upon expansion,
x 2 + 8x + 12 – x 2 – 5x = 0
3x + 12 = 0
(-4 + 2) / (-4 + 5) = -4 / (-4 + 6)
-2/1 = -4 / (2)
18. 2x – (7-5x) / 9x – (3+4x) = 7/6
2x – (7-5x) / 9x – (3+4x) = 7/6
(2x – 7 + 5x) / (9x – 3 – 4x) = 7/6
(7x – 7) / (5x – 3) = 7/6
6(7x – 7) = 7(5x – 3)
42x – 42 = 35x – 21
42x – 35x = -21 + 42
(7(3) -7) / (5(3) – 3) = 7/6
(21-7) / (15-3) = 7/6
14/12 = 7/6
19. (15(2-x) – 5(x+6)) / (1-3x) = 10
15(2-x) – 5(x+6) / (1-3x) = 10
(30-15x) – (5x + 30) / (1-3x) = 10
(30-15x) – (5x + 30) = 10(1- 3x)
30- 15x – 5x – 30 = 10 – 30x
30- 15x – 5x – 30 + 30x = 10
(15(2-x) – 5(x+6)) / (1-3x) = 10
(15(2-1) – 5(1+6)) / (1- 3) = 10
(15 – 5(7))/-2 = 10
(15-35)/-2 = 10
-20/-2 = 10
20. (x+3)/(x-3) + (x+2)/(x-2) = 2
(x+3)/(x-3) + (x+2)/(x-2) = 2
By taking LCM as (x-3) (x-2)
((x+3)(x-2) + (x+2) (x-3)) / (x-3) (x-2) = 2
(x+3)(x-2) + (x+2) (x-3) = 2 ((x-3) (x-2))
x 2 + 3x – 2x – 6 + x 2 – 3x + 2x – 6 = 2(x 2 – 3x – 2x + 6)
2x 2 – 12 = 2x 2 – 10x + 12
2x 2 – 2x 2 + 10x = 12 + 12
(12/5 + 3)/(12/5 – 3) + (12/5 + 2)/(12/5 – 2) = 2
((12+15)/5)/((12-15)/5) + ((12+10)/5)/((12-10)/5) = 2
(27/5)/(-3/5) + (22/5)/(2/5) = 2
-27/3 + 22/2 = 2
((-27×2) + (22×3))/6 = 2
(-54 + 66)/6 = 2
21. ((x+2) (2x-3) – 2x 2 + 6)/(x-5) = 2
((x+2) (2x-3) – 2x 2 + 6)/(x-5) = 2
(x+2) (2x-3) – 2x 2 + 6) = 2(x-5)
2x 2 – 3x + 4x – 6 – 2x 2 + 6 = 2x – 10
x = 2x – 10
x – 2x = -10
((10+2) (2(10) – 3) – 2(10) 2 + 6)/ (10-5) = 2
(12(17) – 200 + 6)/5 = 2
(204 – 194)/5 = 2
22. (x 2 – (x+1) (x+2))/(5x+1) = 6
(x 2 – (x+1) (x+2))/(5x+1) = 6
(x 2 – (x+1) (x+2)) = 6(5x+1)
x 2 – x 2 – 2x – x – 2 = 30x + 6
-3x – 2 = 30x + 6
30x + 3x = -2 – 6
((-8/33) 2 – ((-8/33)+1) (-8/33 + 2))/(5(-8/33)+1) = 6
(64/1089 – ((-8+33)/33) ((-8+66)/33)) / (-40+33)/33) = 6
(64/1089 – (25/33) (58/33)) / (-7/33) = 6
(64/1089 – 1450/1089) / (-7/33) = 6
((64-1450)/1089 / (-7/33)) = 6
-1386/1089 × 33/-7 = 6
1386 × 33 / 1089 × -7 = 6
23. ((2x+3) – (5x-7))/(6x+11) = -8/3
((2x+3) – (5x-7))/(6x+11) = -8/3
3((2x+3) – (5x-7)) = -8(6x+11)
3(2x + 3 – 5x + 7) = -48x – 88
3(-3x + 10) = -48x – 88
-9x + 30 = -48x – 88
-9x + 48x = -88 – 30
x = -118/39
((2(-118/39) + 3) – (5(-118/39) – 7)) / (6(-118/39) + 11) = -8/3
((-336/39 + 3) – (-590/39 – 7)) / (-708/39 + 11) = -8/3
(((-336+117)/39) – ((-590-273)/39)) / ((-708+429)/39) = -8/3
(-219+863)/39 / (-279)/39 = -8/3
644/-279 = -8/3
-8/3 = -8/3
24. Find the positive value of x for which the given equation is satisfied:
(i) (x 2 – 9)/(5+x 2 ) = -5/9
(x 2 – 9)/(5+x 2 ) = -5/9
9(x 2 – 9) = -5(5+x 2 )
9x 2 – 81 = -25 – 5x 2
9x 2 + 5x 2 = -25 + 81
x 2 = 56/14
x = √4
(ii) (y 2 + 4)/(3y 2 + 7) = 1/2
(y 2 + 4)/(3y 2 + 7) = 1/2
2(y 2 + 4) = 1(3y 2 + 7)
2y 2 + 8 = 3y 2 + 7
3y 2 – 2y 2 = 7 – 8
y = √-1
## EXERCISE 9.4 PAGE NO: 9.29
1. Four-fifth of a number is more than three-fourths of the number by 4. Find the number.
Let us consider the number as ‘x’
So, Three-fourth of the number is 3x/4
Fourth-fifth of the number is 4x/5
4x/5 – 3x/4 = 4
By taking LCM of 5 and 4, we get 20
(16x – 15x)/20 = 4
16x – 15x = 4(20)
∴ The number is 80.
2. The difference between the squares of two consecutive numbers is 31. Find the numbers.
Let the two consecutive numbers be x and (x – 1)
x 2 – (x-1) 2 = 31
By using the formula (a-b) 2 = a 2 + b 2 – 2ab
x 2 – (x 2 – 2x + 1) = 31
x 2 – x 2 + 2x – 1 = 31
2x – 1 = 31
Two consecutive numbers are, x and (x-1) : 16 and (16-1) =15
∴ The two consecutive numbers are 16 and 15.
3. Find a number whose double is 45 greater than its half.
2x – x/2 = 45
(4x-x)/2 = 45
∴ The number is 30.
4. Find a number such that when 5 is subtracted from 5 times that number, the result is 4, more than twice the number.
Then, five times the number will be 5x
And, two times, the number will be 2x
5x – 5 = 2x + 4
5x – 2x = 5 + 4
∴ The number is 3.
5. A number whose fifth part increased by 5 is equal to its fourth part diminished by 5. Find the number.
x/5 + 5 = x/4 – 5
x/5 – x/4 = -5 – 5
By taking LCM for 5 and 4, which is 20
(4x-5x)/20 = -10
4x – 5x = -10(20)
∴ The number is 200.
6. A number consists of two digits whose sum is 9. If 27 is subtracted from the number the digits are reversed. Find the number.
We know that one of the digits be ‘x’
The other digit is 9-x
So, the two digit number is 10(9-x) + x
The number obtained after interchanging the digits is 10x + (9-x)
10(9-x) + x – 27 = 10x + (9-x)
90 – 10x + x – 27 = 10x + 9 – x
-10x + x – 10x + x = 9 – 90 + 27
The two-digit number is 10(9-x) + x
Substituting the value of x, we get,
10(9-x) + x
10(9 – 3) + 3
∴ The number is 63.
7. Divide 184 into two parts such that one-third of one part may exceed one-seventh of another part by 8.
Let one of the numbers be ‘x’
The other number is 184 – x
So, One-third of one part may exceed one-seventh of another part by 8.
x/3 – (184-x)/7 = 8
LCM for 3 and 7 is 21
(7x – 552 + 3x)/21 = 8
(7x – 552 + 3x) = 8(21)
10x – 552 = 168
10x = 168 + 552
∴ One of the numbers is 72, and the other number is 184 – x => 184 – 72 = 112.
8. The numerator of a fraction is 6 less than the denominator. If 3 is added to the numerator, the fraction is equal to 2/3. What is the original fraction equal to?
Let us consider the denominator as x and numerator as (x-6)
By using the formula,
Fraction = numerator/denominator = (x-6)/x
(x – 6 + 3)/x = 2/3
(x – 3)/x = 2/3
3(x-3) = 2x
3x – 9 = 2x
3x – 2x = 9
∴ The denominator is x = 9, numerator is (x-6) = (9-6) = 3
And the fraction = numerator/denominator = (x-6)/x = 3/9 = 1/3
9. A sum of Rs 800 is in the form of denominations of Rs 10 and Rs 20. If the total number of notes be 50. Find the number of notes of each type.
Let the number of 10Rs notes be x
Number of 20Rs notes be 50 – x
Amount due to 10Rs notes = 10 × x = 10x
Amount due to 20Rs notes = 20 × (50 – x) = 1000 – 20x
So the total amount is Rs 800
10x + 1000 – 20x = 800
-10x = 800 – 1000
-10x = -200
x = -200/-10
∴ The number of 10Rs notes is 20
Number of 20Rs notes are 50 – 20 = 30
10. Seeta Devi has Rs 9 in fifty-paise and twenty five-paise coins. She has twice as many twenty- five paise coins as she has fifty-paise coins. How many coins of each kind does she have?
Let the number of fifty paise coins be x
The number of twenty-five paise coins be 2x
Amount due to fifty paise coins = (50×x)/100 = 0.50x
Amount due to twenty five paise coins = (25×2x)/100 = 0.50x
So the total amount is Rs 9
0.50x + 0.50x = 9
∴ The number of fifty paise coins is x = 9
Number of twenty-five paise coins, 2x = 2×9 = 18
11. Sunita is twice as old as Ashima. If six years is subtracted from Ashima’s age and four years added to Sunita’s age, then Sunita will be four times Ashima’s age. How old were they two years ago?
Let the present age of Ashima be ‘x’ years
The present age of Sunita is 2x years
Ashima’s new age = (x – 6) years
Sunita’s new age = (2x + 4) years
So, (2x + 4) = 4 (x – 6)
2x + 4 = 4x – 24
2x – 4x = -24 – 4
∴ The age of Ashima is x years = 14 years
Age of Sunita is 2x years = 2(14) = 28 years
Two years ago, age of Ashima is 14 – 2 = 12 years, age of Sunita = 28 – 2 = 26 years
12. The ages of Sonu and Monu are in the ratio 7:5 ten years hence, the ratio of their ages will be 9:7. Find their present ages.
Let the present age of Sonu be 7x years
The present age of Monu is 5x years
Sonu’s age after 10 years = (7x + 10) years
Monu’s age after 10 years = (5x + 10) years
(7x + 10) / (5x + 10) = 9/7
by using cross-multiplication, we get,
7(7x + 10) = 9(5x + 10)
49x + 70 = 45x + 90
49x – 45x = 90 – 70
∴ Present age of Sonu is 7x = 7(5) = 35years
Present age of Monu is 5x = 5(5) = 25years
13. Five years ago, a man was seven times as old as his son. Five years hence, the father will be three times as old as his son. Find their present ages.
Let the age of the son five years ago be x years
The age of man five years ago be 7x years
After five years, the son’s age is x + 5 years
After five years father’s age is 7x + 5 years
So, since five years, the relation in their ages are
7x + 5 + 5 = 3(x + 5 + 5)
7x + 10 = 3x + 15 + 15
7x + 10 = 3x + 30
7x – 3x = 30 – 10
∴ Present father’s age is 7x + 5 = 7(5) + 5 = 35 + 5 = 40years
Present son’s age is x + 5 = 5 + 5 = 10years
14. I am currently 5 times as old as my son. In 6 years time, I will be three times as old as he will be then. What are our ages now?
Let the present son’s age be x years
Present father’s age be 5x years
Son’s age after 6 years = (x + 6) years
Fathers’ age after 6 years = (5x + 6) years
5x + 6 = 3(x + 6)
5x + 6 = 3x + 18
5x – 3x = 18 – 6
∴ present son’s age is x = 6years
Present father’s age is 5x = 5(6) = 30years
15. I have Rs 1000 in ten and five rupee notes. If the number of ten rupee notes that I have is ten more than the number of five rupee notes, how many notes do I have in each denomination?
Let the number of five rupee notes be x
The number of ten rupee notes be (x + 10)
Amount due to five rupee notes = 5 × x = 5x
Amount due to ten rupee notes = 10 (x + 10) = 10x + 100
The total amount = Rs 1000
5x + 10x +100 = 1000
∴ the number of five rupee notes is x = 60
The number of ten rupee notes is x + 10 = 60+10 = 70
16. At a party, colas, squash and fruit juice were offered to guests. A fourth of the guests drank colas, a third drank squash, two-fifths drank fruit juice, and just three did not drink anything. How many guests were in all?
Let the number of guests be x
The given details are the number of guests who drank colas are x/4
The number of guests who drank squash is x/3
The number of guests who drank fruit juice is 2x/5
The number of guests who did not drink anything was 3
x/4 + x/3 + 2x/5 + 3 = x
By taking LCM for 4, 3 and 5, we get 60
(15x+20x+24x-60x)/60 = -3
(15x+20x+24x-60x) = -3(60)
∴ The total number of guests in all was 180
17. There are 180 multiple choice questions in a test. If a candidate gets 4 marks for every correct answer and for every unattempted or wrongly answered question, one mark is deducted from the total score of correct answers. If a candidate scored 450 marks in the test, how many questions did he answer correctly?
Let the number of correct answers be x
The number of questions answered wrong is (180 – x)
Total score when answered right = 4x
Marks deducted when answered wrong = 1(180 – x) = 180 – x
4x – (180 – x) = 450
4x – 180 + x = 450
5x = 450 + 180
∴ 126 questions he answered correctly.
18. A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day he works, and he will be fined Rs 5 for each day he is absent. If he receives Rs 745 in all, how many days he remained absent?
Let us consider the number of absent days as x
So, the number of present days is (20 – x)
The wage for one day of work = Rs 60
Fine for absent day = Rs 5
60(20 – x) – 5x = 745
1200 – 60x – 5x = 744
-65x = 744-1200
-65x = -456
x = -456/-65
∴ For 7 days, the labourer was absent.
19. Ravish has three boxes whose total weight is 60 ½ Kg. Box B weighs 3 ½ kg more than box A, and box C weighs 5 1/3 kg more than box B. Find the weight of box A.
The given details are the total weight of three boxes is 60 ½ kg = 121/2 kg
Let the weight of box A be x kg
Weight of box B be x + 7/2 kg
Weight of box C be x + 7/2 + 16/3 kg
x + x + 7/2 + x + 7/2 + 16/3 = 121/2
3x = 121/2 – 7/2 – 7/2 – 16/3
3x = (363 – 21 – 21 – 32)/6
∴ The weight of box A is 289/18 kg
20. The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2. Find the rational number.
Le the denominator be x and the numerator be (x – 3)
By using the formula
Fraction = numerator/denominator
= (x – 3)/x
So, when the numerator is increased by 2 and Denominator is increased by 5, then the fraction is ½
(x – 3 + 2)/(x + 5) = 1/2
(x – 1)/(x + 5) = 1/2
By using cross-multiplication, we get
2(x – 1) = x + 5
2x – 2 = x + 5
2x – x = 2 + 5
∴ Denominator is x = 7, numerator is (x – 3) = 7 – 3 = 4
And the fraction = numerator/denominator = 4/7
21. In a rational number, twice the numerator is 2 more than the denominator, if 3 is added to each, the numerator and the denominator. The new fraction is 2/3. Find the original number.
Le the numerator be x and the denominator be (2x – 2)
= x / (2x – 2)
So, the numerator and denominator are increased by 3, then the fraction is 2/3
(x + 3)/(2x – 2 + 3) = 2/3
(x + 3)/(2x + 1) = 2/3
3(x + 3) = 2(2x + 1)
3x + 9 = 4x + 2
3x – 4x = 2 – 9
∴ The numerator is x = 7, denominator is (2x – 2) = (2(7) – 2) = 14-2 = 12
And the fraction is numerator/denominator = 7/12
22. The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.
Let the speed of one train be x km/hr.
The speed of the other train be (x + 5) km/hr.
The total distance between the two stations = 340 km
Distance = speed × time
So, the distance covered by one train in 2 hrs. Will be x×2 = 2x km
Distance covered by the other train in 2 hrs. Will be 2(x + 5) = (2x + 10) km
The distance between the trains is 30 km
2x + 2x + 10 + 30 = 340
4x + 40 = 340
4x = 340 – 40
∴ The speed of one train is x = 75 km/hr.
Speed of other train is (x + 5) = 75 + 5 = 80 km/hr.
23. A steamer goes downstream from one point to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream is 1 km/hr., find the speed of the steamer in still water and the distance between the ports.
Let the speed of the steamer be x km/hr.
Speed of stream = 1 km/hr.
Downstream speed = (x + 1) km/hr.
Upstream speed = (x – 1) km/hr.
= (x + 1) × 9 and
= (x – 1) × 10
9x + 9 = 10x – 10
9x – 10x = -10 -9
x = 19 km/hr.
∴ The speed of the steamer in still water is 19 km/hr.
Distance between the ports is 9(x + 1) = 9(19+1) = 9(20) = 180 km.
24. Bhagwanti inherited Rs 12000.00. She invested part of it at 10% and the rest at 12%. Her annual income from these investments is Rs 1280.00 How much did she invest at each rate?
At a rate of 10%, let the investment be Rs x
At the rate of 12%, the investment will be Rs (12000 – x)
At 10% of rate the annual income will be x × (10/100) = 10x/100
At 12% of rate, the annual income will be (12000 – x) × 12/100 = (144000 – 12x)/100
Total investment = 1280
So, 10x/100 + (144000 – 12x)/100 = 1280
(10x + 144000 – 12x)/100 = 1280
(144000 – 2x)/100 = 1280
144000 – 2x = 1280(100)
-2x = 128000 – 144000
-2x = -16000
x = -16000/-2
∴ At 10% of rate, she invested Rs 8000, and at 12% of the rate she invested Rs (12000 – x) = Rs (12000 – 8000) = Rs 4000
25. The length of a rectangle exceeds its breadth by 9 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 84 cm 2 more than that of the given rectangle. Find the length and breadth of the given rectangle.
Let the breadth of the rectangle be x meter
Length of the rectangle be (x + 9) meter
Area of the rectangle length×breadth = x(x +9) m 2
When length and breadth increased by 3cm, then,
New length = x + 9 + 3 = x + 12
New breadth = x + 3
So, the area is
(x + 12) (x + 3) = x (x + 9) + 84
x 2 + 15x + 36 = x 2 + 9x + 84
15x – 9x = 84 – 36
∴ The length of the rectangle (x + 9) = (8 + 9) = 17cm, and the breadth of the rectangle is 8cm.
26. The sum of the ages of Anup and his father is 100. When Anup is as old as his father now, he will be five times as old as his son Anuj is now. Anuj will be eight years older than Anup is now, when Anup is as old as his father. What are their ages now?
Let the age of Anup be x years
So the age of Anup’s father will be (100 – x) years
The age of Anuj is (100-x)/5 years
So, When Anup is as old as his father after (100 – 2x) years,
Then Anuj’s age = present age of his father (Anup) + 8
Present age of Anuj + 100 – 2x = Present age of Anup + 8
(100 – x)/5 + (100 – 2x) = x + 8
(100-x)/5 – 3x = 8 – 100
(100 – x – 15x)/5 = -92
100 – 16x = -460
-16x = -460 – 100
-16x = -560
x = -560/-16
∴ The present age of Anup is 35 years then, the age of Anup’s father will be (100-x) = 100-35 = 65 years
The age of Anuj is (100-x)/5 = (100 – 35)/5 = 65/5 = 13 years
27. A lady went shopping and spent half of what she had on buying hankies and gave a rupee to a beggar waiting outside the shop. She spent half of what was left on lunch and followed that up with a two rupee tip. She spent half of the remaining amount on a book and three rupees on bus fare. When she reached home, she found that she had exactly one rupee left. How much money did she start with?
Amount spent for hankies and given to beggar is x/2 + 1
Remaining amount is x – (x/2 + 1) = x/2 – 1 = (x-2)/2
Amount spent for lunch (x-2)/2×1/2 = (x-2)/4
The amount given as a tip is Rs 2
Remaining amount after lunch = (x-2)/2 – (x-2)/4 – 2 = (2x – 4 – x + 2 – 8)/4 = (x – 10)/4
Amounts spent for books =1/2 × (x-10)/4 = (x-10)/8
The bus fare is Rs 3
Amount left = (x-10)/4 – (x-10)/8 – 3 = (2x – 20 – x + 10 – 24)/8 = (x-34)/8
So from the question, we know that the amount left = Rs 1
(x-34)/8 = 1
x – 34 = 8
∴ the lady started with Rs. 42
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## 3.5 Least squares for simple linear regression, matrix edition
So we’ve managed to figure out the normal equations for the best-fit regression coefficients. Now our job is to solve a rather complicated system of equations. Let’s do the simple regression version (i.e., with one predictor), but with matrices, to illustrate how this works.
First, ask yourself: How many columns are there in $$\boldsymbol{X}$$? What is $$\boldsymbol X' \boldsymbol X$$? ${\bf X'X} = \left(\begin{array}{cccc} 1&1&\ldots&1\\ x_1&x_2&\ldots&x_n \end{array}\right)\left(\begin{array}{cc} 1&x_1\\ 1&x_2\\ \vdots&\vdots\\ 1&x_n\\ \end{array}\right) = \left(\begin{array}{cc} \sum_{i=1}^n (1*1)&\sum_{i=1}^n(1*x_i)\\ \sum_{i=1}^n (x_i*1)&\sum_{i=1}^n(x_i*x_i) \end{array}\right) = \left(\begin{array}{cc} n&\sum_{i=1}^nx_i\\ \sum_{i=1}^n x_i&\sum_{i=1}^nx_i^2 \end{array}\right)$
Okay, what is $$(\boldsymbol X' \boldsymbol X)^{-1}$$?
The esteemed reference “mathisfun.com” says: “To find the Inverse of a 2x2 Matrix: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc).”
Let’s first label the four values.
$a = n,\quad b = \sum_{i=1}^nx_i,\quad c = \sum_{i=1}^nx_i,\quad d = \sum_{i=1}^nx_i^2$
The first step is to find the determinant.
\begin{aligned} det(\boldsymbol X' \boldsymbol X) &= ad-bc \\ &= n\sum_{i=1}^nx_i^2 - (\sum_{i=1}^n x_i)^2 \\ &= n \left(\sum_{i=1}^nx_i^2 - n^{-1}(\sum_{i=1}^n x_i)^2\right)\\ &= n \left(\sum_{i=1}^nx_i^2 - n \bar{x}^2\right)\\ &= n S_{xx} \end{aligned}
Hey look, the sum of squares for x! Knew that would come in handy.
Then the inverse is:
\begin{aligned} (\boldsymbol X' \boldsymbol X)^{-1} &= \frac{1}{nS_{xx}}\left(\begin{array}{cc} \sum_{i=1}^nx_i^2&-\sum_{i=1}^nx_i\\ -\sum_{i=1}^n x_i&n \end{array}\right)\\ &= \frac{1}{S_{xx}}\left(\begin{array}{cc} n^{-1}\sum_{i=1}^nx_i^2&-\bar{x}\\ -\bar{x}&1 \end{array}\right)\\ &= \frac{1}{S_{xx}}\left(\begin{array}{cc} n^{-1}(\sum_{i=1}^nx_i^2 - n\bar{x}^2+ n\bar{x}^2)&-\bar{x}\\ -\bar{x}&1 \end{array}\right)\\ &= \frac{1}{S_{xx}}\left(\begin{array}{cc} n^{-1}S_{xx} + \bar{x}^2&-\bar{x}\\ -\bar{x}&1 \end{array}\right)\\ &= \left(\begin{array}{cc} \frac{1}{n} + \frac{\bar{x}^2}{S_{xx}}&\frac{-\bar{x}}{S_{xx}}\\ \frac{-\bar{x}}{S_{xx}}&\frac{1}{S_{xx}} \end{array}\right)\\ \end{aligned}
So what happens if we keep doing out the multiplication? $\boldsymbol{b} = (\boldsymbol X' \boldsymbol X)^{-1} \boldsymbol X' \boldsymbol y$
I’m focusing on the bottom row because that element is the slope. The first/top element is about the intercept, $$b_0$$; and while $$b_0$$ is mathematically important for specifying where the line is, it’s often not very meaningful in context, and frankly it’s just not as interesting. Take my word for it that we can in fact find a closed-form solution for the best-fit $$b_0$$, and we can all move on with our lives.
Let’s focus on that bottom row. \begin{align*} \begin{pmatrix}b_0 \\ b_1\end{pmatrix} &= \left(\begin{array}{cc} \frac{1}{n} + \frac{\bar{x}^2}{S_{xx}}&\frac{-\bar{x}}{S_{xx}}\\ \frac{-\bar{x}}{S_{xx}}&\frac{1}{S_{xx}} \end{array}\right) \begin{pmatrix} 1 & 1 & \dots \\ x_1 & x_2 & \dots \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{pmatrix} \\ &= \begin{pmatrix} \dots & stuff & \dots \\ \frac{-\bar{x}}{S_{xx}}*1 + \frac{1}{S_{xx}}*x_1 & \frac{-\bar{x}}{S_{xx}}*1 + \frac{1}{S_{xx}}*x_2 & \dots \end{pmatrix} \begin{pmatrix} y_1 \\ y_2 \\ \vdots \\ y_n \end{pmatrix} \\ &= \begin{pmatrix} stuff \\ \frac{x_1-\bar{x}}{S_{xx}}* y_1 + \frac{x_2-\bar{x}}{S_{xx}}*y_2 + \dots \end{pmatrix} \\ &= \begin{pmatrix} stuff \\ \sum_{i} \left( \frac{x_i-\bar{x}}{S_{xx}} \right) y_i \end{pmatrix} \\ \end{align*}
Whoa.
Look at that second element. That’s $$b_1$$, expressed in a very interesting way: as a weighted average of the $$y$$ values. Specifically, they’re weighted by the discrepancy between the point’s $$x$$ value and the average $$x$$ value. This is going to work in a similar way when we have more predictors – that is, more rows in the $$\boldsymbol{b}$$ vector and columns in $$\boldsymbol{X}$$.
This is a hint as to why we always say high-leverage points are of concern. The slope is a weighted average of the $$y$$ values, with weights determined by how far each $$x$$ is from the mean. So for points with extreme $$x$$ values, their $$y$$ values are going to have a major – perhaps disproportionate – effect on our estimate of $$b_1$$. This is still true in multiple regression! |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$-(2x^{n}+1)(10x^n+3)$
Factoring the negative $GCF= -1 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} -20x^{2n}-16x^n-3 \\\\= -1(20x^{2n}+16x^n+3) \\\\= -(20x^{2n}+16x^n+3) .\end{array} Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} -(20x^{2n}+16x^n+3) \end{array} has $ac= 20(3)=60$ and $b= 16 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 10,6 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} -(20x^{2n}+10x^n+6x^n+3) .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} -[(20x^{2n}+10x^n)+(6x^n+3)] .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} -[10x^n(2x^{n}+1)+3(2x^n+1)] .\end{array} Factoring the $GCF= (2x^{n}+1)$ of the entire expression above results to \begin{array}{l}\require{cancel} -[(2x^{n}+1)(10x^n+3)] \\\\= -(2x^{n}+1)(10x^n+3) .\end{array} |
The Surprising Uses of Conic Sections
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Conic sections – the curves made by slicing through cones at various angles – were studied by the ancient Greeks, but because of their useful properties, have many real-world uses. Planets have elliptical orbits, projectiles move in parabolas, and cooling towers have hyperbolic cross-sections. But did you know that one of the most important curves in economics is a hyperbola? Or that ellipses are used to cure kidney stones?
The Surprising Uses of Conic Sections
Professor Sarah Hart
25th April 2022
Conic sections – the curves made by slicing through cones at various angles – were first studied by the ancient Greeks. The ellipse, parabola, and hyperbola have many interesting properties and applications. Today we’ll explore some of these, and along the way will find out how ellipses can be used in medicine, how to make a parabola by paper-folding, and what the hyperbola has to do with economics.
Introduction
When we slice (or “take a section” of) the cone, if the plane is horizontal we get a circle (a special case of an ellipse). As we tilt the plane, we obtain ellipses, until we reach a point where the plane is parallel to the slope of the cone. At this precise angle, the resulting curve is a parabola. Increasing the angle of the plane results in the plane intersecting both halves of the cone, and we then get hyperbolas.
In this lecture we’re going to look at these three curves in turn, and I’ll show you three real-world applications of each one.
The Ellipse
One way to define an ellipse is the following. Given two fixed points F and F’, and fixed distance r , the corresponding ellipse is the set of points P for which FP+PF'=r . This definition gives a nice way to draw an ellipse. You put two pins in a piece of paper, and place a loop of string round them. Then place a pencil against the string and draw the curve formed as the string is kept taut. If the pins are at points F and F’ , and our pencil happens to be at point P , then the total length of the string, which stays constant of course, is |FP| +|PF’| +|F’F| . Since |F’F| is also fixed, what we see is that in this curve the sum of the distances from the two fixed points, which are called foci, is always constant. And so we really do get an ellipse.
It's not obvious that the curve we get when we slice through a cone is an ellipse, with our choice of definition, but it does turn out to be the case. The same is true of the parabola and the hyperbola. These things can be proved with algebra, or using a neat argument nesting spheres between the cone and the intersecting plane, named “Dandelin spheres” after the French mathematician Germinal Pierre Dandelin.
Application 1: Elliptical Gears
In most applications in a gear system we want everything moving at constant speed (though gears of different sizes will move at different constant speeds). However, there are some cases where we want to introduce variations in speed. An example of this might be a production line in a factory where you want items to move quickly between the places where something is being done to them, and then slowly during those points. If you have two elliptical gears of the same size, each fixed with an axle at one focus, and you ensure that those fixed foci are distance r apart, and that the other pair of foci are distance r apart (by means of a rod joining them, for instance), then when you rotate the input gear about its fixed focus, it will cause the output gear to rotate, and the gears will remain touching at all times. But because the distance between the focus and the edge of the moving gear is changing, the speed of the output gear will change even when the input gear is being rotated at constant speed.
One property of the ellipse is really useful and it gives us our second application. If you draw a tangent to the ellipse at a point P , then the angle between FP and the tangent equals the angle between PF’ and the tangent.
Proof
Here’s a proof of this, which we won’t do in the lecture but I thought would be nice to include in the transcript. Let P be a point on the ellipse with foci F and F’ . The first step is to construct a line L which is going to turn out to be a tangent. To do this, extend the line F'P to the point X such that |PX| = |PF|, and then draw the perpendicular bisector of XF . This is our line L . It consists of the set of points equidistant from X and F , and this means in particular that P lies on L . A tangent is a line that meets a curve in exactly one point. So to prove L is the tangent at P , we must show that no other point on L can lie on the ellipse. So, let Q be any point on L other than P . Since L is the perpendicular bisector of XF , we know |QF| = |QX| . Because F'QX is a triangle (since Q is different from P ), we know that |F'Q| + |QX| > |F'X| . But |QX| = |QF| and |F'X| = |F'P| + |PX| = |F'P| + |PF| , because, remember, |PX| = |PF| . But this means that |F'Q| + |QF| > |F'P| + |PF| . But the ellipse consists precisely of those points R for which |F’R| +|RF| is constant. So Q cannot lie on the ellipse. This means that L meets the ellipse at P and only at P . So it’s tangent to the ellipse at P . The angles θ2 and θ3 are corresponding angles in two congruent right-angled triangles. Thus, θ2=θ3 . Also, θ1 and θ3 are vertically opposite angles, so again are equal. Hence θ1=θ2 . In other words, the angle between the tangent and FP is equal to the angle between the tangent and F'P .
QED
The equal angle property means that if we had an elliptical mirror, then any light beam that passed through one focus would reflect to pass through the other focus. By the same token, an elliptical pool table with a pocket at one focus would have the property that if we hit the ball in any direction from the other focus, it would bounce off the cushion and go into the pocket. People have built such things for fun.
Application 2: Lithotripsy
A genuine application of the equal angle property of ellipses is in medicine, with a machine called a lithotripter, which treats kidney stones. You could do an operation to get them out, but that’s risky and invasive. One alternative is to break them up in situ by bombarding them with high energy sound waves. The tiny pieces can then pass harmlessly out of your body. Of course, you don’t want to damage the healthy tissue, so you need a way to focus the waves just on the stones. The solution is to have the waves emitted from a source at one focus of an ellipse, and then have an elliptical reflector, the waves are reflected and converge at the other focus, which is where the kidney stone is.
If we take a point P on the ellipse, and extend the line from F’ to P out to a point X such that |PX| = |PF| , then X is a distance of |F’P| + |PF| from F’ , and this is constant, and independent of the choice of P . Thus these points X lie on a circle centre F’ . If we set |F’P| + |PF| = r , then the points X form a circle, centre F’ and radius r . There’s something else we can say. The point P is equidistant from X and F , so it lies on the perpendicular bisector of the line XF . That is, if we had a piece of paper and folded it to join X and F , then P would lie on the fold line. This gives us another way to create an ellipse – by folding circles. Take a circular piece of paper, and mark a point S a little way in from the edge. Now repeatedly fold in from the edge of the paper to just touch S . Crease the fold, unfold and then repeat at a different point on the circumference. If you work your way round the whole circle, you'll see an ellipse forming. One focus is S , and the other focus is the centre of the circle.
Application 3: Astronomy
Another place where ellipses play an important role is astronomy – the planets have elliptical orbits around the sun, which is at one focus. That's why we used S for our notation just now – it’s for Sun (or Solus, if you write in Latin like Newton). Many years ago when we had pound notes instead of pound coins, they showed Isaac Newton and a picture of the solar system with planets orbiting around the sun in ellipses. Unfortunately, the picture was wrong, because it showed the sun at the centre of the ellipse, not at a focus.
We can use the definition of the ellipse that FP+PF’ is constant to find the equation of an ellipse. If we put the foci equally spaced on the x-axis either side of the origin, then the ellipse has equation x2a2+y2b2=1 , for appropriate a and b with a≥b . (If a<b we still get an ellipse but its foci are on the y -axis. By varying a and b we can vary the size and shape of the ellipse. In the diagram below, the foci are at (c,0) and (-c,0) , and it can be shown that c2=a2-b2 .
The two things that distinguish ellipses from each other are their size, and their degree of “roundness”: are they close to being a circle, or are they very long and narrow? This latter property can be described precisely in terms of a quantity referred to as the eccentricity of an ellipse.
Take an ellipse with foci F and F' . It consists of the points P for which |PF| + |PF'| = r , for some constant r . The eccentricity e of the ellipse is defined to be |FF'|r . On the graph it’s ca . The closer the value of e gets to 0, the more the ellipse resembles a circle; a circle can be thought of as an ellipse with eccentricity 0. As we saw with the “folding circles” construction, the ellipse is contained in a circle centre F’ and radius r . Since the line F'F is part of a radius, we have 0≤e<1 . Now pick r>1 and fix FF’= r-1 . Then e=r-1r=1-1r. If we let r get larger and larger, in the limit we get “eccentricity 1”, which would involve a “circle of infinite radius”. This gives rise to a parabola, and that’s what we’ll look at next.
The Parabola
The parabola is the conic section that results if we cut a cone with a plane that is parallel to the sloped sides of the cone. Thinking about how we made the ellipse by folding a circle, remember that we marked a point F in the circle, and folded points X on the circumference to F , and then for each point P on the ellipse, we had |PF|=|PX| . The other focus was the centre of the circle. Now, imagine making the circle larger and larger. In the limit, the centre of the circle – the other focus – is at infinity; the curve we produce is a parabola, and the circle becomes a straight line. But we still have |PF| =|PX| , so the distance from the focus to each point P on the parabola equals the distance from P to the straight line (called the “directrix”).
We can take this as the definition: a parabola is a set of points in the plane whose distance from some fixed point F (the focus) is equal to their distance from some fixed line L (the directrix) that does not contain F . We can therefore make a parabola by folding, not a circle, but a normal piece of paper. Just mark a point F about an inch from the long left-hand side. Then repeatedly fold the long edge onto the point, creasing the fold and then unfolding.
Suppose the directrix is x=-a and the focus is (a,0) . Let P=(x,y) be a point on the parabola. The distance of P from the directrix is x+a . We use Pythagoras’s Theorem to find the distance from the focus. We get x-a2+y2 . These two must be equal. That is, x+a2=x-a2+y2 . Multiplying out gives x2+2ax+a2=x2-2ax+a2+y2 , and thus y2=4ax . If we choose a=14 , then the equation is y2=x . (We could flip the axes and retrieve the well-known y=x2 for a “vertical” parabola with horizontal directrix.)
It can be shown that the parabola has an equal angle property similar to the ellipse. As we’ve said, essentially a parabola is an ellipse with one focus at infinity. Let L be the directrix, and let’s orient the parabola so that L is vertical, and let F be the focus. If P is a point on the parabola, and we draw a tangent to the parabola at P , then the angle between the tangent and the horizontal line “from infinity” through P is equal to the angle between the tangent and the line PF .
Application 1: Telescopes and Satellite Dishes
The equal angle property of parabolas means if parallel rays of light all perpendicular to the directrix are reflected off the parabola then the reflected beams will pass through the focus. Telescopes exploit this property by using a parabolic mirror to focus the (approximately) parallel beams of light from a distant star. Parabolic antennas do the same thing for radio waves, and that’s why we see them in satellite dishes.
A parabolic mirror can be used to focus light from the sun to a single point, potentially generating enough heat to set fire to a flammable object placed at its focus – and this is where we encounter the Archimedes Death Ray! Archimedes was from Syracuse, which was threatened with invasion by the Romans. Historians recount that one night the fleet was spotted coming into the harbour. The story then goes that this is when Archimedes deployed his genius weapon – parabolic reflectors that focused the rays of the sun onto the ships, causing them to burst into flames. It’s an excellent story, but one or two tiny problems make it likely that it’s not true. Firstly, it’s pretty hard using the technology available to make a reflective enough surface to generate the required amount of heat for combustion. Secondly, no contemporary reports of the invasion say that this happened (the first to mention it was Galen, 350 years later). Thirdly if it was so great why wasn’t it ever used again? And finally, how does the story begin? At night!
Application 2: Lights
We can use the reflector property in reverse. A parabolic mirror will reflect light from a bulb placed at its focus to give a parallel beam of light. We can see this in torches and also the headlamps of vintage cars.
We saw that gravity causes the planets to have elliptical orbits. But when the two bodies involved have extremely different masses, such as a pebble and the entire planet Earth, and the centre of one is vastly far away compared to the distance between them, then the motion is, to a very good approximation, parabolic. We can actually see this using the equations of motion that you might have seen if you studied physics at school. If an object is moving with constant speed v in a given direction, then the displacement after time t is just vt (distance equals speed times time). If an object at rest starts accelerating at a constant acceleration a in a given direction, then at time t its displacement is 12at2 . Now let’s apply this to the situation of a pebble being thrown horizontally. It has a constant horizontal speed, let’s pick a not-very-random value of v=1 metre per second. Its initial vertical speed is zero, and it will accelerate vertically downwards due to gravity, at approximately g=10 metres per second squared. So at time t , if we say it starts at the point (0,0) , it will be at position (x,y) , where x=vt=t , and y=-12gt2=-5t2 and so y=-5x2 . This is just an upside-down parabola (or half of one).
Application 3: Water features
The same mathematics applies to a jet of water flowing horizontally out of a pipe (or indeed a waterfall): it creates a beautiful half-parabola. We can also show that if you instead aim your jet of water upwards at an angle, you still get a parabolic trajectory. If you know the equations you can work out exactly where the water will land, and use this to create pleasing effects, for example a series of fountains where the water comes out in bursts, giving the appearance of jets of water that jump from one place to another.
The Hyperbola
The final type of conic section is the hyperbola. This is what you get when you slice through the cone at a steep enough angle (steeper than the angle of the cone itself) so that you hit both parts of the cone. You may well see hyperbolas every evening because of this. If you have a lamp with a cylindrical lampshade, then the light from the bulb that actually escapes from the lampshade forms a cone. That cone of light intersects the vertical plane of the wall forming a beautiful hyperbolic shadow.
It’s possible to define a hyperbola with a distance rule, just like the ellipse and the parabola. For the ellipse, the sum of the distances from the foci was fixed. For the hyperbola, it’s the difference between these distances that’s a constant.
Application 1: Telescopes (again)
A hyperbolic mirror has the property that a beam of light directed towards (but on the other side of the mirror from) one focus is reflected to pass through the other focus. In a Cassegrain telescope light rays hit a parabolic mirror, and reflect towards the focus, before which they hit a hyperbolic mirror with that same focus. They are then reflected towards the other focus of the hyperbola, where they can be viewed. This makes the telescope more compact, as the focus of a parabola can be quite a long way from the mirror. The same technique is often also used in large satellite dishes.
It can be shown that the graph y=mx , where m is any fixed constant you like, is a hyperbola. The ancient Greek mathematician Manaechmus, who is believed to be the first person to study conics, was doing so to try and solve the problem of doubling the cube. That is, given a cube, to give a geometric construction of one that’s exactly double the volume. It’s not possible if you constrain yourself to “straightedge and compass” constructions, but Menaechmus showed that using the intersections of certain conics does give a way to double the cube. I’m going to show you why that might be, but in a completely anachronistic way – this is not the way Menaechmus would have argued, because algebra didn’t yet exist. Given a cube of volume v , think about where the parabola y=x2 meets the hyperbola y=2vx . It will be where x2=2vx , that is, where x3=2v. To ‘find’ this x , draw the curves and find the point of intersection. Now you can construct a cube of side x . Its volume is x3 , which is 2v , and we’ve doubled the cube!
Application 2: Cooling Towers
If you take a hyperbola and spin it around its axis you create a hyperboloid – typically seen as the shape of cooling towers. Water is often used as a coolant in industry, but of course that warms the water up, and then the water needs to be cooled down before it can be re-used. It is pumped into the tower part-way up its height, then passes down onto what’s called an exchange surface. Some of it evaporates, creating a vacuum which sucks air into the bottom of the tower. That air cools the remaining water, and it drips down through the exchange surface into a reservoir at the bottom of the tower. Only about 2% of the water escapes as steam, so this process is very efficient. The hyperboloid shape was first used in 1918, after being patented by Dutch engineers Frederick Van Iterson and Gerard Kuypers[1]. It is very strong, uses less material than a cylinder of the same height and base, and the shape also helps the air flow efficiently.
The hyperboloid also has the property that although it is a curved shape, it can actually be created with straight lines, which is a useful quality for construction, because they can be made with a lattice of straight beams rather than curved beams. Such things are known as ruled surfaces. There are also “double ruled surfaces” like the hyperbolic paraboloid “pringle” shape.
There’s another fun way to make a hyperbola, and that’s what I’ve christened “tidying rectangles”. Pick a number, say 12, and work out all the rectangles that have area 12, such as 12×1 , 3×4 , or 6×2 . Then organize them nicely by putting one of their corners at the origin, with the two edges sitting on the two axes. You don’t have to stop at whole numbers, you can add in 8×1.5 , and so on. If you keep placing rectangles like this, you start to build up a shape whose boundary is a curve. But which curve? Well, it’s the curve that passes through the top right-hand corner of each rectangle. And if the rectangle has width x and height y , then that will be the point (x,y) . But the area of the rectangle is 12, and so xy=12 . So the curve we have made is y=12x , a hyperbola.
Application 3: The Demand Curve
The law of demand says that demand for goods decreases when the price goes up, as they become less affordable, and increases if the price drops. There are one or two exceptions. For some luxury goods, raising the price gives them even more cachet and demand goes up. These are called Veblen goods. At the other extreme, a very low price-tag signals that the item must be of poor quality. Raising the price can then increase demand. However, in most cases the law of demand holds, that there is an inverse relationship between price and demand. The demand curve for a product plots quantity demanded (i.e. how many you can sell) against price charged. Different goods have different demand curves, but one interesting case is what you might call the “kid’s budget” model. A kid given 50p to spend on sweets is likely going to spend it all. If sweets cost 5p each, you buy 10 sweets. If they cost 2p each, you buy 25 sweets. In this case, the total expenditure is fixed. If we think about a particular price point and quantity demanded, we can put a rectangle under the demand curve, and its area will be the price multiplied by the amount purchased, that is, the total spend. In our pocket money example, we always spend all the money. So the area of all these rectangles is the same (in our example it’s 50). That means we get the “tidying rectangles” curve – a hyperbola. There are lots of cases where the demand curve would be a hyperbola. For instance an adult might have something like this with an annual budget for holidays or meals out.
It’s wonderful to think that conic sections, first studied over two thousand years ago, have such a diverse range of interesting applications. Even if there were no applications at all, they are of course still interesting and beautiful and worthy of study. We can’t know which bits of pure mathematics are going to have applications, but conics are just one instance of a topic that does. They weren’t studied with economics, medicine or astronomy in mind, and yet they have uses in all these areas – amazing!
Image Credits
Unless otherwise specified, the technical diagrams are drawn by me and are ©Sarah Hart, and the photos and videos are either taken/made by me or are public domain images.
• The three conics, by Pbroks13, CC BY 3.0, via Wikimedia Commons https://commons.wikimedia.org/wiki/File:Conic_sections_with_plane.svg
• Lithotripsy diagram from https://medical-dictionary.thefreedictionary.com/lithotripters
• Newton £1 note image from https://www.bankofengland.co.uk/banknotes/withdrawn-banknotes
• Cooling towers image, zootalures (Owen Cliffe), CC BY-SA 3.0, via Wikimedia Commons
https://commons.wikimedia.org/wiki/File:Didcot_power_station_cooling_tower_zootalures.jpg
• Allan H Brooks / New Control Tower Newcastle Airport / Image use permitted under CC BY-SA 2.0 https://commons.wikimedia.org/wiki/File:Newcastle_International_Airport_Control_Tower.jpg
[1] See https://www.engineeringclicks.com/cooling-tower/ for a fuller explanation of how a cooling tower works.
Professor Sarah Hart
Professor of Geometry
Sarah Hart is the first woman Professor of Geometry at Gresham College, and was appointed in 2020. She is Professor of Mathematics and until recently was Head of Mathematics and Statistics at Birkbeck, University of London.
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Math is painful, but math riddles are fun! Try these math riddles. We have also provided the answers at the end.
Math has always been cursed by many students. The theories, formulas, and concepts are too complex to understand for many. Rote learning does not work with math. Study a chapter in the English language, or study a lesson in geography or biography, and your memory may help you in your exams. But math demands practice. Every question in math is different from the other, and thus, different complex formulas may be used for every question.
And oh, the calculations are super tough to handle!
But hey, there is always a silver lining! If math comes with so many problems, it also excites one in the form of riddles. That is where we help you out!
Check out these interesting math riddles. No worries, we have also mentioned the answers at the end!
## Math Riddle 1:
Arrange the numbers 1 to 15 in order so that each pair of neighbors adds up to a square number (for example 11+5=16).
## Math Riddle 2:
Emma has just had four puppies, but her sister Ava has had five. However it’s not the litter size that their owners care about, it’s the premium that they get for female puppies. What’s the chance that Ava has had more female puppies than Emma?
## Math Riddle 4:
In reply to an inquiry about the animals on his farm, the farmer says: “I only ever keep sheep, goats, and horses. In fact, at the moment they are all sheep bar three, all goats bar four, and all horses bar five.” How many does he have of each animal?
## Math Riddle 5:
I add six to eleven and get five. How is this correct?
Math Riddle 1:
Arrange the numbers 1 to 15 in order so that each pair of neighbors adds up to a square number (for example 11+5=16).
### Answer: 9, 7, 2, 14, 11, 5, 4, 12, 13, 3, 6, 10, 15, 1, 8 or the reverse.
Math Riddle 2:
Emma has just had four puppies, but her sister Ava has had five. However it’s not the litter size that their owners care about, it’s the premium that they get for female puppies. What’s the chance that Ava has had more female puppies than Emma?
### Explanation:
Emma either has more female puppies than Emma or she has more male puppies. (Ava can't have more female puppies and more male puppies like Emma.) Since males/females are equally likely, the chance of more females must be 50%.
Math Riddle 3:
### Answer: 888 + 88 + 8 + 8 + 8 = 1,000. This is the kind of number riddle you can work out with times tables, or by simple logic. First, get as close to 1,000 as you can (888).
Math Riddle 4:
In reply to an inquiry about the animals on his farm, the farmer says: “I only ever keep sheep, goats, and horses. In fact, at the moment they are all sheep bar three, all goats bar four, and all horses bar five.” How many does he have of each animal?
### The farmer has 3 sheep, 2 goats, and 1 horse. Take sheep: we know that three animals are goats and horses, so we suppose there are two goats and one horse. Checking this hypothesis gives us three sheep, which works out because there are four non-goats: three sheep, and one horse!
Math Riddle 5:
I add six to eleven and get five. How is this correct? |
# How do you simplify (6x + 3) - 5x?
Nov 30, 2017
(6x+3)-5x=color(blue)(x-3
Refer to the explanation for the process.
#### Explanation:
Simplify:
$\left(6 x + 3\right) - 5 x$
Remove the parentheses.
$6 x + 3 - 5 x$
Gather like terms.
$\left(6 x - 5 x\right) + 3$
Simplify.
$x + 3$
Nov 30, 2017
x = -3
#### Explanation:
Given:
$\left(6 x + 3\right) - 5 x = 0$
Subtract the like terms;
$6 x - 5 x + 3 = 0$
$x + 3 = 0$
$\textcolor{red}{x = - 3}$ |
The Distance Between Notes (Half Steps, Whole Steps, and Intervals)
This lesson covers two basic methods of determining the distance between two notes. The first of these methods is called whole steps and half steps. This method is rather straightforward and pretty easy to learn. The second method we will be looking at is call intervals. Learning intervals is a bit tricky and it helps if you can read music. What we will cover here will be just a basic introduction to the concept of intervals; a more in depth lesson on this topic will find it's way into the Lesson Zone soon.
Each of these techniques lends themselves to different situations. Half steps and whole steps are usually used to tell you how far you have to move from one note to get to another. For example if you were on your 5th fret of your A string and you should be on the seventh I would use whole steps and half steps to describe where you need to be. Intervals, on the other hand, are usually used when trying to determine the relation ship between two notes. So if you were to play the 5th fret on your A string at the same time as you were playing the 2 second fret of your D string I would use intervals to explain the relationship between these two notes.
Understanding how two notes interact with each other is the first step in understanding how chords work.
Half Steps and Whole Steps
Whole steps and half steps are a way of measuring the distance between two notes. One half step up or one have step down would be the movement of one fret in either direction. A whole step is just two half steps, or two frets. You can take this thinking as far as you like. You can say the distance between the 1st fret of your E string and the 4th fret of that same sting as a movement of 3 half steps or 1 1/2 whole steps.
Whole steps are represented by 1
Half steps are represented by 1/2
The little tool on this page should help you get the basic idea of whole and half steps. By hitting the buttons below the fretboard diagram you can move the note (red dot) up or down by whole or half steps. You can also use whole and half steps to tell someone how a scale should be played, for example:
The Major Scale = 1 - 1 - 1/2 - 1 - 1 - 1 - 1/2 |
# Two real numbers
## An Advanced Mathematics Lesson Starter Of The Day
Two numbers cubed add up to four,
Both real gold as you’ll see.
Their reciprocals sum to minus one,
What could those numbers be?
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## Answer
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Clive, Bangkok
Saturday, March 7, 2015
"What a great starter puzzle this is. It revises so many concepts for IB students (Standard and Studies). This is how it went in my class today: The mission is to find the gold, it is in the labyrinth (of your mind!). The first obstacle is to know what kind of Maths topic this is: algebra? OK what kind of algebra is it? Simultaneous equations? Very good, can you write down those simultaneous equations? Can you solve those simultaneous equations? The simultaneous equation solver on the calculator is no good as it only solves for linear equations. How about drawing graphs and seeing where the graphs intersect? You will need to know how to rearrange the equations to make y the subject. Wow, what a surprising answer!"
Mollie Shoe, Georgia
Tuesday, May 5, 2020
"Hello, I've spent a very long time on this question and I still don't know the answer. How would you go about figuring it out?"
Pirenz, Philippines
Wednesday, January 13, 2021
"Let x and y represent the two real numbers.
1/x + 1/y = -1 {Second Given}
y + x = -xy {Multiply by xy}
Let u equal the value of x + y and -xy.
Hence:
4 = x^3 + y^3 {Given}
= (x + y) (x^2 - xy + y^2) {Factorize}
= (x + y) (x^2 + 2xy + y^2 - 3xy) {Rewrite -xy as 2xy - 3xy}
= (x + y) (x + y)^2 + 3(-xy) {Factor x^2 + 2xy + y^2}
= u (u^2 + 3u) {Substitute u = x + y and u = -xy}
= u^3 + 3u^2 {Expand}
So:
u^3 + 3u^2 - 4 = 0
(u - 1)(u + 2)(u + 2) = 0 {Factorize}
Thus, u = 1 or u = -2.
CASE I: u = -2
By definition, this means that x + y = -2 and -xy = -2
So, x + y = -2 and xy = 2
Hence, x ( -2 - x ) = 2 which means 0 = x^2 + 2x + 2.
But, x^2 + 2x + 2 is always at least 1 if x is a real number
So, this case provides no real solutions
CASE II: u = 1
By definition, this means that x + y = 1 and -xy = 1
So, x + y = 1 and xy = -1
Hence, x ( 1 - x ) = -1 which means 0 = x^2 - x - 1
Using the quadratic formula we get these answers:
x = (1 + sqrt(5))/2 and y = (1 - sqrt(5))/2
OR
x = (1 - sqrt(5))/2 and y = (1 + sqrt(5))/2
So, the numbers are (1 + sqrt(5))/2 and (1 - sqrt(5))/2
Note that (1 + sqrt(5))/2 is the golden ratio this is exactly why the second line said "Both real GOLD as you'll see".
Good one, Transum! :)."
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# Search by Topic
#### Resources tagged with Practical Activity similar to Number Chains:
Filter by: Content type:
Stage:
Challenge level:
### There are 28 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Practical Activity
### First Forward Into Logo 5: Pen Up, Pen Down
##### Stage: 2, 3 and 4 Challenge Level:
Learn about Pen Up and Pen Down in Logo
### First Forward Into Logo 8: More about Variables
##### Stage: 3, 4 and 5 Challenge Level:
Write a Logo program, putting in variables, and see the effect when you change the variables.
### First Forward Into Logo 2: Polygons
##### Stage: 2, 3 and 4 Challenge Level:
This is the second in a twelve part introduction to Logo for beginners. In this part you learn to draw polygons.
##### Stage: 3, 4 and 5
Logo helps us to understand gradients of lines and why Muggles Magic is not magic but mathematics. See the problem Muggles magic.
### First Forward Into Logo 9: Stars
##### Stage: 3, 4 and 5 Challenge Level:
Turn through bigger angles and draw stars with Logo.
### Fitting Flat Shapes
##### Stage: 5 Challenge Level:
How efficiently can various flat shapes be fitted together?
### First Forward Into Logo 4: Circles
##### Stage: 2, 3 and 4 Challenge Level:
Learn how to draw circles using Logo. Wait a minute! Are they really circles? If not what are they?
### First Forward Into Logo 6: Variables and Procedures
##### Stage: 3, 4 and 5 Challenge Level:
Learn to write procedures and build them into Logo programs. Learn to use variables.
### First Forward Into Logo 10: Count up - Count Down
##### Stage: 3, 4 and 5 Challenge Level:
What happens when a procedure calls itself?
### First Forward Into Logo 12: Puzzling Sums
##### Stage: 3, 4 and 5 Challenge Level:
Can you puzzle out what sequences these Logo programs will give? Then write your own Logo programs to generate sequences.
### Cubic Conundrum
##### Stage: 2, 3 and 4 Challenge Level:
Which of the following cubes can be made from these nets?
### Proof Sorter - Sum of an AP
##### Stage: 5 Challenge Level:
Use this interactivity to sort out the steps of the proof of the formula for the sum of an arithmetic series. The 'thermometer' will tell you how you are doing
### Making Maths: Celtic Knot Tiles
##### Stage: 2, 3 and 4 Challenge Level:
Make some celtic knot patterns using tiling techniques
### First Forward Into Logo 7: Angles of Polygons
##### Stage: 3, 4 and 5 Challenge Level:
More Logo for beginners. Learn to calculate exterior angles and draw regular polygons using procedures and variables.
### Plaited Origami Polyhedra
##### Stage: 2, 3 and 4 Challenge Level:
These models have appeared around the Centre for Mathematical Sciences. Perhaps you would like to try to make some similar models of your own.
### Well Balanced
##### Stage: 1, 2, 3, 4 and 5 Challenge Level:
Exploring balance and centres of mass can be great fun. The resulting structures can seem impossible. Here are some images to encourage you to experiment with non-breakable objects of your own.
### Turning the Place Over
##### Stage: 3, 4 and 5 Challenge Level:
As part of Liverpool08 European Capital of Culture there were a huge number of events and displays. One of the art installations was called "Turning the Place Over". Can you find our how it works?
### Witch's Hat
##### Stage: 3 and 4 Challenge Level:
What shapes should Elly cut out to make a witch's hat? How can she make a taller hat?
### Whirling Fibonacci Squares
##### Stage: 3 and 4
Draw whirling squares and see how Fibonacci sequences and golden rectangles are connected.
### Cool as Ice
##### Stage: 3 and 4 Challenge Level:
Design and construct a prototype intercooler which will satisfy agreed quality control constraints.
### First Forward Into Logo 11: Sequences
##### Stage: 3, 4 and 5 Challenge Level:
This part introduces the use of Logo for number work. Learn how to use Logo to generate sequences of numbers.
### Ding Dong Bell Interactive
##### Stage: 5 Challenge Level:
Try ringing hand bells for yourself with interactive versions of Diagram 2 (Plain Hunt Minimus) and Diagram 3 described in the article 'Ding Dong Bell'.
### Stonehenge
##### Stage: 5 Challenge Level:
Explain why, when moving heavy objects on rollers, the object moves twice as fast as the rollers. Try a similar experiment yourself.
### The Best Card Trick?
##### Stage: 3 and 4 Challenge Level:
Time for a little mathemagic! Choose any five cards from a pack and show four of them to your partner. How can they work out the fifth?
### Paper Folding - Models of the Platonic Solids
##### Stage: 2, 3 and 4
A description of how to make the five Platonic solids out of paper.
### Gym Bag
##### Stage: 3 and 4 Challenge Level:
Can Jo make a gym bag for her trainers from the piece of fabric she has?
### Factors and Multiples Game
##### Stage: 2, 3 and 4 Challenge Level:
A game in which players take it in turns to choose a number. Can you block your opponent? |
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