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# How do you evaluate ( ( x / (x-1 ) - (1 / lnx ) ) as x approaches 1+?
Nov 16, 2017
$\frac{1}{2}$
#### Explanation:
$\frac{x}{x - 1} - \frac{1}{\ln x} = \frac{x \ln x - \left(x - 1\right)}{\left(x - 1\right) \ln x} = \frac{x \ln x - x + 1}{x \ln x - \ln x}$
Plugging in 1 gives the indeterminate form $\frac{0}{0}$
Using L'Hospital's Rule:
Differentiate numerator and denominator.
$\frac{d}{\mathrm{dx}} \left(x \ln x - x + 1\right) = \ln x$
$\frac{d}{\mathrm{dx}} \left(x \ln x - \ln x\right) = \ln x + \frac{1}{x} \cdot x - \frac{1}{x} = \ln x + 1 - \frac{1}{x}$
We still have the form $\frac{0}{0}$, so differentiate again.
$\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$
$\frac{d}{\mathrm{dx}} \left(\ln x + 1 - \frac{1}{x}\right) = \frac{1}{x} + \frac{1}{x} ^ 2$
$\therefore$
$\frac{\frac{1}{x}}{\frac{1}{x} + \frac{1}{x} ^ 2} = \frac{\frac{x}{x}}{\frac{x}{x} + \frac{x}{x} ^ 2} = \frac{1}{1 + \frac{1}{x}}$
Plugging in 1:
$\frac{1}{1 + \frac{1}{1}} = \frac{1}{1 + 1} = \frac{1}{2}$
$\therefore$
${\lim}_{x \to {1}^{+}} \left(\frac{x \ln x - x + 1}{x \ln x - \ln x}\right) = \frac{1}{2}$ |
# Solve for: {\text{begin}array l-2x+3y=16 } 2x+3y=-16\text{end}array .
## Expression: $\left\{\begin{array} { l } -2x+3y=16 \\ 2x+3y=-16\end{array} \right.$
Solve the equation for $3y$
$\left\{\begin{array} { l } -2x+3y=16 \\ 3y=-16-2x\end{array} \right.$
Substitute the given value of $3y$ into the equation $-2x+3y=16$
$-2x-16-2x=16$
Solve the equation for $x$
$x=-8$
Substitute the given value of $x$ into the equation $3y=-16-2x$
$3y=-16-2 \times \left( -8 \right)$
Solve the equation for $y$
$y=0$
The possible solution of the system is the ordered pair $\left( x, y\right)$
$\left( x, y\right)=\left( -8, 0\right)$
Check if the given ordered pair is the solution of the system of equations
$\left\{\begin{array} { l } -2 \times \left( -8 \right)+3 \times 0=16 \\ 2 \times \left( -8 \right)+3 \times 0=-16\end{array} \right.$
Simplify the equalities
$\left\{\begin{array} { l } 16=16 \\ -16=-16\end{array} \right.$
Since all of the equalities are true, the ordered pair is the solution of the system
$\left( x, y\right)=\left( -8, 0\right)$
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# Math Snap
## MBER is the measure, in degrees, of an angle that is $\frac{92}{360}$ of a turn through a circle?
#### STEP 1
Assumptions1. The total angle in a circle is360 degrees. . The given fraction represents the portion of the circle that the angle covers.
#### STEP 2
We need to find the angle that represents the given fraction of a full turn. We can do this by multiplying the total angle in a circle by the given fraction.
$Angle = Total\, angle\, in\, a\, circle \\times Given\, fraction$
#### STEP 3
Now, plug in the given values for the total angle in a circle and the given fraction to calculate the angle.
$Angle =360\, degrees \\times \frac{92}{360}$
#### STEP 4
implify the fraction in the multiplication.
$\frac{92}{360} = \frac{23}{90}$$Angle =360\, degrees \\times \frac{23}{90}$
##### SOLUTION
Perform the multiplication to find the angle.
$Angle =360\, degrees \\times \frac{23}{90} =92\, degrees$The angle that is $\frac{92}{360}$ of a turn through a circle is92 degrees. |
# Compositions of Functions
Consider functions, f: A → B and g: B → C. The composition of f with g is a function from A into C defined by (gof) (x) = g [f(x)] and is defined by gof.
To find the composition of f and g, first find the image of x under f and then find the image of f (x) under g.
Example1:
Consider the function f = {(1, a), (2, a), (3, b)} and g = {(a, 5), (b, 7)} as in figure. Find the composition of gof.
Solution: The composition function gof is shown in fig:
```(gof) (1) = g [f (1)] = g (a) = 5, (gof) (2) = g [f (2)] = g (a) = 5
(gof) (3) = g [f (3)] = g (b) = 7.
```
Example2: Consider f, g and h, all functions on the integers, by f (n) =n2, g (n) = n + 1 and h (n) = n - 1.
Determine (i) hofog (ii) gofoh (iii) fogoh.
Solution:
```(i) hofog (n) = n + 1,
hofog (n + 1) = (n+1)2
h [(n+1)2 ] = (n+1)2 - 1 = n2 + 1 + 2n - 1 = n2 + 2n.
(ii) gofoh (n) = n - 1, gof (n - 1) = (n-1)2
g [(n-1)2 ] = (n-1)2 + 1 = n2 + 1 - 2n + 1 = n2 - 2n + 2.
(iii) fogoh (n) = n - 1
fog (n - 1) = (n - 1) + 1
f (n) = n2.
```
Note:
• If f and g are one-to-one, then the function (gof) (gof) is also one-to-one.
• If f and g are onto then the function (gof) (gof) is also onto.
• Composition consistently holds associative property but does not hold commutative property. |
# WRITING VARIABLE EXPRESSIONS
Writing variable expressions :
Writing variable expressions is the most important topic in math. Whenever we want to solve word problems, first we have to understand the given statement and convert the words into algebraic expression.
The following table will help us to learn some of the words (phrases) that can be used to indicate mathematical operations:
• The sum of
• increased by
• plus
• more than
## Subtraction
• the difference of
• decreased by
• minus
• subtracted from
• less than
## Multiplication
• the product of
• multiplied by
• times
## Division
• the quotient of
• divided by
• the ratio of
## Writing variable expression - Examples
Let us see some examples based on the above concept.
Example 1 :
Write the algebraic expressions for the following:
Twice the sum of m and n.
Solution :
Since we have the word "sum of m and n", we have to use the binary operator "+"
Sum of m and n = m + n
twice = multiply by 2
Twice the sum of m and n = 2(m + n)
Example 2 :
Write the algebraic expressions for the following:
b decreased by twice a
Solution :
Since we have the word "decreased by", we have to use the binary operator "-"
twice a = multiply 2 by a
b decreased by twice a = b - 2a
Example 3 :
Write the algebraic expressions for the following:
Numbers x and y both squared and added.
Solution :
Since we have the word "added", we have to use the binary operator "+"
square means we have to take power 2 for the variable
x and y both squared and added = x² + y²
Example 4 :
Write the algebraic expressions for the following:
Two times the product of a and b divided by 5
Solution :
Since we have the words "product of a and b", and "divided by", we have to use the binary operators "x" and "/".
Two times the product of a and b = 2ab
Two times the product of a and b divided by 5 = 2ab/5
Example 5 :
Write the algebraic expressions for the following:
Product of p and q added to 7.
Solution :
Since we have the words "product" and "added", we have to use the binary operators "x" and "+".
Product of p and q = pq
Product of p and q added to 7 = pq + 7
Example 6 :
Write the algebraic expressions for the following:
x more than two-third of y
Solution :
Since we have the word "more than", we have to use the binary operator "+"
two-third of y = (2/3)y
x more than two-third of y = x + (2/3)y
Example 7 :
Write the algebraic expressions for the following:
Half a number x decreased by 3.
Solution :
Since we have the word "decreased by", we have to use the binary operator "-"
Half a number x = (1/2) x
decreased by 3 = - 3
Half a number x decreased by 3 = (x/2) - 3
Example 8 :
Write the algebraic expressions for the following:
Sum of numbers m and n decreased by their product
Solution :
Since we have the words "Sum of" and "decreased", we have to use the binary operator "+" and "-"
Sum of numbers m and n = m + n
decreased by their product = -mn
Sum of numbers m and n decreased by their product
= (m + n) - mn
Example 9 :
Write the algebraic expressions for the following:
4 times x less than sum of y and 6.
Solution :
Since we have the words "less than" and "sum of" , we have to use the binary operator "-" and "+"
4 times x = 4x
sum of y and 6 = y + 6
4 times x less than sum of y and 6 = 4x - (y + 6)
Example 10 :
Write the algebraic expressions for the following:
Double the sum of one third of a and m.
Solution :
Since we have the word "sum of", we have to use the binary operator "+"
one third of a and m = (1/3)a + m
Double = multiply by 2
Double the sum of one third of a and m = 2 [(1/3)a + m]
After having gone through the stuff given above, we hope that the students would have understood "Writing variable expressions".
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# Calculus in a Nutshell: the Definite Integral of Monovariate Functions
The definite integral is the key tool in calculus for defining and calculating quantities important to mathematics and science, such as areas, volumes, lengths of curved paths, probabilities, and the weights of various objects, just to mention a few.
The idea behind the integral is that we can effectively compute such quantities by breaking them into small pieces and then summing the contributions from each piece.
### Riemann’s definition of integral
Figure 1 show that the area under a curve $f(x)=25-x^2$ can be approximated using rectangular strips of different size as the sum (also called upper sum):
$\displaystyle { \mathcal{A} \approx S_{upper}^{p=2} = 25\sum\limits_{i=0}^{n} f(x_i)(x_{i+1}-x_i)} \ \ \ \ \ (1)$
By varing the number of strips the approximation of the integral caan be improved. In the example given in the Figure 1a there are two strips therefore the area is given by
${\mathcal{A}\approx S_{upper}^{p=5} = 25\times \frac{5}{2} + 25\times\frac{3}{4} \times \frac{5}{2} = 25\left( \frac{5}{2}+ \frac{15}{8}\right) = 25\frac{35}{8} = 109.375}$.
In the partition in Figure 1b the interval is $\Delta x=1$ and the new sum is given by
${\mathcal{A}\approx S_{upper}^{p=5} = 25 + (25-1) + (25-4) + (25-9)+(25-16)=95 }$
We can also use the lower sums to approximate the function as shown in Figure 2.
It gives us the approximate area of
${\mathcal{A}\approx S_{lower}^{p=5} =(25-1) + (25-4) + (25-9)+(25-16)=70 }$
Therefore the correct area is between the two summations
$S_{lower} < \mathcal{A} < S_{upper}$,
and if we increase the number of rectangles we can define the area as a limit of the value of the two summations. This limit is also very close to the one that we got using rectangles defined using the Middle Point Rule (Figure 3). In this case, the are is given by
${\mathcal{A}\approx S_{lower}^{p=5} =(25-\frac{1}{4}) + (25-\frac{9}{4}) +(25-\frac{25}{4})+(25-\frac{49}{4})+(25-\frac{81}{4})=83.75 }$
that is only 0.5 % different from the analytical value (83.33) of the integral.
Therefore, the definition of the Riemann integral require a passage to the limit of the infinitesimal partition of the area underneath the interval limits. Given the interval ${[a;b]}$ the size of the interval is given by the number ${n}$ of segments ${\Delta x=\frac{b-a}{n}}$, the Riemann integral is defined as
$\displaystyle { \begin{array}{lll} \mathcal{A}&=&\lim_{\Delta x\rightarrow0} \sum\limits_{i=1}^{n} \!f(x_i)\Delta x\\ &=& \int_a^bf(x)dx\\ \end{array}} \ \ \ \ \ (2)$
Rigorously the Riemann integral is defined as follows
Some Definitions
Partition. Let $a be real number. A partition P of the interval $[a,b]$ is a finite subset of real numbers $x_0,\dots,x_n$ such that $a=x_0 < x_1 < \dots x_{n-1}=b$. We write $P=x_0,x_1,\dots,x_n$.
Norm of a Partition. We define the norm of partition P, written$||P||$, to be largest of all the subintervall widths. If $|| P ||$ is a small number then all the subintervalls in the partitionP have a small width.
The Riemann Integral
The $f(x)$ be a function definied on a closed interval $[a,b]$. The number $S$ is the definite integral of $f(x)$ over $[a,b]$ and that S is the limit of the Riemann sum:
$\sum_{i=1}^n f(t_i)\Delta x_j$
if the following condition is satisfied:
Given any number $\epsilon >0$ there is a corresponding number $\delta >0$ such that for every partition $P=x_0,x_1,\dots,x_n$ of $[a,b]$ with $||P||< \delta$ and any choice of $t_k$ in $|x_{k-1}-x_k|$, we have
$\lvert \sum_{i=1}^n f(t_i) \Delta x_i-S \rvert < \epsilon$.
### Definite integral properties
Here a list of some important properties of the definite integral:
Association: ${\int_a^b \left(f(x)+g(x)\right) dx = \int_a^b f(x)dx+\int_a^b g(x)dx} \ \ \ \ \ (3)$
Linearity: ${\int_a^b cf(x)dx = c\int_a^b f(x)dx} \ \ \ \ \ (4)$
Partition: ${\int_a^b cf(x)dx = \int_a^c f(x)dx+\int_c^b f(x)dx } \ \ \ \ \ (5)$
Antisymmetry: ${\int_a^b cf(x)dx = -\int_b^a f(x)dx} \ \ \ \ \ (6)$
### Fundamental Theorem of Calculus
The definite integral is defined as $F(x)=\int \limits_{a}^{x}f(t)dt \ \ \ \ \ (7)$
The function F(x) gives the area under the graph of ${f(t)}$ from a to x when ${f(t)}$ is nonnegative and ${x > a}$.
If we denote the area between ${x}$ and ${x+h}$ as ${F(x+h)-F(x)}$ then
$\displaystyle F(x+h)-F(x) \approx hf(x) \ \ \ \ \ (8)$
Theorem: If ${f}$ is continuous function on ${[a, b]}$, then ${F(x) = \int\limits_{a}^{x}f(t)dt}$ is continuous on ${[a, b]}$ and differentiable on ${(a, b)}$ and its derivative is ${f(x)}$:
$\displaystyle F'(x)=\frac{d}{dx}\int\limits_{a}^{x}f(t)dt=f(x) \ \ \ \ \ (9)$
Proof:
\begin{aligned} \frac{d}{dx}\int_a^x f(s)ds &= lim_{h \rightarrow 0}\frac{\int_a^{x+h} f(s)ds-\int_a^{x} f(s)ds}{h} \\ &= lim_{h \rightarrow 0}\frac{\int_x^{x+h} f(s)ds}{h}\\ &= lim_{h \rightarrow 0}\frac{hf(s)}{h} \\ &= f(x) \end{aligned}
$\Box$
Consider two functions ${F(x)}$ and ${G(x)=F(x)+c}$ with c a constant. Then
$\displaystyle \frac{dF(x)}{dx}=f(x)=\frac{dG(x)}{dx} \ \ \ \ \ (10)$
Thus indefinite integrals are denoted as
$\displaystyle \int f(x)dx=F(x)+c \ \ \ \ \ (11)$
### Definition of Integral as antiderivative
Thus in the limit that ${h \rightarrow 0}$, we obtain
$F'(x)=\frac{dF(x)}{dx} =\lim_{h \rightarrow 0} \frac{F(x+h)-F(x)}{h} =f(x) \ \ \ \ \ (12)$
Therefore, the function of ${F(x)}$ is the inverse of the derivative (or antiderivative) of $f(x)$
\begin{aligned} \frac{d}{dx}\int_a^b f(s)ds &= \frac{d}{dx}\int_a^b F'(s)ds \\ &= lim_{h \rightarrow 0}\sum_{n=1}^N F'(a+(n-1)h)h \\ &= lim_{h \rightarrow 0}\sum_{n=1}^N \frac{\left[F(a+nh)-F(a+(n-1)h)\right] h}{h}\\ &= lim_{h \rightarrow 0}\sum_{n=1}^N {\left[F(a+nh)-F(a+(n-1)h)\right]} \end{aligned}
By examining the last expression it is easy to find that only the values for ${n=1}$, ${-F(a)}$ and the one for n=N, ${F(b)}$ do not cancel out giving the demonstration of the following theorem.
Theorem: If ${f}$ is continuous at every point in ${[a, b]}$ and ${F}$ is any antiderivative of ${f}$ on ${[a, b]}$, then $\displaystyle \int_a^bF(x)dx=\left[F(x)\right]_a^b=F(b)-F(a) \ \ \ \ \ (13)$
#### Some example indefinite integrals
• ${\int adx = ax+c}$
• ${\int x^ndx = \frac{1}{n+1}x^{n+1}+c}$ where ${n\neq-1}$
• ${\int\frac{1}{x}dx=lnx + c}$
• ${\int a^xdx=\frac{1}{\log_e{a}}a^x+c}$
• ${\int e^xdx=e^{x}+c}$
• ${\int \cos(x)dx=\sin{x}+c}$
• ${\int \sin(x)dx=-\cos{x}+c}$
### Substitution method
$\displaystyle \frac{d f(g(x))}{dx}=f'(g(x))g'(x) \ \ \ \ \ (14)$
$\displaystyle \int f'(g(x))g'(x)dx= f(g(x))+c \ \ \ \ \ (15)$
or using the substitution ${y=g(x)}$ and ${dy=g'(x)dx}$
$\displaystyle \int f'(y)dy= f(y)+c \ \ \ \ \ (16)$
### Integral by Parts Method
The derivative of a product of two functions ${(fg)'=f'g+fg'}$, can rearrange be rearranged as ${f'g=(fg)'-fg'}$. By taking the integral of all the terms, we can write
$\int f'(x)g(x)dx= \int (f(x)g(x))'dx - \int f(x)g'(x)dx$
or by introducing the two new functions ${U}$ and ${v}$, as
\begin{aligned} u&= g(x) \\ du&=g'(x)dx\\ v&=f(x)\\ dv&= f'(x)dx \end{aligned}
$\displaystyle \int udv= uv - \int vdu \ \ \ \ \ (31)$
### REFERENCES
[1] Maurice D. Weir, Joel Hass, George B. Thomas. Thomas’s Calculus. 12thEdition, Pearson.
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## Calculating the Slope
To calculate the slope of a line you need only two points from that line, (x1, y1) and (x2, y2).
The equation used to calculate the slope from two points is: On a graph, this can be represented as:
There are three steps in calculating the slope of a straight line when you are not given its equation.
1. Step One: Identify two points on the line.
2. Step Two: Select one to be (x1, y1) and the other to be (x2, y2).
3. Step Three: Use the slope equation to calculate slope.
Take a moment to work through an example where we are given two points.
### Example
Let's say that points (15, 8) and (10, 7) are on a straight line. What is the slope of this line?
1. Step One: Identify two points on the line.
In this example we are given two points, (15, 8) and (10, 7), on a straight line.
2. Step Two: Select one to be (x1, y1) and the other to be (x2, y2).
It doesn't matter which we choose, so let's take (15, 8) to be (x2, y2). Let's take the point (10, 7) to be the point (x1, y1).
3. Step Three: Use the equation to calculate slope.
Once we've completed step 2, we are ready to calculate the slope using the equation for a slope:
We said that it really doesn't matter which point we choose as (x1, y1) and the which to be (x2, y2). Let's show that this is true. Take the same two points (15, 8) and (10, 7), but this time we will calculate the slope using (15, 8) as (x1, y1) and (10, 7) as the point (x2, y2). Then substitute these into the equation for slope:
We get the same answer as before!
Often you will not be given the two points, but will need to identify two points from a graph. In this case the process is the same, the first step being to identify the points from the graph. Below is an example that begins with a graph.
Example
What is the slope of the line given in the graph? The slope of this line is 2.
[detailed solution to example]
Now, take a moment to compare the two lines which are on the same graph.
Notice that the line with the greater slope is the steeper of the two. The greater the slope, the steeper the line. Keep in mind, you can only make this comparison between lines on a graph if: (1) both lines are drawn on the same set of axes, or (2) lines are drawn on different graphs (i.e., using different sets of axes) where both graphs have the same scale.
You are now ready to try a practice problem. If you have already completed the first practice problem for this unit you may wish to try the additional practice. |
# Hard 2 step equations
Apps can be a great way to help learners with their math. Let's try the best Hard 2 step equations. Our website can solving math problem.
## The Best Hard 2 step equations
Absolute value is a concept in mathematics that refers to the distance of a number from zero on a number line. The absolute value of a number can be thought of as its magnitude, or how far it is from zero. For example, the absolute value of 5 is 5, because it is five units away from zero on the number line. The absolute value of -5 is also 5, because it is also five units away from zero, but in the opposite direction. Absolute value can be represented using the symbol "| |", as in "|5| = 5". There are a number of ways to solve problems involving absolute value. One common method is to split the problem into two cases, one for when the number is positive and one for when the number is negative. For example, consider the problem "find the absolute value of -3". This can be split into two cases: when -3 is positive, and when -3 is negative. In the first case, we have "|-3| = 3" (because 3 is three units away from zero on the number line). In the second case, we have "|-3| = -3" (because -3 is three units away from zero in the opposite direction). Thus, the solution to this problem is "|-3| = 3 or |-3| = -3". Another way to solve problems involving absolute value is to use what is known as the "distance formula". This formula allows us to calculate the distance between any two points on a number line. For our purposes, we can think of the two points as being 0 and the number whose absolute value we are trying to find. Using this formula, we can say that "the absolute value of a number x is equal to the distance between 0 and x on a number line". For example, if we want to find the absolute value of 4, we would take 4 units away from 0 on a number line (4 - 0 = 4), which tells us that "the absolute value of 4 is equal to 4". Similarly, if we want to find the absolute value of -5, we would take 5 units away from 0 in the opposite direction (-5 - 0 = -5), which tells us that "the absolute value of -5 is equal to 5". Thus, using the distance formula provides another way to solve problems involving absolute value.
How to solve an equation? There are many ways to solve an equation, but one of the most common methods is by using algebra. Algebra is a branch of mathematics that deals with the solution of equations. In order to solve an equation, you need to find the value of the unknown variable. For example, if you have the equation "x + 3 = 5", then you would need to find the value of "x" that makes the equation true. In this case, "x" would be equal to 2. However, not all equations can be solved using algebra. Some equations may require more advanced methods, such as calculus. But in general, algebra is the method most often used to solve equations.
One way is to graph the function and see where it produces a result. Another way is to look at the definition of the function and see what values of x will produce a result. For example, if we have a function that takes the square root of x, we know that we can only take the square root of positive numbers. Therefore, our domain will be all positive numbers. Once we have found the domain, we can then solve for specific values by plugging in those values and seeing what outputs we get. This process can be helpful in solving problems and understanding how functions work.
Looking for an easy and effective way to solve equations? Look no further than the 3 equation solver! This handy tool can quickly and easily solve any equation with three variables, making it a valuable tool for students, teachers, and professionals alike. Simply enter the equation into the 3 equation solver and press the solve button. The tool will instantly generate a solution, making it easy to check your work or find the correct answer. With its simple and user-friendly interface, the 3 equation solver is a must-have for anyone who needs to solve equations on a regular basis. Give it a try today and see how much time and effort you can save! |
# Texas Go Math Grade 2 Lesson 4.1 Answer Key Equal Parts
Refer to our Texas Go Math Grade 2 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 2 Lesson 4.1 Answer Key Equal Parts.
## Texas Go Math Grade 2 Lesson 4.1 Answer Key Equal Parts
Explore
Put pattern blocks together to match the shape of the hexagon. Trace the shape you made.
FOR THE TEACHER • Have children place a yellow hexagon pattern block on the workspace and make the same shape by using any combination of pattern blocks. Discuss how they know if the outline of the blocks they used is the same shape as the yellow hexagon.
Yes, children make the same shape by using any combination of pattern blocks.
Explanation:
A yellow hexagon pattern block on the workspace and make the same shape by using any combination of 6 triangle pattern blocks are joined to form a same yellow hexagon
Math Talk
Mathematical Processes
Describe how the shapes you used are different from the shapes a classmate used.
Shapes are described depending on the number of its sides and to some degree its angular relationship. Shapes of classmate are described by virtue of the area of space bordered by the lines.
Model and Draw
The green rectangle is the whole. It can be divided into equal parts.
Share and Show
Write how many equal parts there are in the whole. Write halves, fourths, or eighths to name the equal parts.
Question 1.
___________ equal parts
2 halves
two equal parts
Explanation:
On dividing the whole or a object into equal parts, we get 2 halves.
We need to divide an object or a whole number into equal parts to distribute it equally.
These equal parts have to be the same in measurements like weight, volume, dimensions, numbers etc.
Question 2.
___________ equal parts
8 equal parts
Explanation:
A whole can be divided into parts of equal size.
The above rectangle has been divided into eight equal parts.
Question 3.
___________ equal parts
4 equal parts
Explanation:
A whole can be divided into parts of equal size.
The above circle has been divided into 4 equal parts.
Question 4.
___________ equal parts
8 equal parts
Explanation:
A whole can be divided into parts of equal size.
The above rectangle has been divided into eight equal parts.
Question 5.
___________ equal parts
2 equal parts
Explanation:
On dividing the whole or a object into equal parts, we get 2 halves.
We need to divide an object or a whole number into equal parts to distribute it equally.
These equal parts have to be the same in measurements like weight, volume, dimensions, numbers etc.
Question 6.
___________ equal parts
4 equal parts
Explanation:
A whole can be divided into parts of equal size.
The above rectangle has been divided into 4 equal parts.
Problem Solving
Write how many equal parts there are in the whole. Write halves, fourths, or eighths to name the equal parts.
Question 7.
___________ equal parts
2 equal parts
Explanation:
On dividing the whole or a object into equal parts, we get 2 halves.
We need to divide an object or a whole number into equal parts to distribute it equally.
These equal parts have to be the same in measurements like weight, volume, dimensions, numbers etc.
Question 8.
___________ equal parts
4 equal parts
Explanation:
A whole can be divided into parts of equal size.
The above square has been divided into 4 equal parts.
Question 9.
___________ equal parts
8 equal parts
Explanation:
A whole can be divided into parts of equal size.
The above square has been divided into eight equal parts.
Question 10.
H.O.T. Draw to show halves. Explain how you know that the parts are halves.
The complete picture or object is divided in to two equal parts is called halves as shown in below pictures.
Explanation:
On dividing the whole or a object into equal parts, we get 2 halves.
We need to divide an object or a whole number into equal parts to distribute it equally.
These equal parts have to be the same in measurements like weight, volume, dimensions, numbers etc.
Question 11.
H.O.T. Multi-Step Sort the shapes.
• Use red to color the shapes that show eighths.
• Use blue to color the shapes that show fourths.
Explanation:
As, we know from the above exercises about halves, fourths and eighths in detail.
So, from the above pictures circle, rectangle and square are divided into fourths colored with blue and eighths colored with red.
Question 12.
Analyze Joe drew a circle divided into fourths. Which shape has parts that are fourths?
Explanation:
Option (A) shows fourths.
If a whole is divided into four equal parts, each part is a FOURTH.
A fourth is obtained by dividing a whole (1) by 4.
Question 13.
Use Math Language Anna drew a rectangle divided into eighths. Which shape has parts that are eighths?
Explanation:
Option (B) shows eights.
If a whole is divided into eight equal parts, each part is an Eighth.
An eighth is obtained by dividing a whole (1) by 8.
Question 14.
TEXAS Test Prep Which shape has parts that are fourths?
Explanation:
Option (C) shows fourths.
If a whole is divided into four equal parts, each part is a FOURTH.
A fourth is obtained by dividing a whole (1) by 4.
TAKE HOME ACTIVITY • Ask your child to fold one sheet of paper into halves and another sheet of paper into fourths.
Explanation:
When the child folds the paper sheet, he will get the above images.
If a whole is divided into two equal parts, each part is a Halves.
A halves is obtained by dividing a whole (1) by 2.
If a whole is divided into four equal parts, each part is a FOURTH.
A fourth is obtained by dividing a whole (1) by 4.
### Texas Go Math Grade 2 Lesson 4.1 Homework and Practice Answer Key
Write how many equal parts there are in the whole. Write halves, fourths, or eighths to name the equal parts.
Question 1.
Fourths
Explanation:
The above picture is of fourths, it has four equal parts.
A fourth is obtained by dividing a whole (1) by 4.
Question 2.
Halves
Explanation:
The above picture is of halves, it has two equal parts.
A half is obtained by dividing a whole (1) by 2.
Question 3.
Eighths
Explanation:
The above picture is of eighths, it has eight equal parts.
A eighth is obtained by dividing a whole (1) by 8.
Problem Solving
Draw to solve.
Question 4.
Multi-Step Sort the shapes.
• Use red to color the shapes that show eighths.
• Use blue to color the shapes that show fourths.
Explanation:
If a whole is divided into 8 equal parts, each part is an eighth.
An eighth is obtained by dividing a whole (1) by 8.
So, color all the eighths with red after identifying.
If a whole is divided into four equal parts, each part is a FOURTH.
A fourth is obtained by dividing a whole (1) by 4.
So, color all the fourths with blue after identifying.
Lesson Check
Question 5.
Which shape has parts that show fourths?
Explanation:
Option (B) shows the fourths,
If a whole is divided into four equal parts, each part is a FOURTH.
A fourth is obtained by dividing a whole (1) by 4.
Question 6.
Which shape has parts that show halves?
Explanation:
From the above pictures option (C) shows halves.
If a whole is divided into 2 equal parts, each part is a Half.
A Half is obtained by dividing a whole (1) by 2.
Question 7.
Which shape has parts that show eighths?
Explanation:
In the above pictures option (A) shows eighths.
If a whole is divided into 8 equal parts, each part is a Eighth.
A Eighth is obtained by dividing a whole (1) by 8.
Question 8.
The cheese pizza was divided into fourths. Which pizza is divided into fourths? |
# STATS 244.3(01) Elementary Statistical Concepts
## Term Test #1 , September 24, 1998
1. The answer is (g) 10.
(11+5+11+6+11+6)/6=10
2. [back to questions]
3. The answer is (h) 11.
If we put the numbers in order we get
5 6 11 11 11 16
Since there are 6 numbers, and (6+1)/2=3.5, the median is between the third and fourth numbers. Since these are both 11, the median is 11.
[back to questions]
4. The answer is (h) 11.
This is the number occurring the most frequently. 11 appears 3 times, the other numbers only appear once each.
[back to questions]
5. The answer is (i) 16.
The deviations and their squares are
11-10 = 1 and 1² = 1
5-10 = -5 and (-5)² = 25
11-10 = 1 and 1² = 1
6-10 = -4 and (-4)² = 16
11-10 = 1 and 1² = 1
16-10 = 6 and 6² = 36
The sum of the squared deviations add up to 80. For the sample variance we divide by 6-1=5, and 80/5=16.
[back to questions]
6. The answer is (b) 4.
The standard deviation is the square root of the variance, and 4²=16.
[back to questions]
7. The answer is (d) 6.
Putting the numbers in order gives
5 6 11 11 11 16.
The numbers to the left of the median are
5 6 11
and the median of these is 6.
[back to questions]
8. The answer is (h) 11.
Putting the numbers in order gives
5 6 11 11 11 16.
The numbers to the right of the median are
11 11 16
and the median of these is 11
[back to questions]
.
9. The answer is (c) 5.
The lower quartile is 6, and the upper quartile is 11, so the inter-quartile range is 11-6=5.
[back to questions]
10. The answer is (h) 11.
The largest number is 16, the smallest is 5. The difference is 16-5=11.
[back to questions]
11. The answer is (h) 11.
The 75th percentile is the same as the upper quartile.
[back to questions]
12. An attribute of an individual (or "unit") is called a variable. Non-numeric variables are called categorical. So the answer is (i) categorical variable
[back to questions]
13. The answer is (d) parameter. Remember Population Parameter.
[back to questions]
14. (a) Sample. A subset of the population is called a sample.
[back to questions]
15. (f) Outlier.
[back to questions]
16. Mean squared deviation from the mean is the verbal definition of the (c) variance.
[back to questions]
1. Wrong. It is not a histogram--a histogram has bars.
[back to questions]
2. Correct. This is an example of a stemplot or stem-and-leaf diagram.
[back to questions]
3. Wrong. A cumulative distribution would be a line graph.
[back to questions]
4. Wrong. No box, no whiskers.
[back to questions]
5. Wrong. We haven't even discussed scatterplots in this course yet. This choice is put there to catch smart-alecks who figure they must have missed something.
[back to questions]
1. Not 7.
[back to questions]
2. Not 14.
[back to questions]
3. Correct. Each leaf represents an observation ("unit"), and one can count that there are 15 leaves.
[back to questions]
4. Not 53.
[back to questions]
5. Not 77.
[back to questions]
1. Wrong. There is no 11. Only single digits appear in the leaves.
[back to questions]
2. Wrong. Why would you think 13?
[back to questions]
3. Not quite. The smallest first digit is 8, but the diagram has more.
[back to questions]
4. Correct. The smallest first digit is 8, and the smallest second digit is 1. Actually, one cannot tell from the stemplot where the decimal point is, but the choices from the following questions make clear where it must be.
[back to questions]
5. Could be correct. The smallest number could be 811, since the first two digits are still 8 and 1, respectively. However, this choice is not consistent with any of the values from the remaining questions.
[back to questions]
1. Not 96. There are 6 smaller observations, and 8 larger.
[back to questions]
2. Correct.There are 7 smaller observations, and 7 larger.
[back to questions]
3. Not 99. There are 8 smaller observations, and 6 larger.
[back to questions]
4. Wrong. There is no 110.
[back to questions]
5. Not even close. There are 11 smaller observations, and only 3 larger.
[back to questions]
17. The answer is (c) 33. The quartiles would be the 4th observations from either end, thus the lower quartile is 82, and the upper is 115. The interquartile range is the difference of these.
[back to questions]
1. Wrong. This is a stemplot for some other data.
[back to questions]
2. Wrong. The median and the quartiles are in the wrong places.
[back to questions]
3. Correct. Note that median and quartiles are in the correct places.
[back to questions]
4. Wrong. The shape is correct, but note that the numbers along the abcissa do not agree with the values in the stemplot.
[back to questions]
5. Not even close. This looks like a cumulative distribution, but most of the values are greater than 10.
[back to questions]
1. It is not symmetrical, and there don't appear to be outliers.
[back to questions]
2. It is not symmetrical.
[back to questions]
3. Correct. The distribution is positively skewed
[back to questions]
4. It is not negatively skewed.
[back to questions]
5. Wrong. To have a variance of zero, all the values would have to be the same.
[back to questions]
1. Wrong. This histogram would not balance at 10.
[back to questions]
2. Wrong. It would not balance at 20 either
[back to questions]
3. Wrong. This is the "centre" of the distribution in one sense -- in fact it is like the mode, but there is more weight to the right than to the left, so again it would not balance.
[back to questions]
4. Not quite. The median is around 30. But for a skewed distribution like this one, the mean and median are not usually equal. On the other hand, the histogram does not tell you exactly where the values are. If all the values were pushed to the left side of the intervals (as they could be, if the left end-points are included), then it is possible for the median to be less than 30 -- in fact, the median could be as low as 25. Thus, one could argue that a distribution having a mean around 30 is not inconsistent with this histogram. Hence, I will give full marks to those who selected this choice.
[back to questions]
5. Correct. The mean would be to the right of the median, so it would be somewhat greater than 30. If you suppose that the values are concentrated at the mid-points of the intervals, you can figure out that the mean would be 33. Since one could argue that 33 is `around 30' the previous choice is also acceptable.
[back to questions]
18. 20% of the values are less than 20, and 20+30=50% are less than 30, so the 25th percentile is somewhere between these values. Thus the correct answer is (b) between 20 and 30.
[back to questions]
19. All the bars except the last one represent values less than 60. The last bar comprises 5%, so the remaining ones make up 95%. Thus the answer is (d).
[back to questions] |
# C Triangle Check If Triangle Is 90 Degrees With Code Examples
In this article, we will look at how to get the solution for the problem, C Triangle Check If Triangle Is 90 Degrees With Code Examples
## What shows a right triangle?
The first (and easiest) way to identify a right a triangle is if it's already been marked with a 90∘ angle, like the one above. Even if 90∘ isn't written in, if you see a little square drawn in the corner, that's also a way of indicating a right angle.
``````int IsRightTriangle(float a, float b, float c)
{
if((a*a + b*b == c*c) || (b*b + c*c == a) || (a*a + c*c == b*b)){
return 1; //returns 1 if it's a right triangle
}else{
return 0; // returns 0 if it isn't a right triangle
}
}```
```
## Is the given triangle a right-angled triangle?
We can use Pythagoras' Theorem to check if a triangle is right-angled, using the following method: Square the two shorter sides and add the values together. Square the longest side. Check if the results for (1) and (2) are the same.
## How do you identify a right angle?
When two straight lines intersect each other at 90˚ or are perpendicular to each other at the intersection, they form the right angle. A right angle is represented by the symbol ∟.
## How do you find the type of triangle with side lengths in c?
Step 1: Declare three sides of triangle. Step 2: Enter three sides at run time. Step 3: If side1 == side2 && side2 == side3 Go to step 6 Step 4: If side1 == side2 || side2 == side3 || side3 == side1 Go to Step 7 Step 5: Else Go to step 8 Step 6: Print the triangle is equilateral.
## How do you check if a triangle is right angled in C?
Let the sides be A, B, and C. The given triangle is right-angled if and only if A2 + B2 = C2.
## How do you find the C angle of a triangle?
What is the measure of angle C? We know that the sum of the measures of any triangle is 180 degrees. Using the fact that angle A + angle B + angle C = 180 degrees, we can find the measure of angle C.
## How do you verify a triangle?
To determine if 3 side lengths are a triangle, use the triangle inequality theorem, which states that the sum of 2 sides of a triangle must be greater than the third side. Therefore, all you have to do is add together each combination of 2 sides to see if it's greater than the third side.
## Find A Big Length Friend From Array Javascript Finding Longest String In Array In Javascript With Code Examples
In this article, we will look at how to get the solution for the problem, Find A Big Length Friend From Array Javascript Finding Longest String In Array In Javascript With Code Examples How do you find the longest string in an array of strings? java get longest string in array int index = 0; int elementLength = array[0]. length(); for(int i=1; i< array. length(); i++) { if(array[i]. length() > elementLength) { index = i; elementLength = array[i]. length(); } } return array[index]; //find a bi
## Parse Integer Javascript With Code Examples
In this article, we will look at how to get the solution for the problem, Parse Integer Javascript With Code Examples What is the difference between parseInt and number? Number() converts the type whereas parseInt parses the value of input. As you see, parseInt will parse up to the first non-digit character. On the other hand, Number will try to convert the entire string. parseInt(value); let string = "321" console.log(string); // "321" <= string let number = parseInt(string); console.log(numbe
## How To Set Indian Timezone In Django With Code Examples
In this article, we will look at how to get the solution for the problem, How To Set Indian Timezone In Django With Code Examples How do I get current time zone in Python? To get the current time in particular, you can use the strftime() method and pass into it the string ”%H:%M:%S” representing hours, minutes, and seconds. #Tested it's working modify into settings.py file TIME_ZONE = 'Asia/Kolkata' TIME_ZONE = 'Asia/Calcutta' Here is the list of valid timezones: h
## London Poem Analysis With Code Examples
In this article, we will look at how to get the solution for the problem, London Poem Analysis With Code Examples What are the poetic devices used in the poem London? Analysis of Literary Devices Used in “London” Assonance: Assonance is the repetition of vowel sounds in the same line. Consonance: Consonance is the repetition of consonant sounds in the same line. Alliteration: Alliteration is the repetition of consonant sounds in the same line in quick succession. LONDON As we've seen, a l
## Fix 504 Gateway Time Out in Nodejs Applications
Node.js is a popular open-source, cross-platform JavaScript runtime environment that allows developers to run JavaScript code on the server side. However, sometimes when you try to access a website built on Node.js, you might encounter a "504 Gateway Timeout" error message. This error can be frustrating, but it is not uncommon. In this article, we will explain what a 504 Gateway Timeout error is, what causes it, and how you can fix it. What is a 504 Gateway Timeout Error? A 504 Gateway Timeout |
17Calculus Precalculus - Quadratic Partial Fraction Expansion
17Calculus
Partial Fraction Expansion - Quadratic Factors (Single and Repeating)
Okay, now we look at how to handle quadratic factors when they cannot be factored (in the real number system).
Terms with quadratic factors are of the form $$ax^2+bx+c$$ where the highest power on the variable ($$x$$ in this case) is two and $$a \neq 0$$. We do not have the same requirement on $$b$$ and $$c$$, so $$ax^2$$ and $$ax^2+c$$ are also considered quadratic factors. However, the term $$ax^2+bx$$ is NOT considered a quadratic factor, since it can be factored into two simple factors, i.e. $$ax^2 + bx = x(ax+b)$$. In this case, we follow the techniques associated with simple factors.
When we have a quadratic factor, the numerator must be of the form $$Ax+B$$. Notice that we now have an $$x$$ in the numerator, not just constants. Also, notice that the highest power of $$x$$ in the numerator is one less than the highest power in the denominator.
Quadratic Single Factors Example
$$\displaystyle{ \frac{1}{x(x^2+3)} = \frac{A}{x} + \frac{Bx+C}{x^2+3} }$$
Repeated Factors
If you understand how repeated factors work for linear terms discussed on the previous page, you should be able to anticipate how repeated factors work for quadratic factors.
factor in the denominator partial fraction terms $$ax^2+bx+c$$ $$\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} }$$ $$(ax^2+bx+c)^2$$ $$\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} }$$ $$(ax^2+bx+c)^3$$ $$\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + }$$ $$\displaystyle{ + \frac{A_3x+B_3}{(ax^2+bx+c)^3} }$$
Do you see the pattern? For a denominator with the term $$(ax^2+bx+c)^k$$, we would have the factors
$$\displaystyle{ \frac{A_1x+B_1}{ax^2+bx+c} + \frac{A_2x+B_2}{(ax^2+bx+c)^2} + }$$ $$\displaystyle{ \frac{A_3x+B_3}{(ax^2+bx+c)^3} + }$$ $$\displaystyle{ . . . + \frac{A_kx+B_k}{(ax^2+bx+c)^k} }$$
Once these expansions are set up, the steps to find the constants are the same as with linear factors.
Okay, let's work the practice problems.
Practice
Unless otherwise instructed, expand the given fraction using partial fraction expansion. Give your answer in exact terms.
Basic
$$\displaystyle{\frac{5x^2+10+7}{(x^2+2)(x+3)}}$$
Problem Statement
Expand $$\displaystyle{\frac{5x^2+10+7}{(x^2+2)(x+3)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2926 video solution
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$$\displaystyle{\frac{x^4+3x^3+7x-1}{x(x^2+1)^2}}$$
Problem Statement
Expand $$\displaystyle{\frac{x^4+3x^3+7x-1}{x(x^2+1)^2}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2927 video solution
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$$\displaystyle{\frac{6x^2+21x+11}{(x^2+3)(x+5)}}$$
Problem Statement
Expand $$\displaystyle{\frac{6x^2+21x+11}{(x^2+3)(x+5)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2928 video solution
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$$\displaystyle{\frac{3x^2+5x-4}{(x^2-7)(x+1)}}$$
Problem Statement
Expand $$\displaystyle{\frac{3x^2+5x-4}{(x^2-7)(x+1)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2929 video solution
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$$\displaystyle{\frac{3x^4-2x^3+6x^2-3x+3}{(x^2+2)^2(x+3)}}$$
Problem Statement
Expand $$\displaystyle{\frac{3x^4-2x^3+6x^2-3x+3}{(x^2+2)^2(x+3)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2930 video solution
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$$\displaystyle{\frac{x^2 - 1}{x(x^2+1)}}$$
Problem Statement
Expand $$\displaystyle{\frac{x^2 - 1}{x(x^2+1)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2931 video solution
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$$\displaystyle{\frac{2x^2-11x+17}{x^3-7x^2+15x-9}}$$
Problem Statement
Expand $$\displaystyle{\frac{2x^2-11x+17}{x^3-7x^2+15x-9}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2932 video solution
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$$\displaystyle{\frac{7x^2-7x+18}{x^3+8}}$$
Problem Statement
Expand $$\displaystyle{\frac{7x^2-7x+18}{x^3+8}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2933 video solution
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$$\displaystyle{\frac{-x^4+3x^3+6x^2+11x+20}{(x^2+2)^2(x+3)}}$$
Problem Statement
Expand $$\displaystyle{\frac{-x^4+3x^3+6x^2+11x+20}{(x^2+2)^2(x+3)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2934 video solution
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$$\displaystyle{\frac{5x^2+7x+8}{(x+1)(x^2+2x+3)}}$$
Problem Statement
Expand $$\displaystyle{\frac{5x^2+7x+8}{(x+1)(x^2+2x+3)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2935 video solution
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$$\displaystyle{\frac{2x^2-x+4}{x^2+4x}}$$
Problem Statement
Expand $$\displaystyle{\frac{2x^2-x+4}{x^2+4x}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2936 video solution
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$$\displaystyle{\frac{5x-1}{(3x^2-2)(x-1)}}$$
Problem Statement
Expand $$\displaystyle{\frac{5x-1}{(3x^2-2)(x-1)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2937 video solution
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$$\displaystyle{\frac{x^2+4x-2}{x^3+4x}}$$
Problem Statement
Expand $$\displaystyle{\frac{x^2+4x-2}{x^3+4x}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2938 video solution
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$$\displaystyle{\frac{1}{x(x^2+1)}}$$
Problem Statement
Expand $$\displaystyle{\frac{1}{x(x^2+1)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2939 video solution
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$$\displaystyle{\frac{10}{(x-1)(x^2+9)}}$$
Problem Statement
Expand $$\displaystyle{\frac{10}{(x-1)(x^2+9)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2940 video solution
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$$\displaystyle{\frac{x^2+x+1}{(2x+1)(x^2+1)}}$$
Problem Statement
Expand $$\displaystyle{\frac{x^2+x+1}{(2x+1)(x^2+1)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
There is more going on in this problem than partial fraction expansion, which you can ignore. You will get a chance to do this in calculus.
2941 video solution
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$$\displaystyle{\frac{x^4+1}{x(x^2+1)^2}}$$
Problem Statement
Expand $$\displaystyle{\frac{x^4+1}{x(x^2+1)^2}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
Krista King Math - 2942 video solution
video by Krista King Math
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$$\displaystyle{\frac{2x^2-x+9}{x(x^2+9)}}$$
Problem Statement
Expand $$\displaystyle{\frac{2x^2-x+9}{x(x^2+9)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
2943 video solution
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$$\displaystyle{\frac{5x^2-8x+5}{(x-2)(x^2-x+1)}}$$
Problem Statement
Expand $$\displaystyle{\frac{5x^2-8x+5}{(x-2)(x^2-x+1)}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
MIP4U - 1495 video solution
video by MIP4U
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$$\displaystyle{\frac{3x^2-x+4}{x^3+2x^2+6x}}$$
Problem Statement
Expand $$\displaystyle{\frac{3x^2-x+4}{x^3+2x^2+6x}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
PatrickJMT - 1487 video solution
video by PatrickJMT
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$$\displaystyle{\frac{8x^3+13x}{(x^2+2)^2}}$$
Problem Statement
Expand $$\displaystyle{\frac{8x^3+13x}{(x^2+2)^2}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
MIP4U - 1496 video solution
video by MIP4U
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$$\displaystyle{\frac{10x^2+12x+20}{x^3-8}}$$
Problem Statement
Expand $$\displaystyle{\frac{10x^2+12x+20}{x^3-8}}$$ using partial fraction expansion. Give your answer in exact terms.
Solution
Khan Academy - 1489 video solution
video by Khan Academy
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Intermediate
$$\displaystyle{ \frac{4x^2}{4x^4+1} }$$
Problem Statement
Expand $$\displaystyle{ \frac{4x^2}{4x^4+1} }$$ using partial fraction expansion. Give your answer in exact terms.
Hint
To factor $$4x^4+1$$, start by adding and subtracting $$4x^2$$.
Problem Statement
Expand $$\displaystyle{ \frac{4x^2}{4x^4+1} }$$ using partial fraction expansion. Give your answer in exact terms.
Final Answer
$$\displaystyle{ \frac{4x^2}{4x^4+1} }$$ $$\displaystyle{ = \frac{x}{2x^2-2x+1} - }$$ $$\displaystyle{ \frac{x}{2x^2+2x+1} }$$
Problem Statement
Expand $$\displaystyle{ \frac{4x^2}{4x^4+1} }$$ using partial fraction expansion. Give your answer in exact terms.
Hint
To factor $$4x^4+1$$, start by adding and subtracting $$4x^2$$.
Solution
This is part of a improper integral problem. You do not need to know how to work improper integrals to understand how to solve this problem.
Michael Penn - 3856 video solution
video by Michael Penn
Final Answer
$$\displaystyle{ \frac{4x^2}{4x^4+1} }$$ $$\displaystyle{ = \frac{x}{2x^2-2x+1} - }$$ $$\displaystyle{ \frac{x}{2x^2+2x+1} }$$
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# 2009 AMC 8 Problems/Problem 23
## Problem
On the last day of school, Mrs. Wonderful gave jelly beans to her class. She gave each boy as many jelly beans as there were boys in the class. She gave each girl as many jelly beans as there were girls in the class. She brought $400$ jelly beans, and when she finished, she had six jelly beans left. There were two more boys than girls in her class. How many students were in her class?
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 32\qquad\textbf{(E)}\ 34$
## Solution 1
If there are $x$ girls, then there are $x+2$ boys. She gave each girl $x$ jellybeans and each boy $x+2$ jellybeans, for a total of $x^2 + (x+2)^2$ jellybeans. She gave away $400-6=394$ jellybeans.
\begin{align*} x^2+(x+2)^2 &= 394\\ x^2+x^2+4x+4 &= 394\\ 2x^2 + 4x &= 390\\ x^2 + 2x &= 195\\ \end{align*}
From here, we can see that $x = 13$ as $13^2 + 26 = 195$, so there are $13$ girls, $13+2=15$ boys, and $13+15=\boxed{\textbf{(B)}\ 28}$ students.
## Solution 2 (Don't need quadratic equation)
Consider the solutions, there are two more boys than girls, so if there are 26 students, we have 14 boys and 12 girls. \$4^2+2^2 according to bglah blah hkjlhkjhkjlhkjh |
# How to Calculate the Standard Error of Estimate
Co-authored by wikiHow Staff
Updated: March 29, 2019 | References
The standard error of estimate is used to determine how well a straight line can describe values of a data set. When you have a collection of data from some measurement, experiment, survey or other source, you can create a line of regression to estimate additional data. With the standard error of estimate, you get a score that describes how good the regression line is.
### Part One of Two:Tabulating Your Data
1. 1
Create a five column data table. Any statistical work is generally made easier by having your data in a concise format. A simple table serves this purpose very well. To calculate the standard error of estimate, you will be using five different measurements or calculations. Therefore, creating a five-column table is helpful. Label the five columns as follows:[1]
• ${\displaystyle x}$
• ${\displaystyle y}$
• ${\displaystyle y^{\prime }}$
• ${\displaystyle y-y^{\prime }}$
• ${\displaystyle (y-y^{\prime })^{2}}$
• Note that the table shown in the image above performs the opposite subtractions, ${\displaystyle y^{\prime }-y}$. The more standard order, however, is ${\displaystyle y-y^{\prime }}$. Because the values in the final column are squared, the negative is not problematic and will not change the outcome. Nevertheless, you should recognize that the more standard calculation is ${\displaystyle y-y^{\prime }}$.
2. 2
Enter the data values for your measured data. After collecting your data, you will have pairs of data values. For these statistical calculations, the independent variable is labeled ${\displaystyle x}$ and the dependent, or resulting, variable is ${\displaystyle y}$. Enter these values into the first two columns of your data table.
• The order of the data and the pairing is important for these calculations. You need to be careful to keep your paired data points together in order.
• For the sample calculations shown above, the data pairs are as follows:
• (1,2)
• (2,4)
• (3,5)
• (4,4)
• (5,5)
3. 3
Calculate a regression line. Using your data results, you will be able to calculate a regression line. This is also called a line of best fit or the least squares line. The calculation is tedious but can be done by hand. Alternatively, you can use a handheld graphing calculator or some online programs that will quickly calculate a best fit line using your data.[2]
• For this article, it is assumed that you will have the regression line equation available or that it has been predicted by some prior means.
• For the sample data set in the image above, the regression line is ${\displaystyle y^{\prime }=0.6x+2.2}$.
4. 4
Calculate predicted values from the regression line. Using the equation of that line, you can calculate predicted y-values for each x-value in your study, or for other theoretical x-values that you did not measure.
• Using the equation of the regression line, calculate or “predict” values of ${\displaystyle y^{\prime }}$ for each value of x. Insert the x-value into the equation, and find the result for ${\displaystyle y^{\prime }}$ as follows:
• ${\displaystyle y^{\prime }=0.6x+2.2}$
• ${\displaystyle y^{\prime }(1)=0.6(1)+2.2=2.8}$
• ${\displaystyle y^{\prime }(2)=0.6(2)+2.2=3.4}$
• ${\displaystyle y^{\prime }(3)=0.6(3)+2.2=4.0}$
• ${\displaystyle y^{\prime }(4)=0.6(4)+2.2=4.6}$
• ${\displaystyle y^{\prime }(5)=0.6(5)+2.2=5.2}$
### Part Two of Two:Performing the Calculations
1. 1
Calculate the error of each predicted value. In the fourth column of your data table, you will calculate and record the error of each predicted value. Specifically, subtract the predicted value (${\displaystyle y^{\prime }}$) from the actual observed value (${\displaystyle y}$).[3]
• For the data in the sample set, these calculations are as follows:
• ${\displaystyle y(x)-y^{\prime }(x)}$
• ${\displaystyle y(1)-y^{\prime }(1)=2-2.8=-0.8}$
• ${\displaystyle y(2)-y^{\prime }(2)=4-3.4=0.6}$
• ${\displaystyle y(3)-y^{\prime }(3)=5-4=1}$
• ${\displaystyle y(4)-y^{\prime }(4)=4-4.6=-0.6}$
• ${\displaystyle y(5)-y^{\prime }(5)=5-5.2=-0.2}$
2. 2
Calculate the squares of the errors. Take each value in the fourth column and square it by multiplying it by itself. Fill in these results in the final column of your data table.
• For the sample data set, these calculations are as follows:
• ${\displaystyle -0.8^{2}=0.64}$
• ${\displaystyle 0.6^{2}=0.36}$
• ${\displaystyle 1^{2}=1.0}$
• ${\displaystyle -0.6=0.36}$
• ${\displaystyle -0.2=0.04}$
3. 3
Find the sum of the squared errors (SSE). The statistical value known as the sum of squared errors (SSE) is a useful step in finding standard deviation, variance and other measurements. To find the SSE from your data table, add the values in the fifth column of your data table.[4]
• For this sample data set, this calculation is as follows:
• ${\displaystyle 0.64+0.36+1.0+0.36+0.04=2.4}$
4. 4
Finalize your calculations. The Standard Error of the Estimate is the square root of the average of the SSE. It is generally represented with the Greek letter ${\displaystyle \sigma }$. Therefore, the first calculation is to divide the SSE score by the number of measured data points. Then, find the square root of that result.[5]
• If the measured data represents an entire population, then you will find the average by dividing by N, the number of data points. However, if you are working with a smaller sample set of the population, then substitute N-2 in the denominator.
• For the sample data set in this article, we can assume that it is a sample set and not a population, just because there are only 5 data values. Therefore, calculate the Standard Error of the Estimate as follows:
• ${\displaystyle \sigma ={\sqrt {\frac {2.4}{5-2}}}}$
• ${\displaystyle \sigma ={\sqrt {\frac {2.4}{3}}}}$
• ${\displaystyle \sigma ={\sqrt {0.8}}}$
• ${\displaystyle \sigma =0.894}$
5. 5
Interpret your result. The Standard Error of the Estimate is a statistical figure that tells you how well your measured data relates to a theoretical straight line, the line of regression. A score of 0 would mean a perfect match, that every measured data point fell directly on the line. Widely scattered data will have a much higher score.[6]
• With this small sample set, the standard error score of 0.894 is quite low and represents well organized data results.
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Co-Authored By:
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Updated: March 29, 2019
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Article Summary
To calculate the standard error of estimate, create a five-column data table. In the first two columns, enter the values for your measured data, and enter the values from the regression line in the third column. In the fourth column, calculate the predicted values from the regression line using the equation from that line. These are the errors. Fill in the fifth column by multiplying each error by itself. Add together all of the values in column 5, then take the square root of that number to get the standard error of estimate.
Yes
No |
11.1 Graphs, trigonometric identities, and solving trigonometric
Page 1 / 12
History of trigonometry
Work in pairs or groups and investigate the history of the development of trigonometry. Describe the various stages of development and how different cultures used trigonometry to improve their lives.
The works of the following people or cultures can be investigated:
1. Cultures
1. Ancient Egyptians
2. Mesopotamians
3. Ancient Indians of the Indus Valley
2. People
2. Hipparchus (circa 150 BC)
3. Ptolemy (circa 100)
4. Aryabhata (circa 499)
5. Omar Khayyam (1048-1131)
7. Nasir al-Din (13th century)
8. al-Kashi and Ulugh Beg (14th century)
9. Bartholemaeus Pitiscus (1595)
Functions of the form $y=sin\left(k\theta \right)$
In the equation, $y=sin\left(k\theta \right)$ , $k$ is a constant and has different effects on the graph of the function. The general shape of the graph of functions of this form is shown in [link] for the function $f\left(\theta \right)=sin\left(2\theta \right)$ .
Functions of the form $y=sin\left(k\theta \right)$
On the same set of axes, plot the following graphs:
1. $a\left(\theta \right)=sin0,5\theta$
2. $b\left(\theta \right)=sin1\theta$
3. $c\left(\theta \right)=sin1,5\theta$
4. $d\left(\theta \right)=sin2\theta$
5. $e\left(\theta \right)=sin2,5\theta$
Use your results to deduce the effect of $k$ .
You should have found that the value of $k$ affects the period or frequency of the graph. Notice that in the case of the sine graph, the period (length of one wave) is given by $\frac{{360}^{\circ }}{k}$ .
These different properties are summarised in [link] .
$k>0$ $k<0$
Domain and range
For $f\left(\theta \right)=sin\left(k\theta \right)$ , the domain is $\left\{\theta :\theta \in \mathbb{R}\right\}$ because there is no value of $\theta \in \mathbb{R}$ for which $f\left(\theta \right)$ is undefined.
The range of $f\left(\theta \right)=sin\left(k\theta \right)$ is $\left\{f\left(\theta \right):f\left(\theta \right)\in \left[-1,1\right]\right\}$ .
Intercepts
For functions of the form, $y=sin\left(k\theta \right)$ , the details of calculating the intercepts with the $y$ axis are given.
There are many $x$ -intercepts.
The $y$ -intercept is calculated by setting $\theta =0$ :
$\begin{array}{ccc}\hfill y& =& sin\left(k\theta \right)\hfill \\ \hfill {y}_{int}& =& sin\left(0\right)\hfill \\ & =& 0\hfill \end{array}$
Functions of the form $y=cos\left(k\theta \right)$
In the equation, $y=cos\left(k\theta \right)$ , $k$ is a constant and has different effects on the graph of the function. The general shape of the graph of functions of this form is shown in [link] for the function $f\left(\theta \right)=cos\left(2\theta \right)$ .
Functions of the form $y=cos\left(k\theta \right)$
On the same set of axes, plot the following graphs:
1. $a\left(\theta \right)=cos0,5\theta$
2. $b\left(\theta \right)=cos1\theta$
3. $c\left(\theta \right)=cos1,5\theta$
4. $d\left(\theta \right)=cos2\theta$
5. $e\left(\theta \right)=cos2,5\theta$
Use your results to deduce the effect of $k$ .
You should have found that the value of $k$ affects the period or frequency of the graph. The period of the cosine graph is given by $\frac{{360}^{\circ }}{k}$ .
These different properties are summarised in [link] .
$k>0$ $k<0$
Domain and range
For $f\left(\theta \right)=cos\left(k\theta \right)$ , the domain is $\left\{\theta :\theta \in \mathbb{R}\right\}$ because there is no value of $\theta \in \mathbb{R}$ for which $f\left(\theta \right)$ is undefined.
The range of $f\left(\theta \right)=cos\left(k\theta \right)$ is $\left\{f\left(\theta \right):f\left(\theta \right)\in \left[-1,1\right]\right\}$ .
Intercepts
For functions of the form, $y=cos\left(k\theta \right)$ , the details of calculating the intercepts with the $y$ axis are given.
The $y$ -intercept is calculated as follows:
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Exponential Terms Raised to an Exponent
## Multiply to raise exponents to other exponents
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Exponential Terms Raised to an Exponent
What if you had an exponential expression that was raised to a secondary power, like (23)2\begin{align*}(2^3)^2\end{align*}? How could you simplify it? After completing this Concept, you'll be able to use the power of a product property to simplify exponential expressions like this one.
### Watch This
The following video from YourTeacher.com may make it clearer how the power rule works for a variety of exponential expressions:
### Guidance
What happens when we raise a whole expression to a power? Let’s take x\begin{align*}x\end{align*} to the power of 4 and cube it. Again we’ll use the full factored form for each expression:
(x4)3(xxxx)(xxxx)(xxxx)=x4×x4×x43 factors of {x to the power 4}=xxxxxxxxxxxx=x12
So (x4)3=x12\begin{align*}(x^4)^3 = x^{12}\end{align*}. You can see that when we raise a power of x\begin{align*}x\end{align*} to a new power, the powers multiply.
Power Rule for Exponents: (xn)m=x(nm)\begin{align*}(x^n)^m = x^{(n\cdot m)}\end{align*}
If we have a product of more than one term inside the parentheses, then we have to distribute the exponent over all the factors, like distributing multiplication over addition. For example:
(x2y)4=(x2)4(y)4=x8y4.
Or, writing it out the long way:
(x2y)4=(x2y)(x2y)(x2y)(x2y)=(xxy)(xxy)(xxy)(xxy)=xxxxxxxxyyyy=x8y4
Note that this does NOT work if you have a sum or difference inside the parentheses! For example, (x+y)2x2+y2\begin{align*}(x+y)^2 \neq x^2+y^2\end{align*}. This is an easy mistake to make, but you can avoid it if you remember what an exponent means: if you multiply out (x+y)2\begin{align*}(x+y)^2\end{align*} it becomes (x+y)(x+y)\begin{align*}(x+y)(x+y)\end{align*}, and that’s not the same as x2+y2\begin{align*}x^2+y^2\end{align*}. We’ll learn how we can simplify this expression in a later chapter.
#### Example A
Simplify the following expressions.
a) 3537\begin{align*}3^5 \cdot 3^7\end{align*}
b) 262\begin{align*}2^6 \cdot 2\end{align*}
c) (42)3\begin{align*}(4^2)^3\end{align*}
Solution
When we’re just working with numbers instead of variables, we can use the product rule and the power rule, or we can just do the multiplication and then simplify.
a) We can use the product rule first and then evaluate the result: 3537=312=531441\begin{align*}3^5 \cdot 3^7 = 3^{12}=531441\end{align*}.
OR we can evaluate each part separately and then multiply them: 3537=2432187=531441\begin{align*}3^5 \cdot 3^7 = 243 \cdot 2187 = 531441\end{align*}.
b) We can use the product rule first and then evaluate the result: 262=27=128\begin{align*}2^6 \cdot 2 = 2^7 = 128\end{align*}.
OR we can evaluate each part separately and then multiply them: \begin{align*}2^6 \cdot 2 = 64 \cdot 2 = 128\end{align*}.
c) We can use the power rule first and then evaluate the result: \begin{align*}(4^2)^3 = 4^6 = 4096\end{align*}.
OR we can evaluate the expression inside the parentheses first, and then apply the exponent outside the parentheses: \begin{align*}(4^2)^3 = (16)^3 = 4096\end{align*}.
#### Example B
Simplify the following expressions.
a) \begin{align*}x^2 \cdot x^7\end{align*}
b) \begin{align*}(y^3)^5\end{align*}
Solution
When we’re just working with variables, all we can do is simplify as much as possible using the product and power rules.
a) \begin{align*}x^2 \cdot x^7 = x^{2+7} = x^9\end{align*}
b) \begin{align*}(y^3)^5 = y^{3 \times 5}= y^{15}\end{align*}
#### Example C
Simplify the following expressions.
a) \begin{align*}(3x^2 y^3) \cdot (4xy^2)\end{align*}
b) \begin{align*}(4xyz) \cdot (x^2 y^3) \cdot (2y z^4)\end{align*}
c) \begin{align*}(2a^3b^3)^2\end{align*}
Solution
When we have a mix of numbers and variables, we apply the rules to each number and variable separately.
a) First we group like terms together: \begin{align*}(3x^2y^3) \cdot (4xy^2) = (3 \cdot 4) \cdot (x^2 \cdot x) \cdot (y^3 \cdot y^2)\end{align*}
Then we multiply the numbers or apply the product rule on each grouping: \begin{align*}=12 x^3y^5\end{align*}
b) Group like terms together: \begin{align*}(4xy z) \cdot (x^2 y^3) \cdot (2y z^4) = (4 \cdot 2) \cdot (x \cdot x^2) \cdot (y \cdot y^3 \cdot y) \cdot (z \cdot z^4)\end{align*}
Multiply the numbers or apply the product rule on each grouping: \begin{align*}= 8x^3 y^5 z^5\end{align*}
c) Apply the power rule for each separate term in the parentheses: \begin{align*}(2a^3b^3)^2 = 2^2 \cdot (a^3)^2 \cdot (b^3)^2\end{align*}
Multiply the numbers or apply the power rule for each term \begin{align*}=4a^6 b^6\end{align*}
Watch this video for help with the Examples above.
### Vocabulary
• Exponent: An exponent is a power of a number that shows how many times that number is multiplied by itself. An example would be \begin{align*}2^3\end{align*}. You would multiply 2 by itself 3 times: \begin{align*}2 \times 2 \times 2.\end{align*} The number 2 is the base and the number 3 is the exponent. The value \begin{align*}2^3\end{align*} is called the power.
• Product of Powers Property: For all real numbers \begin{align*}\chi\end{align*},
\begin{align*}\chi^n \cdot \chi^m = \chi^{n+m}\end{align*}.
• Power of a Product Property: For all real numbers \begin{align*}\chi\end{align*},
.
### Guided Practice
Simplify the following expressions.
a) \begin{align*}(x^2)^2 \cdot x^3\end{align*}
b) \begin{align*}(2x^2y) \cdot (3xy^2)^3\end{align*}
c) \begin{align*}(4a^2 b^3)^2 \cdot (2ab^4)^3\end{align*}
Solution
In problems where we need to apply the product and power rules together, we must keep in mind the order of operations. Exponent operations take precedence over multiplication.
a) We apply the power rule first: \begin{align*}(x^2)^2 \cdot x^3 = x^4 \cdot x^3\end{align*}
Then apply the product rule to combine the two terms: \begin{align*}x^4 \cdot x^3 = x^7\end{align*}
b) Apply the power rule first: \begin{align*}(2x^2 y) \cdot (3xy^2)^3 = (2x^2y) \cdot (27x^3y^6)\end{align*}
Then apply the product rule to combine the two terms: \begin{align*}(2x^2 y) \cdot (27 x^3 y^6) = 54x^5y^7\end{align*}
c) Apply the power rule on each of the terms separately: \begin{align*}(4a^2 b^3)^2 \cdot (2ab^4)^3 = (16a^4 b^6) \cdot (8a^3 b^{12})\end{align*}
Then apply the product rule to combine the two terms: \begin{align*}(16a^4 b^6) \cdot (8a^3 b^{12}) = 128a^7 b^{18}\end{align*}
### Practice
Simplify:
1. \begin{align*}(a^3)^4\end{align*}
2. \begin{align*}(xy)^2\end{align*}
3. \begin{align*}(-5y)^3\end{align*}
4. \begin{align*}(3a^2b^3)^4\end{align*}
5. \begin{align*}(-2xy^4z^2)^5\end{align*}
6. \begin{align*}(-8x)^3(5x)^2\end{align*}
7. \begin{align*}(-x)^2(xy)^3\end{align*}
8. \begin{align*}(4a^2)(-2a^3)^4\end{align*}
9. \begin{align*}(12xy)(12xy)^2\end{align*}
10. \begin{align*}(2xy^2)(-x^2y)^2(3x^2y^2)\end{align*} |
Unit 16 Normal Distributions
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1 Unit 16 Normal Distributions Objectives: To obtain relative frequencies (probabilities) and percentiles with a population having a normal distribution While there are many different types of distributions that a population may have (see Figures 15-1a to 15-1f), normal distributions are of particular importance in statistics One of the primary reasons normal distributions are so important is because, as we have seen previously, the sampling distribution of x with simple random sampling takes on the properties of a normal distribution, with a sufficiently large sample size n Figure 15-2 gave us a somewhat detailed description of a normal distribution in terms of one, two, and three standard deviations away from the mean We have seen that a normal distribution is symmetric and bell-shaped, and that practically all of the items in a normally distributed population are within three standard deviations of the mean We now want to describe normal distributions in a much more detailed manner than merely in terms of one, two, and three standard deviations away from the mean Table A2 in the appendix provides areas under a normal density curve in terms of z-scores A normal curve described entirely in terms of z-scores is called a standard normal curve By converting raw scores to z-scores, we can use Table A2 to find many different areas under any normal density curve These areas under a normal density curve can be interpreted as relative frequencies or as probabilities In addition to using Table A2, statistical software packages, spreadsheets, programmable calculators, etc can also be used to find areas under a normal density curve The figure at the top of the Table A2 indicates that you can use this table directly to find the area under a normal curve above any value greater than the mean µ Such an area is found by obtaining the z-score of a value greater than the mean and reading the corresponding area in the body of the table The z-scores in the table are all to two decimal place accuracy, with the z-scores to the first decimal place displayed as the row labels, and the column labels providing the second decimal place The areas in the body of the table are all given to four decimal place accuracy To illustrate the use of Table A2, we shall consider several examples involving the population of weights of oranges from a particular grove Let us suppose that the weights of oranges from the grove have a normal distribution with mean µ = 781 oz and standard deviation σ = oz We shall consider finding the percentage of orange weights that lie in a given range, or, in other words, the probability that one randomly 107
3 The probability that one randomly selected orange weighs less than 1066 oz is = (or 9846%) As another illustration, we shall have you find the percentage, or relative frequency, of oranges that weigh more than 653 oz, which can also be interpreted as the probability that one randomly selected orange has a weight more than 653 oz Find the z-score for 653 oz, and label the value 653 oz on the horizontal axis in Figure 16-4 Then, shade the desired area under the normal curve, and use Table A2 to obtain the desired probability (You should find that the z-score is 097, and that the probability that one randomly selected orange weighs more than 653 oz is (or 8340%)) If the desired area under the standard normal curve is in between two given values, then we need to read Table A2 twice Let us obtain the percentage, or relative frequency, of oranges that weigh between 6 and 8 oz, which can also be interpreted as the probability that one randomly selected orange has a weight between 6 and 8 oz To find this probability, we first use µ = 781 and σ = to find the z-score of 6 oz to be = 137, and to find the z-score of 8 oz to be = The proportion of shaded area in Figure 16-5 is the desired probability The unshaded area below 6 oz in Figure 16-5 is a mirror image of the shaded area in the figure at the top of Table A2; also, the unshaded area above 8 oz in Figure 16-5 corresponds exactly to the shaded area in the figure at the top of Table A2 We find the total unshaded area by adding the entry of Table A2 in the row labeled 13 and the column labeled 007 to the entry of Table A2 in the row labeled 01 and the column labeled 004; we then obtain the desired area by subtracting this unshaded area from 1 The probability that one randomly selected orange weighs between 6 and 8 oz is 1 ( ) = (or 4704%) In the illustration just completed, the desired area under the standard normal curve was in between two values, where one was below µ, and the other was above µ We shall now consider instances where the desired area under the standard normal curve is in between two values, where either both values are greater than µ, or both values are less than µ First, we obtain the percentage, or relative frequency, of oranges that weigh between 55 and 65 oz, which can also be interpreted as the probability that one randomly selected orange has a weight between 55 and 65 oz To find this probability, we first use µ = 781 and σ = to find the z-score of 55 oz to be = 175, 109
4 and to find the z-score of 65 oz to be = 099 The proportion of shaded area in Figure 16-6 is the desired probability The unshaded area below 55 oz in Figure 16-6 is a mirror image of the shaded area in the figure at the top of Table A2; also, if the shaded and unshaded areas below 65 oz are combined together in Figure 16-6, this combined area is a mirror image of the shaded area in the figure at the top of Table A2 We find the desired area by subtracting the entry of Table A2 in the row labeled 17 and the column labeled 005 from the entry of Table A2 in the row labeled 09 and the column labeled 009 The probability that one randomly selected orange weighs between 55 and 65 oz is = (or 1210%) Now, we shall let you find the percentage, or relative frequency, of oranges that weigh between 854 and 932 oz, which can also be interpreted as the probability that one randomly selected orange has a weight between 854 and 932 oz Find the z-score for each of 854 and 932 oz, and label the values for 854 and 932 oz on the horizontal axis in Figure 16-7 Then, shade the desired area under the normal curve, and use Table A2 to obtain the desired probability (You should find that the z-scores are +055 and +114, and that the probability that one randomly selected orange weighs between 854 and 932 oz is (or 1641%)) Since practically all of the area under a normal curve is within three standard deviations of the mean, there are no z-scores in Table A2 below 309 or above +309 When one orange weight is randomly selected from the population with mean µ = 781 oz and standard deviation σ = oz, we would consider the probability of observing a weight greater than 20 oz to be practically zero (0), because the z-score of 20 oz is +923 Of course, common sense might suggest to us that the likelihood of finding an orange weighing more than 20 oz is extremely small! In a similar fashion, we would conclude that practically all oranges weigh more than 2 oz (for which the z-score is 440), or in other words, the probability of selecting an orange weighing more than 2 oz is one (1) It is sometimes desirable to obtain percentiles and percentile ranks from a density curve We define the pth percentile to be the value with p% of the total area below it and (100 p)% of the total area above it; also, the percentage of area below a given value x is defined to be the percentile rank of x To illustrate the use of Table B2 in obtaining percentiles and percentile ranks with a normally distributed population, we shall return to the population of orange weights having a normal distribution with mean µ = 781 oz and standard deviation σ = oz Finding the percentile rank for a value x from a normally distributed population simply amounts to calculating the area under the normal density curve which lies below x For instance, earlier we found the percentage of oranges weighing less than 698 oz to be 2643%; we can then say that the percentile rank of an orange weighing 698 oz is 2643 or 26 Often, the percent sign (%) is omitted when stating a percentile rank, 110
5 since a percentile rank is understood to be a percentage As another illustration, recall that we have previously found that the percentage of oranges weighing less than 1066 oz is 9846%; we can then say that the percentile rank of an orange weighing 1066 oz must be 9846 or 98 Finding the pth percentile of a normally distributed population amounts to finding the value below which lies p percent of the area under the normal density curve Since a normal curve is symmetric, the 50th percentile (ie, the median) is equal to the mean µ We determine other percentiles by using Table A2 to find the z-score of the desired percentile and converting this z-score to a raw score To illustrate, we shall obtain the 90th percentile of the orange weights and the 30th percentile of the orange weights To find the 90th percentile, we first realize that the 90th percentile must be greater than the median (which is the mean µ), implying that the z-score of the 90th percentile must be positive We then use Table A2 to determine the desired z-score, that is, the z-score of the orange weight below which lie 90% of all the orange weights and above which lie 10% of all the orange weights Since the areas listed in Table A2 are the areas under the normal curve above a positive z-score, by searching for the area closest to 010 (which is 01003), we find that the desired z-score is +128 In Figure 16-8, label the z-score +128 on the horizontal axis; if you draw a vertical line through the graph at the location of the z-score +128, you should be able to see that roughly 90% of the area under the normal curve lies below this z-score, and 10% of the area under the normal curve lies above this z-score Recall that we use x = µ + zσ to convert a z-score to a raw score Convert the z-score +128 to a raw score to obtain the 90th percentile of the orange weights (You should find that the 90th percentile is approximately 950 oz) To find the 30th percentile of the orange weights, we must first realize that the 30th percentile must be smaller than the median (which is the mean µ), implying that the z-score of the 30th percentile must be negative We then use Table A2 to determine the desired z-score, that is, the z-score of the orange weight below which lie 30% of all the orange weights and above which lie 70% of all the orange weights The areas listed in Table A2 are the areas under the normal curve above a positive z-score, but since normal distributions are symmetric, the areas listed can also be treated as the areas under the normal curve below a negative z-score By searching for the area closest to 030 (which is 03015), we find that the desired z-score is 052 In Figure 16-9, label the z-score 052 on the horizontal axis; if you draw a vertical line through the graph at the location of the z-score 052, you should be able to see that roughly 30% of the area under the normal curve lies below this z-score, and 70% of the area under the normal curve lies above this z-score Use x = µ + zσ to convert the z-score 052 to a raw score to obtain the 30th percentile of the orange weights (You should find that the 30th percentile is approximately 712 oz) We have now illustrated the use of Table A2 in obtaining desired areas under the standard normal curve and in obtaining percentiles Now that we have thoroughly discussed normal distributions, we once again recall how we have previously observed that the sampling distribution of x with simple random samples of sufficiently large size n, takes on the properties of a normal distribution; consequently, the sampling distribution of x can be approximated using a normal distribution, and we shall very shortly begin making heavy use of this fact 111
6 Self-Test Problem 16-1 Suppose the right-hand grip strength for men between the ages of 20 and 40 is normally distributed with mean 863 lbs and standard deviation 78 lbs Draw a sketch illustrating the probability that one randomly selected male between the ages of 20 and 40 will have a right-hand grip strength (a) over 90 lbs, and find this probability; (b) under 96 lbs, and find this probability; (c) over 70 lbs, and find this probability; (d) under 75 lbs, and find this probability; (e) between 85 and 100 lbs, and find this probability; (f) between 88 and 95 lbs, and find this probability; (g) between 75 and 82 lbs, and find this probability (h) Draw a sketch illustrating the probability that the right-hand grip strength for one randomly selected male is within 4 lbs of the population mean, and find this probability (i) Find the percentile rank for a male whose right-hand grip strength is 90 lbs (j) Find the quartiles for the distribution of right-hand grip strengths Answers to Self-Test Problems 16-1 (a) or 3192% (b) or 8925% (c) or 9817% (d) or 735% (e) or 5283% (f) or 2815% (g) or 2177% (h) or 3900% (i) 6808 or 68 (j) The quartiles are approximately 811, 863, and 915 lbs Summary A normal curve described entirely in terms of z-scores is called a standard normal curve Tables of standard normal probabilities (eg, Table A2) provide a detailed description of a normal density curve in terms of z-scores Using such a table together with the fact that a normal distribution is symmetric, we can obtain areas under a normal density curve, which can be interpreted as relative frequencies or as probabilities, and we can obtain percentile ranks and percentiles Statistical software packages, spreadsheets, programmable calculators, etc are often able to supply the same information as a table of standard normal probabilities Since the sampling distribution of x with simple random samples of sufficiently large size n, takes on the properties of a normal distribution, the sampling distribution of x can be approximated using a normal distribution 112
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Week 3&4: Z tables and the Sampling Distribution of X
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Sampling Distributions and the Central Limit Theorem
135 Part 2 / Basic Tools of Research: Sampling, Measurement, Distributions, and Descriptive Statistics Chapter 10 Sampling Distributions and the Central Limit Theorem In the previous chapter we explained
STT315 Chapter 4 Random Variables & Probability Distributions KM. Chapter 4.5, 6, 8 Probability Distributions for Continuous Random Variables
Chapter 4.5, 6, 8 Probability Distributions for Continuous Random Variables Discrete vs. continuous random variables Examples of continuous distributions o Uniform o Exponential o Normal Recall: A random
Math 1011 Homework Set 2
Math 1011 Homework Set 2 Due February 12, 2014 1. Suppose we have two lists: (i) 1, 3, 5, 7, 9, 11; and (ii) 1001, 1003, 1005, 1007, 1009, 1011. (a) Find the average and standard deviation for each of
Chapter 6 Random Variables
Chapter 6 Random Variables Day 1: 6.1 Discrete Random Variables Read 340-344 What is a random variable? Give some examples. A numerical variable that describes the outcomes of a chance process. Examples:
What Does the Normal Distribution Sound Like?
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Chapter 4. Probability and Probability Distributions
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Inferential Statistics
Inferential Statistics Sampling and the normal distribution Z-scores Confidence levels and intervals Hypothesis testing Commonly used statistical methods Inferential Statistics Descriptive statistics are
7. Normal Distributions
7. Normal Distributions A. Introduction B. History C. Areas of Normal Distributions D. Standard Normal E. Exercises Most of the statistical analyses presented in this book are based on the bell-shaped
Def: The standard normal distribution is a normal probability distribution that has a mean of 0 and a standard deviation of 1.
Lecture 6: Chapter 6: Normal Probability Distributions A normal distribution is a continuous probability distribution for a random variable x. The graph of a normal distribution is called the normal curve.
Chapter 5. Section 5.1: Central Tendency. Mode: the number or numbers that occur most often. Median: the number at the midpoint of a ranked data.
Chapter 5 Section 5.1: Central Tendency Mode: the number or numbers that occur most often. Median: the number at the midpoint of a ranked data. Example 1: The test scores for a test were: 78, 81, 82, 76,
TImath.com. Statistics. Areas in Intervals
Areas in Intervals ID: 9472 TImath.com Time required 30 minutes Activity Overview In this activity, students use several methods to determine the probability of a given normally distributed value being
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Unit 7: Normal Curves Summary of Video Histograms of completely unrelated data often exhibit similar shapes. To focus on the overall shape of a distribution and to avoid being distracted by the irregularities
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Cents and the Central Limit Theorem Overview of Lesson GAISE Components Common Core State Standards for Mathematical Practice
Cents and the Central Limit Theorem Overview of Lesson In this lesson, students conduct a hands-on demonstration of the Central Limit Theorem. They construct a distribution of a population and then construct |
Integer Operations
Introduction
In this smore you will learn how to add, subtract, multiply, and divide integers.
1.- First you need to see the signs.
2.- If both signs are the same then you add and stay with the same sign.
3.- If they are different then you will subtract and stay with the sign of the highest number.
EXAMPLES:
-2+(-6)= -8
-4+6= 2
-6+(-12)= -18
Rules for subtracting
First you see the signs.
Then if you have like signs you add the numbers and stay with the sign.
Then if you have different signs you subtract and stay with the highest number sign.
Remember to change all the numbers to addition.
EXAMPLES:
-3-1= -4
-3-(-2)= -5
-5-4= -9
Rules for multiplication
First look at the signs.
Then if they are like signs they stay positive.
Then if they are unlike signs the answer would be negative.
EXAMPLES:
-3*-2= 6
-4*-8= 32
-9*3= -27
Rules for division
First look at the signs.
Then if they are like signs they stay positive.
Then if they are unlike signs the answer would be negative.
EXAMPLES:
-12/-3= 4
-45/5= -9
-25/-5= 5 |
We've updated our
TEXT
# Factoring Special Cases
### Learning Outcomes
• Factor a perfect square trinomial.
• Factor a difference of squares.
• Factor a sum and difference of cubes.
• Factor an expression with negative or fractional exponents.
## Factoring a Perfect Square Trinomial
A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.
$\begin{array}{ccc}\hfill {a}^{2}+2ab+{b}^{2}& =& {\left(a+b\right)}^{2}\hfill \\ & \text{and}& \\ \hfill {a}^{2}-2ab+{b}^{2}& =& {\left(a-b\right)}^{2}\hfill \end{array}$
$\\$
We can use this equation to factor any perfect square trinomial.
### A General Note: Perfect Square Trinomials
A perfect square trinomial can be written as the square of a binomial:
${a}^{2}+2ab+{b}^{2}={\left(a+b\right)}^{2}$
### How To: Given a perfect square trinomial, factor it into the square of a binomial
1. Confirm that the first and last term are perfect squares.
2. Confirm that the middle term is twice the product of $ab$.
3. Write the factored form as ${\left(a+b\right)}^{2}$.
### Example: Factoring a Perfect Square Trinomial
Factor $25{x}^{2}+20x+4$.
Answer: Notice that $25{x}^{2}$ and $4$ are perfect squares because $25{x}^{2}={\left(5x\right)}^{2}$ and $4={2}^{2}$. Then check to see if the middle term is twice the product of $5x$ and $2$. The middle term is, indeed, twice the product: $2\left(5x\right)\left(2\right)=20x$. Therefore, the trinomial is a perfect square trinomial and can be written as ${\left(5x+2\right)}^{2}$.
### Try It
Factor $49{x}^{2}-14x+1$.
Answer: ${\left(7x - 1\right)}^{2}$
### Q & A
Is there a formula to factor the sum of squares? No. A sum of squares cannot be factored.
Watch this video to see another example of how to factor a difference of squares. https://youtu.be/Li9IBp5HrFA
## Factoring the Sum and Difference of Cubes
Now we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.
${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$
[latex-display]\\[/latex-display] Similarly, the sum of cubes can be factored into a binomial and a trinomial but with different signs.
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$
[latex-display]\\[/latex-display] We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following example.
${x}^{3}-{2}^{3}=\left(x - 2\right)\left({x}^{2}+2x+4\right)$
The sign of the first 2 is the same as the sign between ${x}^{3}-{2}^{3}$. The sign of the $2x$ term is opposite the sign between ${x}^{3}-{2}^{3}$. And the sign of the last term, 4, is always positive.
### A General Note: Sum and Difference of Cubes
We can factor the sum of two cubes as
${a}^{3}+{b}^{3}=\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$
We can factor the difference of two cubes as
${a}^{3}-{b}^{3}=\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$
### How To: Given a sum of cubes or difference of cubes, factor it
1. Confirm that the first and last term are cubes, ${a}^{3}+{b}^{3}$ or ${a}^{3}-{b}^{3}$.
2. For a sum of cubes, write the factored form as $\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)$. For a difference of cubes, write the factored form as $\left(a-b\right)\left({a}^{2}+ab+{b}^{2}\right)$.
### Example: Factoring a Sum of Cubes
Factor ${x}^{3}+512$.
Answer: Notice that ${x}^{3}$ and $512$ are cubes because ${8}^{3}=512$. Rewrite the sum of cubes as $\left(x+8\right)\left({x}^{2}-8x+64\right)$.
#### Analysis of the Solution
After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.
### Try It
Factor the sum of cubes $216{a}^{3}+{b}^{3}$.
Answer: $\left(6a+b\right)\left(36{a}^{2}-6ab+{b}^{2}\right)$
### Example: Factoring a Difference of Cubes
Factor $8{x}^{3}-125$.
Answer: Notice that $8{x}^{3}$ and $125$ are cubes because $8{x}^{3}={\left(2x\right)}^{3}$ and $125={5}^{3}$. Write the difference of cubes as $\left(2x - 5\right)\left(4{x}^{2}+10x+25\right)$.
#### Analysis of the Solution
Just as with the sum of cubes, we will not be able to further factor the trinomial portion.
### Try It
Factor the difference of cubes: $1,000{x}^{3}-1$.
Answer: $\left(10x - 1\right)\left(100{x}^{2}+10x+1\right)$
The following video provides more examples of how to factor expressions with fractional exponents. https://youtu.be/R6BzjR2O4z8 |
# Algebraic expression mean
Introduction:
In mathematics, an algebraic expression is a finite combination of symbols that are well-formed according to the rules applicable in the context at hand. Symbols can designate values (constants), variables, operations, relations, or can constitute punctuation or other syntactic entities. The use of expressions can range from simple arithmetic operations. Let us see about algebraic expression mean and examples. (Source: Wikipedia)
Properties Algebraic expression mean:
Commutative property of algebraic expression
Addition: a + b="b" + a
For example: x2 +x= x + x 2
(2) 2 + 2 = 2+ (2) 2 (if we consider x="2)
4+2=2+4
6="6
Multiplication: (if a="5" ,b=3)
For example: a * b = b*a
5*3 = 3*5
15 = 15
The Distributive property of algebraic expression
Addition: a * (b + c) = a * b + a * c
Multiplication: (a + b) * c = a * c + b * c
Associative property of algebraic expression
Addition: (a + b)+ c = a + (b + c)
Multiplication: (a * b) * c = a * (b * c)
The reciprocal of a is `1/a` : a *`(1/a)` = 1
The additive inverse of a is –a: a + (-a) = 0
Example problems for algebraic expression mean:
Example 1: Check that the following expression is satisfy the Commutative property of algebraic expression: x2 +x= x + x 2.
Solution:
(2) 2+2=2 + (2) 2 (if we consider x="2)
4+2=2+4
6="6
Therefore the given expression has the commutative property of addition.
Example 2: Solve the expression: |-4x + 4| + 5 = 5
Solution:
Given |-2x + 2| + 5 = 5
|-2x + 2| = 0 (subtract 5 on both sides)
- 2x + 2 = 0
-2x =-2
x = 1
Example 3: Solve for n: 11m – 11n = 5
Solution:
Given 11m – 11n = 5
-11n = 5 – 11m (subtract 11m on both sides)
- n = `(5 -11m) / 11` (divide 11 on both sides)
n = m - `5/11` (multiply by - sign on both sides)
In this section we have seen about the basic concepts and meaning of algebraic expression. |
# The First Steps in Learning Trigonometry - Trigonometric Functions
Updated on May 31, 2010
## What is Trigonometry?
Trigonometry is the study of triangles, particularly right triangles. It deals with relationships between the sides and angles of the triangles. These relationships are expressed by the functions of sine, cosine and tangent. These functions are also used in describing the motion of waves.
In this article, we will be discussing the basic uses of the trigonometric functions sin, cos and tan.
An introduction of right triangles is found in the article Pythagorean Theorem. Go check it out if you need a bit of a refresher.
To begin with, let us define what sin, cos and tan mean. These three functions are simply ratios of the sides of triangles that help us relate to an angle in the triangle. We'll be using the angle A to compare these.
sin(A) = a / c (sin is the opposite side divided by the hypotenuse)
cos(A) = b / c (cos is the adjacent side divided by the hypotenuse)
tan(A) = a / b (tan is the opposite side divided by the adjacent)
An easy way to remember this is SOHCAHTOA:
SOH, sin = opposite / hypotenuse
CAH, cos = adjacent / hypotenuse
TOA, tan = opposite / adjacent
Note that tan is not an entirely independent definition, it's just used to simplify our math. If you were to divide sin by cos, you would get tan.
sin(A) / cos(A) = tan(A)
(a / c) / (b / c) = tan(A)
a / b = tan(A)
Based on Diagram 1.
In most cases, the notation for the angle is that of the Greek letter theta Θ.
## Solving the Trig Problems
The main trick with trigonometry problems:
If you are given 2 pieces of data in a triangle (i.e. if you're given an angle and a side length or two side lengths) you can solve the entire triangle with the trig ratios and Pythagorean Theorem.
The slight exception to this is if you're given two angles - this would basically be giving you one piece of data since if you have one angle you can find out the other (the two complementary angles add to 90 degrees). You can still find the ratios of the side lengths using the angles that you are given and calculating the trigonometric ratios, but they could be any set of numbers as long as they make up that ratio.
Usually, you'll be given either two side lengths or an angle and a side length though, and you'll be asked to solve the rest of the triangle.
## Example 1
In the first example, you are given two side lengths of 8 and 15, respectively. We are asked to solve for the angle Θ.What function are we supposed to use to solve for Θ?
With respect to Θ, we are given the opposite and the adjacent sides. The hypotenuse is unknown. The easiest way to tackle this is to use tan (opposite / adjacent). We could also use the Pythagorean Theorem to solve for the hypotenuse and then use sin or cos to solve Θ, but why make it harder on yourself?
Now in this we don't use the tan function, we use the inverse tan function (on calculators, it is denoted by tan^-1). Since tanΘ is used to calculate the ratio of opposite / adjacent using Θ, inverse tan is used to calculate Θ using the ratio of opposite / adjacent.
inverse tan (8/15) = 28.07 degrees
It's as simple as that. Make sure your calculator is not in radians - we will talk about that in a future article.
Notice that inverse cos (15/17) = inverse sin (8/17) = inverse tan (8/15) = 28.07 degrees.
## Example 2
In this example, we are given a side length of 12 and Θ, which is equal to 30 degrees. We are asked to find the sides a and b. We know that b is the hypotenuse, a is adjacent to Θ, and the side length of 12 is opposite of the angle Θ.
That's a good indication that we can use the sin function:
sin (30) = 12 / b
Rearrange the equation:
b = 12 / sin (30)
b = 24
Now that we have b, we can solve for a using Pythagoras, or we can use the angle Θ again, this time using the tan ratio.
tan (30) = 12 / a
a = 12 / tan (30)
a = 20.8
You now have the basics to solve trigonometric ratios consisting of sin, cos and tan.
Now go get some practice!
## Popular
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• ### Financial Literacy for Kids Games & Websites
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• aishwarya
4 years ago
Thank you very much it helped me a lot .it is much easier for me now.
• omar21
5 years ago
Your illustration is really good , i loved it ,thank you for your effort.
I am a great fan of tri.
• AUTHOR
mrpopo
• anonymous
5 years ago
Thank you for this!
I need to discuss origami and trigonometry within my research paper (on origami and math) due tomorrow, and idk what trig is (i'm a young 8th grader who is in Algebra 1 right now)
This explained everything to me
• anonymous
5 years ago
0
• manisha
6 years ago
plesase tell me some tricks for solvin questions related to trigonometric identities
• Shamela
6 years ago
Dear mrpopo,
I found Trigonometry easy and refreshing to learn from your hub.
You have given simple steps to remember Sine,Cosine and tangent.
• AUTHOR
mrpopo
Hey mel, I know these look hard at first, but keep practicing and you'll get them!
• mel
6 years ago
i do not get trigonemetric function and they are hard! thanks but this could not help me either. i do not know what i should do!
• AUTHOR
mrpopo
Heheh, I hear ya wilbury. We still gotta know some of these though - or at least how to use the calculator to solve them...
Thanks for the comment!
• wilbury4
8 years ago from England I think?
Memories of school and college.... Before calculators...
• AUTHOR
mrpopo
Haha, thanks for the comment Lamme! Math can be fun, and as easy as pi!
(yeah lame pun, I know)
• Lamme
8 years ago
Great hub. I love it! I could do math for fun LOL
• AUTHOR
mrpopo
Thanks TGS! I hope it helps :)
• TheGlassSpider
8 years ago from On The Web
Whoa!! This is very impressive. Math has never been my strongest suit and trig was always difficult for me, but you've laid this out in a very easy-to-understand, non-threatening manner. Thank you!!
• AUTHOR
mrpopo |
# Video: AQA GCSE Mathematics Foundation Tier Pack 2 • Paper 2 • Question 13
A shop sells two brands of cola, both of which are sold in multipacks. A multipack of brand A cola costs £2.19 and contains 8 cans which are 500 millilitres each. A multipack of brand B cola costs £ 1.86 and contains 6 cans, which are 330 ml each. Which brand is the better value? You must show your working.
03:59
### Video Transcript
A shop sells two brands of cola, both of which are sold in multipacks. A multipack of brand A cola costs two pounds 19 and contains eight cans, which are 500 millilitres each. A multipack of brand B cola costs one pound 86 and contains six cans, which are 330 millilitres each. Which brand is the better value? You must show your working.
What does the question mean by better value? It means the lowest cost per unit. And so we need to consider how much cola each multipack is selling us. To do that, we could calculate the price per millilitre. We already know how many millilitres each can is. However, since we’re buying so many cans, we’re better off calculating the price per litre.
Our first step is to then find out how much cola in litres do we get when we buy brand A. Eight cans times 500 millilitres each tells us that brand A gives us 4000 millilitres of cola. We know that 1000 millilitres equals one litre. If 1000 millilitres equals one litre, 4000 millilitres equals four litres. We divide 4000 by 1000, and we get four. For brand A, you pay two pound 90 and you get four litres of cola.
We follow the same procedure for brand B. There are six cans. And each of them contain 330 millilitres of cola. When we multiply six times 330, we get 1980. To convert 1980 millilitres to litres, we’ll need to divide by 1000. And that means we’ll move our digits three places to the right, which would give us 1.98 litres. You could also enter 1980 divided by 1000 into the calculator. Either way, you find that brand B is 1.98 litres of cola for one pound 86.
To calculate the price per litre, we need to divide. We’ll take the cost of the multipack and divide that by the number of litres you get in that multipack. For brand A, 2.90 per, divided, by four litres. Using our calculator to do this division, we get the result 0.725. Since we’re considering cost, money, we need to round to two decimal places. 0.725 rounds to 0.73. For brand A, you pay 73 pence per litre of cola.
Now for brand B, we’ll take one pound 86 and divide by 1.98 litres. When we do this division in our calculator, we get 0.93 repeating. If we write that out, it could look like this. And again, we’ll need to round to two decimal places. 0.93 repeating rounds to 0.94. For brand B cola, we’re paying 94 pence per litre.
Remember that our question is which brand is the better value. So we need to do a comparison of brand A and brand B. 0.73 is less than 0.94. Brand A cost less per litre than brand B. Therefore, brand A is the better value.
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# Subsets / Power-set Problem
In the Subsets problem, we find the power set of a given input using bitwise operations. In general, if we have n elements then the subsets are 2^n subsets.
Subsets problem, we need to write a program that finds all possible subsets (the power set) of a given input. The solution set must not contain duplicate subsets.
Subsets problem, we find the power set of given input without duplicates. Leetcode, one of the largest tech communities with hundreds of thousands of active users to participate and solve coding problems, listed this problem in their curriculum – the Leetcode Subsets Problem.
### Introduction
Let us discuss the subsets of a given input. This is one of the most popular questions asked in coding interviews.
Companies that have asked this in their coding interview are Apple, Microsoft, Amazon, Facebook, and many more.
### Problem Statement
We need to write a program that finds all possible input subsets ( the power set). The solution set must not contain duplicate subsets.
Example 01:
``````Input: [1, 2, 3]
Output: [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]``````
Example 02:
``````Input: [100]
Output: [[], [100]]``````
Explanation: The subsets of any given input are equal to its power set.
if, input n = 3, then, powerset => 2^n = 2^3 = 8.
Assume input has a length greater than or equal to 1.
Hint: Use the left-shift operator to achieve this.
### Thought Process
In this program, we find the power set of a given input using bitwise operations.
In general, if we have n elements then the subsets are 2^n subsets.
So for every possible case of having at least two elements, we can see that an element is present and not present in the subsets
Think of a solution that is iterative, uses bitwise operators, and generates the powerset.
Here is how we generate each subset using the outer-loop variable counter. Here is a table indicating how the value gets generated based on the counter input.
## HTML Table
Counter (in decimal) Counter (in binary) Subset
0 000 []
1 001 [1]
2 010 [2]
3 011 [1, 2]
4 100 [3]
5 101 [1, 3]
6 110 [2, 3]
7 111 [1, 2, 3]
### Algorithm
We need to consider a counter variable that starts from 0 to 2^n - 1.
For every value, we are considering the binary representation and here we use the set bits in the binary representation to generate corresponding subsets.
• If all set bits are 0, then the corresponding subset is empty [].
• If the last bit is 1, then we put 1 in the subset as [1].
Steps:
We use two loops here, the outer-loop starts from 0 to 2^n - 1, and the inner loop continues to input array length n.
In the inner loop, we conditionally check (counter & (1 << j)) != 0), if yes, then we print the corresponding element from an array.
### Solution Code
``````import java.util.ArrayList;
import java.util.List;
public class Subsets {
public static void main(String[] args) {
int[] input = {1, 2, 3};
System.out.println(subsets(input));
}
public static List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> result = new ArrayList<>();
int n = nums.length;
int powSize = (int) Math.pow(2, n);
for (int i = 0; i < powSize; i++) {
List<Integer> val = new ArrayList<>();
for (int j = 0; j < n; j++) {
if ((i & (1 << j)) != 0) {
}
}
}
return result;
}
}``````
### Complexity Analysis
Time Complexity:`O(n*2^n)`, Time complexity is `n` times the powerset.
Space Complexity:`O(2^n)`, We are storing `2^n` subset elements in an array. So the extra space is directly proportional to `O(2^n)`. |
# Relative Minima and Maxima
Last Updated : 28 Mar, 2024
Relative maxima and minima are the points defined in any function such that at these points the value of the function is either maximum or minimum in their neighborhood. Relative maxima and minima depend on their neighborhood point and are calculated accordingly. We find the relative maxima and minima of any function by using the first derivative test and the second derivative test.
In this article, we have covered Relative Maxima and Minima, methods to find relative maxima and minima, various examples, and others in detail. Before starting with Relative Maxima and Minima, first, learn in brief about Maxima and Minima.
## What is Maxima and Minima?
Maxima and Minima are called critical points of the function. A maxima is a high point and a minima is a low point in any function. In a function, more than one maximum and minimum point can exist. The points at which the function attains the highest and lowest values are called Maxima and Minima.
## What Is Relative Maxima and Minima?
Relative maxima and minima are the points at which the function gives the maximum and minimum values respectively in their neighborhood. Relative maxima and minima of any function are easily found by using the first derivatives and second derivative test respectively. The graph added below the relative maxima and minima of a function in its neighborhood.
## How to Find Points of Maxima and Minima? (First Derivative Test)
In any smoothly changing function, the points where the function flattens out, give us either minima or maxima. Now, this statement gives rise to two questions.Â
1. How to recognize the points at which function flattens out?
2. Suppose we got a point at which function flattens i.eofcritical point. How to tell whether it’s a minimum or a maximum?
To answer the first question let’s look at the slope of the function. The points where the function flattens out have zero slopes. We know that the derivative is nothing but the slope of the function at a particular point. So, we try to find the points where the derivative is zero. Thus, this test is also called the First Derivative Test. Then we equate the differential equation with zero to get the critical points as,
f'(x) = 0
The solution to this equation gives us the position of the critical points. These critical points tells us that these are the points where the tangent to the curve is parallel to x-axis but still we don’t know whether they are points o maxima or minima for that Second Derivative test is used.
## Recognizing Maxima and Minima
As shown in the figure below, it can be seen that if the sign of the derivative is positive before the critical point and negative after the critical point, it is a maximum. Similarly, if it is negative before the critical point and positive after the critical point. It is a minimum. Maxima and minima can also be recognized by the second derivative test.
Notice the figure carefully, and see that the slope of the curve is continuously decreasing, and then it becomes zero and goes further towards a negative value.Â
### Second Derivative Test
When a function’s slope is zero at x, then the second derivative f”(x) at that point is used to tell whether at that point we have maxima or minima.
1. f”(x) < 0, if it is a maxima.
2. f”(x) > 0, if it is a minima.
Note: If the second derivative is zero at the critical points, then the test fails.Â
## Steps to Find Relative Maxima and Minima
To find the relative maxima and minima of a function follow the steps added below, suppose we are given a function,
f(x) = x2 – 4
and we have to find its maximum and minimum value in the interval [-2, 2] then,
### Step 1: Differentiate the Function
Given,
• f(x) = x2 – 4
f'(x) = 2x
### Step 2: Find out Critical Points
Putting f'(x) = 0 gives us the position of the critical points for the function. For the given function,
2x = 0
⇒ x = 0
Thus, x = 0 is a critical point for this function. Now we need to find out whether it is a minimum or a maximum.Â
### Step 3: Second Derivative Test
We use the second derivative test mentioned above to find out whether the given critical point is minima or maxima. In the above case,Â
f”(x) = 2Â
Notice that, f”(x) > 0. Thus, it must be a minimum.
### Step 4: Value at Critical Points
Find the value of the function at critical points get the smallest value(or the minima)
• f(0) = (0)2 – 4 = -4
Thus, the minimum value of f(x) is at x = 0 and the minimum value is -4.
## Applications of Relative Maxima and Minima
Relative Maxima and Minima has various applications. it is used for various purposes such as,
• This concept is used to determine the maximum and minum value of a stock at particular points by equating it to a function that represent the trajectory of the stocks.
• This concept are used in electronic circuits to manage the voltage and current in the system.
• Relative maxima and minima is also used in astrophyscis to find the maximum and minimum trajectory of an abject, etc.
## Examples on Relative Maxima and Minima
Example 1: Find all the critical points of the following function,Â
f(x) = x3 – 6x2 +11x – 6
Solution:
Given Function,
• f(x) = x3 -6x2 +11x – 6
f'(x) = 3x2 -12x +11 = 0Â
⇒ 3x2 -12x +11 = 0Â
x =Â [Tex]\frac{12 \pm \sqrt{12^2 – 4(3)(11)}}{6} [/Tex]
x =Â [Tex]\frac{12 \pm \sqrt{144 – 132}}{6} [/Tex]
x =Â [Tex]\frac{12 \pm \sqrt{12}}{6} [/Tex]
x =Â [Tex]\frac{12 \pm 2\sqrt{3}}{6} [/Tex]
Thus, the critical points of the given function are,
x =Â [Tex]2 \pm \frac{1}{\sqrt{3}} [/Tex]
Example 2: Find the relative minima of the function, f(x) = x2 – 4x + 4.
Solution:
Given function,
• Â x2 – 4x + 4 = 0
f'(x) = 2x – 4 Â = 0
⇒x = 2
Using Second Derivative test,
f”(x) = x
at x = 2
f”(x) = 2 > 0
Thus, at x = 2 is a minima and the minimum value of function is,
f(2) = (2)2 -4(2) + 4 = 4 – 8 + 4 = 0 (Minimum Value)
Example 3: Find the relative maxima and minima of, f(x) =x2 -7x + 12.
Solution:Â
Given function,
• x2 -7x + 12 =0
For finding the critical point of the function,
f'(x) = 2x – 7 = 0
⇒ x = 7/2
Using second derivative test,
f”(x) = x
at x = 7/2
f”(7/2) = 7/2 > 0 (Point of Minima)
Thus, the value of function is minimum at point x = 7/2 and its minimum value is,
f(7/2) = (7/2)2 – 7(7/2) + 12 = -1/4
Example 4: Find relative maxima and minima of, f(x) =ex + e-x.
Solution:Â
Given function,
• f(x) = ex + e-x
Finding the critical points,
f'(x) = ex – e-x = 0
⇒ ex = e-x
Taking log both sides,Â
⇒ x = -xÂ
⇒x = 0Â
Thus, x = 0 is a critical point.
Using Second Derivative Test,
f”(x) = ex + e-x
At, x = 0
f”(2) = 2 > 0Â (Point of Minima)
Finding minimum value of the function,
f(0) = e(0) + e(0) = 1 + 1 = 2
## FAQs on Relative Maxima and Minima
### 1. What is Relative Maxima and Minima of Fucntion?
Relative maxima or minima is the maximum or minmun value of the function for given set of points.
### 2. What is Difference Between Relative Maxima and Absolute Maxima?
Relative maxima or minima is the maximum or minmun value of the function for given set of points whereas, absolute maxima or minima is the maximum or minimum value of the function for the entire domian.
### 3. How to Find Relative Maxima and Minima of Fucntion?
Relative maxima and mimima of the function is found using the steps added below,
Step 1: Find the derivative of the function.
Step 2: Equate the derivative value of the function with zero and get the critical value of the function.
Step 3: Use the second derivative test to check wether we have maxima or minima at critical points.
Step 4: Find the maximum or minimum value of function at critical points.
### 4. What are the Uses of Relative Maxima and Minima?
Relative Maxima and Minima is used to find the maximum or minimum value of the function for the given set of points.
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# Coordinate Geometry: Distance Between Two Points
Do you know how to find the distance between two points? Well, if you want to know, our video below should help you understand this concept.
# Transcript: Distance Between Two Points
Sometimes the test gives two points and asks us for the distance between them. Other times we will need to find distance in the x-y plane in the course of answering a geometric question. Something like is the triangle isosceles? Something like that or is the figure a square?
## Big Idea Number One
Big Idea number 1, in the x-y plane, horizontal and vertical distances are very easy to find.
If two points are on the same horizontal line, all we have to do is subtract the x-coordinates. We subtract (bigger) minus (smaller), because distance is always positive. If two points are on the same vertical line, we subtract the y-coordinates, incredibly easy. We can’t do anything this simple if the two points are separated by a slanted line.
Let’s think about two points diagonally separated. So for example, suppose we have this. So we can’t just do any kind of subtraction but think about it this way. If we draw the slope triangle we were using to find the slope, remember the slope triangle suddenly things become much clearer.
## Distance Between Two Points: Slope Triangle
Okay, so here the same points again, but now we draw a slope triangle.
With very easy to find the rise and the run on that sub-triangle. The rise is 3, we can see there’s a vertical leg has a distance of 3, a length of 3. The horizontal leg, the run, has a length of 4. So if the horizontal leg is 4 and the vertical leg is 3, this is just a 3-4-5 triangle, and so the distance between those two points, the hypotenuse has to be 5.
Now, it won’t always be the case that the three sides of a triangle are nice, neat Pythagorean triplets, although this is quite common on the test. Nevertheless, it will always be the case that the slope triangle is a right triangle. By definition, the slope triangle is a right triangle. The horizontal and vertical legs have lengths that are easy to find, and the length of the diagonal will always be the hypotenuse of the slope triangle.
## Big Idea Number Two
Image by tanewpix
We can always find that length by the Pythagorean theorem, so we’re using the Pythagorean theorem to find the distance between any two points. That’s Big Idea number 2.
## Practice Problem
Here is a practice problem, find the distance between those two points. Pause the video and then we’ll talk about this.
Okay, so we find the x-leg, the run between them.
We get a run of 8 and a rise of 4, a y-leg of 4. You could do the Pythagorean theorem with legs of 4 and 8 but it’s easier if we scale down first by the greatest common factor, a scale factor of 4. We scale down to much smaller triangle with legs of 1 and 2. Well that’s very easy to solve. If we have legs of 1 and 2, we square those, we get c squared equals 5, c equals the square root of 5.
Now we’ll scale back up by a scale factor of 4, And we get a hypotenuse on the larger slope triangle of 4 root 5. And that is the distance between those two points.
## Don’t Use the Distance Formula
So notice that we could use proportional thinking to really simplify that calculation. Now there is a formula known as the distance formula.
I absolutely refuse to teach it, neither even if you know it, I encourage you not to use it. I think sticking to this formula is actually ultimately a more time consuming way that impedes understanding. And you understand much more deeply if you think about it the way I’m showing in this video using the Pythagorean theorem.
So if you use the Pythagorean theorem to find the hypotenuse of a slope triangle, you’re thinking visually. And you can also use the proportional thinking to simplify your calculation, of course, using the distance formula would not allow you to do that. So there are a variety of reasons that using the formula is an exceptionally bad idea, and it’s much better to think about this situation carefully.
When you do think visually about the x-y plane, you understand much more deeply than when you simply plug into a memorized formula. Now we can talk about circles in the x-y plane. Of course, according to its fundamental definition, a circle is the set of all points equidistant from a fixed center. Everything we said about distance comes into play with circles in the x-y plane.
## Practice Problem
Okay, a circle in the x-y plane has a center of 6, 3 and a radius of 5. Find the two x-intercepts. Think about the two radii that go from the center to these two x-intercepts. Each radius has a length of 5, and it’s the hypotenuse of a slope triangle. Also the point 6, 3 is clearly 3 units above the x-axis.
So let’s think about all this visually. We have those two little triangles there. Each one has hypotenuse of 5 and it’s a distance of 3 above the x-axis. So that vertical length is 3, but, of course, what we have there are two 3-4-5 triangles. So the horizontal legs each have a length of 4.
And so starting at 6, we’ve got 4 to the right and 4 to the left. We get two x-intercepts of 2, 0 and 10, 0, and those are the x-intercepts. If the circle is centered at the origin, then the slope triangle of each radius has a horizontal leg of the opposite value of x and a vertical leg of the opposite value of y. This is true regardless of what x and y are. That’s the equation for a circle with center (0, 0) and radius r must be x squared + y squared = r squared.
So notice that we are really just doing the Pythagorean theorem to find the equation of a circle. So for horizontal and vertical distances in the x-y plane we just use subtraction. Very easy.
For slanted distances we draw or imagine the slope triangle and use the Pythagorean theorem, the distance between the two points is the hypotenuse of the slope triangle.
## In Closing
Remember Pythagorean triplets, remember to use scale factors to simplify your calculations, all those things we talked about back in the geometry lessons. For a circle in the x-y plane, each slanted radius forms a slope triangle. And all these slope triangles have equal hypotenuses, because they hypotenuse is the radius. The equation of a circle with radius r centered at the origin is x squared + y squared = r squared.
## Author
• Mike served as a GMAT Expert at Magoosh, helping create hundreds of lesson videos and practice questions to help guide GMAT students to success. He was also featured as "member of the month" for over two years at GMAT Club. Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike's Youtube video explanations and resources like What is a Good GMAT Score? and the GMAT Diagnostic Test. |
We included HMH Into Math Grade 7 Answer Key PDF Module 6 Review to make students experts in learning maths.
Vocabulary
Associative Property of
Multiplication
Inverse Property of
Multiplication
Distributive Property
Order of Operations
Choose the correct term from the Vocabulary box.
Question 1.
Choose the correct term from the Vocabulary box. To find the value of 7 + $$\frac{1}{3}$$ • 6, you first find the product of $$\frac{1}{3}$$ and 6. This is justified by the
This is justified by the Associative Property.
the product of $$\frac{1}{3}$$ and 6.
$$\frac{1}{3}$$ × 6
= 6 ÷ 3
= 2
The value of 7 + $$\frac{1}{3}$$ • 6
= 7 + 2
= 9
Concepts and Skills
Question 2.
A group of hikers starts at an elevation of 2.53 kilometers. When they stop for lunch, their elevation has increased by 1.24 kilometers. When they stop to camp, their elevation has decreased by 0.53 kilometer compared to their lunch stop. Which expressions represent the elevation, in kilometers, of the group’s campsite? Select all that apply.
A. 2.53 – (1.24 – 0.53)
B. 2.53 – (1.24 + 0.53)
C. (2.53 – 0.53) + 1.24
D. 2.53 + (1.24 – 0.53)
E. 2.53 + 1.24 – (-0.53)
F. 1.24 + [2.53 + (-0.53)]
2.53 + (1.24 – 0.53) and 1.24 + [2.53 + (-0.53)
Explanation:
The expressions represent the elevation, in kilometers, of the group’s campsite is 2.53 + (1.24 – 0.53) and 1.24 + [2.53 + (-0.53).
Question 3.
The total change in the water level in a lake over a period of 5 weeks was -5$$\frac{1}{4}$$ inches. Complete the table to show possible changes in the water level in week 2 and week 5.
Question 4.
Use Tools Tessa and Marisol run in a 100-meter race. Tessa runs at an average speed of 8.1 meters per second. Five seconds after the race begins, she is 3.5 meters ahead of Marisol. To the nearest hundredth of a second, how long will it take Marisol to run the entire race if she does not speed up or slow down? State what strategy and tool you will use to answer the question, explain your choice, and then find the answer.
It will take 13.51 seconds.
Explanation:
The speed of Tessa = 8.1 meters
Distance = speed × Time|
Time = 5 seconds
Distance covered by her = 5 × 8.1
= 40.5 meters
Since she is 3.5 m
Distance covered by Marisol in 5 seconds = 40.5 – 3.5 = 37 meters.
Marisol’s speed = 37 ÷ 5
= 7.4 meters per second
The amount of time it will take her = 100 ÷ 7.4 = 13.51 seconds.
Question 5.
When the air conditioner in a building is running, the temperature in the building changes at a rate of -1 °F every 18 minutes. At this rate, how many hours should it take to cool the building from a temperature of 85.5 °F to a temperature of 78 °F?
____ hours
135 min = 2 hours and 15 minutes.
Explanation:
85.5 F – 78 F = 7.5 F
And it takes -7.5 F to cool 85.5 F to 78 F
Therefore it takes 18 minutes to lower the temperature -1
18 × 7.5 = 135
135 minutes = 2 hours and 15 minutes.
Question 6.
Kyle plans to make 15 hamburger patties, each weighing $$\frac{1}{4}$$ pound, for a cookout. At the store, hamburger meat is priced at $3.66 per pound. If Kyle orders the exact amount of meat he needs, how much will it cost, rounded to the nearest cent?$_____
$13.725 Explanation: Given, Production = 15 units Each weight = $$\frac{1}{4}$$ pound Price =$3.66 per pound
Total pound = $$\frac{1}{4}$$ × 15
= 3.75 pounds
The total cost = 3. 75 × $3.66 =$13.725
Question 7.
The percent error of a measurement tells how close the measurement is to the actual value. The label on a mineral sample in a kit gave the weight of the sample as 60.5 ounces. The true weight was found to be 62.5 ounces.
A. To find the percent error, first find the absolute value of the difference between the weight on the label and the true weight.
2.0 ounces
Explanation:
Given,
The weight on the label is 60.5 ounces.
True weight = 62.5 ounces
Difference between True Weight – weight on the label
= 62.5 – 60.5
= 2.0 ounces.
B. Now express the difference from Part A as a percent of the actual value. This is the percent error.
The percent error is 3.2.
Explanation:
Estimated value = 2.0 ounces.
Actual value = 62.5 ounces
Percent error = (2 ÷ 62.5) × 100
= 0.032 × 100
3.2
Question 8.
A rectangular backyard has a length of 68 feet and a width of 40$$\frac{1}{2}$$-feet. The owner wants to plant 75% of the yard with grass seed. The planting rate for the seed is 1.5 pounds per 1,000 square feet of area. To the nearest tenth of a pound, how much grass seed will the owner need to plant?
___ pounds
3.1 pounds of seed.
Explanation:
A = l × w
= 68 × 40$$\frac{1}{2}$$
= 68 × $$\frac{81}{2}$$
= 2754 square feet
=2754 × $$\frac{75}{100}$$
= 2065.5 Square feet.
Planting rate for the seed = 1.5 pounds per 1000 square feet
= (2066.5 ÷ 1000) × 1.5
= 3.1 pounds
Question 9.
Which expressions are equivalent to –$$\frac{3}{4}$$(10 • 8)? Select Equivalent or Not Equivalent for each expression.
Question 10.
Joanna has $25.00 to spend at the zoo. Admission costs$12.25, and she plans to buy a combo meal for $8.92 at the snack bar for lunch. It costs$2.00 to buy a food pellet to feed the giraffes. Which expressions could Joanna use to estimate the number of pellets she can afford if she wants to be certain not to run out of money? Select all that apply.
A. (25 – 10 – 8) ÷ 2
B. (25 – 12 – 8) ÷ 2
C. (25 – 13 – 9) ÷ 2
D. (25 – 13 – 10) ÷ 2
Ê. (30 – 10 – 8) ÷ 2
F. (30 – 12 – 8) ÷ 2 |
# Differential Equations : Matrix Exponentials
## Example Questions
### Example Question #1 : Matrix Exponentials
Use the definition of matrix exponential,
to compute of the following matrix.
Explanation:
Given the matrix,
and using the definition of matrix exponential,
calculate
Therefore
### Example Question #1 : Matrix Exponentials
Given the matrix , calculate the matrix exponential, . You may leave your answer diagonalized: i.e. it may contain matrices multiplied together and inverted.
Explanation:
First we find our eigenvalues by finding the characteristic equation, which is the determinant of (or ). Expansion down column one yields
Simplifying and factoring out a , we have
So our eigenvalues are
To find the eigenvectors, we find the basis for the null space of for each lambda.
lambda = -1
Adding row 1 to row 3 and placing into row 3, dividing row two by 6, and swapping rows two and 1 gives us our reduced row echelon form. For our purposes, it suffices just to do the first step and look at the resulting system.
So that
Which has solutions . Thus, a clean eigenvector here would be
For lambda = 4, we have
Step 1: Add row 3 to row 1.
Step 2: Add 3 row 3 to row 2
Step 3: Add -6/5 row 1 in to row 2. Swap and divide as necessary to get proper pivots.
This gives us
So that
Which has solutions . Thus, a clean eigenvector here would be .
As we only ended up with two eigenvectors, we'll need to grab a generalized eigenvector as well. To do this, we will solve
(for lambda = 1, and we set it equal to the negation of our eigenvector for 1.)
This gives us
And the steps to solve this are identical to the steps to solving for the eigenvector for -1. Following them once more, and further reducing, we get.
Solving the system, our generalized eigenvector is given by . Decomposing into the Jordan matrix gives us
,
When we exponentiate this in the above form, we only need to find the matrix exponential of the Jordan matrix. This is done by exponentiating the entries on the main diagonal, and making the entries on the super diagonals of each Jordan block powers of t over the proper factorials. Thus, the matrix exponential is given by
From here, it would just be a matter of inverting and multiplying together -- daunting algebraically, but conceptually quite easy.
Note: In the final form above, anything with the same entries, but the columns switched is okay. I.e., it's okay to have the first eigenvector in the last column of the last two matrices, and be in the lower right hand corner of the second matrix.
### Example Question #71 : Differential Equations
Given the matrix , calculate the matrix exponential, .
Explanation:
First we find our eigenvalues by finding the characteristic equation, which is the determinant of (or ).
Thus, we have eigenvalues of 4 and 2. Solving for the eigenvectors by finding the bases of the eigenspaces, we have
lambda = 4
Adding Row1 into Row 2, we're left with
So that
And have an eigenvector of .
For lambda = 2, we have
Adding -1 Row 1 into Row 2, we have
So that
and is an eigenvector.
Constructing our diagonalized matrix, we have
Using the formula for calculating the inverses of 2x2 matrices, we have
To calculate the matrix exponential, we can just find the matrix exponential of and multiply and back in. So .
is just found by taking the entries on the diagonal and exponentiating. Thus,
Multiplying together, we get
### Example Question #4 : Matrix Exponentials
Use the definition of matrix exponential,
to compute of the following matrix.
Explanation:
Given the matrix,
and using the definition of matrix exponential,
calculate
Therefore
### Example Question #1 : Matrix Exponentials
Calculate the matrix exponential, , for the following matrix: .
Explanation:
To get the matrix exponential, we will have to diagonalize the matrix, which requires us to find the eigenvalues and eigenvectors. Thus, we have
Using , we then find the eigenvectors by solving for the eigenspace.
This has solutions , or . So a suitable eigenvector is simply .
Repeating for ,
This has solutions , and thus a suitable eigenvector is .
Thus, we have , and using the inverse formula for 2x2 matrices, . Now we just take the matrix exponential of and multiply the three matrices back together. Thus,
Multiplying these out yields
### Example Question #1 : Matrix Exponentials
Find the general solution to |
# Web Puzzles
Showing posts with label vicar. Show all posts
## How Old is the Vicar?
There once was a choirmaster.
One day three people came in and asked to join the choir.
The choirmaster, who believes that there should be age for his choir's members, asks their ages.
To that question, one of them replied: "We can't tell you our ages, but we can tell you the following: the product of our ages is 2450, and the sum of our ages is twice your age."
The choirmaster is puzzled: "That's not enough information!"
Just then, the vicar walked in and said: "But I'm older than all of them"
The choirmaster, who knew the vicar's age, then exlaimed: "Ah! Now I know."
How old is the vicar?
### How Old is the Vicar? Puzzle Solution
The vicar is 50.
The way to solve this puzzle, is to first of all write down all the possible permutations of three numbers whose product is 2450.
Starting Numbers Product Sum Choirmaster
1, 1, 2450 2450 2452 1226
1, 2, 1225 2450 1228 614
1, 5, 490 2450 496 248
1, 7, 350 2450 358 179
1, 10, 245 2450 256 128
1, 14, 175 2450 190 95
1, 25, 98 2450 124 62
1, 35, 70 2450 106 53
1, 49, 50 2450 100 50
2, 5, 245 2450 252 126
2, 7, 175 2450 184 92
2, 25, 49 2450 76 38
2, 35, 35 2450 72 36
5, 5, 98 2450 108 54
5, 7, 70 2450 82 41
5, 10, 49 2450 64 32
5, 14, 35 2450 54 27
7, 7, 50 2450 64 32
7, 10, 35 2450 52 26
7, 14, 25 2450 46 23
Since the choirmaster, after being told that the product of the ages is 2450 and that the sum is twice his age, still can't work out the ages, we can deduce that there are two (or more) combinations with the same sum. Those combinations have been highlighted in the table above.
The vicar then claims to be older than all of them. The oldest of the three is 49 in the first remaining combination, and 50 in the other. The choirmaster knows the vicar's age, and after his claim, he deduces everyone's age. The only way he's able to do so is if the vicar is 50, leaving the combination 7, 7, 50 logically impossible (the vicar has to be older, that is at least 1 year older than the others), and therefore learning that the people's ages are 5, 10, and 49. |
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# How do you factorize the denominator?
How do you factorize the denominator?
The calculation of this factorization can be done as follows:
1. Reduce the coefficients to a common denominator, to obtain the integer quotient of a polynomial with integer coefficients.
2. Collect the GCD of the coefficients of this polynomial as a common factor to obtain the primitive part, being the content.
## What is factorization for?
The decomposition into prime factors, or factoring into prime numbers, is an algebraic procedure that allows you to rewrite a natural number as a product of prime numbers.
## What does prime factorization mean?
The prime factorization, also called prime number factorization, is an algebraic procedure which consists in rewriting a natural number as a product of prime numbers.
## How do you factorize the numbers?
More generally, we can say that to BREAK a number into PRIME FACTORS, it is DIVIDED by the SMALLEST PRIME NUMBER OF ITS DIVIDER, then DIVIDING the QUOTE obtained by the SMALLEST PRIME NUMBER OF ITS DIVIDER, and so on until you get as quoto 1.
## How do you break down the denominator of a fraction?
We divide the numerator and denominator of the fraction by the GCD. A second way to proceed consists in BREAKING THE NUMBER AND NAME into prime factors and REMOVING the FACTORS COMMON to the two terms of the fraction.
## Find 25 related questions
### How do you reduce a fraction to a minimum?
Reducing a fraction to its lowest terms means transforming it into an equivalent fraction having the smallest terms (numerator and denominator). This transformation occurs by simplifying all the common divisors between the numerator and denominator.
### How do you reduce the fractions to a minimum?
A fraction is said to be reduced to its lowest terms (or irreducible) when the numerator and denominator do not have common divisors greater than 1, that is, they are prime to each other.
### What does it mean to factor a number?
factorization or factorization, an operation consisting in the rewriting of a generic numeric or algebraic expression as the product of several factors.
### How does 36 break down?
In fact 6 Β· 6 = 36. Even if mathematically it is correct to say that 36 is divisible by 6, in the prime factorization this way of proceeding is wrong. In fact, in the right column we can only write prime numbers. And 6 is not a prime number.
### How does 15 break down?
Factoring a number into prime factors - is to find the prime numbers that multiply together to form that number.
1. 15 is not a prime number, it is a composite number.
2. 15 can be written as a product of prime numbers. The prime factorization of 15: 15 = 3 Γ 5.
### When can partial collection be applied?
The partial common factor collection, or more briefly partial collection, is the second technique of decomposition of polynomials that we present and is generally applied to polynomials that have an even number of terms.
### What are the prime numbers up to 100?
At the end of the job, the circled numbers are the prime numbers within 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
### What is the applicability condition of the LU factorization?
1. Prove that if A is strongly dominant diagonal then the LU factorization of A exists and is unique. 2. Prove that if A is Hermitian and positive definite then there exists and the LU factorization of A is unique.
### How does 17 break down?
Is the integer 17 a prime number?
1. 17 is a prime number, it cannot be broken down to other prime factors.
2. 17 cannot be written as a product of prime numbers. 17 can be written as the product of positive integers only as: 17 = 1 Γ 17. 1 is neither a prime nor a composite number.
### How do you know if a number is a perfect square?
HOW CAN I KNOW IF A NUMBER IS A PERFECT SQUARE? A natural number is a perfect square if, when decomposed into prime factors, all the exponents of its factors are even numbers. factors are even. 1296 is a perfect square!
### What are the prime numbers from 1 to 1000?
The list with all the digits and prime numbers from 1 to 1000 (from 2 to 997)
• From 1 to 100: ...
• From 101 to 200: 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199.
• From 201 to 300: 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293.
• 301 to 400:
### When do we say that a number is prime?
prime number integer greater than 1 that admits only trivial divisors, that is 1 and itself.
### How to transform limited decimal numbers into fractions reduced to minimum terms?
3,25. DENOMINATOR = 1 followed by AS MANY ZEROS as there are DECIMAL DIGITS of the given number. 3,25 = 325/100. The fraction we have obtained, in this case, can be SIMPLIFIED.
### How is simplification done?
So we can say that to SIMPLIFY a fraction it is enough to DIVIDE both TERMS by the same COMMON DIVIDER. Therefore, a fraction is said to be REDUCED TO THE MINIMUM TERMS when the NUMERATOR and the DENOMINATOR are FIRST OF THEM.
### How do you calculate fractions?
Rule to calculate the fraction of a number
The division is made between the number considered and the denominator of the fraction, and then the result (the quoto) is multiplied by the numerator of the fraction. The number thus calculated will be the quantity expressed by the fraction with respect to the number.
### What term is not common to fractions?
A minimal or irreducible fraction is a fraction whose operators (dividend and divisor) are coprime to each other, that is, they have no common divisors other than unity. To identify this number, a series of simple simplifications are carried out on the numerator and denominator. ...
### What is a fraction called if the numerator is a multiple of the denominator?
Proper fractions, improper fractions and apparent fractions are types of fractions where the numerator is less than the denominator, the numerator is greater than the denominator, the numerator is a multiple of the denominator, respectively. In particular, the apparent fractions can be expressed as whole numbers.
### How are expressions with powers and fractions made?
But how to make the power of an infringement? Even if the base is a fraction it is always a repeated multiplication: you must therefore multiply the fraction by itself as many times as the exponent indicates. Just raise both the numerator and the denominator to that power.
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Chapters
The graph of a linear equation in an x-y plane is a straight line. Linear equations can be written in different forms. One of the forms of linear equations is two-point form also known as point slope form. In this article, we will focus on two-point form of the linear equation.
We will discuss the following topics related to two-point form of the linear equations in detail along with the relevant examples:
• Writing an equation of a straight line in two-point form
• Converting two point-form into the slope intercept form of the linear equation
• Converting two-point form into the standard form of the linear equation
Besides writing the equations into a two-point form from the given set of values and graphs, we will also see how to write the linear equations in two-point form from the given scenarios.
## Write an equation of a straight line in two-point form
### Two-point form formula
The formula of the equation of the straight line passing through two points and in two-point form is given below:
In the above formula, the slope is denoted by m.
### Slope formula
To write the equation of a straight line in a two-point form, first find the slope denoted by “m” of the straight line. The formula for the slope of the straight line is given below:
Some interesting useful facts about slope are given below:
• The slope can be zero or non-zero (positive or negative).
• Zero slope means that no matter what the value of the x-coordinate is, the y-coordinate remains constant. The graph of an equation will be a horizontal line, if slope is zero.
• A positive slope means that there is positive relationship between x and y variables. It means that as the value of x- coordinate increases, the value of y also increases or if the value of x decreases, then the value of y also decreases.
• A negative slope means that the two variables are negatively related to each other. It means that as the value of x increases, the value of y decreases or if the value of x decreases, the value of y increases.
• The slope of two parallel lines is same, whereas the slope of two perpendicular lines are negative reciprocals of each other.
### Example 1
Write the equation of the line passing through the points and .
#### Solution
To write the equation, first find the slope of the equation. From the above two points we get the following values of and intercepts.
and
Substitute these values in the following formula:
Now, put the values of and in the formula to get the equation in two point form passing through the points and :
### Example 2
Now, we will see how can we write an equation of a straight line in point-slope form given the graph of an equation in a coordinate plane. Consider the following graph:
#### Solution
Take two points from the graph. Suppose we take and . In the above graph, and . To write the equation in two-point form, first we need to find the slope by substituting these values in the formula below:
Now, put the values of the , and in the following formula to get the equation in two-point form:
### Example 3
It is not always necessary that two points on the line are given in a question. Sometimes, the slope, and intercepts of one point are given. Consider the example below:
Write the equation of a straight line in two-point form having a slope of passing through the points .
#### Solution
In the above example, , and .
Put these values in the two-point form of the equation:
### Example 4
You can also write the equation in a two-point form from the given scenario. Consider the following situation:
James joined the fitness club to lose an average weight of 7 pounds each month. At the end of the sixth month, his weight was 148 pounds. Write the given scenario in two point form.
#### Solution
Since he lost 7 pounds each month, hence the slope is .
It is given that at the end of the sixth month, his weight was 148 pounds. We will take months as an independent variable and weight as a dependent variable.
Solve the equation by substituting the values of , and in the following formula to get the point slope form of the linear equation.
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## Converting two-point form into slope-intercept form of the linear equation
Now, we know how to write an equation of the line in a two-point form also known as point-slope form. Do you know that besides a two-point form linear equation can also be written in two other forms? These forms are standard and slope-intercept form.
In this section, we will discuss how to convert the two-point form into the slope-intercept form of the linear equation. If you are given two points on the line, you can write the equation in slope-intercept form. But to do so, you first need to write the equation in a two-point form. The formula for a slope-intercept form is given below:
Here, m is the slope and b is y-intercept of the line.
Let’s see how to convert the equation from point-slope form to slope-intercept form from the following examples.
### Example 1
Write the equation of the line in slope-intercept form passing through the points and .
#### Solution
In the above example, and .
First find the slope using the formula below:
Now, write the equation of a line in point-slope form of the linear equation:
We will get the following form by simplifying the right hand side of the equation:
Isolate on the left side by adding 1 to both sides of the equation:
Hence, the slope-intercept form of the line passing through the points and is . [Note that the slope-intercept form of the straight line is . If the equation of the line takes the form of , then it means that the value of b which is the y-intercept of the line is 0.].
### Example 2
Write the equation in slope-intercept form passing through the points and .
#### Solution
First find the slope of the line using the following formula:
In the above example, , and .
Now, put the values of , and into the point-slope form of the equation:
The slope will be multiplied by the term inside the brackets to get the point-slope form:
(Point slope form)
To convert this equation from point-slope form to slope-intercept form, solve the first equation by taking -3 on the right hand side of the equation.
(Slope intercept form)
### Example 3
We can also write the equation in a two-point form or slope-intercept form from the scenario. Sometimes the situation demands writing linear equations in the two-point form first before converting them into the slope intercept form. The example of one such scenario is given below:
Sarah got a new job and decided to save a fixed part of her salary each month. She already had some savings from her previous job. After starting a new job, she had 2110 dollars of savings in two months and in five months her total savings amounted to 3400 dollars.
1. How much salary did she save each month?
2. Write an equation that can be used to determine her salary in number of months.
3. How much savings she already had before she started the new job?
#### Solution
We can write the information in the form of a table. In two months, her savings were 2110 dollars and in five months her total savings amounted to 3400 dollars.
Months (x)Savings (y)
22110
53400
Part 1
To find the fixed amount she saved each month, first find the slope of the linear equation from the above table:
=
=
Hence, she saved 430 dollars each month.
Part 2
To write an equation that can be used to determine her savings after number of months, we need to find the slope-intercept form of the equation. But before finding the slope-intercept form, first, we need to write the equation in a two-point form.
(Point slope form)
Convert the above equation into slope-intercept form by simplifying the right side of the equation.
Isolate on the left hand side by adding 2110 on both sides of the equation:
(Slope-intercept form)
Hence, you can use your calculators to determine her savings after number of months from the equation .
Part 3
In the previous part, we determined the Slope intercept form of the equation from the scenario. In the equation, 1250 represents the -intercept of the equation. The point at which the line intersects the axis is known as -intercept of the line. At this point of intersection the value of intercept is 0.
To find the savings she already had before starting the new job, put in the equation .
Hence, the correct answer is dollars.
## Converting two-point form into the standard form of the linear equation
The standard form of a linear equation is given by the following formula:
### Example 1
We can easily convert the two-point form of the linear equation into a standard form. Consider the following equation in the two-point form:
#### Solution
To convert it into the standard form, the slope of a line which is 2 will be multiplied by the term on the right hand side of the equation:
Isolate y on the right hand side of the equation by adding 3 on both sides:
We will add 11 on both sides of the equation to get the following equation:
We know that the standard form of the equation is , therefore we will isolate the constant by subtracting on both the sides:
Here, and .
### Example 2
Consider the second example below:
Write the equation of the line in the standard form passing through the points and .
#### Solution
To write the equation in standard form, first find the equation in two-point form. We will begin it by finding the slope of the equation.
In the above example, and .
By putting these values in the slope formula, we will get the following value of slope:
Now, we know the value of the slope, we can easily get the two-point form by plugging it in the following formula:
(Two-point form)
By multiplying the slope with the term inside the brackets, we will get the following equation:
Isolate y on the left hand side by subtracting 1 from both sides of the equation.
(Slope intercept form)
The standard form of the equation is . Hence take -16 to the left hand side and y to the right hand side of the equation to get the final equation.
(Standard form)
Here, , and .
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# What is the addition property of equality example?
The addition property of equality is used while solving equations. For example, to solve x – 3 = 5, we have to add 3 to both sides so that only the variable (x) will be left on the left side. This implies, x – 3 + 3 = 5 + 3, which can be simplified as x = 8.
## What is the addition property of equality example?
The addition property of equality is used while solving equations. For example, to solve x – 3 = 5, we have to add 3 to both sides so that only the variable (x) will be left on the left side. This implies, x – 3 + 3 = 5 + 3, which can be simplified as x = 8.
## How do you solve the addition property of equality?
We learned that the addition property of equality tells us that if we add the same quantity to both sides of an equation, then our equation remains the same. The formula is if a = b, then a + c = b + c.
What is the property of addition property of equality?
The addition property of equality tells us that adding the same number to each side of an equation gives us an equivalent equation. ifa−b=c,thena−b+b=c+b,ora=c+b. The same goes with the subtraction property of equality.
### How do you do MPE in math?
By the multiplication property of equality, if you multiply both sides by the same number, the equality still holds true. So, multiply both sides by 4. Therefore, the 300 kids visited the park that day. Another form of the property is if a, b, c, and d are real numbers such that a = b, c = d, then ac = bd.
### What are the 3 properties of addition?
• Closure Property.
• Commutative Property.
• Associative Property.
What is the addition property in geometry?
The property that states that if you add the same number to both sides of an equation, the sides remain equal (i.e., the equation continues to be true.)
#### What are the property of addition?
The 4 main properties of addition are commutative, associative, distributive, and additive identity.
Additive Identity Property On adding zero to any number, the sum remains the original number. Adding 0 to a number does not change the value of the number. |
4.05 Subtraction using algorithm
Lesson
Ideas
How did you find adding numbers, using the standard algorithm ?
Examples
Example 1
Find the value of 41 + 56.
Worked Solution
Create a strategy
Use the addition algorithm.
Apply the idea
Write the addition in a vertical algorithm.\begin{array}{c} & &4 &1 \\ &+ &5 &6 \\ \hline & \\ \hline \end{array}
Add the smallest place value first. So 1 + 6 = 7.\begin{array}{c} & &4 &1 \\ &+ &5 &6 \\ \hline & & &7 \\ \hline \end{array}
Then add the next place value, 4 + 5 = 9.\begin{array}{c} & &4 &1 \\ &+ &5 &6 \\ \hline & &9 &7 \\ \hline \end{array}
So 41 + 56=97.
Idea summary
You might notice that sometimes the standard algorithm is called the 'vertical algorithm'. Let's think about why. When we use the standard algorithm, we line our numbers up in 'vertical' place value columns.
Subtraction without regrouping
Let's try some subtraction problems, using the standard algorithm. We use place value models to help us, so you might like to have some handy as you watch the video.
Examples
Example 2
Find the value of 95 -51.
Worked Solution
Create a strategy
Use the subtraction algorithm.
Apply the idea
Write the subtraction in a vertical algorithm.\begin{array}{c} & &9 &5 \\ &- &5 &1 \\ \hline & \\ \hline \end{array}
Subtract the smallest place value first. So 5 - 1 = 4.\begin{array}{c} & &9 &5 \\ &- &5 &1 \\ \hline & & &4 \\ \hline \end{array}
Then subtract the next place value, 9- 5 = 4.\begin{array}{c} & &9 &5 \\ &- &5 &1 \\ \hline & &4 &4 \\ \hline \end{array}
So, 95- 51 = 44.
Idea summary
When we solve problems using an algorithm, we have to line up our digits by place value. With subtraction, we always put the total at the top, and the number we are subtracting, below.
Outcomes
MA2-5NA
uses mental and written strategies for addition and subtraction involving two-, three-, four and five-digit numbers |
# SSAT Upper Level Math : Number Concepts and Operations
## Example Questions
### Example Question #41 : Number Concepts And Operations
You are asked to fill in both circles in the statement
with the same number from the set
to make a true statement.
How many ways can you do this?
Four
None of the other responses is correct.
Six
Two
None
Four
Explanation:
The problem is asking for a number whose square is a number congruent to 1 in modulo 12 arithmetic - that is, a number whose square, when divided by 12, yields remainder 1. This square must be odd, so the number squared must also be odd. Therefore, we need only test the odd integers. We see that:
Four of the integers have squares congruent to 1 in modulo 12 arithmetic.
### Example Question #1 : How To Multiply
Multiply:
None of the other responses is correct.
Explanation:
We can write 2 pounds, 5 ounces as just ounces as follows:
Multiply:
Divide by 16, noting quotient and remainder, to get pounds and ounces:
Therefore, the correct response is 13 pounds, 14 ounces.
### Example Question #43 : Number Concepts And Operations
Fill in the circle to yield a true statement:
The other answer choices are incorrect.
The other answer choices are incorrect.
Explanation:
The problem is asking for a number whose product with 6 yields a number congruent to 5 in modulo 12 arithmetic - that is, a number which, when divided by 12, yields remainder 6.
However, 6 multiplied by any odd number yields a number which, when divided by 12, yields remainder 6, as is demonstrated using our choices:
Each of the choices yields a product congruent to 6 modulo 12, so none of them is a correct choice.
### Example Question #11 : Perform Operations With Numbers Expressed In Scientific Notation: Ccss.Math.Content.8.Ee.A.4
Write in scientific notation.
Explanation:
Scientific notation is used to simplify exceptionally complex numbers and to quickly present the number of significant figures in a given value. The value is converted to an exponent form using base ten, such that only a single-digit term with any given number of decimal places is used to represent the significant figures of the given value. Non-significant zeroes can be omitted from the leading term, and represented only in the base ten exponent.
The given number has three significant figure () so we write out number as
You must move the decimal place two places to the right, or in other words multiply by . When you move the decimal place to the right, you multiply the number by , so it is .
### Example Question #1 : How To Divide Square Roots
Divide:
Explanation:
We can write 18 pounds, 2 ounces as just ounces as follows:
Divide:
Convert 58 ounces to pounds and ounces as follows:
That is, 3 pounds 10 ounces.
### Example Question #45 : Number Concepts And Operations
Simplify by rationalizing the denominator:
Explanation:
Multiply both numerator and denominator by the conjugate of the denominator, which is :
### Example Question #2 : How To Divide Square Roots
Simplify by rationalizing the denominator:
Explanation:
Multiply both numerator and denominator by the conjugate of the denominator, which is :
### Example Question #41 : Number Concepts And Operations
Simplify by rationalizing the denominator:
None of the other responses is correct.
Explanation:
Multiply both numerator and denominator by the conjugate of the denominator, which is :
### Example Question #41 : Number Concepts And Operations
Simplify by rationalizing the denominator:
The correct answer is not among the other responses.
The correct answer is not among the other responses.
Explanation:
Multiply both numerator and denominator by the conjugate of the denominator, which is :
This is not among the given choices.
### Example Question #2 : How To Divide Square Roots
Simplify by rationalizing the denominator: |
0 energipoint
Studying for a test? Prepare with these 10 lessons on Linear equations & graphs.
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# Horizontal & vertical lines
Video transskription
- [Instructor] What is the equation of the horizontal line through the point negative four comma six? So let's just visualize this. Once you get the hang of it, you might not have to draw a graph, but for explanatory purposes, it might be useful. So negative four comma six, that's going to be in the second quadrant. So if this is my x-axis, That is my y-axis. I'm going to go negative four in the x direction. So one, two, three, four. Negative four. And then one, two, three, four, five, six, in the y direction. So the point that we care about is going to be right over there. Negative four comma six. And what is the equation of the horizontal line? It is a horizontal line. So it's just going to go straight left, right like this. That is what the line would actually look like. So what is that equation? Well, for any x, y is going to be equal to six. This is the equation, y is equal to six. Doesn't matter what x you input here, you're gonna get y equals six. It just stays constant right over there. So the equation is y is equal to six. Let's do another one of these. So here we are asked what is the slope of the line y is equal to negative four? So let's visualize it and then in the future, you might not have to draw it like this. But let's just draw our axis again. X-axis y-axis and the slope of line y equals negative four. So for whatever x you have, y is going to be negative four. Let's say that's negative four right over there. And so, the line is y the line is y equals negative four. So I can draw it like this. So what's the slope of that? Well, slope is change in y for given change in x. And here, no matter what I change my x, y doesn't change. It stays at negative four. My change in y over change in x. Doesn't matter what my change in x is. My change in y is always going to be zero. It's constant. So the slope here is going to be equal to zero. Y doesn't change, no matter how much you change x. Let's do another one of these. This is fun. So now they are asking us, what is the slope of the line x equals negative three? Let me graph that one. So, I'm just going to draw my axis real fast. X-axis y-axis X is equal to negative three. So negative one, negative two, negative three. And so, this line is going to look let me, it's going to look like this. No matter what y or you can say no matter what y is. X is going to be equal to negative three. So, it would look like this. X is equal to negative three. So what's the slope here? Well, it's undefined. A vertical line has an undefined slope. Remember, you want to do what's your change in y or change in x. Change in y or change in x. Well, you can think about what's the slope as you approach this but once again, that could be, some people would say, maybe it's infinite, maybe it's negative infinity. But that's why it's undefined. A vertical line is going to have an undefined slope. So we'll go with undefined. Let's do one more. What is the equation of the vertical line through negative five comma negative two? So, let me do this without even drawing it. I'll draw it right after that. So, if we're talking about a vertical. If we're talking about a vertical line, that means that x doesn't change. X doesn't change. If we were talking about a horizontal line, then we'd say y doesn't change. So if x doesn't change, that means that x is just going to be equal to some constant value. Well, if it contains the points negative five comma negative two, so if it has a point where x is equal to negative five and if x never changes, it's a vertical line, well that means its equation has to be x is equal to negative five. And we can draw that out, if it helps. So let me draw that out. So, I need to make sure that's a straight line. Okay, so we have x, and we have y, so we have the point negative five comma negative two. So negative one, two, three, four, five, negative one, two so we want to go have a vertical line that goes through that point. So a vertical line, well that just goes straight up and down. So it's just going to look like this. And so notice, x never changes. No matter what y is, x is equal to negative five. This has an undefined slope. It is a vertical line. Its equation is x is equal to negative five. |
## What is an example of definition of perpendicular lines?
Two distinct lines intersecting each other at 90° or a right angle are called perpendicular lines. Example: Here, AB is perpendicular to XY because AB and XY intersect each other at 90°. Non-Example: The two lines are parallel and do not intersect each other.
## What does perpendicular segments look like?
We say two lines or line segments are perpendicular if they form a right angle (or several right angles). We can mark a right angle with a little corner .
What is perpendicular line short answer?
In geometry, a branch of mathematics, perpendicular lines are defined as two lines that meet or intersect each other at right angles (90°).
How do you name a perpendicular line segment?
Two lines that intersect and form right angles are called perpendicular lines. The symbol ⊥ is used to denote perpendicular lines.
### How do you show perpendicular lines?
Explanation: If the slopes of two lines can be calculated, an easy way to determine whether they are perpendicular is to multiply their slopes. If the product of the slopes is , then the lines are perpendicular. In this case, the slope of the line is and the slope of the line is .
### What is mean by perpendicular line?
Perpendicular lines are lines that intersect at a right (90 degrees) angle.
What are two perpendicular lines?
Two lines are perpendicular if and only if the product of their slopes is . In other words, the slope of a line that is perpendicular to a given line is the negative reciprocal of that slope. Thus, for a line with a given slope of 3, the line perpendicular to that slope must be the negative reciprocal of 3, or .
How do you know if a line segment is perpendicular?
#### What happens when a line is perpendicular?
Perpendicular lines are lines that intersect at right angles. That is, the slopes of perpendicular lines are opposite reciprocals . (Exception: Horizontal and vertical lines are perpendicular, though you can’t multiply their slopes, since the slope of a vertical line is undefined.)
#### What is the difference between parallel and perpendicular lines?
The word “parallel” refers to two equidistant (having the same distance) lines with the same steepness, whereas the word “perpendicular” refers to something that is positioned at a 90° angle from another thing. If we talk about lines, for example, the points of the parallel lines will never touch each other…
What are examples of perpendicular lines?
Football field
• Railway track crossing
• First aid kit
• Construction of a house in which floor and the wall are perpendiculars
• Television
• Designs in windows
• What does a perpendicular line look like?
A perpendicular angle is formed when the degree of rotation of the terminal side from the initial side is equal to 90. Thus, a perpendicular line looks like a line that has a 90 degree angle between it and another line.
## Do perpendicular lines have the same exact slope?
Perpendicular lines do not have the same slope . The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. |
In science and mathematics, radical is derived from Late Latin radicals “of roots” and Latin radix “root.” A radical is an atom, molecule, or ion in chemistry that is likely to engage in chemical reactions. The square root or nth root is denoted by the radical sign (√).
In general, people tend to be lazy. And with the mathematicians who do some mathematical tasks that is especially pronounced.
When a clever scientist had to add the same number several times, he decided he couldn’t stand it anymore and created multiplication to make writing prolonged expressions much faster.
Then, a generation or two later, their descendants were presented with a similar challenge: multiplying the same number several times. They remembered their ancestor’s wit at that moment and boldly declared that they would have none of it.
Inverse exponents are known as radicals (also known as roots). When we have the number 390,625, and we know it’s to the eighth degree, the (8th) root of those 390,625 will be 5.
What method did we use to arrive at this conclusion?
Calculating the roots isn’t always straightforward. We frequently turn to other resources for assistance in more complicated circumstances, such as our multiplication calculator. However, a few strategies and principles might come in handy, and we’ll demonstrate how to multiply square roots to see whether they work.
## Multiplying square roots
We’ll look at the statement a√b * c√d to see how to multiply square roots (note that an analogous equation is at the top of the multiplying radicals calculator). The underlying concept is that numbers outside of the roots and those within belong to different categories. To be more specific, the following identification is correct:
a√b * c√d = (a * c) * √(b * d)
Multiplying square roots are, to some extent, all there is to it. However, the answer we acquire in this manner is frequently unsatisfactory, i.e., the resultant radical statement may be simplified.
Take, for example, 3√30 * 5√6. As a result of the above, we have:
3√30 * 5√6 = (3 * 5) * √ (30 * 6) = 15√180
Let’s take a deeper look at the radical before we declare this to be our final response. The number has become something we can write in a neater way by multiplying the roots. We’ll reduce the radical phrase to be more precise.
To do so, we begin by determining the number’s initial factorization below the root:
180 = 2 * 2 * 3 * 3 * 5 = 2^2 * 3^2 * 5
We’re looking for pairs of the same prime numbers in the list above. We have two in our case: a pair of 2s and a pair of 3s. The numbers that represent each pair are extracted from the radicals, while those that did not locate the pair are kept under it.
3\sqrt<b style="mso-bidi-font-weight:normal">{</b>30} * 5\sqrt{6} = 15\sqrt{180} = 15 * 2 * 3 * \sqrt{5} = 90\sqrt{5}
Only now can we be positive that we have finished the square root multiplication and have arrived at the simplest radical form. A similar step-by-step approach may be also in Omni’s multiplier calculator.
Let’s establish what we mean by the term “radical” before we delve into the mathematics underpinning radicals. Simply explained, a radical is a number called the radicand that is included within a root – such as a square root, cube root, or other roots. The radical sign is a term used to describe these roots.
It’s also worth noting that anything, even variables, has the potential to be in origin! Radicals are commonly encountered as you move through arithmetic.As a result, understanding how to collaborate with them is crucial. We’ll look at the mathematics underlying radical simplification and radical multiplication, also known as square root simplification and multiplication, in this article.
Despite its frightening appearance, multiplying radicals is a rather straightforward procedure! However, before we begin multiplying radicals directly, we must first study how to simplify radicals.
To simplify a radical, all we have to do is pull the radicand’s words out of the root, if that is feasible.
Of course, like with any profession or discipline, there are certain duties that are simpler and others that are more challenging. Here are some measures you may take to start the mediation preparation process.
If you follow these two guidelines while multiplying radicals, you’ll never have any problems:
1) Multiply the radicands and store the result in the root.
2) Simplify the radical if feasible, either before or after multiplication.
## Example: using the multiplying radicals
We’ll use the example of multiplying 2 *\sqrt[3]{12} and 5*\sqrt{6}.
The multiplication roots are symbolically represented at the top of our tool: a * ⁿ√b * c * ᵐ√d.
Because we want to finde 2 *\sqrt[3]{12} * 5 * \sqrt{18}– we have to input:
a = 2, b = 12, n = 3 for the first factor, and, c = 5, d = 18, m = 2 for seconed factor.
The last step is just to read the last value. It’s also worth noting that the calculator walks you through the process of arriving at an answer in straightforward steps. |
CBSE Class 12 Sample Paper for 2023 Boards
Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards
If the length and the breadth of the rectangular field be 2x and 2y respectively, then find the area function in terms of x.
Transcript
Question 37 [Case Based] In an elliptical sport field the authority wants to design a rectangular soccer field with the maximum possible area. The sport field is given by the graph of π₯^2/π^2 +π¦^2/π^2 =1Based on the above information answer the following questions. Question 37 (i) If the length and the breadth of the rectangular field be 2x and 2y respectively, then find the area function in terms of x.Given equation of ellipse is π₯^2/π^2 +π¦^2/π^2 =1 where Major axis of ellipse is along x-axis Let ABCD be the rectangle Where Length of rectangle = 2x Breadth of rectangle = 2yBy Symmetry CP = BP = 1/2 BC CP = BP = 1/2 Γ 2π¦=π¦ Thus, y-coordinate of point C = π Similarly, By Symmetry OP = 1/2 AB OP = 1/2 Γ 2π₯=π₯ Thus, x-coordinate of point C = π Thus, coordinates of point C = (x, y) Since point C lies in ellipse, it will satisfy itβs equation π₯^2/π^2 +π¦^2/π^2 =1 π¦^2/π^2 =1β" " π₯^2/π^2 π¦^2/π^2 =" " (π^2 β π₯^2)/π^2 π¦^2=" " π^2/π^2 (π^2 β π₯^2 ) Taking square root by sides π¦=" " β(π^2/π^2 (π^2 β π₯^2 ) ) π=(" " π)/π β((π^π β π^π ) ) Now, Area of Rectangle = Length Γ Breadth = 2x Γ 2y = 4xy Putting π¦=(" " π)/π β((π^2 β π₯^2 ) ) = 4x Γ (" " π)/π β((π^2 β π₯^2 ) ) = (" " ππ)/π πβ((π^π β π^π ) ) |
# If f(x)= 2 x^2 + x and g(x) = sqrtx + 1 , how do you differentiate f(g(x)) using the chain rule?
Feb 23, 2016
$f ' \left(g \left(x\right)\right) = 2 + \frac{5}{2 \sqrt{x}}$
#### Explanation:
Find f(g(x)) $= f \left(\sqrt{x} + 1\right)$
substitute $x = \sqrt{x} + 1 \text{ in f(x) to obtain }$
f(g(x)) = $2 {\left(\sqrt{x} + 1\right)}^{2} + \left(\sqrt{x} + 1\right)$
distribute ${\left(\sqrt{x} + 1\right)}^{2} = x + 2 \sqrt{x} + 1$
$\Rightarrow f \left(g \left(x\right)\right) = 2 \left(x + 2 \sqrt{x} + 1\right) + \sqrt{x} + 1$
$= 2 x + 4 \sqrt{x} + 2 + \sqrt{x} + 1 = 2 x + 5 \sqrt{x} + 3$
and writing $2 x + 5 \sqrt{x} + 3 \text{ as } 2 x + 5 {x}^{\frac{1}{2}} + 3$
now differentiate using$\textcolor{b l u e}{\text{ power rule }}$
ie $\frac{d}{\mathrm{dx}} \left(a {x}^{n}\right) = n a {x}^{n - 1} \text{ term by term }$
$\Rightarrow f ' \left(g \left(x\right)\right) = 2 + \frac{5}{2} {x}^{- \frac{1}{2}} = 2 + \frac{5}{2 \sqrt{x}}$ |
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# Segments from Secants
## Relationships of products of sections versus sums of sections of secants.
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Segments from Secants
What if you wanted to figure out the distance from the orbiting moon to different locations on Earth? At a particular time, the moon is 238,857 miles from Beijing, China. On the same line, Yukon is 12,451 miles from Beijing. Drawing another line from the moon to Cape Horn (the southernmost point of South America), we see that Jakarta, Indonesia is collinear. If the distance from Cape Horn to Jakarta is 9849 miles, what is the distance from the moon to Jakarta? After completing this Concept, you'll be able to solve problems like this.
### Watch This
CK-12 Foundation: Chapter9SegmentsfromSecantsA
### Guidance
In addition to forming an angle outside of a circle, the circle can divide the secants into segments that are proportional with each other.
If we draw in the intersecting chords, we will have two similar triangles.
From the inscribed angles and the Reflexive Property \begin{align*}( \angle R \cong \angle R), \triangle PRS \sim \triangle TRQ\end{align*}. Because the two triangles are similar, we can set up a proportion between the corresponding sides. Then, cross-multiply. \begin{align*}\frac{a}{c+d}=\frac{c}{a+b} \Rightarrow a(a+b)=c(c+d)\end{align*}
Two Secants Segments Theorem: If two secants are drawn from a common point outside a circle and the segments are labeled as above, then \begin{align*}a(a+b)=c(c+d)\end{align*}. In other words, the product of the outer segment and the whole of one secant is equal to the product of the outer segment and the whole of the other secant.
#### Example A
Find the value of the missing variable.
Use the Two Secants Segments Theorem to set up an equation. For both secants, you multiply the outer portion of the secant by the whole.
\begin{align*}18 \cdot (18+x)=16 \cdot (16+24)\\ 324+18x=256+384\\ 18x=316\\ x=17 \frac{5}{9}\end{align*}
#### Example B
Find the value of the missing variable.
Use the Two Secants Segments Theorem to set up an equation. For both secants, you multiply the outer portion of the secant by the whole.
\begin{align*}x \cdot (x+x)=9 \cdot 32\\ 2x^2=288\\ x^2=144\\ x=12\end{align*}
\begin{align*}x \neq -12\end{align*} because length cannot be negative.
#### Example C
True or False: Two secants will always intersect outside of a circle.
This is false. If the two secants are parallel, they will never intersect. It's also possible for two secants to intersect inside a circle.
Watch this video for help with the Examples above.
CK-12 Foundation: Chapter9SegmentsfromSecantsB
#### Concept Problem Revisited
The given information is to the left. Let’s set up an equation using the Two Secants Segments Theorem.
\begin{align*}238857 \cdot 251308 &= x \cdot (x+9849)\\ 60026674956 &= x^2+9849x\\ 0 &= x^2+9849x-60026674956\\ Use \ the \ Quadratic \ Formula \ x & \approx \frac{-9849 \pm \sqrt{9849^2-4(-60026674956)}}{2}\\ x & \approx 240128.4 \ miles\end{align*}
### Guided Practice
Find \begin{align*}x\end{align*} in each diagram below. Simplify any radicals.
1.
2.
3.
Use the Two Secants Segments Theorem.
1.
\begin{align*}8(8+x)&=6(6+18)\\64+8x&=144\\8x&=80\\x&=10\end{align*}
2.
\begin{align*}4(4+x)&=3(3+13)\\16+4x&=48\\4x&=32\\x&=8\end{align*}
3.
\begin{align*}15(15+27)&=x\cdot 45\\630&=45x\\x&=14\end{align*}
### Explore More
Solve for the missing segment.
Find \begin{align*}x\end{align*} in each diagram below. Simplify any radicals.
1. Prove the Two Secants Segments Theorem.
Given: Secants \begin{align*}\overline{PR}\end{align*} and \begin{align*}\overline{RT}\end{align*}
Prove: \begin{align*}a(a+b)=c(c+d)\end{align*}
Solve for the unknown variable.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 9.10.
### My Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
central angle
An angle formed by two radii and whose vertex is at the center of the circle.
chord
A line segment whose endpoints are on a circle.
diameter
A chord that passes through the center of the circle. The length of a diameter is two times the length of a radius.
Inscribed Angle
An inscribed angle is an angle with its vertex on the circle. The measure of an inscribed angle is half the measure of its intercepted arc.
intercepted arc
The arc that is inside an inscribed angle and whose endpoints are on the angle.
point of tangency
The point where the tangent line touches the circle.
AA Similarity Postulate
If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar.
Congruent
Congruent figures are identical in size, shape and measure.
Reflexive Property of Congruence
$\overline{AB} \cong \overline{AB}$ or $\angle B \cong \angle B$
Secant
The secant of an angle in a right triangle is the value found by dividing length of the hypotenuse by the length of the side adjacent the given angle. The secant ratio is the reciprocal of the cosine ratio.
secant line
A secant line is a line that joins two points on a curve.
Tangent line
A tangent line is a line that "just touches" a curve at a single point and no others.
Two Secants Segments Theorem
Two secants segments theorem states that if you have a point outside a circle and draw two secant lines from it, there is a relationship between the line segments formed. |
## Cone
Definition:
Cone is the solid that is generated by rotating a line segment which is passing through a fixed point and making a constant angle with a fixed line.
In the above figure here , VO is a fixed line and VA is a rotating line which is making constant angle with VO.The point A would describe a circle with center O such that the line segment VO is perpendicular to the base.
VO is the height "h" and OA is the base radius "r" VA is the slant height "l".
It is very clear VAO is right angled triangle and the right angle is at at O.
Since VAO is the right angled triangle by Pythagorean theorem
we have l² = h² + r²
from l² = h² + r² , we can get the value of one of the measurements if we know the value of the other two measurements. For example, if we have the height "h", and the radius "r" , we can easily determine the value of the slant height "l".
Example:
If the height and radius of a right circular conoid are 4cm and 3cm respectively,find the slant height.
Solution:
Let us plug the known values in to the equation.
l² = h² + r²
l² = 4² + 3²
l² = 16 + 9
l² = 25
so l = 5
that is slant height = 5 cm
Role of radius,slant height and height in finding area and volume:
Measurement of radius,height and slant height plays a vital role in finding curved surface area, total surface area and volumes.
Example
A heap of paddy is in the form of a conoid whose diameter is 4.2 m and height is 2.8 m. If the heap is to be covered exactly by a canvas to protect it from rain,then find the area of the canvas needed.
Solution:
Diameter of heap of paddy = 4.2 m
r = 4.2/2
= 2.1 m
height of paddy (h) = 2.8 m
L² = r² + h²
L = √(2.1)² + (2.8)²
L = √4.41 + 7.84
L = √12.25
L = √ 3.5 x 3.5
L = 3.5 cm
Curved surface area of heap of paddy = Π r l
= (22/7) x (2.1) x (3.5)
= 22 x (2.1) x (0.5)
= 23.1 cm²
Curved surface area of paddy = 23.1 cm²
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# Compound Interest Problems for Bank Po with Solution for SBI PO 2021 and LIC AAO 2021, CET, NIACL at Smartkeeda
Direction : Study the following questions carefully and choose the right answer.
Important for :
1
Shantanu borrowed Rs. 2.5 lakh from a bank to purchase one car. If the rate of interest be 6% per annum compounded annually, what payment (approximately) will he have to make after 2 years 6 months?
» Explain it
D
CI for 2 years 6 months at the rate of 6, applying the net% effect for first 2 years
= 6 + 6 + 6 × 6 = 12.36% 100
Rate of interest for 6 months = 6 × 6 = 3% 12
For next 6 months = 12.36 + 3 + 12.36 × 3 = 15.36 + 0.37% = 15.73% 100
Here, we can see that in 2 years 6 months the given compound rate of interest is approximate 15.73%.
Now, 115.73% of 250000 = 115.73 × 250000 = 289,325. 100
Hence, option D is correct.
2
A certain amount of money is lent out at compound interest at the rate of 20% per annum for two years, compounded annually. It would give Rs. 482 more if the amount is compounded half yearly. Find the principle.
» Explain it
E
Approach I: To solve this question, we can apply the net % effect formula
x + y + xy % 100
Compounded annually at rate 20% per annum for 2 years, we get
= 20 + 20 + 20 × 20 = 44% 100
Similarly, compounded half yearly at rate 10%, we get
= 10 + 10 + 10 × 10 = 21% 100
And, 21 + 10 + 21 × 10 = 33.1% 100
And, 33.1 + 10 + 33.1 × 10 = 46.41% 100
Now as per the question,
Difference between compound interest yearly and half yearly = 46.41 – 44 = 2.41%
Given, 2.41% ≡ 482
100% ≡ x
⇒ x = 482 × 100 = 20,000 2.41
Approach II:
When compounded annually, the amount received at the end of the period is
A = P [ 1 + r ] n 100
When compounded half yearly, the amount received at the end of the period is
A = P [ 1 + r/2 ] 2n 100
Let the principle be P.
Interest on this amount when compounded annually at the rate of 20% per annum = P [(1.20)2 − 1]
Interest on this amount when compounded half yearly = P [(1.10)4 − 1]
The difference between the two is Rs. 482
∴ P [(1.10)4 − 1] – P [(1.20)2 −1] = 482
∴ P [1.4641 – 1.44] = 482
∴ P = Rs. 20,000
Hence, option E is correct.
3
A man gave 50% of his savings of Rs 67,280 to his wife and divided the remaining sum between his two sons A and B of 14 and 12 years of age respectively. He divided it in such a way that each of his sons, when they attain the age of 18 years, would receive the same amount at 5% compound interest per annum. The share of B was
» Explain it
D
Total Income = 67,280
After giving 50% salary to his wife the man is left with an amount = 33,640
Let's assume the man gave Rs. x to A. Therefore B will get Rs. (33640 – x).
↙ 33640 ↘ 14 years A 12 years B x (33640 – x)
Now, as per the question A & B will be getting an equal amount with CI at 5% rate per year at the 18th year.
⇒ x ( 1 + 5 ) 4 = (33640 – x) [ 1+ 5 ] 6 100 100
⇒
x (33640 – x)
=
( 1 + 5 ) 6 100
( 1 + 5 ) 4 100
⇒ x = ( 21 × 21 ) (33640 – x) 20 20
⇒ 400 x = 33640 × 441 – 441x
⇒ 841x = 33640 × 441
x = 33640 × 441 = 40 × 441 = 17640/- 841
Therefore, at the time of divison of money, B would have got a sum = (33640 – 17640) = Rs. 16000
Hence, option D is correct.
4
Aditya and Bhushan invested 10000 each in scheme A and scheme B respectively for 3 years. Scheme A offers Simple interest @ 12% per annum and scheme B offers compound interest @ 10%. After 3 years, who will have larger amount and by how much?
» Explain it
C
Lets first calculate the total rate % that Aditya will have after 3 years:
As per the question Aditya invested at rate of 12% pa simple interst
So, for 3 years tenure he will get = 12 × 3 = 36%
And the amount that Bhushan invested at rate of 10% pa compound interest
By net% effect formula, we can calculate the total perecntage for 3 years tenure = 33.1% (sub details)
So, the difference between SI and CI = 36% – 33.1% = 2.9% (SI is more)
Here Aditya will get, 2.9% of 10000 = 290
So Aditya will have Rs. 290 more than Bhushan.
---------------------------------------------------------------------------------
Sub-details:-
Net% effect = x + y = xy % 100
For the first 2 years: Here, x = y = 10%
= 10 + 10 = 10 × 10 = 21% 100
And for the next year: Here x = 21% and y = 10%
= 21 + 10 = 21 × 10 = 33.1% 100
Hence, option C is correct.
5
A sum of Rs. 9960 was borrowed at 15/2% per annum compound interest and paid back in two years in two equal annual installments. What was the amount of each installment?
» Explain it
B
Let the each instalment be x.
x + x = 9960 ( 1 + 15 ) ( 1 + 15 ) 2 2 × 100 2 × 100
x + x = 9960 ( 1 + 3 ) ( 1 + 3 ) 2 40 40
⇒ 40 x + 1600 x = 9960 43 1849
⇒ 1720 x + 1600 x = 9960 1849
⇒ 3320 x = 9960 × 1849 ⇒ x = Rs. 5547
Hence, option B is correct. |
Question
# Let $A = \left\{ {\left( {x,{\text{ }}y} \right):{y^2} \leqslant 4x,{\text{ }}y - 2x \geqslant - 4} \right\}$. Then the area (in square units) of the region A is (A) 8 (B) 9 (C) 10 (D) 11
Hint: Here first we will consider the equations of given curves and then draw their diagram and find the point of intersection of two curves and then find the area of the shaded region using the intersection points as the lower and upper limit of integration.
The given curves are:-
${y^2} = 4x$ …………………..(1)
$y - 2x = - 4$……………………………(2)
The given curves can be drawn as:-
Now we will find the intersection points of the given curves.
Evaluating the value of x from equation 2 we get:-
$2x = y + 4 \\ \Rightarrow x = \dfrac{{y + 4}}{2} \\$
Putting this value in equation 1 we get:-
${y^2} = 4\left( {\dfrac{{y + 4}}{2}} \right)$
Simplifying it further we get:-
$\Rightarrow {y^2} = 2\left( {y + 4} \right) \\ \Rightarrow {y^2} - 2y - 8 = 0 \\$
For a standard quadratic equation $a{x^2} + bx + c = 0$ the quadratic formula for roots is given by:-
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Applying this formula for above quadratic equation we get:-
$y = \dfrac{{ - \left( { - 2} \right) \pm \sqrt {{{\left( { - 2} \right)}^2} - 4\left( 1 \right)\left( { - 8} \right)} }}{{2\left( 1 \right)}}$
Simplifying it further we get:-
$y = \dfrac{{2 \pm \sqrt {4 + 32} }}{2} \\ \Rightarrow y = \dfrac{{2 \pm \sqrt {36} }}{2} \\ \Rightarrow y = \dfrac{{2 \pm 6}}{2} \\$
Evaluating the value of y we get:-
$\Rightarrow y = \dfrac{{2 + 6}}{2};y = \dfrac{{2 - 6}}{2} \\ \Rightarrow y = \dfrac{8}{2};y = \dfrac{{ - 4}}{2} \\ \Rightarrow y = 4;y = - 2 \\$
Now putting these values in equation 1 and evaluating the respective values of x we get:-
When $y = 4$
${\left( 4 \right)^2} = 4x$
Evaluating for x we get:-
$16 = 4x \\ \Rightarrow x = \dfrac{{16}}{4} \\ \Rightarrow x = 4 \\$
When $y = - 2$
${\left( { - 2} \right)^2} = 4x$
Evaluating for x we get:-
$4 = 4x \\ \Rightarrow x = \dfrac{4}{4} \\ \Rightarrow x = 1 \\$
Hence we got the intersection points.
Let the intersection points of the two graphs be P and Q such that:
$P = \left( {1, - 2} \right)$ and $Q\left( {4,4} \right)$
Now we will evaluate the area under the curve formed by the intersection of two curves.
Hence,
${\text{Area under the curve}} = {\text{area under the line with equation }}(y - 2x = - 4) - {\text{area under the parabola with equation }}({y^2} = 4x)$
Therefore, evaluating the area with respect to y we get:-
${\text{Area under the curve}} = \int_{ - 2}^4 {\left( {\dfrac{{y + 4}}{2}} \right) - \left( {\dfrac{{{y^2}}}{4}} \right)dy}$
Now taking the LCM and solving it further we get:-
${\text{Area under the curve}} = \int_{ - 2}^4 {\left( {\dfrac{{2\left( {y + 4} \right) - {y^2}}}{4}} \right)dy} \\ {\text{ }} = \int_{ - 2}^4 {\left( {\dfrac{{2y + 8 - {y^2}}}{4}} \right)dy} \\$
Now distributing the integral we get:-
${\text{Area under the curve}} = \int_{ - 2}^4 {\left( {\dfrac{{2y}}{4}} \right)dy} + \int_{ - 2}^4 {\left( {\dfrac{8}{4}} \right)dy} - \int_{ - 2}^4 {\left( {\dfrac{{{y^2}}}{4}} \right)dy}$
Simplifying it further we get:-
${\text{Area under the curve}} = \dfrac{1}{2}\int_{ - 2}^4 {ydy} + 2\int_{ - 2}^4 {dy} - \dfrac{1}{4}\int_{ - 2}^4 {{y^2}dy}$
Now using the following formula of integration:-
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C$
We get:-
${\text{Area under the curve}} = \dfrac{1}{2}\left[ {\dfrac{{{y^2}}}{2}} \right]_{ - 2}^4 + 2\left[ y \right]_{ - 2}^4 - \dfrac{1}{4}\left[ {\dfrac{{{y^3}}}{3}} \right]_{ - 2}^4$
Simplifying it further we get:-
${\text{Area under the curve}} = \dfrac{1}{4}\left[ {{y^2}} \right]_{ - 2}^4 + 2\left[ y \right]_{ - 2}^4 - \dfrac{1}{{12}}\left[ {{y^3}} \right]_{ - 2}^4$
Now evaluating the limits we get:-
${\text{Area under the curve}} = \dfrac{1}{4}\left[ {{{\left( 4 \right)}^2} - {{\left( { - 2} \right)}^2}} \right] + 2\left[ {4 - \left( { - 2} \right)} \right] - \dfrac{1}{{12}}\left[ {{{\left( 4 \right)}^3} - {{\left( { - 2} \right)}^3}} \right]$
Simplifying it further we get:-
${\text{Area under the curve}} = \dfrac{1}{4}\left[ {16 - 4} \right] + 2\left[ {4 + 2} \right] - \dfrac{1}{{12}}\left[ {64 + 8} \right] \\ {\text{ }} = \dfrac{1}{4}\left[ {12} \right] + 2\left[ 6 \right] - \dfrac{1}{{12}}\left[ {72} \right] \\ {\text{ }} = 3 + 12 - 6 \\ {\text{ }} = 15 - 6 \\ {\text{ }} = 9{\text {sq units}} \\$
Hence the required area is 9 sq units
So, the correct answer is “Option B”.
Note: Students might make mistakes while evaluating the limits so the correct value of the limits should be should and substituted in order to get the correct answer.
Also, students should note that the general formula for integration is:
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C$ |
# Multiples of N Between A and B
Count numbers that are multiples of AND multiples of OR multiples of AND NOT multiples of (no extra conditions) Between and
A common counting problem is to determine the number of multiples of an integer n in a given range from a to b (inclusive) where a < b. There may be other conditions on the problem, such as counting the numbers that are multiples of both n AND m, multiples of EITHER n OR m, or multiples of n but NOT of m.
For small ranges it is easy to list them by hand, but for larger ranges it is more efficient to use a formula. Each condition scenario is analyzed below. You can also use the convenient calculator on the left to quickly count multiples and generate them as a list.
#### Case I: Multiples of n
To count the number of multiples of n between a and b, you need to use the floor and ceiling functions, ⌊⌋ and ⌈⌉ respectively. The number of multiples is given by the expression
⌊b/n⌋ - ⌈a/n⌉ + 1
For example, the multiples of 4 between 0 and 24 are {0, 4, 8, 12, 16, 20, 24}. To count them more quickly without having to list them, you compute
⌊24/4⌋ - ⌈0/4⌉ + 1
= 6 - 0 + 1
= 7
#### Case II: Multiples of both n AND m
To count integers that are multiples of both n and m, it suffices to count numbers that are multiples of the least common multiple (LCM) of n and m. The formula is
⌊b/y⌋ - ⌈a/y⌉ + 1
where y = LCM(n, m). For example, suppose you want to count numbers that are multiples of both 4 and 6 between 100 and 200. Since the LCM of 4 and 6 is 12, you compute
⌊200/12⌋ - ⌈100/12⌉ + 1
= 16 - 9 + 1
= 8
The exact set is {108, 120, 132, 144, 156, 168, 180, 192}.
#### Case III: Multiples of Either n OR m
To determine the number of integers that are multiples of either n or m (or both), you calculate the multiples of just n, then add the multiples of just m, and subtract the multiples of LCM(n, m). If you don't subtract this quantity, the multiples of LCM(n, m) will be counted twice. The full expression is
⌊b/n⌋ - ⌈a/n⌉ + ⌊b/m⌋ - ⌈a/m⌉ - ⌊b/y⌋ + ⌈a/y⌉ + 1
where y = LCM(n, m). For instance, suppose you need to find the number of integers between 20 and 40 that are multiples of either 4 or 6. Since LCM(4, 6) = 12, the answer is
⌊40/4⌋ - ⌈20/4⌉ + ⌊40/6⌋ - ⌈20/6⌉ - ⌊40/12⌋ + ⌈20/12⌉ + 1
= 10 - 5 + 6 - 4 - 3 + 2 + 1
= 7
The exact set is {20, 24, 28, 30, 32, 36, 40}.
#### Case IV: Multiples of n but NOT of m
The number of multiples of n that are NOT multiples of m is given by the expression
⌊b/n⌋ - ⌈a/n⌉ - ⌊b/y⌋ + ⌈a/y⌉
where y = LCM(n, m). For example how many multiples of 4 between are there between 0 and 60 that are not multiples of 6? To calculate the answer, evaluate the expression
⌊60/4⌋ - ⌈0/4⌉ - ⌊60/12⌋ + ⌈0/12⌉
= 15 - 0 - 5 + 0
= 10.
The full list is {4, 8, 16, 20, 28, 32, 40, 44, 52, 56}. |
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### Section 2.1 : Trig Function Evaluation
One of the problems with most trig classes is that they tend to concentrate on right triangle trig and do everything in terms of degrees. Then you get to a calculus course where almost everything is done in radians and the unit circle is a very useful tool.
So first off let’s look at the following table to relate degrees and radians.
Degree 0 30 45 60 90 180 270 360 Radians 0 $$\displaystyle \frac{\pi }{6}$$ $$\displaystyle \frac{\pi }{4}$$ $$\displaystyle \frac{\pi }{3}$$ $$\displaystyle \frac{\pi }{2}$$ $$\displaystyle \pi$$ $$\displaystyle \frac{{3\pi }}{2}$$ $$\displaystyle 2\pi$$
Know this table! There are, of course, many other angles in radians that we’ll see during this class, but most will relate back to these few angles. So, if you can deal with these angles you will be able to deal with most of the others.
Be forewarned, everything in most calculus classes will be done in radians!
Now, let’s look at the unit circle. Below is the unit circle with just the first quadrant filled in. The way the unit circle works is to draw a line from the center of the circle outwards corresponding to a given angle. Then look at the coordinates of the point where the line and the circle intersect. The first coordinate is the cosine of that angle and the second coordinate is the sine of that angle. There are a couple of basic angles that are commonly used. These are $$0,\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{2},\pi ,\frac{{3\pi }}{2}$$ and $$2\pi$$ and are shown below along with the coordinates of the intersections. So, from the unit circle below we can see that $$\cos \left( {\frac{\pi }{6}} \right) = \frac{{\sqrt 3 }}{2}$$ and $$\sin \left( {\frac{\pi }{6}} \right) = \frac{1}{2}$$.
Remember how the signs of angles work. If you rotate in a counter clockwise direction the angle is positive and if you rotate in a clockwise direction the angle is negative.
Recall as well that one complete revolution is $$2\pi$$, so the positive $$x$$-axis can correspond to either an angle of 0 or $$2\pi$$ (or $$4\pi$$, or $$6\pi$$, or $$- 2\pi$$, or $$- 4\pi$$, etc. depending on the direction of rotation). Likewise, the angle $$\frac{\pi }{6}$$ (to pick an angle completely at random) can also be any of the following angles:
$$\displaystyle \frac{\pi }{6} + 2\pi = \frac{{13\pi }}{6}$$ (start at $$\displaystyle \frac{\pi }{6}$$ then rotate once around counter clockwise)
$$\displaystyle \frac{\pi }{6} + 4\pi = \frac{{25\pi }}{6}$$ (start at $$\displaystyle \frac{\pi }{6}$$ then rotate around twice counter clockwise)
$$\displaystyle \frac{\pi }{6} - 2\pi = - \frac{{11\pi }}{6}$$ (start at $$\displaystyle \frac{\pi }{6}$$ then rotate once around clockwise)
$$\displaystyle \frac{\pi }{6} - 4\pi = - \frac{{23\pi }}{6}$$ (start at $$\displaystyle \frac{\pi }{6}$$ then rotate around twice clockwise)
etc.
In fact, $$\frac{\pi }{6}$$ can be any of the following angles $$\frac{\pi }{6} + 2\pi \,n\,,\;\;n = 0,\, \pm 1,\, \pm 2,\, \pm 3,\, \ldots$$ In this case $$n$$ is the number of complete revolutions you make around the unit circle starting at $$\frac{\pi }{6}$$. Positive values of $$n$$ correspond to counter clockwise rotations and negative values of $$n$$ correspond to clockwise rotations.
So, why did I only put in the first quadrant? The answer is simple. If you know the first quadrant then you can get all the other quadrants from the first. You’ll see this in the following examples.
Find the exact value of each of the following. In other words, don’t use a calculator. Show All SolutionsHide All Solutions
1. $$\sin \left( {\frac{{2\pi }}{3}} \right)$$ and $$\sin \left( { - \frac{{2\pi }}{3}} \right)$$
Show Solution
The first evaluation here uses the angle $$\frac{{2\pi }}{3}$$. Notice that $$\frac{{2\pi }}{3} = \pi - \frac{\pi }{3}$$. So $$\frac{{2\pi }}{3}$$ is found by rotating up $$\frac{\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$\frac{{2\pi }}{3}$$ will be a mirror image of the line for $$\frac{\pi }{3}$$ only in the second quadrant. The coordinates for $$\frac{{2\pi }}{3}$$ will be the coordinates for $$\frac{\pi }{3}$$ except the $$x$$ coordinate will be negative.
Likewise, for $$- \frac{{2\pi }}{3}$$we can notice that $$- \frac{{2\pi }}{3} = - \pi + \frac{\pi }{3}$$, so this angle can be found by rotating down $$\frac{\pi }{3}$$ from the negative $$x$$-axis. This means that the line for $$- \frac{{2\pi }}{3}$$ will be a mirror image of the line for $$\frac{\pi }{3}$$ only in the third quadrant and the coordinates will be the same as the coordinates for $$\frac{\pi }{3}$$ except both will be negative.
Both of these angles along with their coordinates are shown on the following unit circle.
From this unit circle we can see that $$\sin \left( {\frac{{2\pi }}{3}} \right) = \frac{{\sqrt 3 }}{2}$$and $$\sin \left( { - \frac{{2\pi }}{3}} \right) = - \frac{{\sqrt 3 }}{2}$$.
This leads to a nice fact about the sine function. The sine function is called an odd function and so for ANY angle we have
$\sin \left( { - \theta } \right) = - \sin \left( \theta \right)$
2. $$\cos \left( {\frac{{7\pi }}{6}} \right)$$ and $$\cos \left( { - \frac{{7\pi }}{6}} \right)$$
Show Solution
For this example, notice that $$\frac{{7\pi }}{6} = \pi + \frac{\pi }{6}$$ so this means we would rotate down $$\frac{\pi }{6}$$ from the negative $$x$$-axis to get to this angle. Also $$- \frac{{7\pi }}{6} = - \pi - \frac{\pi }{6}$$ so this means we would rotate up $$\frac{\pi }{6}$$ from the negative $$x$$-axis to get to this angle. These are both shown on the following unit circle along with appropriate coordinates for the intersection points.
From this unit circle we can see that $$\cos \left( {\frac{{7\pi }}{6}} \right) = - \frac{{\sqrt 3 }}{2}$$and $$\cos \left( { - \frac{{7\pi }}{6}} \right) = - \frac{{\sqrt 3 }}{2}$$. In this case the cosine function is called an even function and so for ANY angle we have
$\cos \left( { - \theta } \right) = \cos \left( \theta \right)$
3. $$\tan \left( { - \frac{\pi }{4}} \right)$$ and $$\tan \left( {\frac{{7\pi }}{4}} \right)$$
Show Solution
Here we should note that $$\frac{{7\pi }}{4} = 2\pi - \frac{\pi }{4}$$ so $$\frac{{7\pi }}{4}$$ and $$- \frac{\pi }{4}$$ are in fact the same angle! The unit circle for this angle is
Now, if we remember that $$\tan \left( x \right) = \frac{{\sin \left( x \right)}}{{\cos \left( x \right)}}$$ we can use the unit circle to find the values the tangent function. So,
$\tan \left( \frac{7\pi }{4} \right)=\tan \left( -\frac{\pi }{4} \right)=\frac{\sin \left( -{\pi }/{4}\; \right)}{\cos \left( -{\pi }/{4}\; \right)}=\frac{-{\sqrt{2}}/{2}\;}{{\sqrt{2}}/{2}\;}=-1$
On a side note, notice that $$\tan \left( {\frac{\pi }{4}} \right) = 1$$ and we se can see that the tangent function is also called an odd function and so for ANY angle we will have
$\tan \left( { - \theta } \right) = - \tan \left( \theta \right)$
4. $$\sin \left( {\frac{{9\pi }}{4}} \right)$$
Show Solution
For this problem let’s notice that $$\frac{{9\pi }}{4} = 2\pi + \frac{\pi }{4}$$. Now, recall that one complete revolution is $$2\pi$$. So, this means that $$\frac{{9\pi }}{4}$$ and $$\frac{\pi }{4}$$ are at the same point on the unit circle. Therefore,
$\sin \left( {\frac{{9\pi }}{4}} \right) = \sin \left( {2\pi + \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}$
This leads us to a very nice fact about the sine function. The sine function is an example of a periodic function. Periodic function are functions that will repeat themselves over and over. The “distance” that you need to move to the right or left before the function starts repeating itself is called the period of the function.
In the case of sine the period is $$2\pi$$. This means the sine function will repeat itself every $$2\pi$$. This leads to a nice formula for the sine function.
$\sin \left( {x + 2\pi n} \right) = \sin \left( x \right)\hspace{0.25in}\,\,\,\,\,\,\,\,n = 0,\, \pm 1,\,\, \pm 2,\, \ldots$
Notice as well that because
$csc \left( x \right) = \frac{1}{{\sin \left( x \right)}}$
we can say the same thing about cosecant.
$\csc \left( {x + 2\pi n} \right) = \csc \left( x \right)\hspace{0.25in}\,\,\,\,\,\,\,\,n = 0,\, \pm 1,\,\, \pm 2,\, \ldots$
Well, actually we should be careful here. We can say this provided $$x \ne n\pi$$ since sine will be zero at these points and so cosecant won’t exist there!
5. $$\sec \left( {\frac{{25\pi }}{6}} \right)$$
Show Solution
Here we need to notice that $$\frac{{25\pi }}{6} = 4\pi + \frac{\pi }{6}$$. In other words, we’ve started at $$\frac{\pi }{6}$$ and rotated around twice to end back up at the same point on the unit circle. This means that
$\sec \left( {\frac{{25\pi }}{6}} \right) = \sec \left( {4\pi + \frac{\pi }{6}} \right) = \sec \left( {\frac{\pi }{6}} \right)$
Now, let’s also not get excited about the secant here. Just recall that
$\sec \left( x \right) = \frac{1}{{\cos \left( x \right)}}$
and so all we need to do here is evaluate a cosine! Therefore,
$\sec \left( \frac{25\pi }{6} \right)=\sec \left( \frac{\pi }{6} \right)=\frac{1}{\cos \left( \frac{\pi }{6} \right)}=\frac{1}{{}^{\sqrt{3}}/{}_{2}}=\frac{2}{\sqrt{3}}$
We should also note that cosine and secant are periodic functions with a period of $$2\pi$$. So,
\begin{align*}\begin{aligned}\cos \left( {x + 2\pi n} \right) & = \cos \left( x \right)\\ \sec \left( {x + 2\pi n} \right) & = \sec \left( x \right) \end{aligned}& \hspace{0.5in}n = 0, \pm 1, \pm 2,\, \ldots \end{align*}
6. $$\tan \left( {\frac{{4\pi }}{3}} \right)$$
Show Solution
To do this problem it will help to know that tangent (and hence cotangent) is also a periodic function, but unlike sine and cosine it has a period of $$\pi$$.
\begin{align*}\begin{aligned}\tan \left( {x + \pi n} \right) & = \tan \left( x \right)\\ \cot \left( {x + \pi n} \right) & = \cot \left( x \right) \end{aligned}& \hspace{0.5in}n = 0, \pm 1, \pm 2,\, \ldots \end{align*}
So, to do this problem let’s note that $$\frac{{4\pi }}{3} = \pi + \frac{\pi }{3}$$. Therefore,
$\tan \left( {\frac{{4\pi }}{3}} \right) = \tan \left( {\pi + \frac{\pi }{3}} \right) = \tan \left( {\frac{\pi }{3}} \right) = \sqrt 3$
#### Trig Evaluation Final Thoughts
As we saw in the previous examples if you know the first quadrant of the unit circle you can find the value of ANY trig function (not just sine and cosine) for ANY angle that can be related back to one of those shown in the first quadrant. This is a nice idea to remember as it means that you only need to memorize the first quadrant and how to get the angles in the remaining three quadrants!
In these problems I used only “basic” angles, but many of the ideas here can also be applied to angles other than these “basic” angles as we’ll see in Solving Trig Equations. |
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# A sum of Rs.1000 is invested at 8% simple interest per annum. Calculate the interest at the end of 1,2,3 etc years. Is the sequence of interests an AP? Find the interest at the end of 30 years.
Last updated date: 21st Jul 2024
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Hint: calculate the interest at the end of each year using the simple interest formula and then try to generalise the interests obtained in order to show it as a sequence of AP or not. Then using the formula for the nth term of AP series find the interest at the end of 30 years.
Formula for simple interest is
$\text{SI=}\dfrac{P\times T\times R}{100}............\left( 1 \right)$
Where SI = simple interest
P= principal
R= rate of interest
T= time
Let us assume that the interests at the end of the years as
$\text{S}{{\text{I}}_{1}},\text{S}{{\text{I}}_{2}},\text{S}{{\text{I}}_{3}},\text{etc}$
Given in the question that
P=1000
R=8
Now to calculate the interest at the end of the 1 year we need to take
T=1
By substituting the values of P,R,T in the above simple interest formula (1) we get,
\begin{align} & \text{S}{{\text{I}}_{1}}\text{=}\frac{P\times T\times R}{100} \\ & \Rightarrow \text{S}{{\text{I}}_{1}}=\frac{1000\times 8\times 1}{100} \\ & \Rightarrow \text{S}{{\text{I}}_{1}}=10\times 8 \\ & \therefore \text{S}{{\text{I}}_{1}}=80 \\ \end{align}
Now the interest at the end of the 2 year can be calculated similarly by substituting the values
P=1000
R=8
T=2 in the above simple interest formula (1)
\begin{align} & \text{S}{{\text{I}}_{2}}\text{=}\frac{P\times T\times R}{100} \\ & \Rightarrow \text{S}{{\text{I}}_{2}}=\frac{1000\times 8\times 2}{100} \\ & \Rightarrow \text{S}{{\text{I}}_{2}}=10\times 8\times 2 \\ & \Rightarrow \text{S}{{\text{I}}_{2}}=80\times 2 \\ & \Rightarrow \text{S}{{\text{I}}_{2}}=160 \\ & \therefore \text{S}{{\text{I}}_{2}}=2\times \text{S}{{\text{I}}_{1}}\text{ }\left[ \because \text{S}{{\text{I}}_{1}}=80 \right] \\ \end{align}
Similarly interest at the end of 3 years can be calculated by substituting the respective values in simple interest formula (1)
P=1000
R=8
T=3
\begin{align} & \text{S}{{\text{I}}_{3}}\text{=}\dfrac{P\times T\times R}{100} \\ & \Rightarrow \text{S}{{\text{I}}_{3}}=\dfrac{1000\times 8\times 3}{100} \\ & \Rightarrow \text{S}{{\text{I}}_{3}}=10\times 8\times 3 \\ & \Rightarrow \text{S}{{\text{I}}_{3}}=80\times 3 \\ & \Rightarrow \text{S}{{\text{I}}_{3}}=240 \\ & \therefore \text{S}{{\text{I}}_{3}}=3\times \text{S}{{\text{I}}_{1}}\text{ }\left[ \because \text{S}{{\text{I}}_{1}}=80 \right] \\ \end{align}
Similarly for all other SI also we can write in terms of SI1 [SI1=80]
Now by considering the interests at the end of 1,2,3 years SI1 ,SI2 , SI3 we can observe that
$\text{S}{{\text{I}}_{2}}=2\times \text{S}{{\text{I}}_{1}}$
$\text{S}{{\text{I}}_{3}}=3\times \text{S}{{\text{I}}_{1}}$
etc
Hence , interest at the end of any year let’s suppose be n can be generalised as
$\text{S}{{\text{I}}_{n}}=n\times \text{S}{{\text{I}}_{1}}$
Here we can also observe that
$\text{S}{{\text{I}}_{2}}-\text{S}{{\text{I}}_{1}}=\text{S}{{\text{I}}_{1}}$
$\text{S}{{\text{I}}_{3}}-\text{S}{{\text{I}}_{2}}=\text{S}{{\text{I}}_{1}}$
etc
$\text{S}{{\text{I}}_{n}}-\text{S}{{\text{I}}_{n-1}}=\text{S}{{\text{I}}_{1}}$
As we already know that from the definition,
A sequence in which the difference of two consecutive terms is constant, is called Arithmetic Progression (AP)
Hence the given series of interests at the end of the years 1,2,3…. forms an AP of common difference SI1 i.e., 80
As we know that in an AP series the $nth\text{ term}$ can be calculated by using the formula
${{\text{a}}_{n}}=a+\left( n-1 \right)d..........(2)$
Where a= first term
d=common difference
now interest at the end of 30 years can be calculated by using the above formula of AP (2)
\begin{align} & {{\text{a}}_{n}}=a+\left( n-1 \right)d \\ & a=\text{S}{{\text{I}}_{1}}=80 \\ & d=\text{S}{{\text{I}}_{1}}=80 \\ & n=30 \\ \end{align}
Now by substituting the values in the above formula by assuming interest at ath end of 30 years as $\text{S}{{\text{I}}_{30}}$ we get,
\begin{align} & \text{S}{{\text{I}}_{30}}=\text{S}{{\text{I}}_{1}}+\left( n-1 \right)\text{S}{{\text{I}}_{1}} \\ & \Rightarrow \text{S}{{\text{I}}_{30}}=80+(30-1)80 \\ & \Rightarrow \text{S}{{\text{I}}_{30}}=80+29\times 80 \\ & \Rightarrow \text{S}{{\text{I}}_{30}}=(1+29)\times 80 \\ & \Rightarrow \text{S}{{\text{I}}_{30}}=30\times 80 \\ & \therefore \text{S}{{\text{I}}_{30}}=2400 \\ \end{align}
Hence, the interest at the end of 30 years is Rs.2400
Note:
We need to write the interests at the end of 2,3 years in terms of interest at the end of 1 year so that we can get a generalised form to calculate the interest at the end of any year easily.
Interest at the end of 30 years can also be calculated by using the simple interest formula instead of using the $nth\text{ term}$ of an AP series formula
\begin{align} & \text{SI=}\dfrac{P\times T\times R}{100} \\ & \Rightarrow \text{S}{{\text{I}}_{30}}=\frac{1000\times 30\times 8}{100} \\ & \Rightarrow \text{S}{{\text{I}}_{30}}=10\times 30\times 8 \\ & \Rightarrow \text{S}{{\text{I}}_{30}}=300\times 8 \\ & \therefore \text{S}{{\text{I}}_{30}}=2400 \\ \end{align}
Which gives the same result in either ways.
By writing the generalised form for the interests at the end of each year helps in finding whether the sequence is in AP or not. |
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# Excluded Values for Rational Expressions
## Excluding values that result in division by zero
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Excluded Values for Rational Expressions
What if you had a rational expression like $\frac{x + 2}{x^2 + 3x + 2}$ ? How could you simplify it? After completing this Concept, you'll be able to reduce rational expressions like this one to their simplest terms and find their excluded values.
### Watch This
Watch this video for more examples of how to simplify rational expressions.
### Guidance
A simplified rational expression is one where the numerator and denominator have no common factors. In order to simplify an expression to lowest terms , we factor the numerator and denominator as much as we can and cancel common factors from the numerator and the denominator.
Simplify Rational Expressions
#### Example A
Reduce each rational expression to simplest terms.
a) $\frac{4x-2}{2x^2+x-1}$
b) $\frac{x^2-2x+1}{8x-8}$
c) $\frac{x^2-4}{x^2-5x+6}$
Solution
a) $\text{Factor the numerator and denominator completely:} \qquad \frac{2(2x-1)}{(2x-1)(x+1)}\!\\\\\text{Cancel the common factor} \ (2x - 1): \qquad \qquad \qquad \qquad \qquad \frac{2}{x+1}$
b) $\text{Factor the numerator and denominator completely:} \qquad \frac{(x-1)(x-1)}{8(x-1)}\!\\\\\text{Cancel the common factor}\ (x - 1): \qquad \qquad \qquad \qquad \qquad \ \ \frac{x-1}{8}$
c) $\text{Factor the numerator and denominator completely:} \qquad \frac{(x-2)(x+2)}{(x-2)(x-3)}\!\\\\\text{Cancel the common factor} (x - 2): \qquad \qquad \qquad \qquad \qquad \quad \frac{x+2}{x-3}$
When reducing fractions, you are only allowed to cancel common factors from the denominator but NOT common terms. For example, in the expression $\frac{(x+1) \cdot (x-3)}{(x+2) \cdot (x-3)}$ , we can cross out the $(x - 3)$ factor because $\frac{(x-3)}{(x-3)}=1$ . But in the expression $\frac{x^2+1}{x^2-5}$ we can’t just cross out the $x^2$ terms.
Why can’t we do that? When we cross out terms that are part of a sum or a difference, we’re violating the order of operations (PEMDAS). Remember, the fraction bar means division. When we perform the operation $\frac{x^2+1}{x^2-5}$ , we’re really performing the division $(x^2+1) \div (x^2-5)$ — and the order of operations says that we must perform the operations inside the parentheses before we can perform the division.
Using numbers instead of variables makes it more obvious that canceling individual terms doesn’t work. You can see that $\frac{9+1}{9-5}=\frac{10}{4}=2.5$ — but if we canceled out the 9’s first, we’d get $\frac{1}{-5}$ , or -0.2, instead.
Find Excluded Values of Rational Expressions
Whenever there’s a variable expression in the denominator of a fraction, we must remember that the denominator could be zero when the independent variable takes on certain values. Those values, corresponding to the vertical asymptotes of the function, are called excluded values. To find the excluded values, we simply set the denominator equal to zero and solve the resulting equation.
#### Example B
Find the excluded values of the following expressions.
a) $\frac{x}{x+4}$
b) $\frac{2x+1}{x^2-x-6}$
Solution
a) $\text{When we set the denominator equal to zero we obtain:} \quad \ \ x+4=0 \Rightarrow x=-4\!\\\\\text{So} \ \mathbf{-4} \ \text{is the excluded value.}$
b) $\text{When we set the denominator equal to zero we obtain:} \qquad x^2-x-6=0\!\\\\\text{Solve by factoring:} \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \qquad (x-3)(x+2)=0\!\\\\{\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \Rightarrow x=3 \ \text{and}\ x = -2\!\\\\\text{So}\ \mathbf{3}\ \mathbf{and}\ \mathbf{-2} \ \text{are the excluded values.}$
Removable Zeros
Removable zeros are those zeros from the original expression, but is not a zero for the simplified version of the expression. However, we have to keep track of them, because they were zeros in the original expression. This is illustrated in the following examples.
#### Example C
Determine the removable values of $\frac{4x-2}{2x^2+x-1}$ .
Solution:
Notice that in the expressions in Example A, we removed a division by zero when we simplified the problem. For instance, we rewrote $\frac{4x-2}{2x^2+x-1}$ as $\frac{2(2x-1)}{(2x-1)(x+1)}$ . The denominator of this expression is zero when $x = \frac{1}{2}$ or when $x = -1$ .
However, when we cancel common factors, we simplify the expression to $\frac{2}{x+1}$ . This reduced form allows the value $x = \frac{1}{2}$ , so $x = -1$ is its only excluded value.
Technically the original expression and the simplified expression are not the same. When we reduce a radical expression to its simplest form, we should specify the removed excluded value. In other words, we should write our final answer as $\frac{4x-2}{2x^2+x-1}=\frac{2}{x+1}, x \neq \frac{1}{2}$ .
#### Example D
Determine the removable values of the expressions from Example A parts b and c.
Solution:
We should write the answer from Example A, part b as $\frac{x^2-2x+1}{8x-8}=\frac{x-1}{8}, x \neq 1$ .
The answer from Example A, part c as $\frac{x^2-4}{x^2-5x+6}=\frac{x+2}{x-3}, x \neq 2$ .
Watch this video for help with the Examples above.
### Vocabulary
• Whenever there’s a variable expression in the denominator of a fraction, we must remember that the denominator could be zero when the independent variable takes on certain values. Those values, corresponding to the vertical asymptotes of the function, are called excluded values.
• Removable zeros are those zeros from the original expression, but is not a zero for the simplified version of the expression.
### Guided Practice
Find the excluded values of $\frac{4}{x^2-5x}$ .
Solution
$\text{When we set the denominator equal to zero we obtain:} \quad \ \ x^2-5x=0\!\\\\\text{Solve by factoring:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ x(x-5)=0\!\\\\{\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \Rightarrow x=0 \ \text{and} \ x = 5\!\\\\\text{So} \ \mathbf{0 \ and \ 5}\ \text{are the excluded values.}$
### Practice
Reduce each fraction to lowest terms.
1. $\frac{4}{2x-8}$
2. $\frac{x^2+2x}{x}$
3. $\frac{9x+3}{12x+4}$
4. $\frac{6x^2+2x}{4x}$
5. $\frac{x-2}{x^2-4x+4}$
6. $\frac{x^2-9}{5x+15}$
7. $\frac{x^2+6x+8}{x^2+4x}$
8. $\frac{2x^2+10x}{x^2+10x+25}$
9. $\frac{x^2+6x+5}{x^2-x-2}$
10. $\frac{x^2-16}{x^2+2x-8}$
11. $\frac{3x^2+3x-18}{2x^2+5x-3}$
12. $\frac{x^3+x^2-20x}{6x^2+6x-120}$
Find the excluded values for each rational expression.
1. $\frac{2}{x}$
2. $\frac{4}{x+2}$
3. $\frac{2x-1}{(x-1)^2}$
4. $\frac{3x+1}{x^2-4}$
5. $\frac{x^2}{x^2+9}$
6. $\frac{2x^2+3x-1}{x^2-3x-28}$
7. $\frac{5x^3-4}{x^2+3x}$
8. $\frac{9}{x^3+11x^2+30x}$
9. $\frac{4x-1}{x^2+3x-5}$
10. $\frac{5x+11}{3x^2-2x-4}$
11. $\frac{x^2-1}{2x^2+x+3}$
12. $\frac{12}{x^2+6x+1}$
13. In an electrical circuit with resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance. $\frac{1}{R_c}=\frac{1}{R_1}+\frac{1}{R_2}$ . If $R_1 = 25 \ \Omega$ and the total resistance is $R_c = 10 \ \Omega$ , what is the resistance $R_2$ ?
14. Suppose that two objects attract each other with a gravitational force of 20 Newtons. If the distance between the two objects is doubled, what is the new force of attraction between the two objects?
15. Suppose that two objects attract each other with a gravitational force of 36 Newtons. If the mass of both objects was doubled, and if the distance between the objects was doubled, then what would be the new force of attraction between the two objects?
16. A sphere with radius $R$ has a volume of $\frac{4}{3} \pi R^3$ and a surface area of $4 \pi R^2$ . Find the ratio the surface area to the volume of a sphere.
17. The side of a cube is increased by a factor of 2. Find the ratio of the old volume to the new volume.
18. The radius of a sphere is decreased by 4 units. Find the ratio of the old volume to the new volume.
### Vocabulary Language: English
Oblique Asymptote
Oblique Asymptote
An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Rational Expression
Rational Expression
A rational expression is a fraction with polynomials in the numerator and the denominator.
Vertical Asymptote
Vertical Asymptote
A vertical asymptote is a vertical line marking a specific value toward which the graph of a function may approach, but will never reach. |
Unlike geometric series and p-series, a power series often converges or diverges based on its x value. This leads to a new concept when dealing with power series: the interval of convergence.
The interval of convergence for a power series is the set of x values for which that series converges.
## The interval of convergence is never empty
Every power series converges for some value of x. That is, the interval of convergence for a power series is never the empty set.
Although this fact has useful implications, it’s actually pretty much a no-brainer. For example, take a look at the following power series:
When x = 0, this series evaluates to 1 + 0 + 0 + 0 + ..., so it obviously converges to 1. Similarly, take a peek at this power series:
This time, when x = –5, the series converges to 0, just as trivially as the last example.
Note that in both of these examples, the series converges trivially at x = a for a power series centered at a.
## Three possibilities for the interval of convergence
Three possibilities exist for the interval of convergence of any power series:
• The series converges only when x = a.
• The series converges on some interval (open or closed at either end) centered at a.
• The series converges for all real values of x.
For example, suppose that you want to find the interval of convergence for:
This power series is centered at 0, so it converges when x = 0. Using the ratio test, you can find out whether it converges for any other values of x. To start out, set up the following limit:
To evaluate this limit, start out by canceling xn in the numerator and denominator:
Next, distribute to remove the parentheses in the numerator:
As it stands, this limit is of the form
so apply L’Hopital’s Rule, differentiating over the variable n:
From this result, the ratio test tells you that the series:
• Converges when –1 < x < 1
• Diverges when x < –1 and x > 1
• May converge or diverge when x = 1 and x = –1
Fortunately, it’s easy to see what happens in these two remaining cases. Here’s what the series looks like when x = 1:
Clearly, the series diverges. Similarly, here’s what it looks like when x = –1:
This alternating series swings wildly between negative and positive values, so it also diverges.
As a final example, suppose that you want to find the interval of convergence for the following series:
This series is centered at 0, so it converges when x = 0. The real question is whether it converges for other values of x. Because this is an alternating series, you apply the ratio test to the positive version of it to see whether you can show that it’s absolutely convergent:
First off, you want to simplify this a bit:
Next, you expand out the exponents and factorials:
At this point, a lot of canceling is possible:
This time, the limit falls between –1 and 1 for all values of x. This result tells you that the series converges absolutely for all values of x, so the alternating series also converges for all values of x. |
Q:
# What is 15/6 Divided by 1/12?
Accepted Solution
A:
What is 15/6 Divided by 1/12? Methods Breaking down the problem: First, let’s break down each piece of the problem, as each piece is important. We have the first fraction, or the dividend, which can be divided by its numerator, 15, and its denominator, 6. We also have the first fraction, or the divisor, which can be broken down into its numerator, 1, and its denominator, 12: Numerator of the dividend: 15 Denominator of the dividend: 6 Numerator of the divisor: 1 Denominator of the divisor: 12 So, what is 15/6 divided by 1/12? Let’s work through the problem, and find the answer in both fraction and decimal forms. What is 15/6 Divided by 1/12, Step-by-step First let’s set up the problem: 15 6 ÷ 1 12 \frac{15}{6} ÷ \frac{1}{12} 6 15 ÷ 12 1 Step 1: Interestingly, the first step to solving a division problem between two fractions is to multiply. First, you multiply the numerator of the dividend, 15, by the denominator of the divisor, 12. 15 x 12 = 180 Step 2: Then, multiply the denominator of the dividend, 6, by the numerator of the divisor, 1: 6 x 1 = 6 This will be the denominator of the answer. Step 3: Put the two answers together into one fraction, and this will be the answer to the problem in fraction form: 180 6 \frac{180}{6} 6 180 = 30 1 \frac{30}{1} 1 30 A fraction that has 1 as its denominator is an improper fraction. So, we should simplify this to just be 30 Because 30 is a whole number, there is no reason to write the answer in decimal form. So, 15/6 divided by 1/12 = 30. Practice Other Division Problems Like This One If this problem was a little difficult or you want to practice your skills on another one, give it a go on any one of these too! What is 4/12 divided by 42? 84 divided by what equals 26? What divided by 23 equals 11? What is 5/6 divided by 10/19? What is 80 divided by 11/20? |
#### EQUATIONS CONTAINING VARIABLES UNDER ONE OR MORE RADICALS
Note:
• In order to solve for x, you must isolate x.
• In order to isolate x, you must remove it from under the radical.
• If there are three radicals in the equation, isolate one of the radicals.
• Then raise both sides of the equation to a power equal to the index of the isolated radical.
• Raise both sides of the equation to a power equal to the index of the isolated radical.
• You should now have a polynomial equation. Solve it.
• Remember that you did not start out with a polynomial; therefore, there may be extraneous solutions. Therefore, you must check your answers.
Example 3:
First make a note of the fact that you cannot take the square root of a negative number. Therefore, the term is valid only if the second term is valid if , and the term is valid only if The equation is valid if all three terms are valid, therefore the domain is restricted to the common domain of the three terms or the set of real numbers .
Square both sides of the equation and simplify.
Isolate the term and simplify.
Square both sides of the equation and simplify.
Check the solution by substituting 3 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
• Left Side:
• Right Side:
Since the left side of the original equation equals the right side of the original equation after we substituted 3 for x, then x=3 is a solution.
You can also check the answer by graphing the equation:
The graph represents the right side of the original equation minus the left side of the original equation. The x-intercept(s) of this graph is(are) the solution(s). Since there is just one x-intercept at 3, then the only solution is x=3.
### If you would like to test yourself by working some problems similar to this example, click on Problem
. If you would like to go back to the equation table of contents, click on Contents.
[Algebra] [Trigonometry]
[Geometry] [Differential Equations]
[Calculus] [Complex Variables] [Matrix Algebra] |
Posted on
09
Feb 2021
## Triangle Inequality Rule
One of the less-common but still need-to-know rules tested on the GMAT is the “triangle inequality” rule, which allows you to draw conclusions about the length of the third side of a triangle given information about the lengths of the other two sides.
Often times, this rule is presented in two parts, but I find it is easiest to condense it into one, simple part that concerns a sum and a difference. Here’s what I mean, and we’ll use a SCENARIO:
Suppose we have a triangle that has two sides of length 3 and 5:
What can we say about the length of the third side? Of course, we can’t nail down a single definitive value for that length, but we can actually put a limit on its range. That range is simply the difference and the sum of the lengths of the other two sides, non-inclusive.
So, in this case, since the difference between the lengths of the other two sides is 2, and their sum is 8, we can say for sure that the third side of this triangle must have a length between 2 and 8, non-inclusive. [Algebraically, this reads as (5-3) < x < (5+3) OR 2 < x < 8.]
If you’d like to see that put into words:
**The length of any side of a triangle must be shorter than the sum of the other two side lengths and longer than the difference of the other two side lengths.**
It’s important to note that this works for any triangle. But why did we say non-inclusive? Well, let’s look at what would happen if we included the 8 in the above example. Imagine a “triangle” with lengths 3, 5, and 8. Can you see the problem? (Think about it before reading the next paragraph.)
Imagine a twig of length 3 inches and another of length 5 inches. How would you form a geometric figure of length 8 inches? You’d simply join the two twigs in a straight line to form a longer, single twig of 8 inches. It would be impossible to form a triangle with a side of 8 inches with the original two twigs.
If you wanted to form a triangle with the twigs of 3 and 5, you’d have to “break” the longer twig of 8 inches and bend the two twigs at an angle for an opportunity to have a third side, guaranteed to be shorter than 8 inches:
The same logic would hold for the other end of the range (we couldn’t have a triangle of 3, 5, and 2, as the only way to form a length of 5 from lengths of 2 and 3 would be to form a longer line segment of 5.)
Now that we’ve covered the basics, let’s dive into a few problems, starting with this Official Guide problem:
If k is an integer and 2 < k < 7, for how many different values of k is there a triangle with sides of lengths 2, 7, and k?
(A) one
(B) two
(C) three
(D) four
(E) five
As usual with the GMAT, it’s one thing to know the rule, but it’s another when you’re presented with a carefully worded question that tests your ability to pay close attention to detail. First, we are told that two of the lengths of the triangle are 2 and 7. What does that mean for the third side, given the triangle inequality rule? We know the third side must have a length between 5 (the difference between the two sides) and 9 (the sum of the two sides).
Here, you can actually use the answer choices to your advantage, at least to eliminate some answers. Notice that k is specified as an integer. How many integers do we know now are possible? Well, if k must be between 5 and 9 (and remember, it’s non-inclusive), the only options possibly available to us are 6, 7, and 8. That means a maximum of three possible values of k, thus eliminating answers D and E.
Since the GMAT is a time-intensive test, you might have to end up guessing now and then, so if you can strategically eliminate answers, it increases your chances of guessing correctly.
Now for this problem, there’s another condition given, namely that 2 < k < 7. We already determined that k must be 6, 7, or 8. However, of those numbers, only 6 fits in the given range 2 < k < 7. This means that 6 is the only legal value that fits for k. The correct answer is A.
## Note
It’s important to emphasize that the eliminate answers strategy is not a mandate. We’re simply presenting it as an option that works here because it is useful on many GMAT problems and should be explored and practiced as often as possible.
Check out the following links for our other articles on triangles and their properties:
By: Rich Zwelling, Apex GMAT Instructor
Date: 9th February, 2021 |
# 24 solver with fractions
Here, we debate how 24 solver with fractions can help students learn Algebra. Our website can solve math problems for you.
## The Best 24 solver with fractions
This 24 solver with fractions supplies step-by-step instructions for solving all math troubles. To solve a right triangle, the Pythagorean theorem is used. This states that if a triangle has a hypotenuse of , the triangles legs are in the ratio . And so by substituting the values into this theorem, we can find out the lengths of the two legs of our triangle: The hypotenuse can also be found by using similar triangles. For example, if we know one leg is and want to find the hypotenuse, we would use the formula: Taking these steps shows that we have solved our right triangle. To see why this works, let’s use an example: We start with a right triangle with legs of and . We know that since the angle between them is 60°. We also know that since the angle between them is 60° and since they both bisect angles. This means that is equal to one-half times . Therefore, . Therefore, . Therefore, . Therefore, . Therefore, , which is what we wanted to find!
Vertical asymptote is the line that shows the maximum value of a variable. When you have to do a calculation, you may need to find the value of vertical asymptote, so here are some tips on how to solve vertical asymptote: In order to solve vertical asymptote, you have to know what is the variable you are dealing with and its units. For example, if your variable is temperature in Celsius and you want to find out how hot it is, take the temperature in Celsius, convert it into Fahrenheit and then find out how high it goes. Keep on doing this until you reach 100 °F. If this works, then your answer is close to vertical asymptote. On the other hand, if this does not work out, then you will have to do more calculations until you get closer to vertical asymptote. Not only that, but there are formulas that can help you find out exactly how much hotter or cooler it is than the maximum value of your unit.
The best log problem solver can be described as the person who can take care of log problems in the most efficient way possible. This person should always have a good understanding of how to deal with log problems and be able to apply various techniques and strategies when needed. They should also be able to interpret the information from logs, which are often presented in a complex manner. This ability is vital for anyone who wants to become a successful log problem solver. The best log problem solver will know how to find the root cause of any given log problem, which is crucial in order to get rid of it as soon as possible. It is also important for them to be able to resolve any given log problem quickly and efficiently. If you are interested in learning more about this topic, read this article about why you should become a better log problem solver:
Solving for x is a technique used to determine the value of a variable that has been defined in terms of another variable or expression. Solving for x is also known as substitution or elimination. It can be performed by isolating the variable and replacing it with its value. If the variable is a constant (i.e. a number or a letter), then its value can be substituted directly into the problem at hand to obtain the desired result. However, if the variable is an expression (i.e. a mathematical operation), then it must be rewritten using its value in place of each operation (i.e. "2" becomes "2", "4" becomes "4", etc.). After all of the operations have been replaced, the expression can then be simplified by removing any variables that have already been accounted for. Once this process has been completed, it may be necessary to perform some simple arithmetic operations to make sure that the final result is correct.
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### Xuxa Green
College math questions Math problem search Synthetic division problem solver Factor a binomial Help on math Secant line solver |
Linear equations
Linear equations are all equations that have the following form: y = ax + b, a and b are real numbers and are called constants.
In y = ax + b, x is called independent variable and y is called dependent variable.
Some examples of linear equations are y = 2x + 5 with a = 2 and b = 5, y = -3x + 2 with a = -3 and b = 2, and y = 4x + - 1 with a = 4 and b = -1
Linear and non-linear equations
Notice that all linear equations are equations of degree 1, meaning that the variable cannot be raised to a power other than 1. If the power of the equation is not 1, you are dealing with non-linear equations.
For example, y = 3x, y = x2 - 3x + 2, y = √x, and y = (x - 3) / x + 2 are non-linear equations.
It may not be obvious why y = (x - 3) / x + 2 is not a linear equation. Let us rewrite the equation by multiplying each side of the equation by x + 2.
y(x + 2) = (x - 3)
xy + 2y = x - 3
xy + 2y - x + 3 = 0
The term that is raised to a power other than 1 is xy. This term can also give us the degree of this equation. Just add the exponents for each variable to get the degree. Therefore the degree is 2 since xy = x1y1
y = √x is not a linear equation either since it can be written as y = x1/2 and x1/2 is not an equation of degree 1.
Forms of linear equations
A linear equation can have different forms. There are three major forms. Then, the other forms, that are less common, are to write linear equations as functions and the intercept form.
Here are the three major forms of linear equations
Slope-intercept form of a linear equation
y = mx + b, where m is the slopeb is the y-intercept, x is the coordinate of the x-axis, and y is the coordinate of the y-axis.
The name "slope-intercept form" came from the fact that you can clearly and quickly identify the slope and the y-intercept in the equation. This can help to quickly graph the line of the equation.
For example, y = 5x - 6 is in slope-intercept form where 5 is the slope and -6 is the y-intercept.
Point-slope form of a linear equation
y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
The name "point-slope form" also came from the fact that you can clearly and quickly identify the slope and a point just by looking at the equation. This can help you quickly graph the line of the equation. You do this by putting the point first on the coordinate plane. Then, use the slope to find one more point. Finally draw a straight line between the two points. See how this is done in the lesson about graphing slope.
For example, y = 5x - 6 can be written in point-slope form as y + 1 = 5(x - 1) or y - -1 = 5(x - 1)
5 is the slope and (-1, 1) is a point on the line.
General form of a linear equation
Ax + By = C, where A, B, and C are real numbers and A and B are not both zero.
The name "general form" most likely came from the fact that you cannot quickly identify anything specific in the equation such as the slope. However, the general form can be used to quickly find the x-intercept and the y-intercept.
For example, the general of y = 5x - 6 is 5x - y = 6
The standard form of linear equations with two variables can be written as Ax + By = C where A, B, and C are real numbers and A and B are not both zero.
When either the variable x or the variable y is equal to zero, we get Ax = C or By = C.
An equation of the form Ax = C or By = C is called linear equation with one variable. A and B cannot be zeros.
Less common forms of linear equations
As a function
The equation y = 5x - 6 can also be written as a function with f(x), g(x), h(x) ... instead of y.
f(x) = 5x - 6
g(x) = 5x - 6
h(x) = 5x - 6
The intercept form
x/a + y/b = 1, where a is the x-intercept and b is the y-intercept
For example, y = 5x - 6 can be written in intercept form as (x / 1.2) + (y / -6) = 1
An interesting math problem leading to a linear equation
I am thinking of a number. If I add 2 to that number, I will get 5. What is the number?
Although it may be fairly easy to guess that the number is 3, you can model the situation above with a linear equation.
Let x be the number in my mind.
Add 2 to x to get 5.
Adding 2 to x to get 5 means that whatever x is, when I add 2 to x, it has to equal to 5.
The equation is 2 + x = 5
A real-life example leading to a linear equation
Soon or later, we need the service of a taxi driver. Taxi drivers usually charge an initial fixed fee as part of using their services. Then, for each mileage, they charge a certain amount.
Say for instance the initial fee is 4 dollars and each mileage cost 2 dollars.
The total cost can be modeled with an equation that is linear.
Let y be the total cost.
Let n be number of mileage.
Total cost = 4 + cost for n miles.
Notice that cost for n miles = N × 2.
Therefore, y = 4 + n × 2.
Say for instance, a taxi driver takes you to a distance of 20 miles, how much money do you have to pay using y = 4 + n × 2 ?
When n = 20, y = 4 + 20 × 2 = 4 + 40 = 44 dollars
Now, let's ask the question the other way around!
If you pay 60 dollars, how far did the taxi driver took you?
This time = 60
Replacing 60 into the equation gives you the following equation:
60 = 4 + n × 2
It is not obvious to see that n = 28.
That is why it is important to learn to solve linear equations!
Real life examples or word problems on linear equations are numerous.
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# Misc Solution: Inscribed Squares
A square is inscribed in a circle. A smaller square is drawn within the circle such that it shares a side with the inscribed square and its corners touch the circle. What is the ratio of the large square's area to the small square's area?
## Solution
Let's focus on the large square first. In order to get it's size we say the circle has radius $$r$$. The diameter and as such also the diameter of the square is thus $$2r$$. The resulting triangles inside the big square have its hypotenuse being the diameter and the catheti being the side length $$a$$ of the square. Using Pythagorean theorem we know $$(2r)^2 = 2a^2 \Leftrightarrow 4r^2=2a^2\Leftrightarrow a=r\sqrt{2}$$:
Now to the small square. Lets say the sides of the smaller square is $$b$$. We can now draw a line with length $$r$$ from the circle center to one of the corners of the small square sitting on the circle. We can then also draw a line from the middle of the circle to the middle of the small square, which is perpendicular to the side of the squares:
The length of the perpendicular line is $$\frac{a}{2}$$ plus $$b$$. We can use Pythagorean theorem again and state
$\begin{array}{rrl} & r^2 =& \left(\frac{b}{2}\right)^2 + \left(\frac{a}{2}+b\right)^2\\ & =& \left(\frac{b}{2}\right)^2 + \left(\frac{r\sqrt{2}}{2}+b\right)^2\\ & 0 =& \frac{1}{4}b^2+\frac{2}{4}r^2+rb\sqrt{2}+b^2-r^2\\ & = & \frac{5}{4}b^2 +rb\sqrt{2} -\frac{1}{2}r^2 \end{array}$
This can be used for the Quadratic equation:
$\begin{array}{rl} b &= \frac{-r\sqrt{2}- \sqrt{2r^2 + 4\frac{5}{8}r^2}}{2\frac{5}{4}}\\ \end{array}$
It follows that $$b=\frac{r\sqrt{2}}{5}$$ or $$b=-r\sqrt{2}$$. Since we are looking for the length of a side, the second solution isn't possible. The ratio of the areas is thus $$\frac{a^2}{b^2}=\frac{(r\sqrt{2})^2}{(\frac{r\sqrt{2}}{5})^2} \Rightarrow 25:1$$.
« Back to problem overview |
GCD Subtraction Method
Our Objective
To implement a program to find the Greatest Common Divisor (GCD) between two positive integer numbers using subtraction method.
The Theory
The GCD (Greatest Common Divisor) of two numbers is the largest positive integer that divides both numbers without leaving a remainder. There are many ways to find the GCD of two numbers, including the Subtraction Method.
The Subtraction Method is a simple and iterative algorithm for finding the GCD of two numbers. The basic idea is to subtract the smaller number from the larger number until both numbers become equal. The resulting number is the GCD of the original two numbers.
Here is the step-by-step process for finding the GCD of two numbers using the Subtraction Method:
1. Take two numbers "a" and "b", where "a" is greater than or equal to "b".
2. Subtract "b" from "a". If the result is greater than or equal to "b", replace "a" with the result and go back to step 2. Otherwise, let "a" be the smaller number and let "b" be the difference.
3. Repeat step 2 until both numbers are equal.
4. The resulting number is the GCD of the original two numbers.
For example, let's find the GCD of 42 and 28 using the Subtraction Method:
a = 42 and b = 28.
Subtract "b" from "a": 42 - 28 = 14. Since 14 is less than 28, let "a = 28" and "b = 14".
Subtract "b" from "a": 28 - 14 = 14. Since both numbers are equal, stop.
The GCD of 42 and 28 is 14.
The Subtraction Method is a simple and easy-to-understand algorithm for finding the GCD of two numbers. However, it can be inefficient for large numbers or when the numbers are relatively prime (have no common factors). Other algorithms, such as the Euclidean Algorithm, are more efficient for these cases.
Learning Outcomes
• Understanding of basic arithmetic concepts: Using the Subtraction Method to find the GCD of two numbers requires knowledge of arithmetic operations such as subtraction, division, and remainders.
• Development of problem-solving skills: Using the Subtraction Method requires breaking down the problem of finding the GCD of two numbers into smaller, more manageable steps. This can help individuals develop their problem-solving skills.
• Improved logical thinking: The Subtraction Method involves creating a logical sequence of steps to find the GCD of two numbers. This can help individuals improve their logical thinking skills.
• Understanding of number theory: The Subtraction Method is one of several algorithms for finding the GCD of two numbers. Learning about different algorithms for finding the GCD can help individuals develop a deeper understanding of number theory.
• Practice with calculation and computation: Using the Subtraction Method to find the GCD of two numbers requires practice with calculation and computation. This can help individuals develop their math skills. |
# Lea Michele Under The Microscope (03/26/2020)
How will Lea Michele do on 03/26/2020 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is of questionable accuracy – take it with a grain of salt. I will first find the destiny number for Lea Michele, and then something similar to the life path number, which we will calculate for today (03/26/2020). By comparing the difference of these two numbers, we may have an indication of how smoothly their day will go, at least according to some astrology people.
PATH NUMBER FOR 03/26/2020: We will take the month (03), the day (26) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. This is how it’s calculated. First, for the month, we take the current month of 03 and add the digits together: 0 + 3 = 3 (super simple). Then do the day: from 26 we do 2 + 6 = 8. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 3 + 8 + 4 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 03/26/2020.
DESTINY NUMBER FOR Lea Michele: The destiny number will consider the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Lea Michele we have the letters L (3), e (5), a (1), M (4), i (9), c (3), h (8), e (5), l (3) and e (5). Adding all of that up (yes, this can get tedious) gives 46. This still isn’t a single-digit number, so we will add its digits together again: 4 + 6 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the destiny number for Lea Michele.
CONCLUSION: The difference between the path number for today (6) and destiny number for Lea Michele (1) is 5. That is greater than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t get too concerned! As mentioned earlier, this is for entertainment purposes only. If you want to see something that people really do vouch for, check out your cosmic energy profile here. Check it out now – what it returns may blow your mind.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
Geometry/Chapter 2/Lesson 5
Introduction
Chapter 2, Lesson 5 will introduce you to the concept of deductive reasoning. We will also review these two laws: The Law of Detachment and the Law of Syllogism.
Deductive Reasoning
Deductive reasoning is different from inductive reasoning---Deductive Reasoning is reaching logical and sensible conclusions by the means of facts, rules, definitions, and properties. Deductive reasoning is what doctors use to reach a conclusion on how much medicine a patient must take. Doctors don't use inductive reasoning (examples) to reach conclusions!
An example of deductive reasoning is the father of this concept, Aristotle. He concluded Socrates as a mortal from these two facts:
• Humans are mortals
• Socrates is a human
This is known as the Law of Syllogism, which we will get to in section 1.1.2.
Law of Detachment
The Law of Detachment is a form of deductive reasoning that is used to draw conclusions. This law has a specific format that distinguishes itself as a law.
Refer back to Lesson 3 if you do not remember logic symbols
• p → q
• p
• q
If the order is switched, such as q being first instead of p, it is no longer known as the Law of Detachment. Instead, it is invalid. In order for it to be the Law of Detachment, it must follow the format described above. Here, we will look at this with an example:
If the measure of an angle is greater than 90, then it is obtuse.
m ∠T is greater than 90.
∠T is obtuse.
This example exactly follows the format of the Law of Detachment:
1. p → greater than 90 || q → it is obtuse
2. p → greater than 90
3. q → is obtuse
But on the other hand, this example does not follow the Law of Detachment:
If Pedro is taking history, then he will study about WWII.
Pedro is taking history.
Closer evaluation
1. p → taking history || q → will study about WWII
2. q → will study about WWII
3. p → taking history
As you can see, this statement switched the orders of the p and q. Thus, making this statement invalid and not Law of Detachment.
Law of Syllogism
The Law of Syllogism is another law of logic and is similar to the Transitive Property. This law, like the Law of Detachment, has a specific format to it.
• p → q
• q → r
• p → r
If any part of the format is changed, then it is no longer the Law of Syllogism, and thus, is invalid. Here, we will look at an example here that follows the Law of Syllogism correctly:
Statement 1: If it continues to rain (p), then the soccer field will become wet and muddy (q). This becomes if p, then q.
Statement 2: If the soccer field becomes wet and muddy (q), then the game will be canceled (r). This becomes if q, then r.
Statement 3: If it continues to rain (p), then the game will be canceled (r). This final statement is the conclusion and becomes p, then r.
(Source of problem is on the back)
As you can see, these statements follow the p to q, q to r, and finally p to r format, and thus, follows the Law of Syllogism. Now, we will look at an example that does not follow the Law of Syllogism, and thus, is invalid.
Statement 1: If I take this test, then I will get a better grade than 0.
Statement 2: If I get a better grade than 0, then I will make my parents happy.
Statement 3: If I take this test, then I will get a better grade than 0.
This does not follow the Law of Syllogism because it follows the p to q, q to r, and p to q format--and is invalid. |
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# wikiHow to Find the Nth Term for an Arithmetic Sequence of High Order
An arithmetic sequence of high order is a sequence of numbers where the difference of terms get larger and forms an arithmetic sequence of their own. It may seem confusing at first, but with Newton's little formula for nth term, it will be a piece of cake.
## Steps
1. 1
Write down the formula. It's ${\displaystyle a_{n}=A[{}_{(n-1)}C_{0}]+B[{}_{(n-1)}C_{1}]+C[{}_{(n-1)}C_{2}]}$.
• ${\displaystyle A={\text{first term in the sequence}}}$
• ${\displaystyle B={\text{first difference}}}$
• ${\displaystyle C={\text{second difference}}}$
• ${\displaystyle n={\text{the n-th term in an arithmetic sequence that you need to find}}}$
2. 2
Write down the first four terms of the sequence. Then, draw a straight line leading toward the first difference or ${\displaystyle B}$. Make sure to leave some space apart each other. Look at the diagram for help.
3. 3
Find the positive difference between each pair of terms. Then, write each difference at the end of two lines that lead to the numbers which it came from. Draw a straight line leading toward the second difference or ${\displaystyle C}$. Make sure the second difference is the same for all four terms.
4. 4
Find the positive difference between the pair of numbers in the second row. Then, write each difference at the end of two lines that lead to the numbers which it came from.
5. 5
Plug in the values in the formula ${\displaystyle a_{n}=A[{}_{(n-1)}C_{0}]+B[{}_{(n-1)}C_{1}]+C[{}_{(n-1)}C_{2}].}$${\displaystyle }$
• Let us pretend we are searching for the 28th term.
• In the diagram, we have ${\displaystyle a_{28}=5[{}_{(28-1)}C_{0}]+5[{}_{(28-1)}C_{1}]+5[{}_{(28-1)}C_{2}].}$
• In the formula, ${\displaystyle {}_{(n-1)}C_{0}}$ means ${\displaystyle {\frac {(n-1)!}{[(n-1)-0]!*0!}}}$
6. 6
• ${\displaystyle a_{28}=5[{}_{(28-1)}C_{0}]+5[{}_{(28-1)}C_{1}]+5[{}_{(28-1)}C_{2}]}$
• ${\displaystyle a_{28}=5[{}_{(27)}C_{0}]+5[{}_{(27)}C_{1}]+5[{}_{(27)}C_{2}].}$
• ${\displaystyle a_{28}=5(1)+5(27)+5(351)}$
• ${\displaystyle a_{28}=5+135+1755}$
• ${\displaystyle a_{28}=1895}$ |
What is 10/12 as a percentage?
To convert a fraction such as 10/12 into a percentage, one can follow these procedures. It is crucial to bear in mind that percentages consistently possess a denominator of 100 during the process of converting fractions into percentages.
• Divide the top number (10) by the bottom number (12) to obtain the decimal representation.
• Multiply the decimal by 100 in order to obtain the percentage.
• To arrive at the correct solution, you can reverse these steps and keep in mind that you will multiply the numerator by 100 first, then divide the result by the denominator, giving you a percentage correct of 83.33%.
When are fractions useful?
You may want to express a percentage as a fraction. You often describe the problem using fractions when you are trying to test a score or splitting a bill. Fractions are commonly used in everyday life.
Fraction Conversion Table
The Oceanography Institute in Nha Trang is an intriguing destination for people of different age ranges.
Find the Denominator
A percentage is a numerical value out of 100, therefore we must ensure that our denominator is set at 100!
If the initial denominator is 12, we must determine how to change the denominator to 100.
To convert this fraction, we would divide 100 by 12, which results in 8.33.
Find the Numerator
Now, we multiply 8.33 by 10, our initial numerator, which is equivalent to 83.33.
Understanding Fractions and Percentages
Converting between percentages and fractions can help us better understand and compare different proportions or quantities. Both ways represent a whole as a part, with the numerator representing the top part (part of the whole) and the denominator representing the bottom part. In contrast, a fraction A represents the numerator with the part representing the whole as the denominator, while a percentage represents a part out of 100.
When solving problems or interpreting data, it can be useful to know that both percentages and fractions are equivalent. In this example, the fraction 12/10 is equivalent to 100 out of 83.33 parts. This means that we have converted the percentage 83.33% into the fraction 12/10.
Fraction to Percent Conversion Summary
Here’s a possible solution to the problems you have in your head or in hand: we recommend using a calculator to solve these. Luckily, this page will cover the division and multiplication bit of this problem for us. Sometimes, you may want to express a fraction in the form of a percentage. You can also learn about fractions in the fractions section of our website. We encourage you to check out our introduction page for a little recap on what percentage is.
• Step 1: Divide 10 by 12 to obtain the number as a decimal. 10 / 12 = 0.83.
• Step 2: Multiply 0.83 by 100. 0.83 multiplied by 100 equals 83.33. That’s all there is to it! |
# RATIONAL EXPRESSIONS. EVALUATING RATIONAL EXPRESSIONS Evaluate the rational expression (if possible) for the given values of x: X = 0 X = 1 X = -3 X =
## Presentation on theme: "RATIONAL EXPRESSIONS. EVALUATING RATIONAL EXPRESSIONS Evaluate the rational expression (if possible) for the given values of x: X = 0 X = 1 X = -3 X ="— Presentation transcript:
RATIONAL EXPRESSIONS
EVALUATING RATIONAL EXPRESSIONS Evaluate the rational expression (if possible) for the given values of x: X = 0 X = 1 X = -3 X = 3
DOMAIN OF A FUNCTION The domain of a function is the set of all real numbers that when substituted into the function produces a real number Steps to find the Domain of a Rational Expression 1.Set the denominator equal to zero and solve the resulting equation. 2.The domain is the set of real numbers excluding the values found in Step 1.
2y = -7 Find the Domain of Rational Expressions 2y + 7 = 0 Set the denominator equal to zero. 2y + 7 - = 0 - 7 Solve the equation
Find the Domain of these Expressions
Simplifying a Rational Expression to Lowest Terms 1.Factor the numerator and denominator 2.Determine the domain of the expression 3.Simplify the expression to lowest terms 1.Factor the numerator and denominator 2.Determine the domain of the expression 3.Simplify the expression to lowest terms 2.(p + 7)(p – 7) = 0 p = -7p = 7 Domain: {p | p is a real number and p ≠ -7 and p ≠ 7 2.(p + 7)(p – 7) = 0 p = -7p = 7 Domain: {p | p is a real number and p ≠ -7 and p ≠ 7
Factor out the GCF in the numerator. Factor the denominator as a difference of squares. SOLUTION
2.(p + 7)(p – 7) = 0 p + 7 = 0 p = -7p = 7 Domain: {p | p is a real number and p ≠ -7 and p ≠ 7 2.(p + 7)(p – 7) = 0 p + 7 = 0 p = -7p = 7 Domain: {p | p is a real number and p ≠ -7 and p ≠ 7 To find the domain restrictions, set the denominator equal to zero. The equation is quadratic. Set each factor equal to 0.
Reduce common factors whose ratio is 1. (provided p ≠ 7 or p ≠ -7)
Simplifying Rational Expressions to Lowest Terms SOLVE
To reduce this rational expression, first factor the numerator and the denominator. CAUTION: Remember when you have more than one term, you cannot cancel with one term. You can cancel factors only. REDUCING RATIONAL EXPRESSIONS There is a common factor so we can reduce.
Multiplication and Division of Rational Expressions Factor into prime factors. Simplify
To multiply rational expressions we multiply the numerators and then the denominators. However, if we can reduce, we’ll want to do that before combining so we’ll again factor first. MULTIPLYING RATIONAL EXPRESSIONS Now cancel any like factors on top with any like factors on bottom. Now multiply numerator and multiply denominator
To divide rational expressions remember that we multiply by the reciprocal of the divisor (invert and multiply). Then the problem becomes a multiplying rational expressions problem. DIVIDING RATIONAL EXPRESSIONS Multiply by reciprocal of bottom fraction.
To add rational expressions, you must have a common denominator. Factor any denominators to help in determining the lowest common denominator. ADDING RATIONAL EXPRESSIONS So the common denominator needs each of these factors. This fraction needs (x + 2) This fraction needs (x + 6) FOIL distribute
Subtracting rational expressions is much like adding, you must have a common denominator. The important thing to remember is that you must subtract each term of the second rational function. SUBTRACTING RATIONAL EXPRESSIONS So a common denominator needs each of these factors. This fraction needs (x + 2) This fraction needs (x + 6) Distribute the negative to each term. FOIL
For complex rational expressions you can combine the numerator with a common denominator and then do the same with the denominator. After doing this you multiply by the reciprocal of the denominator. COMPLEX RATIONAL EXPRESSIONS For complex rational expressions you can combine the numerator with a common denominator and then do the same with the denominator. After doing this you multiply by the reciprocal of the denominator.
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# Population of state X increased by x% and the population of state Y increased by y% from 2001 to 2011. Assume that x is greater than y. Let P be the ratio of the population of state X to state Y in a given year. The percentage increase in P from 2001 to 2011 is _______.
Free Practice With Testbook Mock Tests
## Options:
1. $$\frac{x}{y}$$
2. x – y
3. $$\frac{{100\left( {x - y} \right)}}{{100 + x}}$$
4. $$\frac{{100\left( {x - y} \right)}}{{100 + y}}$$
### Correct Answer: Option 4 (Solution Below)
This question was previously asked in
GATE CE 2019 Official Paper: Shift 2
## Solution:
Let,
Population of state X in 2001 = a
∴ Population of state X in 2011 = a + 0.01 × x × a
Population of state Y in 2001 = b
∴ Population of state Y in 2011 = b + 0.01 × y × b
∵ P is the ratio of the population of state X to state Y in a given year
$${P_{2001}} = \frac{a}{b}$$
$${P_{2011}} = \frac{{\left( {1{\rm{\;}} + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;x\;}}} \right) \times {\rm{\;a}}}}{{\left( {1 + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;y}}} \right) \times {\rm{\;b}}}}$$
$${\rm{\Delta }}P\left( \% \right) = \frac{{{P_2} - {P_1}}}{{{P_1}}} \times 100 = \frac{{\frac{{\left( {1{\rm{\;}} + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;x\;}}} \right) \times {\rm{\;a}}}}{{\left( {1 + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;y}}} \right) \times {\rm{\;b}}}} - \frac{a}{b}}}{{\frac{a}{b}}} \times 100$$
$${\rm{\Delta }}P\left( \% \right) = \left( {\frac{{\left( {1{\rm{\;}} + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;x\;}}} \right)}}{{\left( {1 + {\rm{\;}}0.01{\rm{\;}} \times {\rm{\;y}}} \right)}} - 1} \right) \times 100 = \frac{{100\left( {{\rm{x}} - {\rm{y}}} \right)}}{{100 + {\rm{y}}}}$$ |
# 2012 AMC 10A Problems/Problem 16
## Problem
Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
$\textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000$
## Solution 1
First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let $x$ be the time these runners run in seconds.
$$4.8x-4.4x=500 \Rightarrow x=1250$$
Because $4.4(1250)=5500$ is a multiple of 500, it turns out they just meet back at the start line.
Now we must find a time that is a multiple of $1250$ and results in the 5.0 m/s runner to end up on the start line. Every $1250$ seconds, that fastest runner goes $5.0(1250)=6250$ meters. In $2(1250)=2500$ seconds, he goes $5.0(2500)=12500$ meters. Therefore the runners run $\boxed{\textbf{(C)}\ 2,500}$ seconds.
## Solution 2
Working backwards from the answers starting with the smallest answer, if they had run $1000$ seconds, they would have run $4400, 4800, 5000$ meters, respectively. The first two runners have a difference of $400$ meters, which is not a multiple of $500$ (one lap), so they are not in the same place.
If they had run $1250$ seconds, the runners would have run $5500, 6000, 6250$ meters, respectively. The last two runners have a difference of $250$ meters, which is not a multiple of $500$.
If they had run $2500$ seconds, the runners would have run $11000, 12000, 12500$ meters, respectively. The distance separating each pair of runners is a multiple of $500$, so the answer is $\boxed{\textbf{(C)}\ 2,500}$ seconds.
## Solution 3
Let $t$ be the time run in seconds, then the difference in meters run between the three runners is $0.2t, 0.4t, 0.6t$. For them to be at the same location all of them need to be multiples of 500. It is now easy to see that $0.2t=500, 0.4t=1000, 0.6t=1500$, so $t=\boxed{\textbf{(C)}\ 2,500}$.
## Solution 4
After $t$ seconds, respectively the runners would've ran $4.4t, 4.8t,$ and $5t$ meters. Their current positions on the track are these values $\pmod{500}$. We're trying to find the value of $t$ such that $$4.4t \equiv 4.8t \equiv 5t \pmod{500}$$ Subtracting $4.4t$ on all sides, we get $$0 \equiv 0.4t \equiv 0.6t \pmod{500}$$ Now, we must find a value for $t$ such that both $0.6t$ and $0.4t$ are simultaneously multiples of $500$.
Plugging in $500$ for $0.4t$ we get $t=1250$, but this does not work for $0.6t$ ($750$ isn't a multiple of $500$). Plugging in $0.4t=1000$, we get $t=2500$, and this does work for $0.6t$.
Therefore, $t=2500$ and the answer is $\boxed{\textbf{(C) } 2500}$.
• Note: Modular Arithmetic works only for integral values, so my usage of decimals is technically incorrect but the intuition leads to the right answer
## Solution 5
Similar to the solution above, but is much quicker and does not involve trial and error. This uses decimal mod arithmetic, which can be justified by intuition... After $t$ seconds, respectively the runners would've ran $4.4t, 4.8t,$ and $5t$ meters. These three values are congruent $\pmod{500}$, so $$4.4t \equiv 4.8t \equiv 5t \pmod{500}$$. Subtract $4.4t$ from all three sides to get $0, 0.4t,$ and $0.6t$ are congruent. Now all we need to find is a value of $t$ for which $0.4t$ and $0.6t$ are congruent $\pmod{500}$. Subtract $0.4t$ from both sides to get $0.2t$ and $0$ are congruent mod $500$, or that $0.2t=\dfrac{t}{5}$ is a multiple of $500$. Let $t=500k$, so we want $100k$ to be a multiple of $500$, or $k$ to be a multiple of $5$. Therefore, the smallest value of $t$ is when $k=5$, and when $t=500k=500(5)=2500 \space \boxed{(\text{C})}$
- Solution by mathchampion1 |
CATEGORIES:
# Decomposition of vectors.
Theorem 1. An arbitrary vector in the plane can be decomposed into two noncollinear vectors:
.
Theorem 2. An arbitrary vector in space can be decomposed into three noncoplanar vectors
.
Let be noncoplanar vectors.
The Cartesian system of coordinates. Consider the following coordinate system: take mutually perpendicular unit vectors and , draw coordinate axes x,y and z along them, and fix a unit on metric scale:
,
.
Definition. The triple of vectors is called right if, looking from the endpoint of the last vector, we see that the shorter rotation from the first vector to the second is anticlockwise.
From the triangle ÎÌÌ1, we obtain
,
Since the vector is collinear to the unit vector , it follows that
.
From the triangle ÎÀÌ1, we obtain
,
because, by analogy, the vectors and are collinear to the unit vectors and . Substituting the vector thus obtained, we see that
. (2)
Thus, the radius vector is represented as the sum of and multiplied by the corresponding coordinates of the point Ì.
Consider the vectors
and
and their sum
.
Under addition the respective coordinates are added
Let us multiply the vector by a number l:
.
When a vector is multiplied by a number l, each coordinate of this vector is multiplied by this number.
Example. Find the vector if
; .
Let us find the required vector in vector notation:
.
To find the same vector in vector notation, we multiply the first vector by 4 and the second by –3 and sum their coordinates:
.
Given two points and in space, find the vector .
Thus, we have found the required vector in the coordinate notation:
. (2¢¢)
To find the coordinates of a vector, we must subtract the coordinates of its tail from the coordinates of the head.
For example, let us find vectors with given coordinates of heads and tails:
Ì1(7;4;–3); Ì2(1;–2;–2);
={–6; –6; 1}; ={6; 6; –1}.
Find the length of a vector :
.
From the right triangle ÎÌ1Ì2 , we find the hypotenuse
,
z
where .
M1
From the other right triangle ÎÀÌ2 , 0 z1 y
we find the hypotenuse .
Substituting it into À x1
the first hypotenuse, we obtain x y1 M2
.
Thus, the length of a vector is defined by the formula
. (3)
Date: 2015-01-02; view: 1400
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# If the side of a square is increased by $$16\frac{2}{3}\%$$ of its original value, then find the percentage change in area of square.
1. $$22\frac{1}{9}\%$$
2. $$36\frac{1}{9}\%$$
3. $$33\frac{1}{9}\%$$
4. $$27\frac{1}{9}\%$$
Option 2 : $$36\frac{1}{9}\%$$
## Detailed Solution
Given:
the side of a square is increased by $$16\frac{2}{3}\%$$ of its original value
Concept used:
Area of square = (side)2
As we know that $$16\frac{2}{3}\% = \frac{1}{6}$$
Calculations:
Let side of square is x m.
⇒ Area of square = x × x = x2
Now side increased by $$16\frac{2}{3}\%$$
⇒ increased in side = x + x × 1 / 6 = 7x / 6
Now new area = (7x / 6)2 = (49x2 / 36)
⇒ increased area = (49x2 / 36) - x2 = (49x2 - 36x2) / 36 = 13x2 / 36.
⇒ percentage increased in area = (13x2 / 36) / x2 × 100
∴ percentage increased in area = $$36\frac{1}{9}\%$$
Short trick / Topper’s Approach:
Let initially side of the square be 6 unit
Then, increased side of square = 7 unit
⇒ Initial area = 6 × 6 = 36 sq. unit
⇒ New area = 7 × 7 = 49 sq. unit
⇒ increased in area = 49 - 36 = 13 sq. unit
∴ percentage change in area = 13 × 100 / 36 = $$36\frac{1}{9}\%$$ |
### Half-Life Problems #26 - 40
Problem #26: The half life in two different samples, A and B, of radio-active nuclei are related according to T(1/2,B) = T(1/2,A)/2. In a certain period the number of radio-active nuclei in sample A decreases to one-fourth the number present initially. In the same period the number of radio-active nuclei in sample B decreases to a fraction f of the number present initially. Find f.
Solution:
a) Sample A underwent two half-lives:
1 ---> 1/2 ---> 1/4
b) Let us set the length of one half-life of A equal to 1. Therefore the total amount of elapsed time for A was 2.
T(1/2,B) = T(1/2,A)/2
Since T(1/2,A) = 1,
we now know that T(1/2,B) = 1/2
c) Allow B to go through several half-lives such that the total amount of time = 2. This is 4 half-lives (1/2 + 1/2 + 1/2 + 1/2 = 2):
Four half-lives:
1 ---> 1/2 ---> 1/4 ---> 1/8 ---> 1/16
f = 1/16
Problem #27: You have 20.0 grams of P-32 that decays 5% daily. How long will it take for half the original to decay?
Solution:
In 24 hours, the sample goes from 100% to 95%
(1/2)n = 0.95
n log 0.5 = log 0.95
n = 0.074
24 hrs / 0.074 = 324 hrs (one half-life)
Problem #28: A sample of radioactive isotopes contains two different nuclides, labeled A and B. Initially, the sample composition is 1:1, i.e., the same number of nuclei A as nuclei B. The half-life of A is 3 hours and, that of B, 6 hours. What is the expected ratio A/B after 18 hours?
Solution:
A has a half-life of 3 hrs, so 18 hrs = 6 half-lives.
B has a half-life of 6 hrs, so 18 hrs = 3 half-lives.
After 6 half lives, the fraction of A left is 1/(26) = 1/64
The fraction of B left is 1/(23) = 1/8.
Since A/B started out at 1/1, A/B at 18 hrs = (1/64) / (1/8) = 1/8.
This could also be expressed as: 0.56 / 0.53 = 0.56-3 = 1/8
Problem #29: The ratio of tritium, H-3, to hydrogen, H-1, in a sample of water was 1:1x1019. If the half life of tritium is 12.25 years, calculate the actual number of tritium atoms remaining in 10.0 g water after a period of 49 years.
Solution:
1) Moles, then molecules of water:
10.0 g / 18.0148 g/mol = 0.555099 mol
(0.555099 mol) (6.02214 x 1023 molecules / mol) = 3.342884 x 1023 molecules
2) Atoms of hydrogen in 10.0 g of water:
(2 atoms/molecules) (3.342884 x 1023 molecules) = 6.685768 x 1023 atoms
3) Atoms of H-3:
6.685768 x 1023 / 1 x 1019 = 66858 atoms of H-3 (to the nearest whole number)
4) Half-lives elapsed:
49 yr / 12.25 yr = 4
5) Amount remaining after 4 half-lives:
(1/2)4 = 1/16 = 0.0625
6) atoms remaining after 4 half-lives:
(66858 atoms) (0.0625) = 4179 atoms (to the nearest whole number)
Problem #30: The isotope Ra-226 decays to Pb-206 in a number of stages which have a combined half-life of 1640 years. Chemical analysis of a certain chunk of concrete from an atomic-bombed city, preformed by an archaeologist in the year 6264 AD, indicated that it contained 2.50 g of Ra-226 and 6.80 g of Pb-206. What was the year of the nuclear war?
Solution:
Start by ignoring a few chemical realities and assume all the Ra-226 ends up as lead.
a) Calculate moles of Ra-226 decayed:
6.80 g / 205.974465 g/mol = 0.033013801 mol of Pb-206 decayed
∴ 0.033013801 mol of Ra-226 decayed
b) Calculate grams of Ra-226 initially present:
(0.033013801 mol) (226.02541 g/mol) = 7.462 g of Ra-226 decayed
7.462 + 2.50 = 9.962 g of Ra-226 initially present
c) Calculate decimal fraction of Ra-226 remaining:
2.50 g / 9.962 g = 0.251
d) Calculate number of half-lives elapsed:
(1/2)n = 0.251
n = 2
e) Calculate year of war:
1640 x 2 = 3280 y elapsed since war
6264 − 3280 = 2984 AD
Problem #31: A radioactive sample contains 3.25 x 1018 atoms of a nuclide that decays at a rate of 3.4 x 1013 disintegrations per 26 min.
(a) What percentage of the nuclide will have decayed after 159 days?
(b) What is the half-life of the nuclide?
Solution to a:
159 days x 24 hrs/day x 60 min/hour = 228960 min
228960 min x (3.4 x 1013 disintegrations per 26 min) = 2.994 x 10 x 1017 total dis in 159 days
2.994 x 10 x 1017 / 3.25 x 1018 = 0.0921
9.21% has disintegrated
Solution to b:
0.9079 is the decimal fraction of the substance remaining since 0.0921 has gone away
(1/2)n = 0.9079
n log 0.5 = log 0.9079
n = 0.139 half-lives
159 day / 0.139 = 1144 days
Problem #32: The radioisotope potassium-40 decays to argon-40 by positron emission with a half life of 1.27 x 109 yr. A sample of moon rock was found to contain 78 argon-40 atoms for every 22 potassium-40 atoms. What is the age of the rock?
Solution:
Assume the sample was 100% K-40 at start. In the present day, the sample contains 78% Ar-40 and 22% K-40. We will use 0.22, the decimal percent of K-40 remaining:
(1/2)n = 0.22
where n is the number of half-lives.
n log 0.5 = log 0.22
n = 2.18
What is the total elapsed time?
(2.18) (1.27 x 109) = 2.77 x 109 yrs
Problem #33: What is the age of a rock in which the mass ratio of Ar-40 to K-40 is 3.8? K-40 decays to Ar-40 with a half-life of 1.27 x 109 yr.
Solution:
Since, the sample is 3.8 parts by mass Ar and 1 part K, the orginal sample contained 4.8 parts K and zero parts Ar.
What is the decimal amount of K-40 that remains?
1 part divided by 4.8 parts = 0.20833
How many half-lives are required to reach 0.20833 remaining?
(1/2)n = 0.20833
where n is the number of half-lives.
n log 0.5 = log 0.20833
n = 2.263
What is the total elapsed time?
(2.263) (1.27 x 109) = 2.87 x 109 yrs
Problem #34: A radioactive isotope has a half-life of 4.5 days. What fractions of the sample will exist after 9 and 18 days?
(a) 1/2 and 1/4 of the original amount
(b) 1/9 and 1/18 of the original amount
(c) 1/4 and 1/16 of the original amount
(d) 1/4 and 1/8 of the original amount
Solution:
Half-lives elapsed zero one two three four Days elapsed 0 4.5 9 13.5 8 Fraction remaining 1 1/2 1/4 1/8 1/16
Answer: (c) 1/4 and 1/16 of the original
Problem #35: Selenium-75 has a half-life of 120 days and is used medically for pancreas scans. (a) Approximately how much selenium-75 would remain of a 0.050 g sample that has been stored for one year? (b) How long would it take for a sample of selenium-75 to lose 99% of its radioactivity?
Solution to (a):
365 day / 120 day = 3.0417 half-lives
(1/2)3.0417 = 0.12144 (this is the decimal amount that remains)
0.12144 x 0.050 g = 0.006072 g
to two sig figs = 0.0061 g
Solution to (b):
99% gone means 1% remaining, which is 0.01 as a decimal
(1/2)n = 0.01
n log 0.5 = log 0.01
n = 6.643856 (this is the number of half-lives elapsed)
120 day times 6.643856 = 797 days
Problem #36: Natural samarium (average atomic mass 150.36) contains 14.99% of the radioactive isotope Sm-147. A 1 g sample of natural Sm has an activity of 89 decays per second. Estimate the half life of Sm-147.
Solution:
1 g times 0.1499 = 0.1499 g of Sm-147
0.1499 g / 146.915 g/mol = 0.00102032 mol
0.00102032 mol times 6.022 x 1023 mol¯1 = 6.144367 x 1020 atoms
6.144367 x 1022 atoms / 2 = 3.0721835 x 1020 atoms
3.0721835 x 1020 atoms / 89 decays/sec = 3.45189 x 1018 sec
3.45189 x 1018 sec = 1.094 x 1011 yr.
The Wiki article for isotopes of samarium gives 1.06 x 1011 yr.
Implicit in this solution is that the decays rate of 89 decays/second remains constant for the entire half-life. The decay rate actually becomes lesser over time but, for purposes of making an estimate, we ignore this.
Problem #37: You measure the radioactivity of a substance, then when measuring it 120 days later, you find that it only has 54.821% of the radioactivity it had when you first measured it. What is the half life of that substance?
Solution:
1) How many half-lives have elapsed in the 120 days?
(1/2)n = 0.54821
n log 0.5 = log 0.54821
n = 0.8672
2) Determine the half-life.
120 day / 0.8672 = 138.4 day
Problem #38: Arsenic-74 is a medical radioisotopes with a half-life of 18 days. If the initial amount of arsenic-74 injected is 2.30 mCi, how much arsenic-74 is left in the body after 54 days?
Solution:
54 days / 18 days = 3 half-lives elapsed
(1/2)3 = 0.125 <--- amount remaining after three half-lives
2.30 mCi x 0.125 = 0.2875 mCi
to three sig figs, 0.288 mCi
Problem #39: Assume that today there is 10 grams of substance, while 1000 years ago there was 100 grams of it. If there is 15 grams of substance today, how much will there be 600 years from now?
Solution:
1) What is needed is the length of the half-life.
10 g / 100 g = 0.1 <--- this is the decimal amount remaining after 1000 years
(1/2)n = 0.1 <--- where n is the number of half-lives needed to elapse to give 0.1 remaining
n log 0.5 = log 0.1
n = 3.321928 half-lives
1000 yr / 3.321928 = 301.03 yr
2) Now, we turn to the 15 g sample. We need to know how many half-lives have elapsed
600 yr / 301.03 yr = 1.99316 half-lives
(1/2)1.99316 = 0.251188 <--- that's the decimal amount remaining after 600 years.
15 g is to 1 as x is to 0.251188
x = 3.8 g
Problem #40: A radionuclide, cobalt-60, has a half-life of 5.27 years. How many hours would it take for the activity to diminish to one-thirty-second (3.125%) of its original value?
Solution:
(1/2)n = 0.03125
n log 0.5 = log 0.03125
n = 5 (this is how many half-lives have elapsed)
5.27 x 5 = 26.35 years
You may convert years to hours. |
The Principal Ideal Domain of Polynomials over a Field
# The Principal Ideal Domain of Polynomials over a Field
Recall from the Principal Ideals and Principal Ideal Domains (PIDs) page that if $(R, +, *)$ is a ring then an ideal of the form $I = aR = \{ ar : r \in R \}$ is called a principal ideal, and if every ideal in $R$ is a principal ideal, then $R$ is said to be a principal ideal domain.
We will now look at an important example of a principal ideal domain.
Theorem 1: Let $(F, +, *)$ be a field. Then the ring of polynomials $F[x]$ is a principal ideal domain.
• Proof: Let $I$ be an ideal of $F[x]$. There are two cases to consider.
• Case 1: If $I = \{ 0 \}$ then trivially, $I$ is a principal ideal since $I = 0F[x]$.
• Case 2: If $I \neq \{ 0 \}$ then there exists a nonzero polynomial $f \in I$ with minimal degree. Let $g \in I$. By the division algorithm there exists polynomials $q, r \in R[x]$ such that:
(1)
\begin{align} \quad g(x) = f(x)q(x) + r(x) \end{align}
• where $r(x) = 0$ or $0 \leq \deg r \leq \deg f$. We rewrite the equation above as:
(2)
\begin{align} \quad r(x) = g(x) - f(x)q(x) \end{align}
• Since $f \in I$ and $q \in R[x]$ we have by definition of $I$ being an ideal that $fq \in I$. Since $g \in I$ we have that $(g - fq) \in I$. So $r \in I$. But by minimality of the degree of $f$ in $I$ we must have that $r(x) = 0$. Therefore:
(3)
\begin{align} \quad g(x) = f(x)q(x) \end{align}
• Since $g \in I$ was arbitrary we have that $I = f(x)F[x]$. So $I$ is a principal ideal.
• Hence every ideal in $F[x]$ is a principal ideal. So $F[x]$ is a principal ideal domain. $\blacksquare$ |
# How do you write a quadratic function in standard form whose graph passes through points (1/2, -3/10), (1, 6/5), (1/4, -3/10)?
Apr 18, 2017
Start with the vertex form and use symmetry to find the x coordinate.
Use the points to find the other 2 two values of the vertex form.
Expand the vertex form into the standard form.
#### Explanation:
Because you specified a function we discard the vertex form $x = a {\left(y - k\right)}^{2} + h$ and use only the form:
$y = a {\left(x - h\right)}^{2} + k \text{ [1]}$
Because the we know the 2 x values corresponding to y value, $- \frac{3}{10}$ we know know that the x coordinate of the vertex is halfway between $\frac{1}{4} \mathmr{and} \frac{1}{2}$
$h = \frac{\frac{1}{2} + \frac{1}{4}}{2}$
$h = \frac{3}{8}$
Substitute $\frac{3}{8}$ for h into equation [1]:
$y = a {\left(x - \frac{3}{8}\right)}^{2} + k \text{ [2]}$
Substitute the point $\left(\frac{1}{2} , - \frac{3}{10}\right)$ into equation [2]:
$- \frac{3}{10} = a {\left(\frac{1}{2} - \frac{3}{8}\right)}^{2} + k$
$- \frac{3}{10} = a {\left(\frac{1}{8}\right)}^{2} + k$
$- \frac{3}{10} = \frac{a}{64} + k \text{ [3]}$
Substitute the point $\left(1 , \frac{6}{5}\right)$ into equation [2]:
$\frac{6}{5} = a {\left(1 - \frac{3}{8}\right)}^{2} + k$
$\frac{6}{5} = a {\left(\frac{5}{8}\right)}^{2} + k$
$\frac{6}{5} = \frac{25 a}{64} + k \text{ [4]}$
Subtract equation [3] from equation [4]:
$\frac{6}{5} + \frac{3}{10} = \frac{24 a}{64}$
$\frac{15}{10} = \frac{24}{64} a$
$a = \frac{3}{2} \left(\frac{64}{24}\right)$
$a = \frac{1}{2} \left(\frac{64}{8}\right)$
$a = 4$
Use equation [3] to find the value of k:
$- \frac{3}{10} = \frac{4}{64} + k$
$- \frac{3}{10} = \frac{1}{16} + k$
$k = - \frac{3}{10} - \frac{1}{16}$
$k = - \frac{29}{80}$
Substitute the value of "a" and "k" into equation [2]:
$y = 4 {\left(x - \frac{3}{8}\right)}^{2} - \frac{29}{80}$
Expand the square:
$y = 4 \left({x}^{2} - \frac{6}{8} x + \frac{9}{64}\right) - \frac{29}{80}$
Distribute the 4:
$y = 4 {x}^{2} - 3 x + \frac{9}{16} - \frac{29}{80}$
This is the standard form:
$y = 4 {x}^{2} - 3 x + \frac{1}{5}$
The following is a graph of the equation and the 3 points: |
## Saturday, October 8, 2016
### Unizor - Derivatives - Notation
Notes to a video lecture on http://www.unizor.com
Derivatives - Notation
We have introduced a concept of derivative at point x0 of a real function f(x) defined in some neighborhood of this point as the following limit (if it exists):
limΔx→0[f(x0+Δx)−f(x)]/Δx
Let's emphasize once more that this definition is valid only if the limit above exists.
For some functions, like f(x)=2x this limit exists at any point of their domain. For some others, like f(x)=|x|, there are points (in this case x=0) where this limit does not exist, while at others (x≠0) it does.
The limit defining a derivative of function f(x) at point x0, if it exists, is some real value defined by both the function itself and a point x0 in its domain. Therefore, for all these points of a domain of function f(x) where this limit exists the derivative is a new function defined at all these points.
This new function, a derivative of function f(x), defined at all points of a domain of function f(x) where the above limit exists, is traditionally denoted as f'(x) and called the first derivative of function f(x).
The domain of the derivative f'(x) is a subset of a domain of function f(x) since, in theory, the above limit might not exist at all points of domain of f(x). If this limit exists at each point of domain of f(x), the domain of a derivative f'(x) coincides with the domain of original function f(x).
The definition of a derivative as a limit of ratio of two infinitesimals, increment of function and increment of argument at some point, prompts us to use the following notation:
Δf(x) = f(x+Δx)−f(x)
Then the expression of a derivative of function f(x) at point x would look like this:
f'(x) = limΔx→0Δf(x)/Δx
Using the symbol f' to denote a derivative is attributed to Italian mathematician Lagrange (1736-1813).
To make a notation even shorter, mathematicians have abbreviated the above expression based on limits even further:
f'(x) = df(x)/dx
or, sometimes, (d/dx)f(x).
In these cases expression dx is called a differential of argument x - an infinitesimal variable, and expression df(x) is called a differential of function f(x) - also an infinitesimal variable.
These notations are attributed to German mathematician Leibniz (1646-1716). They are as popular as the Lagrange's notation above. They have certain advantage since they remind us that a derivative is a ratio of two infinitesimals - increment of function df and increment of its argument dx.
In particular, since it implies that these two infinitesimals are proportional to each other with a ratio f'(x) (a function of argument x), it makes sense to use the following notation that relates differentials with a derivative:
df(x) = f'(x)·dx
Newton (England, 1643-1727) used a dot above the function symbol to denote its derivative, which we will rarely use.
Euler used Df(x) as a notation of a derivative.
The process of deriving a derivative from a function is called differentiation. We will go through this process for major functions in the Examples. |
Class 11 NCERT Math Solution
TOPICS
Miscellaneous
Question-1 :- The relation f is defined by The relation g is defined by Show that f is a function and g is not a function.
Solution :-
``` It is observed that for 0 ≤ x < 3,
f(x) = x²
3 < x ≤ 10,
f(x) = 3x
Also, at x = 3,
f(x) = 3² = 9 or
f(x) = 3 × 3 = 9 i.e.,
at x = 3, f(x) = 9.
Therefore, for 0 ≤ x ≤ 10, the images of f(x) are unique.
Thus, the given relation is a function.
It can be observed that for x = 2, g(x) = 2² = 4 and g(x) = 3 × 2 = 6
Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6.
Hence, this relation is not a function.
```
Question-2 :- If f(x) = x², find = [f(1.1)-f(1)]/(1.1 - 1).
Solution :-
``` Given that f(x) = x².
[f(1.1) - f(1)]/(1.1 - 1)
= [(1.1)² - (1)²]/(1.1 - 1)
= (1.21 - 1)/(0.1)
= 0.21/0.1
= 2.1
```
Question-3 :- Find the domain of the function f(x) = (x² + 2x + 1)/(x² - 8x + 12).
Solution :-
``` Given that f(x) = (x² + 2x + 1)/(x² - 8x + 12).
(x² - 8x + 12) = x² - 6x - 2x + 12
= x(x - 6) - 2(x - 6)
= (x - 2)(x - 6)
f(x) = (x² + 2x + 1)/(x - 2)(x - 6)
It can be seen that function f is defined for all real numbers except at x = 6 and x = 2.
Hence, the domain of f is R – {2, 6}.
```
Question-4 :- Find the domain and the range of the real function f defined by f (x) = √x - 1
Solution :-
``` Given that f(x) = f (x) = √x - 1.
It can be seen that is defined for (x – 1) ≥ 0.
i.e., is defined for x ≥ 1.
Therefore, the domain of f is the set of all real numbers greater than or equal to 1 i.e., the domain of f = [1, ∞).
As x ≥ 1 ⇒ (x – 1) ≥ 0 ⇒ √x - 1 ≥ 0
Therefore, the range of f is the set of all real numbers greater than or equal to 0 i.e., the range of f = [0, ∞).
```
Question-5 :- Find the domain and the range of the real function f defined by f (x) = |x - 1|.
Solution :-
``` Given that f(x) = |x - 1|.
It is clear that |x – 1| is defined for all real numbers.
Domain of f = R
Also, for x ∈ R, |x – 1| assumes all real numbers.
Hence, the Range of f is the set of all non-negative real numbers.
```
Question-6 :- Let f = {(x, x²/(1 + x²)) : x ∈ R} be a function from R into R. Determine the range of f.
Solution :-
``` Given that f(x) = {(x, x²/(1 + x²)) : x ∈ R}.
f(x) = {(0, 0), (±0.5, 1/5), (±1, 1/2), (±1.3, 9/13), (±2, 4/5), (3, 9/10),.....}
The range of f is the set of all second elements.
It can be observed that all these elements are greater than or equal to 0 but less than 1.
Range of f = [0, 1)
```
Question-7 :- Let f, g : R → R be defined, respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and f/g.
Solution :-
``` Given that f(x) = x + 1, g(x) = 2x – 3.
f + g = x + 1 + 2x - 3 = 3x - 2
f - g = x + 1 - 2x + 3 = -x + 4
f/g = (x + 1)/(2x - 3)
```
Question-8 :- Let f = {(1,1), (2,3), (0,–1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.
Solution :-
``` Given that f = {(1,1), (2,3), (0,–1), (–1, –3)} and
f(x) = ax + b
(1, 1) ∈ f
⇒ f(1) = 1
⇒ a × 1 + b = 1
⇒ a + b = 1
(0, –1) ∈ f
⇒ f(0) = –1
⇒ a × 0 + b = –1
⇒ b = –1
On substituting b = –1 in a + b = 1, we obtain a + (–1) = 1 ⇒ a = 1 + 1 = 2.
Thus, the respective values of a and b are 2 and –1.
```
Question-9 :- Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b²}. Are the following true?
(i) (a,a) ∈ R, for all a ∈ N
(ii) (a,b) ∈ R, implies (b,a) ∈ R
(iii) (a,b) ∈ R, (b,c) ∈ R implies (a,c) ∈ R.
Solution :-
``` Given that R = {(a, b) : a, b ∈ N and a = b²}.
(i) (a,a) ∈ R, for all a ∈ N
It can be seen that 2 ∈ N;
However, 2 ≠ 2² = 4.
Therefore, the statement '(a, a) ∈ R, for all a ∈ N' is not true.
(ii) (a,b) ∈ R, implies (b, a) ∈ R
It can be seen that (9, 3) ∈ N because 9, 3 ∈ N and 9 = 3².
Now, 3 ≠ 9² = 81;
Therefore, (3, 9) ∉ N.
Therefore, the statement '(a, b) ∈ R, implies (b, a) ∈ R' is not true.
(iii) (a,b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R.
It can be seen that (9, 3) ∈ R, (16, 4) ∈ R because 9, 3, 16, 4 ∈ N and 9 = 3² and 16 = 4².
Now, 9 ≠ 4² = 16;
Therefore, (9, 4) ∉ N
Therefore, the statement '(a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R' is not true.
```
Question-10 :- Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)} Are the following true?
(i) f is a relation from A to B
(ii) f is a function from A to B.
Solution :-
``` Given that A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}.
A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11),
(2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5),
(4, 9), (4, 11), (4, 15), (4, 16)}
(i) f is a relation from A to B
A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.
It is observed that f is a subset of A × B.
Thus, f is a relation from A to B.
(ii) f is a function from A to B.
Since the same first element i.e., 2 corresponds to two different images
i.e., 9 and 11, relation f is not a function.
```
Question-11 :- Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.
Solution :-
``` Given that f = {(ab, a + b) : a, b ∈ Z}.
The relation f is defined as f = {(ab, a + b): a, b ∈ Z}
We know that a relation f from a set A to a set B is said to be a function
if every element of set A has unique images in set B.
Since 2, 6, –2, –6 ∈ Z, (2 × 6, 2 + 6), [(–2) × (–6), (–2) + (–6)) ∈ f i.e.,
(12, 8), (12, –8) ∈ f
It can be seen that the same first element i.e.,
12 corresponds to two different images i.e., 8 and –8.
Thus, relation f is not a function.
```
Question-12 :- Let A = {9, 10 ,11 ,12 ,13} and let f : A → N be defined by f (n) = the highest prime factor of n. Find the range of f.
Solution :-
``` Given that A = {9,10,11,12,13} & f(n) = the highest prime factor of n.
Prime factor of 9 = 3
Prime factors of 10 = 2, 5
Prime factor of 11 = 11
Prime factors of 12 = 2, 3
Prime factor of 13 = 13
f(9) = The highest prime factor of 9 = 3
```
CLASSES |
## Engage NY Eureka Math 5th Grade Module 2 Lesson 7 Answer Key
### Eureka Math Grade 5 Module 2 Lesson 7 Sprint Answer Key
A
Multiply by Multiples of 10 and 100
2 x 10 = 20, 12 x 10 = 120, 12 x 100 = 1200, 4 x 10 = 40, 34 x 10 = 340, 34 x 100 = 3400, 7 x 10 = 70, 27 x 10 = 270, 27 x 100 = 2700, 3 x 10 = 30, 3 x 2 = 6, 3 x 20 = 60, 13 x 10 = 130, 13 x 2 = 26, 13 x 20 = 260, 13 x 100 = 1300, 13 x 200 = 2600, 2 x 4 = 8, 22 x 4 = 88, 22 x 40 = 880, 22 x 400 = 8800, 33 x 2 = 66.
Explanation:
In the above-given question,
given that,
multiply by multiples of 10 and 100.
2 x 10 = 20, 12 x 10 = 120, 12 x 100 = 1200, 4 x 10 = 40, 34 x 10 = 340, 34 x 100 = 3400, 7 x 10 = 70, 27 x 10 = 270, 27 x 100 = 2700, 3 x 10 = 30, 3 x 2 = 6, 3 x 20 = 60, 13 x 10 = 130, 13 x 2 = 26, 13 x 20 = 260, 13 x 100 = 1300, 13 x 200 = 2600, 2 x 4 = 8, 22 x 4 = 88, 22 x 40 = 880, 22 x 400 = 8800, 33 x 2 = 66.
33 x 20 = 660, 33 x 200 = 6600, 24 x 10 = 240, 24 x 20 = 480, 24 x 100 = 2400, 24 x 200 = 4800, 23 x 30 = 690, 23 x 300 = 6900, 71 x 2 = 142, 71 x 20 = 1420, 14 x 2 = 28, 14 x 3 = 42, 14 x 30 = 420, 14 x 300 = 4200, 82 x 20 = 1640, 15 x 300 = 4500, 71 x 600 = 42,600, 18 x 40 = 720, 75 x 30 = 2250, 84 x 300 = 25200, 87 x 60 = 5220, 79 x 800 = 63200.
Explanation:
In the above-given question,
given that,
multiply by multiples of 10 and 100.
33 x 20 = 660, 33 x 200 = 6600, 24 x 10 = 240, 24 x 20 = 480, 24 x 100 = 2400, 24 x 200 = 4800, 23 x 30 = 690, 23 x 300 = 6900, 71 x 2 = 142, 71 x 20 = 1420, 14 x 2 = 28, 14 x 3 = 42, 14 x 30 = 420, 14 x 300 = 4200, 82 x 20 = 1640, 15 x 300 = 4500, 71 x 600 = 42,600, 18 x 40 = 720, 75 x 30 = 2250, 84 x 300 = 25200, 87 x 60 = 5220, 79 x 800 = 63200.
Question 1.
2 × 10 = 20.
2 x 10 = 20.
Explanation:
In the above-given question,
given that,
2 x 10.
20.
Question 2.
12 × 10 = 120.
12 x 10 = 120.
Explanation:
In the above-given question,
given that,
12 x 10.
120.
Question 3.
12 × 100 = 1200.
12 x 100 = 1200.
Explanation:
In the above-given question,
given that,
12 x 100.
1200.
Question 4.
4 × 10 = 40.
4 x 10 = 40.
Explanation:
In the above-given question,
given that,
4 x 10.
40.
Question 5.
34 × 10 = 340
34 x 10 = 340.
Explanation:
In the above-given question,
given that,
34 x 10.
340.
Question 6.
34 × 100 = 3400
34 x 100 = 3400.
Explanation:
In the above-given question,
given that,
34 x 100.
3400.
Question 7.
7 × 10 = 70
7 x 10 = 70.
Explanation:
In the above-given question,
given that,
7 x 10.
70.
Question 8.
27 × 10 = 270
27 x 10 = 270.
Explanation:
In the above-given question,
given that,
27 x 10.
270.
Question 9.
27 × 100 = 2700
27 x 100 = 2700.
Explanation:
In the above-given question,
given that,
27 x 100.
2700.
Question 10.
3 × 10 = 30
3 x 10 = 30.
Explanation:
In the above-given question,
given that,
3 x 10.
30.
Question 11.
3 × 2 = 6
2 x 3 = 6.
Explanation:
In the above-given question,
given that,
2 x 3.
6.
Question 12.
3 × 20 = 60
20 x 3 = 60.
Explanation:
In the above-given question,
given that,
20 x 3.
60.
Question 13.
13 × 10 = 130
13 x 10 = 130.
Explanation:
In the above-given question,
given that,
13 x 10.
130.
Question 14.
13 × 2 = 26
2 x 13 = 26.
Explanation:
In the above-given question,
given that,
2 x 13.
26.
Question 15.
13 × 20 = 260
13 x 20 = 260.
Explanation:
In the above-given question,
given that,
13 x 20.
260.
Question 16.
13 × 100 = 1300
13 x 100 = 1300.
Explanation:
In the above-given question,
given that,
13 x 100.
1300.
Question 17.
13 × 200 = 2600
13 x 200 = 2600.
Explanation:
In the above-given question,
given that,
13 x 200.
2600.
Question 18.
2 × 4 = 8
2 x 4 = 8.
Explanation:
In the above-given question,
given that,
2 x 4.
8.
Question 19.
22 × 4 = 88.
22 x 4 = 88.
Explanation:
In the above-given question,
given that,
22 x 4.
88.
Question 20.
22 × 40 = 880
22 x 40 = 880.
Explanation:
In the above-given question,
given that,
22 x 40.
880.
Question 21.
22 × 400 = 8800
22 x 40 = 8800.
Explanation:
In the above-given question,
given that,
22 x 40.
8800.
Question 22.
33 × 2 = 66
2 x 33 = 66.
Explanation:
In the above-given question,
given that,
2 x 33.
66.
Question 23.
33 × 20 = 660.
33 x 20 = 660.
Explanation:
In the above-given question,
given that,
33 x 20.
660.
Question 24.
33 × 200 = 6600.
33 x 200 = 6600.
Explanation:
In the above-given question,
given that,
33 x 200.
6600.
Question 25.
24 × 10 = 240
24 x 10 = 240.
Explanation:
In the above-given question,
given that,
24 x 10.
240.
Question 26.
24 × 20 = 480
24 x 20 = 480.
Explanation:
In the above-given question,
given that,
24 x 20.
480.
Question 27.
24 × 100 = 2400
24 x 100 = 2400.
Explanation:
In the above-given question,
given that,
24 x 100.
2400.
Question 28.
24 × 200 = 4800
24 x 200 = 4800.
Explanation:
In the above-given question,
given that,
24 x 200.
4800.
Question 29.
23 × 30 = 690
23 x 30 = 690.
Explanation:
In the above-given question,
given that,
23 x 30.
690.
Question 30.
23 × 300 = 6900
23 x 300 = 6900.
Explanation:
In the above-given question,
given that,
23 x 300.
6900.
Question 31.
71 × 2 = 142.
2 x 71 = 142.
Explanation:
In the above-given question,
given that,
2 x 71.
142.
Question 32.
71 × 20 = 1420
20 x 71 = 1420.
Explanation:
In the above-given question,
given that,
20 x 71.
1420.
Question 33.
14 × 2= 28.
2 x 14 = 28.
Explanation:
In the above-given question,
given that,
2 x 14.
28.
Question 34.
14 × 3 = 42.
14 x 3 = 42.
Explanation:
In the above-given question,
given that,
14 x 3.
42.
Question 35.
14 × 30 = 420.
14 x 30 = 420.
Explanation:
In the above-given question,
given that,
14 x 30.
420.
Question 36.
14 × 300 = 4200
14 x 300 = 4200.
Explanation:
In the above-given question,
given that,
14 x 300.
4200.
Question 37.
82 × 20 = 1640
82 x 20 = 1640.
Explanation:
In the above-given question,
given that,
82 x 20.
1640.
Question 38.
15 × 300 = 4500
15 x 300 = 4500.
Explanation:
In the above-given question,
given that,
15 x 300.
4500.
Question 39.
71 × 600 = 42600.
71 x 600 = 42600.
Explanation:
In the above-given question,
given that,
71 x 600.
42600.
Question 40.
18 × 40 = 720
18 x 40 = 720.
Explanation:
In the above-given question,
given that,
18 x 40.
720.
Question 41.
75 × 30 = 2250
75 x 30 = 2250.
Explanation:
In the above-given question,
given that,
75 x 30.
2250.
Question 42.
84 × 300 = 25200
84 x 300 = 25200.
Explanation:
In the above-given question,
given that,
84 x 300.
25200.
Question 43.
87 × 60 = 5220
87 x 60 = 5220.
Explanation:
In the above-given question,
given that,
87 x 60.
5220.
Question 44.
79 × 800 = 63200
79 x 800 = 63200.
Explanation:
In the above-given question,
given that,
79 x 800.
63200.
B
Multiply by Multiples of 10 and 100
3 x 10 = 30, 13 x 10 = 130, 13 x 100 = 1300, 5 x 10 = 50, 35 x 10 = 350, 35 x 100 = 3500, 8 x 10 = 80, 28 x 10 = 280, 28 x 100 = 2800, 4 x 10 = 40, 4 x 2 = 8, 4 x 20 = 80, 14 x 10 = 140, 14 x 2 = 28, 14 x 20 = 280, 14 x 100 = 1400, 2 x 3 = 6, 22 x 3 = 66, 22 x 30 = 660, 22 x 300 = 6600, 44 x 2 = 88.
Explanation:
In the above-given question,
given that,
multiply by multiples of 10 and 100.
3 x 10 = 30, 13 x 10 = 130, 13 x 100 = 1300, 5 x 10 = 50, 35 x 10 = 350, 35 x 100 = 3500, 8 x 10 = 80, 28 x 10 = 280, 28 x 100 = 2800, 4 x 10 = 40, 4 x 2 = 8, 4 x 20 = 80, 14 x 10 = 140, 14 x 2 = 28, 14 x 20 = 280, 14 x 100 = 1400, 2 x 3 = 6, 22 x 3 = 66, 22 x 30 = 660, 22 x 300 = 6600, 44 x 2 = 88.
44 x 20 = 800, 44 x 200 = 8000, 42 x 10 = 420, 42 x 20 = 840, 42 x 100 = 4200, 42 x 200 = 8400, 32 x 30 = 960, 32 x 300 = 9600, 81 x 2 = 162, 81 x 20 = 1620, 13 x 3 = 39, 13 x 4 = 52, 13 x 40 = 520, 13 x 400 = 5200, 72 x 30 = 2160, 15 x 300 = 4500, 81 x 600 = 48600, 16 x 40 = 640, 65 x 30 = 1950, 48 x 300 = 14400, 89 x 60 = 5340, 76 x 800 = 60,800.
Explanation:
In the above-given question,
given that,
multiply by multiples of 10 and 100.
44 x 20 = 800, 44 x 200 = 8000, 42 x 10 = 420, 42 x 20 = 840, 42 x 100 = 4200, 42 x 200 = 8400, 32 x 30 = 960, 32 x 300 = 9600, 81 x 2 = 162, 81 x 20 = 1620, 13 x 3 = 39, 13 x 4 = 52, 13 x 40 = 520, 13 x 400 = 5200, 72 x 30 = 2160, 15 x 300 = 4500, 81 x 600 = 48600, 16 x 40 = 640, 65 x 30 = 1950, 48 x 300 = 14400, 89 x 60 = 5340, 76 x 800 = 60,800.
Question 1.
3 × 10 = 30
3 x 10 = 30.
Explanation:
In the above-given question,
given that,
3 x 10.
30.
Question 2.
13 × 10 = 130.
13 x 10 = 130.
Explanation:
In the above-given question,
given that,
13 x 10.
130.
Question 3.
13 × 100 = 1300
13 x 10 = 1300.
Explanation:
In the above-given question,
given that,
13 x 10.
1300.
Question 4.
5 × 10 = 50
5 x 10 = 50.
Explanation:
In the above-given question,
given that,
5 x 10.
50.
Question 5.
35 × 10 = 350
35 x 10 = 350.
Explanation:
In the above-given question,
given that,
35 x 10.
350.
Question 6.
35 × 100 = 3500
35 x 100 = 3500.
Explanation:
In the above-given question,
given that,
35 x 100.
3500.
Question 7.
8 × 10 = 80.
8 x 10 = 80.
Explanation:
In the above-given question,
given that,
8 x 10.
80.
Question 8.
28 × 10 = 280
28 x 10 = 280.
Explanation:
In the above-given question,
given that,
28 x 10.
280.
Question 9.
28 × 100 = 2800
28 x 100 = 2800.
Explanation:
In the above-given question,
given that,
28 x 100.
2800.
Question 10.
4 × 10 = 40
4 x 10 = 40.
Explanation:
In the above-given question,
given that,
4 x 10.
40.
Question 11.
4 × 2 = 8
4 x 2 = 8.
Explanation:
In the above-given question,
given that,
4 x 2.
8.
Question 12.
4 × 20 = 80.
4 x 20 = 80.
Explanation:
In the above-given question,
given that,
4 x 20.
80.
Question 13.
14 × 10 = 140.
14 x 10 = 140.
Explanation:
In the above-given question,
given that,
14 x 10.
140.
Question 14.
14 × 2 = 28
14 x 2 = 28.
Explanation:
In the above-given question,
given that,
14 x 2.
28.
Question 15.
14 × 20 = 280
14 x 20 = 280.
Explanation:
In the above-given question,
given that,
14 x 20.
280.
Question 16.
14 × 100 = 1400
14 x 100 = 1400.
Explanation:
In the above-given question,
given that,
14 x 100.
1400.
Question 17.
14 × 200 = 2800
14 x 200 = 2800.
Explanation:
In the above-given question,
given that,
14 x 200.
2800.
Question 18.
2 × 3 = 6
3 x 2 = 6.
Explanation:
In the above-given question,
given that,
3 x 2.
6.
Question 19.
22 × 3 = 66.
3 x 22 = 66.
Explanation:
In the above-given question,
given that,
3 x 22.
66.
Question 20.
22 × 30 = 660
22 x 30 = 660.
Explanation:
In the above-given question,
given that,
22 x 30.
660.
Question 21.
22 × 300 = 6600
22 x 300 = 6600.
Explanation:
In the above-given question,
given that,
22 x 300.
6600.
Question 22.
44 × 2 = 88
42 x 2 = 88.
Explanation:
In the above-given question,
given that,
42 x 2.
88.
Question 23.
44 × 20 = 880
44 x 20 = 880.
Explanation:
In the above-given question,
given that,
44 x 20.
880.
Question 24.
44 × 200 = 8800
44 x 200 = 8800.
Explanation:
In the above-given question,
given that,
44 x 200.
8800.
Question 25.
42 × 10 = 420
42 x 10 = 420.
Explanation:
In the above-given question,
given that,
42 x 10.
420.
Question 26.
42 × 20 = 840
42 x 20 = 840.
Explanation:
In the above-given question,
given that,
42 x 20.
840.
Question 27.
42 × 100 = 4200
42 x 100 = 4200.
Explanation:
In the above-given question,
given that,
42 x 100.
4200.
Question 28.
42 × 200 = 8400
42 x 200 = 8400.
Explanation:
In the above-given question,
given that,
42 x 200.
8400.
Question 29.
32 × 30 = 960.
32 x 30 = 960.
Explanation:
In the above-given question,
given that,
32 x 30.
9600.
Question 30.
32 × 300 = 9600
32 x 300 = 9600.
Explanation:
In the above-given question,
given that,
32 x 300.
9600.
Question 31.
81 × 2 = 162.
81 x 2 = 162.
Explanation:
In the above-given question,
given that,
81 x 2.
162.
Question 32.
81 × 20 = 1620.
81 x 20 = 1620.
Explanation:
In the above-given question,
given that,
81 x 20.
1620.
Question 33.
13 × 3 = 39
3 x 13 = 39.
Explanation:
In the above-given question,
given that,
3 x 13.
39.
Question 34.
13 × 4 = 52
13 x 4 = 52.
Explanation:
In the above-given question,
given that,
13 x 4.
52.
Question 35.
13 × 40 = 520
13 x 40 = 520.
Explanation:
In the above-given question,
given that,
13 x 40.
520.
Question 36.
13 × 400 = 5200
13 x 400 = 5200.
Explanation:
In the above-given question,
given that,
13 x 400.
5200.
Question 37.
72 × 30 = 2160.
72 x 30 = 2160.
Explanation:
In the above-given question,
given that,
72 x 30.
2160.
Question 38.
15 × 300 = 4500
15 x 300 = 4500.
Explanation:
In the above-given question,
given that,
15 x 300.
4500.
Question 39.
81 × 600 = 48600.
81 x 600 = 48600.
Explanation:
In the above-given question,
given that,
81 x 600.
48600.
Question 40.
16 × 40 = 640
16 x 40 = 640.
Explanation:
In the above-given question,
given that,
16 x 40.
640.
Question 41.
65 × 30 = 1950.
65 x 30 = 1950.
Explanation:
In the above-given question,
given that,
65 x 30.
1950.
Question 42.
48 × 300 = 14400
48 x 300 = 14400.
Explanation:
In the above-given question,
given that,
48 x 300.
14400.
Question 43.
89 × 60 = 5340
89 x 60 = 5340.
Explanation:
In the above-given question,
given that,
89 x 60.
5340.
Question 44.
76 × 800 = 60800
76 x 800 = 60800.
Explanation:
In the above-given question,
given that,
76 x 800.
60800.
### Eureka Math Grade 5 Module 2 Lesson 7 Problem Set Answer Key
Question 1.
Draw an area model. Then, solve using the standard algorithm. Use arrows to match the partial products from the area model to the partial products in the algorithm.
a. 481 × 352
481 x 352 = 169312.
Explanation:
In the above-given question,
given that,
481 x 352.
169312.
b. 481 × 302
481 x 302 = 145262.
Explanation:
In the above-given question,
given that,
481 x 302.
145262.
c. Why are there three partial products in 1(a) and only two partial products in 1(b)?
Question 2.
Solve by drawing the area model and using the standard algorithm.
a. 8,401 × 305
8401 x 305 = 25,62,305.
Explanation:
In the above-given question,
given that,
8401 x 305.
25,62,305.
b. 7,481 × 350
7481 x 350 = 26,18,350.
Explanation:
In the above-given question,
given that,
7481 x 350.
2618350.
Question 3.
Solve using the standard algorithm.
a. 346 × 27
346 x 27 = 9342.
Explanation:
In the above-given question,
given that,
346 x 27.
9342.
b. 1,346 × 297
1346 x 297 = 399762.
Explanation:
In the above-given question,
given that,
1346 x 297.
399762.
c. 346 × 207
346 x 207 = 71622.
Explanation:
In the above-given question,
given that,
346 x 207.
71622.
d. 1,346 × 207
1346 x 207 = 278622.
Explanation:
In the above-given question,
given that,
1346 x 207.
278622.
Question 4.
A school district purchased 615 new laptops for their mobile labs. Each computer cost $409. What is the total cost for all of the laptops? Answer: The total cost for all of the laptops = 251535. Explanation: In the above-given question, given that, A school didtrict purchased 615 new laptops for their mobile labs. Each computer cost$409.
The total cost for all the laptops = 615 x 409.
251535.
Question 5.
A publisher prints 1,512 copies of a book in each print run. If they print 305 runs, how many books will be printed?
The books will be printed = 461160.
Explanation:
In the above-given question,
given that,
A publisher prints 1,512 copies of a book in each print run.
If they print 305 runs.
1512 x 305 = 461160.
Question 6.
As of the 2010 census, there were 3,669 people living in Marlboro, New York. Brooklyn, New York, has 681 times as many people. How many more people live in Brooklyn than in Marlboro?
The more people live in Brooklyn than in Marlboro = 24, 98, 589.
Explanation:
In the above-given question,
given that,
As of the 2010 census, there were 3669 people living in Marlboro, New York.
Brooklyn, New York, has 681 times as many people.
3669 x 681 = 24,98,589.
### Eureka Math Grade 5 Module 2 Lesson 7 Exit Ticket Answer Key
Draw an area model. Then, solve using the standard algorithm.
a. 642 × 257
642 x 257 = 158574.
Explanation:
In the above-given question,
given that,
642 x 257.
158574.
b. 642 × 207
642 x 207 = 132894.
Explanation:
In the above-given question,
given that,
642 x 207.
132894.
### Eureka Math Grade 5 Module 2 Lesson 7 Homework Answer Key
Question 1.
Draw an area model. Then, solve using the standard algorithm. Use arrows to match the partial products from your area model to the partial products in your algorithm.
a. 273 × 346
273 x 346 = 94458.
Explanation:
In the above-given question,
given that,
273 x 346.
94458.
b. 273 × 306
273 x 306 = 83538.
Explanation:
In the above-given question,
given that,
273 x 306.
83538.
c. Both Parts (a) and (b) have three-digit multipliers. Why are there three partial products in Part (a) and only two partial products in Part (b)?
Question 2.
Solve by drawing the area model and using the standard algorithm.
a. 7,481 × 290
7481 x 290 = 2169490.
Explanation:
In the above-given question,
given that,
7481 x 290.
2169490.
b. 7,018 × 209
7018 x 209 = 14,66,762.
Explanation:
In the above-given question,
given that,
7018 x 209.
14,66,762.
Question 3.
Solve using the standard algorithm.
a. 426 × 357
426 x 375 = 159750.
Explanation:
In the above-given question,
given that,
426 x 375.
159750.
b. 1,426 × 357
1426 x 357 = 509082.
Explanation:
In the above-given question,
given that,
1426 x 357.
509082.
c. 426 × 307
426 x 307 = 130782.
Explanation:
In the above-given question,
given that,
426 x 307.
130782.
d. 1,426 × 307
1426 x 307 = 437782.
Explanation:
In the above-given question,
given that,
1426 x 307.
437782.
Question 4.
The Hudson Valley Renegades Stadium holds a maximum of 4,505 people. During the height of their popularity, they sold out 219 consecutive games. How many tickets were sold during this time?
The tickets were sold during this time = 9,86, 595.
Explanation:
In the above-given question,
given that,
One Saturday at the farmer’s market, each of the 94 vendors made $502 in profit. How much profit did all vendors make that Saturday? Answer: The profit did all vendors make that saturday =$47188.
One Saturday at the farmer’s market, each of the 94 vendors made $502 in profit. 94 x$502 = 47188. |
### N - Piazza
```Chapter 6
With Question/Answer Animations
Chapter Summary
The Basics of Counting
The Pigeonhole Principle
Permutations and Combinations
Binomial Coefficients and Identities
Generalized Permutations and Combinations
Generating Permutations and Combinations (not yet
included in overheads)
Section 6.1
Section Summary
The Product Rule
The Sum Rule
The Subtraction Rule
The Division Rule
Examples, Examples, and Examples
Tree Diagrams
Basic Counting Principles: The Product
Rule
The Product Rule: A procedure can be broken down
into a sequence of two tasks. There are n1 ways to do
the first task and n2 ways to do the second task. Then
there are n1∙n2 ways to do the procedure.
Example: How many bit strings of length seven are
there?
Solution: Since each of the seven bits is either a 0 or a
1, the answer is 27 = 128.
The Product Rule
Example: How many different license plates can be
made if each plate contains a sequence of three
uppercase English letters followed by three digits?
Solution: By the product rule,
there are 26 ∙ 26 ∙ 26 ∙ 10 ∙ 10 ∙ 10 = 17,576,000
different possible license plates.
Counting Functions
Counting Functions: How many functions are there from a set
with m elements to a set with n elements?
Solution: Since a function represents a choice of one of the n
elements of the codomain for each of the m elements in the
domain, the product rule tells us that there are n ∙ n ∙ ∙ ∙ n = nm
such functions.
Counting One-to-One Functions: How many one-to-one
functions are there from a set with m elements to one with n
elements?
Solution: Suppose the elements in the domain are
a1, a2,…, am. There are n ways to choose the value of a1 and n−1
ways to choose a2, etc. The product rule tells us that there are
n(n−1) (n−2)∙∙∙(n−m +1) such functions.
Telephone Numbering Plan
Example: The North American numbering plan (NANP) specifies that a telephone
number consists of 10 digits, consisting of a three-digit area code, a three-digit office
code, and a four-digit station code. There are some restrictions on the digits.
Let X denote a digit from 0 through 9.
Let N denote a digit from 2 through 9.
Let Y denote a digit that is 0 or 1.
In the old plan (in use in the 1960s) the format was NYX-NNX-XXX.
In the new plan, the format is NXX-NXX-XXX.
How many different telephone numbers are possible under the old plan and the new
plan?
Solution: Use the Product Rule.
There are 8 ∙2 ∙10 = 160 area codes with the format NYX.
There are 8 ∙10 ∙10 = 800 area codes with the format NXX.
There are 8 ∙8 ∙10 = 640 office codes with the format NNX.
There are 10 ∙10 ∙10 ∙10 = 10,000 station codes with the format XXXX.
Number of old plan telephone numbers: 160 ∙640 ∙10,000 = 1,024,000,000.
Number of new plan telephone numbers: 800 ∙800 ∙10,000 = 6,400,000,000.
Counting Subsets of a Finite Set
Counting Subsets of a Finite Set: Use the product rule to
show that the number of different subsets of a finite set S is
2|S|. (In Section 5.1, mathematical induction was used
to prove this same result.)
Solution: When the elements of S are listed in an
arbitrary order, there is a one-to-one correspondence
between subsets of S and bit strings of length |S|. When
the ith element is in the subset, the bit string has a 1 in the
ith position and a 0 otherwise.
By the product rule, there are 2|S| such bit strings, and
therefore 2|S| subsets.
Product Rule in Terms of Sets
If A1, A2, … , Am are finite sets, then the number of
elements in the Cartesian product of these sets is the
product of the number of elements of each set.
The task of choosing an element in the Cartesian
product A1 ⨉ A2 ⨉ ∙∙∙ ⨉ Am is done by choosing an
element in A1, an element in A2 , …, and an element
in Am.
By the product rule, it follows that:
|A1 ⨉ A2 ⨉ ∙∙∙ ⨉ Am |= |A1| ∙ |A2| ∙ ∙∙∙ ∙ |Am|.
DNA and Genomes
A gene is a segment of a DNA molecule that encodes a particular
protein and the entirety of genetic information of an organism is called
its genome.
DNA molecules consist of two strands of blocks known as nucleotides.
Each nucleotide is composed of bases: adenine (A), cytosine (C),
guanine (G), or thymine (T).
The DNA of bacteria has between 105 and 107 links (one of the four
bases). Mammals have between 108 and 1010 links. So, by the product
rule there are at least 4105 different sequences of bases in the DNA of
bacteria and 4108 different sequences of bases in the DNA of mammals.
The human genome includes approximately 23,000 genes, each with
1,000 or more links.
Biologists, mathematicians, and computer scientists all work on
determining the DNA sequence (genome) of different organisms.
Basic Counting Principles: The Sum Rule
The Sum Rule: If a task can be done either in one of n1
ways or in one of n2 ways to do the second task, where
none of the set of n1 ways is the same as any of the n2 ways,
then there are n1 + n2 ways to do the task.
Example: The mathematics department must choose
either a student or a faculty member as a representative for
a university committee. How many choices are there for
this representative if there are 37 members of the
mathematics faculty and 83 mathematics majors and no
one is both a faculty member and a student.
Solution: By the sum rule it follows that there are
37 + 83 = 120 possible ways to pick a representative.
The Sum Rule in terms of sets.
The sum rule can be phrased in terms of sets.
|A ∪ B|= |A| + |B| as long as A and B are disjoint
sets.
Or more generally,
|A1 ∪ A2 ∪ ∙∙∙ ∪ Am |= |A1| + |A2| + ∙∙∙ + |Am|
when Ai ∩ Aj = ∅ for all i, j.
The case where the sets have elements in common will
be discussed when we consider the subtraction rule
and taken up fully in Chapter 8.
Combining the Sum and Product
Rule
Example: Suppose statement labels in a programming
language can be either a single letter or a letter
followed by a digit. Find the number of possible labels.
Solution: Use the product rule.
26 + 26 ∙ 10 = 286
Counting Passwords
Combining the sum and product rule allows us to solve more complex problems.
Example: Each user on a computer system has a password, which is six to eight
characters long, where each character is an uppercase letter or a digit. Each password
must contain at least one digit. How many possible passwords are there?
Solution: Let P be the total number of passwords, and let P6, P7, and P8 be the
passwords of length 6, 7, and 8.
By the sum rule P = P6 + P7 +P8.
To find each of P6, P7, and P8 , we find the number of passwords of the specified length
composed of letters and digits and subtract the number composed only of letters. We find
that:
P6 = 366 − 266 =2,176,782,336 − 308,915,776 =1,867,866,560.
P7 = 367 − 267 =
78,364,164,096 − 8,031,810,176 = 70,332,353,920.
P8 = 368 − 268 =
2,821,109,907,456 − 208,827,064,576 =2,612,282,842,880.
Consequently, P = P6 + P7 +P8 = 2,684,483,063,360.
Internet Addresses
Version 4 of the Internet Protocol (IPv4) uses 32 bits.
Class A Addresses: used for the largest networks, a 0,followed by a 7-bit netid
and a 24-bit hostid.
Class B Addresses: used for the medium-sized networks, a 10,followed by a
14-bit netid and a 16-bit hostid.
Class C Addresses: used for the smallest networks, a 110,followed by a 21-bit
netid and a 8-bit hostid.
Neither Class D nor Class E addresses are assigned as the address of a computer
on the internet. Only Classes A, B, and C are available.
1111111 is not available as the netid of a Class A network.
Hostids consisting of all 0s and all 1s are not available in any network.
Counting Internet Addresses
Example: How many different IPv4 addresses are available for
computers on the internet?
Solution: Use both the sum and the product rule. Let x be the number
of available addresses, and let xA, xB, and xC denote the number of
addresses for the respective classes.
To find, xA: 27 − 1 = 127 netids. 224 − 2 = 16,777,214 hostids.
xA = 127∙ 16,777,214 = 2,130,706,178.
To find, xB: 214 = 16,384 netids. 216 − 2 = 16,534 hostids.
xB = 16,384 ∙ 16, 534 = 1,073,709,056.
To find, xC: 221 = 2,097,152 netids. 28 − 2 = 254 hostids.
xC = 2,097,152 ∙ 254 = 532,676,608.
Hence, the total number of available IPv4 addresses is
x = xA + xB + xC
= 2,130,706,178 + 1,073,709,056 + 532,676,608
= 3, 737,091,842.
Not Enough Today !!
The newer IPv6 protocol solves the problem
of too few addresses.
Basic Counting Principles:
Subtraction Rule
Subtraction Rule: If a task can be done either in one
of n1 ways or in one of n2 ways, then the total number
of ways to do the task is n1 + n2 minus the number of
ways to do the task that are common to the two
different ways.
Also known as, the principle of inclusion-exclusion:
Counting Bit Strings
Example: How many bit strings of length eight either
start with a 1 bit or end with the two bits 00?
Solution: Use the subtraction rule.
Number of bit strings of length eight
that start with a 1 bit: 27 = 128
Number of bit strings of length eight
that start with bits 00: 26 = 64
Number of bit strings of length eight
that start with a 1 bit and end with bits 00 : 25 = 32
Hence, the number is 128 + 64 − 32 = 160.
Basic Counting Principles: Division
Rule
Division Rule: There are n/d ways to do a task if it can be done using a procedure that can
be carried out in n ways, and for every way w, exactly d of the n ways correspond to way
w.
Restated in terms of sets: If the finite set A is the union of n pairwise disjoint subsets
each with d elements, then n = |A|/d.
In terms of functions: If f is a function from A to B, where both are finite sets, and for
every value y ∈ B there are exactly d values x ∈ A such that f(x) = y, then |B| = |A|/d.
Example: How many ways are there to seat four people around a circular table, where two
seatings are considered the same when each person has the same left and right
neighbor?
Solution: Number the seats around the table from 1 to 4 proceeding clockwise. There are
four ways to select the person for seat 1, 3 for seat 2, 2, for seat 3, and one way for seat 4.
Thus there are 4! = 24 ways to order the four people. But since two seatings are the same
when each person has the same left and right neighbor, for every choice for seat 1, we get
the same seating.
Therefore, by the division rule, there are 24/4 = 6 different seating arrangements.
Tree Diagrams
Tree Diagrams: We can solve many counting problems through the
use of tree diagrams, where a branch represents a possible choice and
the leaves represent possible outcomes.
Example: Suppose that “I Love Discrete Math” T-shirts come in five
different sizes: S,M,L,XL, and XXL. Each size comes in four colors
(white, red, green, and black), except XL, which comes only in red,
green, and black, and XXL, which comes only in green and black. What
is the minimum number of stores that the campus book store needs to
stock to have one of each size and color available?
Solution: Draw the tree diagram.
The store must stock 17 T-shirts.
Section 6.2
Section Summary
The Pigeonhole Principle
The Generalized Pigeonhole Principle
The Pigeonhole Principle
If a flock of 20 pigeons roosts in a set of 19 pigeonholes, one of
the pigeonholes must have more than 1 pigeon.
Pigeonhole Principle: If k is a positive integer and k + 1 objects
are placed into k boxes, then at least one box contains two or
more objects.
Proof: We use a proof by contraposition. Suppose none of the k
boxes has more than one object. Then the total number of
objects would be at most k. This contradicts the statement that
we have k + 1 objects.
The Pigeonhole Principle
Corollary 1: A function f from a set with k + 1
elements to a set with k elements is not one-to-one.
Proof: Use the pigeonhole principle.
Create a box for each element y in the codomain of f .
Put in the box for y all of the elements x from the
domain such that f(x) = y.
Because there are k + 1 elements and only k boxes, at
least one box has two or more elements.
Hence, f can’t be one-to-one.
Pigeonhole Principle
Example: Among any group of 367 people, there must be at least
two with the same birthday, because there are only 366 possible
birthdays.
Example (optional): Show that for every integer n there is a
multiple of n that has only 0s and 1s in its decimal expansion.
Solution: Let n be a positive integer. Consider the n + 1 integers
1, 11, 111, …., 11…1 (where the last has n + 1 1s). There are n
possible remainders when an integer is divided by n. By the
pigeonhole principle, when each of the n + 1 integers is divided
by n, at least two must have the same remainder. Subtract the
smaller from the larger and the result is a multiple of n that has
only 0s and 1s in its decimal expansion.
The Generalized Pigeonhole Principle
The Generalized Pigeonhole Principle: If N objects are
placed into k boxes, then there is at least one box
containing at least ⌈N/k⌉ objects.
Proof: We use a proof by contraposition. Suppose that
none of the boxes contains more than ⌈N/k⌉ − 1 objects.
Then the total number of objects is at most
where the inequality ⌈N/k⌉ < ⌈N/k⌉ + 1 has been used. This
is a contradiction because there are a total of n objects.
Example: Among 100 people there are at least
⌈100/12⌉ = 9 who were born in the same month.
The Generalized Pigeonhole Principle
Example: a) How many cards must be selected from a standard
deck of 52 cards to guarantee that at least three cards of the
same suit are chosen?
b) How many must be selected to guarantee that at least three
hearts are selected?
Solution: a) We assume four boxes; one for each suit. Using the
generalized pigeonhole principle, at least one box contains at
least ⌈N/4⌉ cards. At least three cards of one suit are selected if
⌈N/4⌉ ≥3. The smallest integer N such that ⌈N/4⌉ ≥3 is
N = 2 ∙ 4 + 1 = 9.
b) A deck contains 13 hearts and 39 cards which are not hearts.
So, if we select 41 cards, we may have 39 cards which are not
hearts along with 2 hearts. However, when we select 42 cards, we
must have at least three hearts. (Note that the generalized
pigeonhole principle is not used here.)
Section 6.3
Section Summary
Permutations
Combinations
Combinatorial Proofs
Permutations
Definition: A permutation of a set of distinct objects
is an ordered arrangement of these objects. An ordered
arrangement of r elements of a set is called an
r-permuation.
Example: Let S = {1,2,3}.
The ordered arrangement 3,1,2 is a permutation of S.
The ordered arrangement 3,2 is a 2-permutation of S.
The number of r-permuatations of a set with n
elements is denoted by P(n,r).
The 2-permutations of S = {1,2,3} are 1,2; 1,3; 2,1; 2,3;
3,1; and 3,2. Hence, P(3,2) = 6.
A Formula for the Number of
Permutations
Theorem 1: If n is a positive integer and r is an integer with
1 ≤ r ≤ n, then there are
P(n, r) = n(n − 1)(n − 2) ∙∙∙ (n − r + 1)
r-permutations of a set with n distinct elements.
Proof: Use the product rule. The first element can be chosen in n
ways. The second in n − 1 ways, and so on until there are
(n − ( r − 1)) ways to choose the last element.
Note that P(n,0) = 1, since there is only one way to order zero
elements.
Corollary 1: If n and r are integers with 1 ≤ r ≤ n, then
Solving Counting Problems by
Counting Permutations
Example: How many ways are there to select a firstprize winner, a second prize winner, and a third-prize
winner from 100 different people who have entered a
contest?
Solution:
P(100,3) = 100 ∙ 99 ∙ 98 = 970,200
Solving Counting Problems by
Counting Permutations (continued)
Example: Suppose that a saleswoman has to visit eight
different cities. She must begin her trip in a specified
city, but she can visit the other seven cities in any order
she wishes. How many possible orders can the
saleswoman use when visiting these cities?
Solution: The first city is chosen, and the rest are
ordered arbitrarily. Hence the orders are:
7! = 7 ∙ 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 5040
If she wants to find the tour with the shortest path that
visits all the cities, she must consider 5040 paths!
Solving Counting Problems by
Counting Permutations (continued)
Example: How many permutations of the letters
ABCDEFGH contain the string ABC ?
Solution: We solve this problem by counting the
permutations of six objects, ABC, D, E, F, G, and H.
6! = 6 ∙ 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 720
Combinations
Definition: An r-combination of elements of a set is an
unordered selection of r elements from the set. Thus, an
r-combination is simply a subset of the set with r elements.
The number of r-combinations of a set with n distinct
elements is denoted by C(n, r). The notation
is also
used and is called a binomial coefficient. (We will see the
notation again in the binomial theorem in Section 6.4.)
Example: Let S be the set {a, b, c, d}. Then {a, c, d} is a 3combination from S. It is the same as {d, c, a} since the
order listed does not matter.
C(4,2) = 6 because the 2-combinations of {a, b, c, d} are the
six subsets {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, and {c, d}.
Combinations
Theorem 2: The number of r-combinations of a set
with n elements, where n ≥ r ≥ 0, equals
Proof: By the product rule P(n, r) = C(n,r) ∙ P(r,r).
Therefore,
Combinations
Example: How many poker hands of five cards can be dealt
from a standard deck of 52 cards? Also, how many ways are
there to select 47 cards from a deck of 52 cards?
Solution: Since the order in which the cards are dealt does
not matter, the number of five card hands is:
The different ways to select 47 cards from 52 is
This is a special case of a general result. →
Combinations
Corollary 2: Let n and r be nonnegative integers with
r ≤ n. Then C(n, r) = C(n, n − r).
Proof: From Theorem 2, it follows that
and
Hence, C(n, r) = C(n, n − r).
This result can be proved without using algebraic manipulation. →
Combinatorial Proofs
Definition 1: A combinatorial proof of an identity is a
proof that uses one of the following methods.
A double counting proof uses counting arguments to
prove that both sides of an identity count the same
objects, but in different ways.
A bijective proof shows that there is a bijection between
the sets of objects counted by the two sides of the
identity.
Combinatorial Proofs
Here are two combinatorial proofs that
C(n, r) = C(n, n − r)
when r and n are nonnegative integers with r < n:
Bijective Proof: Suppose that S is a set with n elements. The
function that maps a subset A of S to is a bijection between
the subsets of S with r elements and the subsets with n − r
elements. Since there is a bijection between the two sets, they
must have the same number of elements.
Double Counting Proof: By definition the number of subsets
of S with r elements is C(n, r). Each subset A of S can also be
described by specifying which elements are not in A, i.e.,
those which are in . Since the complement of a subset of S
with r elements has n − r elements, there are also C(n, n − r)
subsets of S with r elements.
Combinations
Example: How many ways are there to select five players
from a 10-member tennis team to make a trip to a match at
another school.
Solution: By Theorem 2, the number of combinations is
Example: A group of 30 people have been trained as
astronauts to go on the first mission to Mars. How many
ways are there to select a crew of six people to go on this
mission?
Solution: By Theorem 2, the number of possible crews is
Section 6.4
Section Summary
The Binomial Theorem
Pascal’s Identity and Triangle
Other Identities Involving Binomial Coefficients (not
currently included in overheads)
Powers of Binomial Expressions
Definition: A binomial expression is the sum of two terms, such as x + y. (More
generally, these terms can be products of constants and variables.)
We can use counting principles to find the coefficients in the expansion of (x + y)n where
n is a positive integer.
To illustrate this idea, we first look at the process of expanding (x + y)3.
(x + y) (x + y) (x + y) expands into a sum of terms that are the product of a term from
each of the three sums.
Terms of the form x3, x2y, x y2, y3 arise. The question is what are the coefficients?
To obtain x3 , an x must be chosen from each of the sums. There is only one way to do this.
So, the coefficient of x3 is 1.
To obtain x2y, an x must be chosen from two of the sums and a y from the other. There
are
ways to do this and so the coefficient of x2y is 3.
To obtain xy2, an x must be chosen from of the sums and a y from the other two . There
are
ways to do this and so the coefficient of xy2 is 3.
To obtain y3 , a y must be chosen from each of the sums. There is only one way to do this. So,
the coefficient of y3 is 1.
We have used a counting argument to show that (x + y)3 = x3 + 3x2y + 3x y2 + y3 .
Next we present the binomial theorem gives the coefficients of the terms in the expansion
of (x + y)n .
Binomial Theorem
Binomial Theorem: Let x and y be variables, and n a
nonnegative integer. Then:
Proof: We use combinatorial reasoning . The terms in
the expansion of (x + y)n are of the form xn−jyj for
j = 0,1,2,…,n. To form the term xn−jyj, it is necessary to
choose n−j xs from the n sums. Therefore, the
coefficient of xn−jyj is
which equals
.
Using the Binomial Theorem
Example: What is the coefficient of x12y13 in the
expansion of (2x − 3y)25?
Solution: We view the expression as (2x +(−3y))25.
By the binomial theorem
Consequently, the coefficient of x12y13 in the expansion
is obtained when j = 13.
A Useful Identity
Corollary 1: With n ≥0,
Proof (using binomial theorem): With x = 1 and y = 1, from the
binomial theorem we see that:
Proof (combinatorial): Consider the subsets of a set with n
elements. There are
subsets with zero elements,
with one
element,
with two elements, …, and
with n elements.
Therefore the total is
Since, we know that a set with n elements has 2n subsets, we
conclude:
Blaise Pascal
(1623-1662)
Pascal’s Identity
Pascal’s Identity: If n and k are integers with n ≥ k ≥ 0, then
Proof (combinatorial): Let T be a set where |T| = n + 1, a ∊T, and
S = T − {a}. There are
subsets of T containing k elements.
Each of these subsets either:
contains a with k − 1 other elements, or
contains k elements of S and not a.
There are
subsets of k elements that contain a, since there are
subsets of k − 1 elements of S,
subsets of k elements of T that do not contain a, because there
are
subsets of k elements of S.
Hence,
See Exercise 19
for an algebraic
proof.
Pascal’s Triangle
The nth row in
the triangle
consists of the
binomial
coefficients
,
k = 0,1,….,n.
By Pascal’s identity, adding two adjacent bionomial coefficients results is the
binomial coefficient in the next row between these two coefficients.
Section 6.5
Section Summary
Permutations with Repetition
Combinations with Repetition
Permutations with Indistinguishable Objects
Distributing Objects into Boxes
Permutations with Repetition
Theorem 1: The number of r-permutations of a set of n
objects with repetition allowed is nr.
Proof: There are n ways to select an element of the set for
each of the r positions in the r-permutation when
repetition is allowed. Hence, by the product rule there are
nr r-permutations with repetition.
Example: How many strings of length r can be formed
from the uppercase letters of the English alphabet?
Solution: The number of such strings is 26r, which is the
number of r-permutations of a set with 26 elements.
Combinations with Repetition
Example: How many ways are there to select five bills
from a box containing at least five of each of the
following denominations: \$1, \$2, \$5, \$10, \$20, \$50,
and \$100?
Solution: Place the selected bills in the appropriate
position of a cash box illustrated below:
continued →
Combinations with Repetition
Some possible ways of
placing the five bills:
The number of ways to select five bills corresponds to the
number of ways to arrange six bars and five stars in a row.
This is the number of unordered selections of 5 objects from a
set of 11. Hence, there are
ways to choose five bills with seven types of bills.
Combinations with Repetition
Theorem 2: The number 0f r-combinations from a set with n
elements when repetition of elements is allowed is
C(n + r – 1,r) = C(n + r – 1, n –1).
Proof: Each r-combination of a set with n elements with
repetition allowed can be represented by a list of n –1 bars and r
stars. The bars mark the n cells containing a star for each time
the ith element of the set occurs in the combination.
The number of such lists is C(n + r – 1, r), because each list is a
choice of the r positions to place the stars, from the total of
n + r – 1 positions to place the stars and the bars. This is also
equal to C(n + r – 1, n –1), which is the number of ways to place
the n –1 bars.
Combinations with Repetition
Example: How many solutions does the equation
x1 + x2 + x3 = 11
have, where x1 , x2 and x3 are nonnegative integers?
Solution: Each solution corresponds to a way to select
11 items from a set with three elements; x1 elements of
type one, x2 of type two, and x3 of type three.
By Theorem 2 it follows that there are
solutions.
Combinations with Repetition
Example: Suppose that a cookie shop has four
different kinds of cookies. How many different ways
can six cookies be chosen?
Solution: The number of ways to choose six cookies is
the number of 6-combinations of a set with four
elements. By Theorem 2
is the number of ways to choose six cookies from the
four kinds.
Summarizing the Formulas for Counting Permutations
and Combinations with and without Repetition
Permutations with
Indistinguishable Objects
Example: How many different strings can be made by reordering the
letters of the word SUCCESS.
Solution: There are seven possible positions for the three Ss, two Cs,
one U, and one E.
The three Ss can be placed in C(7,3) different ways, leaving four
positions free.
The two Cs can be placed in C(4,2) different ways, leaving two
positions free.
The U can be placed in C(2,1) different ways, leaving one position free.
The E can be placed in C(1,1) way.
By the product rule, the number of different strings is:
The reasoning can be generalized to the following theorem. →
Permutations with
Indistinguishable Objects
Theorem 3: The number of different permutations of n objects, where there are
n1 indistinguishable objects of type 1, n2 indistinguishable objects of
type 2, …., and nk indistinguishable objects of type k, is:
Proof: By the product rule the total number of permutations is:
C(n, n1 ) C(n − n1, n2 ) ∙∙∙ C(n − n1 − n2 − ∙∙∙ − nk, nk) since:
The n1 objects of type one can be placed in the n positions in C(n, n1 ) ways,
leaving n − n1 positions.
Then the n2 objects of type two can be placed in the n − n1 positions in
C(n − n1, n2 ) ways, leaving n − n1 − n2 positions.
Continue in this fashion, until nk objects of type k are placed in
C(n − n1 − n2 − ∙∙∙ − nk, nk) ways.
The product can be manipulated into the desired result as follows:
Distributing Objects into Boxes
Many counting problems can be solved by counting
the ways objects can be placed in boxes.
The objects may be either different from each other
(distinguishable) or identical (indistinguishable).
The boxes may be labeled (distinguishable) or unlabeled
(indistinguishable).
Distributing Objects into Boxes
Distinguishable objects and distinguishable boxes.
There are n!/(n1!n2! ∙∙∙nk!) ways to distribute n distinguishable
objects into k distinguishable boxes.
(See Exercises 47 and 48 for two different proofs.)
Example: There are 52!/(5!5!5!5!32!) ways to distribute hands of 5
cards each to four players.
Indistinguishable objects and distinguishable boxes.
There are C(n + r − 1, n − 1) ways to place r indistinguishable
objects into n distinguishable boxes.
Proof based on one-to-one correspondence between
n-combinations from a set with k-elements when repetition is
allowed and the ways to place n indistinguishable objects into k
distinguishable boxes.
Example: There are C(8 + 10 − 1, 10) = C(17,10) = 19,448 ways to
place 10 indistinguishable objects into 8 distinguishable boxes.
Distributing Objects into Boxes
Distinguishable objects and indistinguishable boxes.
Example: There are 14 ways to put four employees into three
indistinguishable offices (see Example 10).
There is no simple closed formula for the number of ways to
distribute n distinguishable objects into j indistinguishable boxes.
See the text for a formula involving Stirling numbers of the second
kind.
Indistinguishable objects and indistinguishable boxes.
Example: There are 9 ways to pack six copies of the same book into
four identical boxes (see Example 11).
The number of ways of distributing n indistinguishable objects into
k indistinguishable boxes equals pk(n), the number of ways to write
n as the sum of at most k positive integers in increasing order.
No simple closed formula exists for this number.
``` |
# Triangular Series
A triangular series is a series of numbers where each number could be the row of an equilateral triangle.
So 1, 2, 3, 4, 5 is a triangular series, because you could stack the numbers like this:
Their sum is 15, which makes 15 a triangular number.
A triangular series always starts with 1 and increases by 1 with each number.
Since the only thing that changes in triangular series is the value of the highest number, it’s helpful to give that a name. Let’s call it n.
# n is 8 1, 2, 3, 4, 5, 6, 7, 8
Triangular series are nice because no matter how large n is, it’s always easy to find the total sum of all the numbers.
Take the example above. Notice that if we add the first and last numbers together, and then add the second and second-to-last numbers together, they have the same sum! This happens with every pair of numbers until we reach the middle. If we add up all the pairs of numbers, we get:
1 + 8 = 9 2 + 7 = 9 3 + 6 = 9 4 + 5 = 9
This is true for every triangular series:
1. Pairs of numbers on each side will always add up to the same value.
2. That value will always be 1 more than the series’ n.
This gives us a pattern. Each pair's sum is n+1, and there are \frac{n}{2} pairs. So our total sum is: (n + 1) * \frac{n}{2}
Or:
\frac{n^2 + n}{2}
Ok, but does this work with triangular series with an odd number of elements? Yes. Let's say n = 5. So if we calculate the sum by hand:
1+2+3+4+5=15
And if we use the formula, we get the same answer:
\frac{5^2 + 5}{2}=15
One more thing:
What if we know the total sum, but we don't know the value of n?
Let’s say we have:
1 + 2 + 3 + \ldots + (n - 2) + (n - 1) + n = 78
No problem. We just use our formula and set it equal to the sum!
\frac{n^2 + n}{2}=78
Now, we can rearrange our equation to get a quadratic equation (remember those?)
n^2 + n = 156 n^2 + n - 156 = 0
Here's the quadratic formula:
\frac{-b\pm\sqrt{b^2-4ac}}{2a}
If you don't really remember how to use it, that's cool. You can just use an online calculator. We don't judge.
Taking the positive solution, we get n = 12.
So for a triangular series, remember—the total sum is:
\frac{n^2 + n}{2}
. . . |
# Trigonometry - Solve Equations
## Trigonometry - Solve Equations - How it Works - Video
### Unit Circle
Unit Circle:
The unit circle is vital to answer any exact answer. Do you have to memorize every bit? That is up to. It would be faster if you had it encoded in your brain, but you can memorized just the first quadrant and use symmetry to fill out the unit circle.
### Example 1
Example 1:
Since we have tangent by itself on the left side, we inverse both sides to get phi by itself. We know the angle is sqrt(3). Since we have tangent, we want sqrt(3) to be in the y coordinate in our ordered pair. Remember tangent = y/x. We want this so we don't have to radicalized the denominator.
Here we have two options one in the first quadrant and one in the third quadrant. So phi equals, π /3 or 4π /3. Since the period of tangent is π , every solution is going to be π /3 + π k and other 4π /3 + π k. If k = 1 for π /3 + π k then we get 4π /3. So we can combine the two answers into one, π /3 + π k.
Here we have graphed our equation. You can either graph it as two equations and see where they intersect or set the equation equal to zero, and graph one equation. We set the equation equal to zero, f(φ) = tan(φ) - sqrt(3). So we are looking for the equation crosses the x-axis. The first positive one is π /3. The next one is 4π /3. And the next one is 7π /3. Each time we are adding π .
### Example 2 Part 1
Example 2 Part 1:
Since we have zero on the right side, we can solve for theta. Since both terms involve square numbers, we have a difference of square numbers. Let's factor.
csc4(2θ) - 4 = 0
[csc2(2θ)]2 - 22 = 0
(csc2(2θ) - 2) * (csc2(2θ) + 2)
Rewrote the terms to the be powers.
Factored.
Now we have two factors, so let's set each to zero and solve.
(csc2(2θ) - 2) = 0
csc2(2θ) = 2
csc(2θ) = ± sqrt(2)
Rewrote the terms to the be powers.
Factored.
Let's flip that to sin() = ± sqrt(2)/2.
### Example 2 Part 2
Example 2 Part 2:
Once we have two numbers we substitute the numbers into the formula, cos(u - v) = cos u * cos v + sin u * sin v.
So we get cos(5π /6 - π /4) = cos 5π /6 * cos π /4 - sin 5π /6 * sin π /4, where 5π /6 = u and π /4 = v.
=> cos(5π /6 - π /4) = cos 5π /6 * cos π /4 - sin 5π /6 * sin π /4
=> -sqrt(3) / 2 * sqrt(2) / 2- 1 / 2 * sqrt(2) / 2
=> -sqrt(6) / 4 + sqrt(2) / 4
=> [-sqrt(6) + sqrt(2)] / 4
Substituted u and v.
Use the unit chart to find the corresponding x and y values.
Multiplied numerators and multiplied denominators.
Combined fractions.
So our exact is [-sqrt(6) + sqrt(2)] / 4.
Sine and tangent work the same way.
### Example 2 Part 3
Example 2 Part 3:
We have four options where sin() = ± sqrt(2)/2, one in Q I, one in Q II, one in Q III, and one in Q IV.
Solving for θ for each one:
2θ = π/4
2θ = π/4 + 2πk
θ = π/8 + πk
Q I
Divided by 2.
2θ = 3π/4
2θ = π/4 + 2πk
θ = 3π/8 + πk
Q II
Divided by 2.
2θ = 5π/4
2θ = 5π/4 + 2πk
θ = c + πk
Q III
Divided by 2.
2θ = 7π/4
2θ = 7π/4 + 2πk
θ = 7π/8 + πk
Q IV |
### Chapter 4 – section 1
```College Algebra
Acosta/Karwowski
Chapter 4
CHAPTER 4 – SECTION 1
• A quadratic is any equation with a 2nd degree
term and no higher degree terms
• In standard form ax2 + bx + c = y
• Solving an equation means to isolate the x
variable. Since there is an x2 term as well as
an x term this presents a challenge
algebraically.
•
Using roots to isolate the x
• If the x appears in the problem only once you can
isolate the x using inverses but first note
• In general the inverse of a power is its root
• But as seen in ch 5 square root is not the inverse
of square power unless we modify the functions
• you can cancel square powers with square roots
if you remember that
2
2 ≠ for every x
2
2 = ||
• thus: if 2 = 25 ℎ = ±5
Reminders : working with square roots
• If the square root is irrational – leave the radical form of
called the exact form of the number
ex: 16 ans: 4
21 ans: 21
21 : 21
80 : 4 5
• Square root of a negative is an IMAGINARY number
−4 = 2 not - no solution
−12 = 2 3
Examples: solve
• 5x2 – 3 = 77
• 3x2 + 7 = 25
• (x – 4)2 = 16
• 24 + (x – 3)2 = 15
• x2 +8x = 15
Draw back to inversing
• If x does not appear in the problem only once
using square root to isolate the x becomes
problematic since:
• A. You cannot combine unlike terms
3x2 + 5x
• B. You cannot do the square root on an
expression that contains addition outside of ()
−5 2 = −5
2 − 4 ≠ − 2
• ex: x2 – 7x + 12 =
3x2 - 19x + 20 =
9x2 – 121 =
15x2 + 35x – 90 =
• Mathematical fact if ab = 0 either
a= 0 or b = 0
Thus: zero product rule is born
• Given a quadratic equation, you can
SOMETIMES isolate the x by
1. Making the equation = 0
3. Splitting the one equation into 2
and
4. Solving each equation
•
Examples
•
•
•
•
x2 + 6x – 16 = 0
x2 – 7x = -12
(x - 3)(x – 8) = 66
6x2 + 29x = 5
• x2 – 10 = 0
Bridging the gap
• Some problems cannot be done by factoring
because they do not have integral factors.
• These problems must be re written so that the
x only appears in the equation once.
• The process is called completing the square
Completing the square
• Concept - write a quadratic using the variable x only
once • tools :
x2 + 2hx + h2 = (x – h)2
• so - a quadratic of this form can be written with only
one x
• Given the first 2 terms you can determine the 3rd term
•
Fill in the blank
x2 + 8x + ___
x2 – 10x + ___
x2 + 5x + ____
Filling in this blank is called completing the square
Completing the square to solve
equations
• 4x2 – 40x - 20 = 0
• Prepare for completing the square by dividing by 4 and
moving the 20 ( which will be a 5)
• x2 – 10x
= 5
• Completing the square CHANGES the number so add
the same amount to BOTH sides of the equation
• x2 – 10x + 25 = 5 + 25
• Factor and solve
•
(x – 5)2 = 30
x = 5 ± 30
Examples : solve the following
• 3x2 – 24x + 6 = 0
• x2 – 7x – 3 = 0
• 3x2 – 11x – 4 = 0
• x2 + 5x + 15
CHAPTER 4 – SECTION 3
• The solutions of inequalities in one variable
are intervals on the number line
• The solutions to inequalities are directly
related to the solution of equations
• There are 2 solutions to quadratic equations.
What the graph of the quadratic
function tells us
y
•
x
find f(x) = 0
find f(x)>0
find f(x)<0
So quadratic inequalities work like absolute
value inequalities
Examples
• find x2 – 9x >0
• Find 2x2 + 7x – 30 < 0
• Find -2(x + 3)2 + 2 > 0
Assignment
• P 370(6-16)
Guaranteed test question
CHAPTER 4 -
Guaranteed test question – worth 1020 pts
• Mathematics finds the formulas –
• You find a formula by solving without using
any known numbers
• Solve ax2 + bx + c =y
for x thereby
deriving a formula for x
Solve ax2 + bx + c = 0 by completing the square
• Divide by a :
• Move constant to the
right
• Divide middle coefficient
• Get a common
denominator and
combine fraction
• Write as a square
• Isolate the x
x
2
b
c
x
a
x
2
b
a
x
a
x
2
x
2
b
0
c
a
b
x
2
a
4a
b
b
x
a
b
a
2
4a
2
2
c
4a
b 4 ac
2
2
4a
2
b
b 4 ac
x
2
2a
4a
x
b
2
b 4 ac
2
2a
2
2
CHAPTER 4 – SECTION 2
• f(x) = ax2 + bx + c = y is a quadratic function
called standard form
• f(x) = a(x – h)2 + k = y can be the SAME
quadratic function - called vertex form
• f(x) = (x – x1)(x – x2) can be the same function
also- called factored form
Standard form to vertex form
• Given f(x) = 2x2 – 12x + 10
Related topic for circles
• Standard form of circle (transformation form)
• Get standard form from simplified form.
• homework p 19(30-35)
Function concepts- what the equation
and graph tell us
•
•
•
•
•
•
x and y intercepts
f(x) = 0 is the x- intercept
f(0) = y is the y - intercept
maximum/minimum of a parabola
called vertex
related to line of symmetry
(a vertical line through the vertex)
intervals of increase or
decrease
f(#) = y (evaluate for y)
f(x) = # (solve for x)
Concavity – (also related to max/min)
Sketching a graph from the equation
Concavity
• Oriented up/ down
• Width
Examples: find x and y intercepts
• f(x) = x2 – 12x +32
also find f(3) and f(x) = 12
• g(x) = (x + 2)(x – 9) also find g(7) and g(x) = 2
• k(x) = 3(x – 6)2 - 15 also find k(4) and k(x) = 9
Find vertex(ie: max/min) / line of
symmetry/ intervals of inc. or dec.
•
g(x) = 3(x + 5)2 + 3
Finding vertex etc.
• f(x) = -2x2 + 4x – 7
• method one – easiest – uses information
• Method two – change to vertex form -
Example: word problem
• A ball is thrown into the air has a height given
by the function h(x) = -16x2 + 34x + 15
•
•
•
•
•
Find the y intercept and interpret its meaning?
When will the ball hit the ground?
Where will the ball be in 1.2 seconds?
When will the ball be 17 feet above ground?
How high did the ball go?
Sketching a graph for a quadratic
Assignment
•
•
•
•
sec 4.1 P326(1-33)(56 – 61) all
P 354 (5-38) odd add 355(46-50)
Worksheet
CHAPTER 4 - SUPPLEMENT
Writing equations – word problems
• To write an equation for a quadratic you need either
• 3 random points
•
or the vertex and one other solution point
• Or factors and one other point
• or a formula which has already been derived by
someone else.
Physics formula
•
h(x) = ax2 + bx + c
• has a physics application where
•
a = acceleration of an object after it is released (usually
this is gravity)
•
b = its initial velocity (force with which it is released)
•
c = the original location of the object
• Write a function for an object with gravitational
acceleration (known to be -16 ft/sec2) an initial velocity
force of 12 ft/sec and an initial location of 29 feet.
Banking formula- compound interest
• P(r) = p0 (1 + r)t
• where p0 is original amount in bank
•
t is a set amount of time
•
r is the rate of interest
•
note: t must be the same units as r
if r is 6% per month then t = 7 is 7 months
Ex. Mark invests \$4500 for 3 years. Write a
function for his account as a function of r.
Given 3 point
• (2,13) (-3, 38) ( 1,6)
•
Given vertex and one point
• (2,4) is the vertex and the graph goes through
the point (3,10)
Given factors/x intercepts
• x – intercepts can be found from factors so
factors can be found from x- intercepts - even
if those intercepts are irrational
• Factors can also be found given solutions that
are not intercepts – ie solutions that are
imaginary – note that imaginary and irrational
solutions come in pairs (conjuagates)
• Note that a third point is necessary to find the
stretch factor
examples
• Given the points (5,0) (-2,0) and the point (3,-2) write
an equation – you could do a system of equations since
this is 3 points
• Or •
a(x – 5) (x + 2)= 0 must be true
• Thus a(x -5)(x +2) = y is the basis for the equation
and
•
a(3 – 5)(3 +2) = -2
so -10a = -2 and a = 1/5
answer g(x) = 1/5 (x – 5)(x + 2)
Examples
•
•
•
•
given intercept 2 + 3, 0 and point (4, -3)
You know there is another intercept –
You have factors
You can simplify the quadratic and find the
stretch factor
example
•
•
•
•
Given solution - x = 3i and point (3, - 2)
You know another solution is x = - 3i
You have factors
You simplify and find the stretch factor
assignment
• P328 (62,63,66,67)
• Worksheet – writing equations for quadratics.
Solving root equations
CHAPTER 4 – SECTION 4
The inverse of a root is it’s
corresponding power
• Algebraic solving involves inversing an
operator or function to determine the input
value.
•
Examples
3
•
=9
=4
•
2 − 7 = 3
•
10 − 2 = 2 +
•
+2+1=
3 − 5
4
=5
assignment
• P377(1-31)odd
Other equations – quadratic in nature
SUPPLEMENT F
Powers/roots generalized
• Rational exponents
• The inverse of x a/b is x b/a
• Quadratic like: a trinomial in which the degree
of the leading term is twice the degree of the
middle term can be solved like a quadratic
Examples
5
3
• = −32
2
3
• 5 − 8 = 117
substitution
• If the equation has 3 terms where the middle
exponent is half of the leading exponent then
it is like a quadratic equation and can be
solved using a temporary substitution
ex
x4 – 5x2 = 6
2 3 + 1 3 = 12
• Or If the equation is written as three “terms”
with this same condition
ex (x – 7)6 + 2(x - 7)3 - 15 = 0
Examples: solve
•
•
•
7
x6
2
x5
−3
+4
=
9
5 x 30
7
x2
2
5
4x
x 7 x 10
2+1
3+60
2
Assignment
• Supplement - p69 (1-9)(13-26)odd (30- 33)all
``` |
# 773 and Level 6
• 773 is a prime number.
• Prime factorization: 773 is prime.
• The exponent of prime number 773 is 1. Adding 1 to that exponent we get (1 + 1) = 2. Therefore 773 has exactly 2 factors.
• Factors of 773: 1, 773
• Factor pairs: 773 = 1 x 773
• 773 has no square factors that allow its square root to be simplified. √773 ≈ 27.8028775.
How do we know that 773 is a prime number? If 773 were not a prime number, then it would be divisible by at least one prime number less than or equal to √773 ≈ 27.8. Since 773 cannot be divided evenly by 2, 3, 5, 7, 11, 13, 17, 19, or 23, we know that 773 is a prime number.
Here is today’s puzzle for you to try to solve:
Print the puzzles or type the solution on this excel file: 12 Factors 2016-02-25
————————————-
What else is special about the number 773?
22² + 17² = 773 so 773 is the hypotenuse of the primitive Pythagorean triple 195-748-773 which was calculated using 22² – 17², 2(17)(22), 22² + 17².
Thus 195² + 748² + 773².
773 is also the sum of three squares six different ways:
• 26² + 9² + 4² = 773
• 25² + 12² + 2² = 773
• 24² + 14² + 1² = 773
• 23² + 12² + 10² = 773
• 22² + 15² + 8² = 773
• 20² + 18² + 7² = 773
773 is a palindrome in two other bases:
• 545 BASE 12, note that 5(144) + 4(12) + 5(1) = 773
• 3D3 BASE 14 (D = 13 base 10); note that 3(196) + 13(14) + 3(1) = 773
Here’s another way we know that 773 is a prime number: Since its last two digits divided by 4 leave a remainder of 1, and 22² + 17² = 773 with 22 and 17 having no common prime factors, 773 will be prime unless it is divisible by a prime number Pythagorean triple hypotenuse less than or equal to √773 ≈ 27.8. Since 773 is not divisible by 5, 13, or 17, we know that 773 is a prime number.
————————————-
## 2 thoughts on “773 and Level 6”
1. I’ve done a lattice labyrinth for the sufficiently massive prime 773, based on number pair (17,22), 17^2 + 22^2 being equal to 773, as you point out, but don’t know how to send it. Best wishes, Dave Mitchell
• If you give me the link to the lattice labyrinth on your blog, I can cut and paste it onto this blog. Thanks.
This site uses Akismet to reduce spam. Learn how your comment data is processed. |
How do you integrate {x/(sqrt(4+4x^2))} dx from 0 to 2?
Feb 1, 2017
The answer is $= \frac{1}{2} \left(\sqrt{5} - 1\right)$
Explanation:
We need
$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$
Let $u = 4 + 4 {x}^{2}$
$\mathrm{du} = 8 x \mathrm{dx}$,
$x \mathrm{dx} = \frac{1}{8} \mathrm{du}$
Therefore,
$\int \frac{x \mathrm{dx}}{\sqrt{4 + 4 {x}^{2}}} = \frac{1}{8} \int \frac{\mathrm{du}}{\sqrt{u}} = \frac{1}{8} {u}^{- \frac{1}{2} + 1} / \left(\frac{1}{2}\right)$
$= \frac{1}{4} \sqrt{u} = \frac{1}{4} \sqrt{4 + 4 {x}^{2}} = \frac{1}{2} \sqrt{1 + {x}^{2}}$
so,
${\int}_{0}^{2} \frac{x \mathrm{dx}}{\sqrt{4 + 4 {x}^{2}}} = {\left[\frac{1}{2} \sqrt{1 + {x}^{2}}\right]}_{0}^{2}$
$= \left(\frac{\sqrt{5}}{2}\right) - \left(\frac{1}{2}\right)$
$= \frac{1}{2} \left(\sqrt{5} - 1\right)$ |
Convert indigenous decimal come fraction. Convert 0.78 come Fraction. Decimal to portion chart and also calculator. Writes any type of decimal number as a fraction.
You are watching: What is 78 as a fraction
## How to convert a Decimal to a portion - Steps
Step 1: write down the decimal together a portion of one (decimal/1);Step 2: If the decimal is no a whole number, multiply both top and bottom by 10 till you get an interger at the numerator.
Learn more reading the examples listed below or usage our self-explaining calculator above
## Convert decimal 0.05 to a fraction
0.05 = 1/20 as a fraction
### Step by step Solution
To transform the decimal 0.05 come a portion follow these steps:
Step 1: write down the number as a fraction of one:
0.05 = 0.05/1
Step 2: main point both top and bottom through 10 for every number after ~ the decimal point:
As we have actually 2 number after the decimal point, us multiply both numerator and denominator by 100. So,
0.05/1 = (0.05 × 100)/(1 × 100) = 5/100.
Step 3: simplify (or reduce) the fraction:
5/100 = 1/20 when lessened to the most basic form.
## What is 0.45 together a fraction?
0.45 = 9/20 together a fraction
### Step by action Solution
To convert the decimal 0.45 come a portion follow these steps:
Step 1: create down the number as a portion of one:
0.45 = 0.45/1
Step 2: multiply both top and also bottom through 10 for every number ~ the decimal point:
As we have actually 2 number after the decimal point, we multiply both numerator and denominator through 100. So,
0.45/1 = (0.45 × 100)/(1 × 100) = 45/100.
Step 3: simplify (or reduce) the fraction:
45/100 = 9/20 when reduced to the simplest form.
## Equivalent portion for 1.3 percent
1.3 = 13/10 = 13/10 as a fraction
### Step by step Solution
To transform the decimal 1.3 to a portion follow these steps:
Step 1: create down the number together a portion of one:
1.3 = 1.3/1
Step 2: multiply both top and also bottom by 10 because that every number after ~ the decimal point:
As we have actually 1 numbers after the decimal point, us multiply both numerator and also denominator through 10. So,
1.3/1 = (1.3 × 10)/(1 × 10) = 13/10.
(This portion is alread reduced, we can"t alleviate it any kind of further).
As the molecule is greater than the denominator, we have an wrong fraction, so us can likewise express it as a blended NUMBER, hence 13/10 is additionally equal to 1 3/10 once expressed as a mixed number.
## Conversion table: portion to decimal inches and also millimeter equivalence
To transform fractions to decimals and also millimeters and also vice-versa use this formula: 1 customs = 25.4 mm exactly, so ...To transform from inch to millimeter multiply inch value by 25.4.To convert from millimeter inch division millimeter value by 25.4.
one easier way to execute it is to usage the table below. How?
### Example 1
Convert 1 1/32" come mm: uncover 1 1/32 and also read come the ideal under mm column! friend will find 26.1938.
### Example 2
Convert 0.875 decimal inches come inches (fraction form).Look under the decimal obelisk until you discover 0.875, then review to the left to discover 7/8 inchesor relocate to the right tower to uncover the mm value!
fractioninchesmm
1/640.01560.3969
1/320.03130.7938
3/640.04691.1906
1/160.06251.5875
5/640.07811.9844
3/320.09382.3813
7/640.10942.7781
1/80.12503.1750
9/640.14063.5719
5/320.15633.9688
11/640.17194.3656
3/160.18754.7625
13/640.20315.1594
7/320.21885.5563
15/640.23445.9531
1/40.25006.3500
17/640.26566.7469
9/320.28137.1438
19/640.29697.5406
5/160.31257.9375
21/640.32818.3344
11/320.34388.7313
23/640.35949.1281
3/80.37509.5250
25/640.39069.9219
13/320.406310.3188
27/640.421910.7156
7/160.437511.1125
29/640.453111.5094
15/320.468811.9063
31/640.484412.3031
1/20.500012.7000
33/640.515613.0969
17/320.531313.4938
35/640.546913.8906
9/160.562514.2875
37/640.578114.6844
19/320.593815.0813
39/640.609415.4781
5/80.625015.8750
41/640.640616.2719
21/320.656316.6688
43/640.671917.0656
11/160.687517.4625
45/640.703117.8594
23/320.718818.2563
47/640.734418.6531
3/40.750019.0500
49/640.765619.4469
25/320.781319.8438
51/640.796920.2406
13/160.812520.6375
53/640.828121.0344
27/320.843821.4313
55/640.859421.8281
7/80.875022.2250
57/640.890622.6219
29/320.906323.0188
59/640.921923.4156
15/160.937523.8125
61/640.953124.2094
31/320.968824.6063
63/640.984425.0031
11.000025.4000
fractioninchesmm
1 1/641.015625.7969
1 1/321.031326.1938
1 3/641.046926.5906
1 1/161.062526.9875
1 5/641.078127.3844
1 3/321.093827.7813
1 7/641.109428.1781
1 1/81.125028.5750
1 9/641.140628.9719
1 5/321.156329.3688
1 11/641.171929.7656
1 3/161.187530.1625
1 13/641.203130.5594
1 7/321.218830.9563
1 15/641.234431.3531
1 1/41.250031.7500
1 17/641.265632.1469
1 9/321.281332.5438
1 19/641.296932.9406
1 5/161.312533.3375
1 21/641.328133.7344
1 11/321.343834.1313
1 23/641.359434.5281
1 3/81.375034.9250
1 25/641.390635.3219
1 13/321.406335.7188
1 27/641.421936.1156
1 7/161.437536.5125
1 29/641.453136.9094
1 15/321.468837.3063
1 31/641.484437.7031
1 1/21.500038.1000
1 33/641.515638.4969
1 17/321.531338.8938
1 35/641.546939.2906
1 9/161.562539.6875
1 37/641.578140.0844
1 19/321.593840.4813
1 39/641.609440.8781
1 5/81.625041.2750
1 41/641.640641.6719
1 21/321.656342.0688
1 43/641.671942.4656
1 11/161.687542.8625
1 45/641.703143.2594
1 23/321.718843.6563
1 47/641.734444.0531
1 3/41.750044.4500
1 49/641.765644.8469
1 25/321.781345.2438
1 51/641.796945.6406
1 13/161.812546.0375
1 53/641.828146.4344
1 27/321.843846.8313
1 55/641.859447.2281
1 7/81.875047.6250
1 57/641.890648.0219
1 29/321.906348.4188
1 59/641.921948.8156
1 15/161.937549.2125
1 61/641.953149.6094
1 31/321.968850.0063
1 63/641.984450.4031
22.000050.8000
fractioninchesmm
2 1/642.015651.1969
2 1/322.031351.5938
2 3/642.046951.9906
2 1/162.062552.3875
2 5/642.078152.7844
2 3/322.093853.1813
2 7/642.109453.5781
2 1/82.125053.9750
2 9/642.140654.3719
2 5/322.156354.7688
2 11/642.171955.1656
2 3/162.187555.5625
2 13/642.203155.9594
2 7/322.218856.3563
2 15/642.234456.7531
2 1/42.250057.1500
2 17/642.265657.5469
2 9/322.281357.9438
2 19/642.296958.3406
2 5/162.312558.7375
2 21/642.328159.1344
2 11/322.343859.5313
2 23/642.359459.9281
2 3/82.375060.3250
2 25/642.390660.7219
2 13/322.406361.1188
2 27/642.421961.5156
2 7/162.437561.9125
2 29/642.453162.3094
2 15/322.468862.7063
2 31/642.484463.1031
2 1/22.500063.5000
2 33/642.515663.8969
2 17/322.531364.2938
2 35/642.546964.6906
2 9/162.562565.0875
2 37/642.578165.4844
2 19/322.593865.8813
2 39/642.609466.2781
2 5/82.625066.6750
2 41/642.640667.0719
2 21/322.656367.4688
2 43/642.671967.8656
2 11/162.687568.2625
2 45/642.703168.6594
2 23/322.718869.0563
2 47/642.734469.4531
2 3/42.750069.8500
2 49/642.765670.2469
2 25/322.781370.6438
2 51/642.796971.0406
2 13/162.812571.4375
2 53/642.828171.8344
2 27/322.843872.2313
2 55/642.859472.6281
2 7/82.875073.0250
2 57/642.890673.4219
2 29/322.906373.8188
2 59/642.921974.2156
2 15/162.937574.6125
2 61/642.953175.0094
2 31/322.968875.4063
2 63/642.984475.8031
33.000076.2000
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# Derivative of tan x, sec x & tan x More
The derivative of tan x, sec x & tan x – The derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). Derivatives are a fundamental tool of calculus. For example, the derivative of the position of a moving object with respect to time is the object’s velocity: this measures how quickly the position of the object changes when time advances.
The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The tangent line is the best linear approximation of the function near that input value. For this reason, the derivative is often described as the “instantaneous rate of change”, the ratio of the instantaneous change in the dependent variable to that of the independent variable.
## The derivative of tan x
The derivative of tan x is sec2x.
To see why you’ll need to know a few results. First, you need to know that the derivative of sinx is cosx. Here’s a proof of that result from first principles
Once you know this, it also implies that the derivative of cosx is sinx (which you’ll also need later). You need to know one more thing, which is the Quotient Rule for differentiation:
Once all those pieces are in place, the differentiation goes as follows:
## The derivative of sec x tan x
Suppose we have y=f(x)g(x)
Then, using Product Rule, y=f(x)g(x)+f(x)g(x)
In simple language, keep the first term as it is and differentiate the second term, then differentiate the first term and keep the second term as it is or vice-versa.
## Derivative of tanx^2
In differential calculus, we use the Chain Rule when we have a composite function. It states that The derivative will be equal to the derivative of the outside function with respect to the inside, times the derivative of the inside function. Let’s see what that looks like mathematically
Chain Rule
f(g(x))g(x)
Let’s say we have the composite function sin(5x). We know:
f(x)=sinxf(x)=cosx
g(x)=5xg(x)=5
So the derivative will be equal to
cos(5x)5
=5cos(5x)
We just have to find our two functions, find their derivatives and input into the Chain Rule expression.
using the chain rule for derivative of tanx^2
Use the chain rule…
## What’s the derivative of SEC 2x?
The derivative of sec2 (x) is 2sec2 (x) tan (x). The chain rule states that the derivative of f(g(x)) is equal to f ‘ (g(x)) ⋅ g ‘ (x)
## What is the differentiation of theta?
It depends on the derivative of the variable you are taking with respect to the variable. For example, if you are taking the derivative of θ with respect to θ, you would get one. Generally, however, the derivative is taken with respect to x, because x is the most commonly used variable.
## What is Sec x?
The secondary trig functions are cosecant, secant, and cotangent [csc, sec, cot]. They are ratios that relate side lengths (opposite, adjacent, hypotenuse) to an angle in a right triangle. So secX is just the ratio of the length of a hypotenuse to the length of an adjacent side.
## What are the derivatives of the 6 trig functions?
Derivatives of Trigonometric Functions. The basic trigonometric functions include the following 6 functions: sine (sinx), cosine (cosx), tangent (tanx), cotangent (cotx), secant (secx) and cosecant (cscx). All these functions are continuous and differentiable in their domains.
## Is TANX continuous?
The function tan x is not continuous but is continuous on for example the interval −π/2 <x<π/2. It has infinitely many points of discontinuity, at ±π/2, ±3π/2, etc.; all are infinite discontinuities.
## How do you find a derivative?
Basically, we can compute the derivative of f(x) using the limit definition of derivatives with the following steps:
1. Find f(x + h).
2. Plug f(x + h), f(x), and h into the limit definition of a derivative.
3. Simplify the difference quotient.
4. Take the limit, as h approaches 0, of the simplified difference quotient.
## Is Tan Sin over COS?
So far in this course, the only trigonometric functions which we have studied are sine and cosine. … The tangent of x is defined to be its sine divided by its cosinetan x = sin x cos x . The cotangent of x is defined to be the cosine of x divided by the sine of x: cot x = cos x sin x .
## What is COTX?
What is ‘cotx‘? cot is a short way to write ‘cotangent’. This is the reciprocal of the trigonometric function ‘tangent’ or tan(x). Therefore, cot(x) can be simplified to 1/tan(x). Using trigonometric rules, an alternative way to write 1/tan(x) is cos(x)/sin(x). |
# How to Find Average Value of a Function – A Simple Guide for Beginners
To find the average value of a function, I start by considering the function over a specific interval. In calculus, this concept is important because it gives insight into the function’s overall behavior across that interval rather than at just a single point.
To calculate it accurately, I use the formula for the average value of a function $f(x)$ over an interval $[a,b]$ given by $\frac{1}{b-a}\int_{a}^{b} f(x) dx$. This formula represents the integral of the function over the interval divided by the width of the interval.
Applying this to a real-world scenario, like finding the average temperature over a day or the average cost over some time, helps me understand average values as they apply to everyday situations.
Stick around to discover how simple it is to apply this powerful tool from calculus to solve practical problems and gain insights into various functions.
## Calculating the Average Value
When I talk about the average value of a function, I’m referring to the mean value a function takes on an interval. This is an essential concept in understanding the overall behavior of a function on a given range.
### Applying Definite Integrals
To calculate the average value of a continuous function on a certain interval, I use definite integrals. This approach relies on the idea that the average value represents a certain “balance point” of the function values over the interval. The definite integral of a function gives me the total area under the curve between two points, which is key to figuring out the average value.
### Using Integration to Find Average Value
The mean value theorem for integrals provides a handy formula for this calculation. For a continuous function ( f(x) ) over the interval ([a, b]), the average value $f_{avg}$ is:
$$f_{avg} = \frac{1}{b-a} \int_{a}^{b} f(x) , dx$$
This formula asserts that I can find the average by integrating the function over the interval and then dividing by the length of that interval.
### Practical Examples and Calculations
To solidify my understanding, I like to work through specific examples. If I have a function $f(x) = x^2$, and I’m interested in finding its average value on the interval ([1, 3]), I would set up the integral:
$$f_{avg} = \frac{1}{3-1} \int_{1}^{3} x^2 , dx = \frac{1}{2} [ \frac{x^3}{3} ]_{1}^{3} = \frac{1}{2} [9 – \frac{1}{3}] = 4$$
Calculations like this help me understand the process and see integration in action.
### Interpreting Negative and Positive Areas
When integrating to find an average, I keep in mind that areas above the x-axis contribute positively, while those below contribute negatively to the total integral value. This reflects how the function behaves:
If it spends more time below the x-axis on an interval, I can expect a negative average value.
The area can be visually represented as the space under a function’s curve, with positive values pointing upwards and negative values extending downwards from the x-axis.
This attribute is significant when calculating the average value as it affects the sign of my final result, representing the net “effect” of the function over the interval.
## Conclusion
In my exploration of the average value of a function, I’ve discussed that this concept is essential to understanding the overall behavior of a function over a given interval.
When I calculate the average value, I am essentially finding the mean of the function’s outputs over that interval—much like finding the average of a set of numbers.
To determine the average value, I use the formula $f_{\text{ave}} = \frac{1}{b-a} \int_{a}^{b} f(x) dx$, where $[a, b]$ is the interval over which I’m averaging the function $f(x)$.
For example, if examining a linear function over an interval $[1, 5]$, my goal is to integrate the function from $1$ to $5$ and then divide it by the length of the interval, which in this case is $4$.
I’ve learned that the Mean Value Theorem for Integrals plays a crucial role, guaranteeing that for continuous functions, there is at least one point $c$ in the interval where the function’s value at $c$ represents the average value over the entire interval.
In practical applications, this computation aids in predicting trends and making informed decisions. By considering the average value, I get a single, representative number that reflects the collective behavior of the function over the interval, rather than getting lost in the infinite possible values the function may take.
Remember, mastering the step-by-step process to find the average value of a function is a powerful tool in both mathematics and various real-life scenarios. It simplifies complex data, providing clarity and insight into the nature of functions. |
Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 9th
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# Samacheer Kalvi solutions for Mathematics Class 9th Tamil Nadu State Board chapter 7 - Mensuration [Latest edition]
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## Chapter 7: Mensuration
Exercise 7.1Exercise 7.2Exercise 7.3Exercise 7.4
Exercise 7.1 [Pages 252 - 253]
### Samacheer Kalvi solutions for Mathematics Class 9th Tamil Nadu State Board Chapter 7 Mensuration Exercise 7.1 [Pages 252 - 253]
Exercise 7.1 | Q 1. (i) | Page 252
Using Heron’s formula, find the area of a triangle whose sides are 10 cm, 24 cm, 26 cm
Exercise 7.1 | Q 1. (ii) | Page 252
Using Heron’s formula, find the area of a triangle whose sides are 1.8 m, 8 m, 8.2 m
Exercise 7.1 | Q 2 | Page 252
The sides of the triangular ground are 22 m, 120 m and 122 m. Find the area and cost of levelling the ground at the rate of ₹ 20 per m2
Exercise 7.1 | Q 3 | Page 252
The perimeter of a triangular plot is 600 m. If the sides are in the ratio 5 : 12 : 13, then find the area of the plot
Exercise 7.1 | Q 4 | Page 252
Find the area of an equilateral triangle whose perimeter is 180 cm
Exercise 7.1 | Q 5 | Page 252
An advertisement board is in the form of an isosceles triangle with perimeter 36 m and each of the equal sides are 13 m. Find the cost of painting it at ₹ 17.50 per square metre.
Exercise 7.1 | Q 6 | Page 252
Find the area of the unshaded region
Exercise 7.1 | Q 7 | Page 253
Find the area of a quadrilateral ABCD whose sides are AB = 13 cm, BC = 12 cm, CD = 9 cm, AD = 14 cm and diagonal BD = 15 cm
Exercise 7.1 | Q 8 | Page 253
A park is in the shape of a quadrilateral. The sides of the park are 15 m, 20 m, 26 m and 17 m and the angle between the first two sides is a right angle. Find the area of the park
Exercise 7.1 | Q 9 | Page 253
A land is in the shape of rhombus. The perimeter of the land is 160 m and one of the diagonal is 48 m. Find the area of the land.
Exercise 7.1 | Q 10 | Page 253
The adjacent sides of a parallelogram measures 34 m, 20 m and the measure of the diagonal is 42 m. Find the area of parallelogram
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Exercise 7.2 [Page 257]
### Samacheer Kalvi solutions for Mathematics Class 9th Tamil Nadu State Board Chapter 7 Mensuration Exercise 7.2 [Page 257]
Exercise 7.2 | Q 1 | Page 257
Find the Total Surface Area and the Lateral Surface Area of a cuboid whose dimensions are: length = 20 cm, breadth = 15 cm, height = 8 cm
Exercise 7.2 | Q 2 | Page 257
The dimensions of a cuboidal box are 6 m × 400 cm × 1.5 m. Find the cost of painting its entire outer surface at the rate of ₹ 22 per m2.
Exercise 7.2 | Q 3 | Page 257
The dimensions of a hall is 10 m × 9 m × 8 m. Find the cost of white washing the walls and ceiling at the rate of ₹ 8.50 per m2
Exercise 7.2 | Q 4. (i) | Page 257
Find the T.S.A and L.S.A of the cube whose side is 8 m
Exercise 7.2 | Q 4. (ii) | Page 257
Find the T.S.A and L.S.A of the cube whose side is 21 cm
Exercise 7.2 | Q 4. (iii) | Page 257
Find the T.S.A and L.S.A of the cube whose side is 7.5 cm
Exercise 7.2 | Q 5 | Page 257
If the total surface area of a cube is 2400 cm2 then, find its lateral surface area
Exercise 7.2 | Q 6 | Page 257
A cubical container of side 6.5 m is to be painted on the entire outer surface. Find the area to be painted and the total cost of painting it at the rate of ₹ 24 per m2
Exercise 7.2 | Q 7 | Page 257
Three identical cubes of side 4 cm are joined end to end. Find the total surface area and lateral surface area of the new resulting cuboid
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Exercise 7.3 [Page 263]
### Samacheer Kalvi solutions for Mathematics Class 9th Tamil Nadu State Board Chapter 7 Mensuration Exercise 7.3 [Page 263]
Exercise 7.3 | Q 1. (i) | Page 263
Find the volume of a cuboid whose dimensions are length = 12 cm, breadth = 8 cm, height = 6 cm
Exercise 7.3 | Q 1. (ii) | Page 263
Find the volume of a cuboid whose dimensions are length = 60 m, breadth = 25 m, height = 1.5 m
Exercise 7.3 | Q 2 | Page 263
The dimensions of a match box are 6 cm × 3.5 cm × 2.5 cm. Find the volume of a packet containing 12 such match boxes
Exercise 7.3 | Q 3 | Page 263
The length, breadth and height of a chocolate box are in the ratio 5 : 4 : 3. If its volume is 7500 cm3, then find its dimensions
Exercise 7.3 | Q 4 | Page 263
The length, breadth and depth of a pond are 20.5 m, 16 m and 8 m respectively. Find the capacity of the pond in litres
Exercise 7.3 | Q 5 | Page 263
The dimensions of a brick are 24 cm × 12 cm × 8 cm. How many such bricks will be required to build a wall of 20 m length, 48 cm breadth and 6 m height?
Exercise 7.3 | Q 6 | Page 263
The volume of a container is 1440 m3. The length and breadth of the container are 15 m and 8 m respectively. Find its height
Exercise 7.3 | Q 7. (i) | Page 263
Find the volume of a cube whose side is 5 cm
Exercise 7.3 | Q 7. (ii) | Page 263
Find the volume of a cube whose side is 3.5 m
Exercise 7.3 | Q 7. (iii) | Page 263
Find the volume of a cube whose side is 21 cm
Exercise 7.3 | Q 8 | Page 263
A cubical milk tank can hold 125000 litres of milk. Find the length of its side in metres
Exercise 7.3 | Q 9 | Page 263
A metallic cube with side 15 cm is melted and formed into a cuboid. If the length and height of the cuboid is 25 cm and 9 cm respectively then find the breadth of the cuboid
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Exercise 7.4 [Pages 263 - 264]
### Samacheer Kalvi solutions for Mathematics Class 9th Tamil Nadu State Board Chapter 7 Mensuration Exercise 7.4 [Pages 263 - 264]
#### Multiple choice questions
Exercise 7.4 | Q 1 | Page 263
The semi-perimeter of a triangle having sides 15 cm, 20 cm and 25 cm is
• 60 cm
• 45 cm
• 30 cm
• 15 cm
Exercise 7.4 | Q 2 | Page 263
If the sides of a triangle are 3 cm, 4 cm and 5 cm, then the area is
• 3 cm2
• 6 cm2
• 9 cm2
• 12 cm2
Exercise 7.4 | Q 3 | Page 263
The perimeter of an equilateral triangle is 30 cm. The area is
• 10sqrt(3) "cm"^2
• 12sqrt(3) "cm"^2
• 15sqrt(3) "cm"^2
• 25sqrt(3) "cm"^2
Exercise 7.4 | Q 4 | Page 263
The lateral surface area of a cube of side 12 cm is
• 144 cm2
• 196 cm2
• 576 cm2
• 664 cm2
Exercise 7.4 | Q 5 | Page 264
If the lateral surface area of a cube is 600 cm2, then the total surface area is
• 150 cm
• 400 cm
• 900 cm
• 1350 cm
Exercise 7.4 | Q 6 | Page 264
The total surface area of a cuboid with dimension 10 cm × 6 cm × 5 cm is
• 280 cm2
• 300 cm2
• 360 cm2
• 600 cm2
Exercise 7.4 | Q 7 | Page 264
If the ratio of the sides of two cubes are 2 : 3, then ratio of their surface areas will be
• 4 : 6
• 4 : 9
• 6 : 9
• 16 : 36
Exercise 7.4 | Q 8 | Page 264
The volume of a cuboid is 660 cm3 and the area of the base is 33 cm2. Its height is
• 10 cm
• 12 cm
• 20 cm
• 22 cm
Exercise 7.4 | Q 9 | Page 264
The capacity of a water tank of dimensions 10 m × 5 m × 1.5 m is
• 75 litres
• 750 litres
• 7500 litres
• 75000 litres
Exercise 7.4 | Q 10 | Page 264
The number of bricks each measuring 50 cm × 30 cm × 20 cm that will be required to build a wall whose dimensions are 5 m × 3 m × 2 m is
• 1000
• 2000
• 3000
• 5000
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## Chapter 7: Mensuration
Exercise 7.1Exercise 7.2Exercise 7.3Exercise 7.4
## Samacheer Kalvi solutions for Mathematics Class 9th Tamil Nadu State Board chapter 7 - Mensuration
Samacheer Kalvi solutions for Mathematics Class 9th Tamil Nadu State Board chapter 7 (Mensuration) 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 Tamil Nadu Board of Secondary Education Mathematics Class 9th Tamil Nadu State Board solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. Samacheer Kalvi textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Mathematics Class 9th Tamil Nadu State Board chapter 7 Mensuration are Mensuration, Area of a Triangle, Area of a Triangle by Heron's Formula, Application of Heron’s Formula in Finding Areas of Quadrilaterals, Surface Area of a Cuboid, Surface Area of a Cube, Volume of a Cuboid, Volume of Cube.
Using Samacheer Kalvi Class 9th solutions Mensuration exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in Samacheer Kalvi Solutions are important questions that can be asked in the final exam. Maximum students of Tamil Nadu Board of Secondary Education Class 9th prefer Samacheer Kalvi Textbook Solutions to score more in exam.
Get the free view of chapter 7 Mensuration Class 9th extra questions for Mathematics Class 9th Tamil Nadu State Board and can use Shaalaa.com to keep it handy for your exam preparation
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# ML Aggarwal CBSE Solutions Class 6 Math Twelveth Chapter Algebra Exercise 12.4
### ML Aggarwal CBSE Solutions Class 6 Math 12th Chapter Algebra Exercise 12.4
Ex 12.4
(1) State which of the following are equations with a variable. In case of an equation with a variable, identify the variable.
(i) 17 + x = 5
Solution: It is an equation with variable x.
(ii) 2b – 3 = 7
Solution: It is an equation with variable b.
(iii) (y -7) > 5
Solution: it is not an equation.
(iv) 9/3 = 3
Solution: It is a numerical equation. It has no variable.
(v) 7 x 3 – 19 = 2
Solution: It is a numerical equation. It has no variable.
(vi) 5 x 4 – 8 = 3t
Solution: It is an equation with variable t.
(vii) 2p < 15
Solution: It is not an equation.
(viii) 7 = 11 x 5 – 12 x 4
Solution: It is a numerical equation. It has no variable.
(ix) 3/2 q = 5
Solution: It is an equation with variable q.
(2) (i) x + 12 = 20 (12, 8, 20, 0)
Solution: The equation is x + 12 = 20
For x = 12, LHS = 12 + 12 = 24 ≠ RHS Therefore, x = 12 is not a solution
For x = 8, LHS 8 + 12 = 20 = RHS Therefore, x = 8 is a solution
For x = 20, LHS 20 + 12 = 32 ≠ RHS Therefore x = 20 is not a solution
For x = 0, LHS 0 + 12 = 12 ≠ RHS Therefore x = 0 is not a solution.
Hence, x = 8 is the solution of the given equation.
(ii) y/2 = 7 (7, 2, 10, 14)
Solution: The given equation y/2 = 7
For y = 7, LHS = 7/2 = 3.5 ≠ RHS Therefore y = 7 is not a solution.
For y = 2, LHS = 2/2 = 1 ≠ RHS Therefore, y = 2 is not a solution
For y = 10, LHS = 10/2 = 5 ≠ RHS Therefore y = 10 is not a solution
For y = 14, LHS = 14 / 2 = 7 = RHS Therefore y = 14 is a solution
Hence, y = 14 is the solution of the given equation.
(iii) p – 4 = 0 (4. -4, 8, 0)
Solution: The given equation p – 4 = 0
For p = 4, LHS = 4 – 4 = 0 = RHS Therefore p = 4 is the solution
For p = -4, LHS = -4 – 4 = -8 ≠ RHS Therefore, p = -4 is not a solution
For p = 8, LHS = 8 – 4 = 4 ≠ RHS Therefore p = 8 is not a solution
For p = 0, LHS = 0 – 4 = -4 ≠ RHS Therefore p = o is not a solution
Hence, p = 4 is the solution of the given equation
(iv) n + 4 = 2 (-2, 0, 2, 4)
Solution: The given equation n + 4 = 2
For n = -2, LHS = -2 + 4 = 2 = RHS Therefore n = -2 is the solution
For n = 0, LHS = 0 + 4 = 4 ≠ RHS Therefore n = 0 is not a solution
For n = 2, LHS = 2 + 4 = 6 ≠ RHS Therefore n = 2 is not a solution
For n = 4, LHS = 4 + 4 = 8 ≠ RHS Therefore n = 4 is not a solution
Hence, n = -2 is the solution of the given equation.
(v) 2y – 3 = 7 (0,3,5,7)
Solution: For y = 0, LHS = 2×0 – 3 = 0 – 3 = -3 ≠ RHS Therefore y = 0 is not a solution
For y = 3, LHS = 2×3 – 3 = 6 – 3 = 3 ≠ RHS Therefore, y = 3 is not a solution.
For y = 5, LHS = 2 x 5 – 3 = 10 – 3 = 7 = RHS Therefore, y = 5 is a solution.
For y = 7, LHS = 7 x 5 – 3 = 35 – 3 = 32 ≠ RHS Therefore, y = 7 is not a solution.
(3) Complete the entries in the fourth of the following table:
Solution:
S. No. Equation Value of Variable Equation Satisfied Yes / No (i) 5n = 40 10 No (ii) 5n = 40 8 Yes (iii) 5n = 40 5 No (iv) X + 10 = 30 10 No (v) X + 10 20 Yes (vi) X + 20 = 30 30 No (vii) 2p – 3 = 7 5 Yes (viii) 2p – 3 = 7 10 No (ix) 2p – 3 = 7 15 No (x) Y + 3 = 1 1 No (xi) Y + 3 = 1 0 No (xii) Y + 3 = 1 -1 No (xiii) Y + 3 = 1 -2 yes
(5) Complete the following table and by inspection of the table, find the solution of the equation 2x + 3 = 17
(i) Solution: The given equation is 2x + 3 = 17. We complete the table as under:
x 1 2 3 4 5 6 7 8 9 …. 2x + 3 5 7 9 11 13 15 17 19 21 ….
(ii) Solution:
x 3 4 5 6 7 8 9 10 11 …. 25 – 3x 16 13 10 7 4 1 -2 -5 -8 ….
Updated: December 4, 2019 — 2:32 pm |
# Evaluate an exponential function
This lesson will show how to evaluate an exponential function and how to solve a real-world problem by evaluating an exponential function.
## An example showing how to evaluate an exponential function
Example #1
Evaluate the following function for x = 0, 1, 2, -1, -2.
Notice that this time, we used a table to organize our calculations. It is always good practice to use a table when evaluating exponential functions.
f(x) = 4 × 5x
x 4 × 5x f(x) 0 4 × 50 = 4 × 1 = 4 4 1 4 × 51 = 4 × 5 = 20 20 2 4 × 52 = 4 × 5 × 5 = 4 × 25 = 100 100 -1 4 × 5-1 = 4 × 1/5 = 4/1 × 1/5 = (4 × 1) / (4 × 1) = 4/5 0.8 -2 4 × 5-2 = 4 × 1/25 = 4/1 × 1/25 = (4 × 1)/ (1 × 25) = 4/25 0.16
Notice that you can also write the answers as shown below:
f(0) = 4
f(1) = 20
f(2) = 100
f(-1) = 0.8
f(-2) = 0.16
## Evaluate an exponential function to solve a real-world problem.
Example #2
Suppose 30 toads are taken to an island. The toad population quintuple every 3 months. Model this situation with an exponential function and then evaluate the function to find how many toads would be there after 4 years.
Solution
There are two Important
keys words to understand in this problem and these are 'quintuple' and 'every 3 months'.
Quintuple:
The word quintuple means that the population was increased 5 times.
Every 3 months:
Every 3 months is the same as every one-fourth of a year or 1 quarter.
Let x be the number of quarters, then f(x) = 30 × 5x
For this problem, there are 4 quarters in a year since the growth is happening every 3 months. For two years then, we get 8 quarters.
f(8) = 30 × 58
f(8) = 30 × 390625
f(8) = 11718750
After two years, there will 11,718,750 toads in that island.
Notice that when x = 0, f(0) = 30×50
f(0) = 30×1
f(0) = 30.
x = 0 refers to the starting point or the day the toads were taken to the island.
## A little summary on how to evaluate an exponential function
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Recommended |
# Difference between revisions of "stat341 / CM 361"
Computational Statistics and Data Analysis is a course offered at the University of Waterloo
Spring 2009
Instructor: Ali Ghodsi
## Sampling (Generating Random numbers)
### Lecture of May 12th, 2009
In order to study statistics computationally, we need a good way to generate random numbers from various distributions using computational methods, or at least numbers whose distribution appears to be random (pseudo-random). Outside a computational setting, this is fairly easy (at least for the uniform distribution). Rolling a die, for example, produces numbers with a uniform distribution very well.
We begin by considering the simplest case: the uniform distribution.
#### Multiplicative Congruential Method
One way to generate pseudorandom numbers from the uniform distribution is using the Multiplicative Congruential Method. This involves three integer parameters a, b, and n, and a seed variable x0. This method deterministically (based on the seed) generates a sequence of numbers with a seemingly random distribution (with some caveats). It proceeds as follows:
$x_{i+1} = ax_{i} + b \mod{n}$
For example, with a = 13, b = 0, m = 31, x0 = 1, we have:
$x_{i+1} = 13x_{i} \mod{31}$
So,
\begin{align} x_{0} &{}= 1 \end{align}
\begin{align} x_{1} &{}= 13*1 + 0 \mod{31} \\ &{}= 13 \end{align}
\begin{align} x_{2} &{}= 13*13 + 0 \mod{31} \\ &{}= 14 \end{align}
etc.
This method generates numbers between 0 and m - 1, and by scaling this output by dividing the terms of the resulting sequence by m - 1, we create a sequence of numbers between 0 and 1. For correctly chosen values of a, b, and m, this method will generate a sequence of integers including all integers between 0 and m - 1.
Of course, not all values of a, b, and m will behave in this way, and will not be suitable for use in generating pseudorandom numbers. In practice, it has been found by a paper published in 1988 by Park and Miller, that a = 75, b = 0, and m = 231 - 1 = 2147483647 (the maximum size of a signed integer in a 32-bit system) are good values for the Multiplicative Congruential Method.
There are however some drawbacks to this method:
• No integer will ever be generated twice in a row (otherwise the method would generate that integer forever from that point onwards)
• Can only be used to generate pseudorandom numbers from the uniform distribution
#### General Methods
Since the Multiplicative Congruential Method can only be used for the uniform distribution, other methods must be developed in order to generate pseudorandom numbers from other distributions.
##### Inverse Transform Method
This method uses the fact that when the inverse of a distribution cumulative density function for a given distribution is applied to the uniform distribution, the resulting distribution is the distribution of the cdf. This is shown by this theorem:
Theorem:
If $U \sim~ Unif[0, 1]$ is a random variable and $X = F^{-1}(U)$, where F is continuous, and is the cumulative density function for some distribution, then the distribution of the random variable X is given by F(X).
Proof:
Recall that, if f is the pdf corresponding to F, then,
$F(x) = P(X \leq x) = \int_{-\infty}^x f(x)$
So F is monotonically increasing, since the probability that X is less than a greater number must be greater than the probability that X is less than a lesser number.
Note also that in the uniform distribution on [0, 1], we have for all a within [0, 1], $P(U \leq a) = a$.
So,
\begin{align} P(F^{-1}(U) \leq x) &{}= P(F(F{-1}(U)) \leq F(x)) \\ &{}= P(U \leq F(x)) \\ &{}= F(x) \end{align}
Completing the proof.
Procedure (Continuous Case)
This method then gives us the following procedure for finding pseudorandom numbers from a continuous distribution:
• Step 1: Draw $U~ \sim~ Unif [0,1]$.
• Step 2: Compute $X = F^{-1}(U)$.
Example:
Suppose we want to draw a sample from $f(x) = \lambda e^{-\lambda x}$ where $x \gt 0$ (the exponential distribution).
We need to first find $F(x)$ and then its inverse, $F^{-1}$.
$F(x) = \int^x_0 \theta e^{-\theta u} du = 1 - e^{-\theta x}$
$F^{-1}(x) = \frac{-log(1-y)}{\theta}$
Now we can generate our random sample $i=1\dots n$ from $f(x)$ by:
$1)\ u_i \sim UNIF(0,1)$
$2)\ x_i = \frac{-log(1-u_i)}{\theta}$
The $x_i$ are now a random sample from $f(x)$.
The major problem with this approach is that we have to find $F^{-1}$ and for many distributions it is too difficult (or impossible) to find the inverse of $F(x)$. Further, for some distributions it is not even possible to find $F(x)$
Procedure (Discrete Case)
The above method can be easily adapted to work on discrete distributions as well.
In general in the discrete case, we have $x_0, \dots , x_n$ where:
\begin{align}P(X = x_i) &{}= p_i \end{align}
$x_0 \leq x_1 \leq x_2 \dots \leq x_n$
$\sum p_i = 1$
Thus we can define the following method to find pseudorandom numbers in the discrete case:
• Step 1: Draw $U~ \sim~ Unif [0,1]$.
• Step 2:
• If $U \lt p_0$, return $X = x_0$
• If $U \lt p_0 + p_1$, return $X = x_1$
• ...
• In general, if $U \lt p_0 + \dots + p_k$, return $X = x_k$
Example:
Suppose we have the following discrete distribution:
\begin{align} P(X = 0) &{}= 0.3 \\ P(X = 1) &{}= 0.2 \\ P(X = 2) &{}= 0.5 \end{align}
The cumulative density function for this distribution is then:
$F(x) = \begin{cases} 0, & \text{if } x \lt 0 \\ 0.3, & \text{if } 0 \leq x \lt 1 \\ 0.5, & \text{if } 1 \leq x \lt 2 \\ 1, & \text{if } 2 \leq x \end{cases}$ |
# HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5
Haryana State Board HBSE 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Ex 2.5 Textbook Exercise Questions and Answers.
## Haryana Board 8th Class Maths Solutions Chapter 2 Linear Equations in One Variable Exercise 2.5
Solve the following linear equations
Question 1.
$$\frac{x}{2}-\frac{1}{5}$$ = $$\frac{x}{3}+\frac{1}{4}$$
Solution:
Question 2.
$$\frac{n}{2}$$ – $$\frac{3n}{4}$$ + $$\frac{5n}{6}$$ = 21
Solution:
Question 3.
x + 7 – $$\frac{8x}{3}$$ = $$\frac{17}{6}$$ – $$\frac{5x}{2}$$
Solution:
Question 4.
$$\frac{x-5}{3}$$ = $$\frac{x-3}{5}$$
Solution:
$$\frac{x-5}{3}$$ = $$\frac{x-3}{5}$$
or, 5(x – 5) = 3 (x – 3)
or, 5x – 25 = 3x – 9
or, 5x – 3x = -9 + 25
or, 2x = 16
or, x = $$\frac{16}{2}$$
∴ x = 8
Question 5.
$$\frac{3t-2}{4}$$ – $$\frac{2t-3}{3}$$ = $$\frac{2}{3}$$ – t
Solution:
or, 3 (13t – 18) = 24
or, 39t – 54 = 24
or, 39t = 24 + 54 = 78
or, 39t = 78
or, t = 2
Question 6.
m – $$\frac{m-1}{2}$$ = 1 – $$\frac{m-2}{3}$$
Solution:
or, 5m – 1 = 6
or, 5m = 6 + 1
or, 5m = 7
∴ m = $$\frac{7}{5}$$
Simplify and solve the following linear equations:
Question 7.
3 (t – 3) = 5 (2t + 1)
Solution:
3 (t – 3) = 5 (2t + 1)
or, 3t – 9 = 10t + 5
or, 3t – 10t = 5 + 9
or, -7t = 14
or, t = $$-\frac{14}{7}$$ =-2
t = -2
Question 8.
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Solution:
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
or, 15y – 60 – 2y + 18 + 5y + 30 = 0
or, 20y – 2y – 60 + 48 = 0
or, 18y – 12 = 0
or, 18y = 12
or, y = $$\frac{12}{18}$$ = $$\frac{2}{3}$$
∴ y = $$\frac{2}{3}$$
Question 9.
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Solution:
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
or, 15z – 21 – 18z + 22 = 32z – 52 – 17
or, 15z – 18z – 32z = -52 – 17 + 21 – 22
or, 15z – 50z = 21 – 91
or, -35z = -70
or, 35z = 70
or, z = 2
∴ z = 2
Question 10.
0.25 (4f – 3) = 0.05 (10f – 9)
Solution:
0.25 (4f – 3) = 0.05 (10f – 9)
or, f – 0.75 = 0.5f – 0.45
or, f – 0.5f = -0.45 + 0.75
or, 0.5f = 0.3
f = $$\frac{0.3}{0.5}$$ = $$\frac{3}{5}$$ = 0.6
∴ f = 0.6 |
# How do you find the limit of (sin (4x)) / (tan(5x)) as x approaches 0?
Mar 16, 2016
Use Algebra, trigonometry and the fundamental trigonometric limit.
#### Explanation:
${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$
We can use this to find ${\lim}_{x \rightarrow 0} \sin \frac{7 x}{x}$ because
${\lim}_{x \rightarrow 0} \sin \frac{7 x}{x} = {\lim}_{x \rightarrow 0} \frac{7}{1} \sin \frac{7 x}{7 x}$
$= 7 {\lim}_{x \rightarrow 0} \sin \frac{7 x}{7 x}$
Now, with $7 x = \theta$ we see that ${\lim}_{x \rightarrow 0} \sin \frac{7 x}{7 x} = 1$
So we finish with
$7 {\lim}_{x \rightarrow 0} \sin \frac{7 x}{7 x} = 7 \left(1\right) = 7$#
We'll also need ${\lim}_{\theta \rightarrow 0} \cos \theta = 1$ (Cosine is continuous at $0$.)
$\tan \theta = \sin \frac{\theta}{\cos} \theta$
Here is the solution
$\sin \frac{4 x}{\tan} \left(5 x\right) = \sin \frac{4 x}{\sin \frac{5 x}{\cos} \left(5 x\right)}$
$= \sin \frac{4 x}{1} \cos \left(5 x\right) \frac{1}{\sin} \left(5 x\right)$
$= \sin \frac{4 x}{4} x \cdot \frac{4}{1} \cos \left(5 x\right) 5 \frac{x}{\sin} \left(5 x\right) \cdot \frac{1}{5}$
$= \frac{4}{5} \left[\sin \frac{4 x}{4} x\right] \left[\cos \left(5 x\right)\right] \left[5 \frac{x}{\sin} \left(5 x\right)\right]$
Taking limits as $x \rightarrow 0$ we get,
$= \frac{4}{5} \left[1\right] \left[1\right] \left[1\right] = \frac{4}{5}$ |
Grade 1 Mathematics Module 2, Topic D | EngageNY
## Grade 1 Mathematics Module 2, Topic D
Topic D closes the module with students renaming ten as a unit: a ten (1.NBT.2a). This is the very first time students are introduced to this language of ten as a unit, so this is exciting! The unit of ten is the foundation for our whole number system, wherein all units are comprised of ten of the adjacent unit on the place value chart. In Lesson 26, students take a walk down memory lane to revisit representations of 10 ones worked with in the past. They rename their Rekenrek bracelet, the ten-frame, the fingers on two hands, and two 5-groups as 1 ten. They connect teen numbers to the unit form, e.g., 1 ten and 1 one, 1 ten and 2 ones, and represent the numbers with Hide Zero cards. They now analyze the digit 1 in the tens place as representing both 10 ones and 1 unit of ten, further setting the foundation for later work with place value in Module 4 and beyond. Students use their very own Magic Counting Sticks (their fingers) to help them bundle 1 ten. By bundling 1 ten, students realize that some ones are left over, which clarifies the meaning of the ones unit (1.NBT.2b). In Lesson 27, students solve both abstract and contextualized result unknown problems (1.OA.1). The lesson takes them through a progression from problems with teens decomposed or composed using 1 ten and some ones to problems wherein they find the hidden ten, e.g., 8 + 6 or 12 – 5. In Lesson 28 students solve familiar problems such as, “Maria had 8 snowballs on the ground and 5 in her arms. How many snowballs did Maria have?” As students write their solutions, they break apart an addend to make a ten with another addend, and write two equations leading to the solution (see the bond and equations to the right). This movement forward in their ability to record the two steps allows them to own the structure of the ten they have been using for the entire module, on a new level (MP.7). Topic D closes with Lesson 29, where students solve add to with change unknown and take apart/put together with addend unknown problems. Similarly to Lesson 28, students write both equations leading to solution, as they take from the ten (see bond and equation to the right). |
Binary number representation
In this lab we are going to learn about binary numbers and its representation.
Binary numbers
The binary numeral system is a way to write numbers using only two digits: 0 and 1. These are used in computers as a series of “off” and “on” switches. In binary, each digit’s place value is twice as much as that of the next digit to the right (since each digit holds two values). In decimal - the system that humans normally use - each digit holds ten values, and the place value increases by a power of ten (ones, tens, hundreds place, etc.). The place value of the rightmost digit in either case is 1.
Example: 10110011
• The place value of the last 1 (rightmost position) is 20=1.
• The place value of the 1 before that is 21=2.
• The place value of the 0 before that is 22=4.
• The place value of the 0 before that is 23=8.
• The place value of the 1 before that is 24=16.
• The place value of the 1 before that is 25=32.
• The place value of the 0 before that is 26=64.
• The place value of the 1 before that is 27=128.
Adding together all the place values that have 1s, it would be 1+2+16+32+128 = 179. For convenience, binary digits (bits, for short) are usually grouped together in 8 bits, or a byte.
To convert from a decimal integer numeral to its binary equivalent, the number is divided by two, and the remainder is the least-significant bit. The (integer) result is again divided by two, its remainder is the next least significant bit. This process repeats until the quotient becomes zero.
Example: 118
• 118 divided by 2 is equal to 59 with remainder 0, so the rightmost position is a 0.
• 59 divided by 2 is equal to 29 with remainder 1, so the value before that is a 1.
• 29 divided by 2 is equal to 14 with remainder 1, so the value before that is a 1.
• 17 divided by 2 is equal to 7 with remainder 0, so the value before that is a 0.
• 7 divided by 2 is equal to 3 with remainder 1, so the value before that is a 1.
• 3 divided by 2 is equal to 1 with remainder 1, so the value before that is a 1.
• 1 divided by 2 is equal to 0 with remainder 1, so the value before that is a 1.
The binary representation of the number 118 is 1110110.
Bitwise operations
Bitwise operations are those that operates on binary numbers at the level of their individual bits.
In the explanations below, any indication of a bit’s position is counted from the right (least significant) side, advancing left. For example, the binary value 0001 (decimal 1) has zeroes at every position but the first one.
Operator NOT
The bitwise NOT also called complement is an unary operation (only operates on one binary number). This operation perform a logical negation on each bit: bits that are 0 becomes 1, and those that are 1 becomes 0. For example:
NOT 0111
= 1000
Operator AND
The bitwise AND takes two binary representation of equal lenght and performs the logical AND on each pair of corresponding bits. The result in each position is one if both bits are 1, otherwise the result is 0. For example:
0101
AND 0011
= 0001
Operator OR
The bitwise OR takes two bit patterns of equal length and performs the logical inclusive OR operation on each pair of corresponding bits. The result in each position is 1 if at least one of the bits is 1, otherwise the result is 0. For example:
0101
OR 0011
= 0111
Operator XOR
The bitwise XOR takes two bit patterns of equal length and performs the logical exclusive OR operation on each pair of corresponding bits. The result in each position if 1 if only one of the bits is one, otherwise the result is 0. For example:
0101
XOR 0011
= 0110
The XOR operator can be seen as a comparison of two bits patterns, given the result in each position 1 if both bits differs and 0 if they are the same.
Bitwise operations in python
In python the bitwise operators has the following symbols:
• NOT: ~
• AND: &
• OR: |
• XOR: ^
Bit shift
The bit shifts are sometimes considered bitwise operations, because they treat a value as a series of bits rather than as a numerical quantity. In these operations the digits are moved, or shifted, to the left or right. Examples:
SHIFT-LEFT 0101
= 1010
SHIFT-RIGHT 0101
= 0010
Bit shift in python
In python the left and right shift operators are << and >>, respectively. The number of places to shift is given as the second argument to the shift operators. For example:
y = x >> 2
Now y contains the value of x shifted two positions to the right.
Testing for bits in python
To test for different bits in python you can use the combination of the shift operator and the and operator. If you want to test if the bit in the i-th position (from right to left counting form 0) you can do:
(x >> i) & 1
The result of the above operation will be 1 if the bit in the i-th position has value 1, zero otherwise.
Representing binary numbers with LED on the Raspberry Pi
We are going to do a fun exercise. Lets create a circuit with 3 LEDs representing each one a bit in a binary number of size 3, and increment that number every second. The circuit would be the following:
Notice that we are using GPIO pins 17, 27 and 22 as output pins for the LEDs. The following program will show the different binary numbers from 0 to 7 displayed using the LEDs:
import RPi.GPIO as GPIO
import time
GPIO.setmode(GPIO.BOARD)
GPIO.setup(11, GPIO.OUT)
GPIO.setup(13, GPIO.OUT)
GPIO.setup(15, GPIO.OUT)
c = 0
try:
while True:
c = (c + 1) % 8
GPIO.output(11, (c >> 0) & 1)
GPIO.output(13, (c >> 1) & 1)
GPIO.output(15, (c >> 2) & 1)
time.sleep(1)
except KeyboardInterrupt:
GPIO.cleanup()
Exercise
Add a button to the circuit and extend the above program to increase the binary number when the button is pressed. |
Question Video: Determining Total Distance Traveled | Nagwa Question Video: Determining Total Distance Traveled | Nagwa
# Question Video: Determining Total Distance Traveled Physics • First Year of Secondary School
## Join Nagwa Classes
What is the total distance covered by someone who walks along the lines shown in the diagram?
02:35
### Video Transcript
What is the total distance covered by someone who walks along the lines shown in the diagram?
Alright, so in this question, we see that we’re presented with a diagram that contains three lines, and we’re asked to find the total distance covered by someone who walks along these lines. Let’s begin by recalling our definition of distance. Distance is defined as the length of the path between two positions. The question tells us that our person walks along the lines, so we see that the path taken by the person in this question must consist of walking along each of these lines in turn. In other words, if they start here, the first part of their path consists of walking along this line, ending up here. We’ll label this line one as it’s the first segment of the path.
The next part of the path that our person follows is this line here. And at the end of it, they end up at this position here. We’ll label this line two. Finally, our person’s path follows this remaining line here, and their final position is here at the end of this line. We’ll label this final line line three. Remember that the question asked us for the total distance covered and that our definition of distance tells us that it is the length of the path that is taken between two positions.
Now we’ve drawn the path that the person from this question takes, and we can see that it consists of three straight line segments which we’ve labeled one, two, and three. If we label the total distance as 𝑑, then we can see that 𝑑, which we know is the length of the path taken, must be equal to the length of line one plus the length of line two plus the length of line three. We are given each of these three lengths in the diagram. So we can substitute these values into our expression for 𝑑. The length of line one is five meters, the length of line two is eight meters, and finally the length of line three is seven meters. And so we have that the total distance covered 𝑑 is given by the sum five meters plus eight meters plus seven meters. If we evaluate this sum, we find that 𝑑 equals 20 meters.
And so we have our answer to the question that the total distance covered by someone who walks along the lines shown in the diagram is equal to 20 meters.
## Join Nagwa Classes
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## Day 81 - Lesson 8.2
##### Learning Targets
• State and check the Random, 10%, and Large Counts conditions for constructing a confidence interval for a population proportion.
• Determine the critical value for calculating a C% confidence interval for a population proportion using a table or technology.
• Construct and interpret a confidence interval for a population proportion.
##### Activity:
This is always one of the best Activities of the year! Students are trying to figure out the proportion of Hershey’s Kisses that will land flat when dropped onto a table. This Activity is also exciting because students are able to put together their learning from the past few chapters into one goal: constructing a confidence interval.
This Activity is a bit longer than others, so students need to work efficiently. We suggest taking the students through question #8 as a whole group in order to avoid a common student error (see below).
##### The Story of Inference - Making Connections
In Chapter 2, students learned about the 68-95-99.7 rule for Normal Distributions. They learned that 95% of the data in a Normal Distribution is within 2 standard deviations of the mean (yes, we lied to them…we know better now to use 1.96). This idea becomes the critical value for a confidence interval.
In Chapter 4, students learned that the purpose of a random sample is so that we can generalize to the larger population. In this chapter, this idea becomes the Random condition for inference.
In Chapter 6, students learned to check the 10% condition in the binomial setting to be sure that there is independence between trials. In Chapter 7, students used the 10% condition to be sure they could use certain formulas for the standard deviation of sampling distributions. In this chapter, this idea becomes the 10% condition for inference.
In Chapter 6, students learned to check the Large Counts condition in the binomial setting to be sure that the binomial distribution could be modeled with a Normal distribution. In Chapter 7, students extended this reasoning to apply to the sampling distribution of a sample proportion. In this chapter, this idea becomes the Large Counts condition for inference.
##### Common Student Error
When calculating critical values from Table A, students will mistakenly look for the wrong area. For example, when finding the critical value for an 80% confidence interval, naturally students want to look for an area of 0.80. But we know that Table A gives the area to the left of a specific z-score, which includes the 10% area in the left tail. So students should be looking for an area of 0.90. This will only matter for a few days, because once students get familiar with Table B (take a look at the last row), they will no longer want to use Table A for critical values. |
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# Multiplicative Identity Property (Number Multiplied by 1)
Last updated date: 16th Apr 2024
Total views: 154.5k
Views today: 4.54k
## Overview of Multiplication
Have you ever heard about the answer obtained when any number is multiplied by 1? When any number is multiplied by 1 the product is always the number itself. In the below-discussed article, children would gain knowledge about the steps involved in solving multiplication problems, which is considered one of the basic and most widely used mathematical and arithmetical operations. Generally, it is represented using the symbol '×' or '*' in between the two numbers. Now, let us start with our topic here:
## What is Multiplication?
Multiplication is a process of calculating the product of two quantities or numbers by multiplying them together. '×' is used to depict the product of two quantities. The number multiplied by the multiplier is called the multiplicand and the number by which the multiplicand is multiplied is called the multiplier. It can also be stated as a way of adding a number repeatedly to acquire their product.
Showing Terms Used in the Multiplication of Two Numbers
## Steps Involved in Solving Multiplication Queries
The steps required to follow in the multiplication of two numbers are given below:
• Write the given multiplicand and multiplier in the column form by taking into consideration their place values
• Then put the multiplication sign (×) preceding the multiplier
• Start the multiplication from the right-hand side of the multiplier and move to the left side.
• Count the number of digits in multiplier quantity and place the zeroes 1 less than the number of digits in the multiplier from the right side
• Finally, add up all the products respective to each digit of the multiplier.
Multiplication of 8 and 5
## Multiplicative Identity Property
Multiplicative identity property states that "Any number multiplied by 1 is the number itself". It means that when any number is multiplied by 1 the answer is always the number. It serves as a user identity in solving various problems. 1 is the identity of itself i.e. 1 × 1 = 1. It is also called the Identity Property of Multiplication because of the reason, here the identity of a number remains unaltered.
Showing Identity Property of Multiplication
## Solved Examples
Q 1. 12 × 1
Ans: Steps to be followed to calculate the product using Multiplicative Identity
• Align the given numbers in columns or rows to calculate the product
• Multiplying 1 with the given number and noting the result
Thus, we can conclude that when any number is multiplied by 1 the answer is the number itself.
Showing Any Number Multiplied by 1
Q 2. 65 × 31
Ans: Steps to calculate the result of this statement are given below:
• Write the given multiplier and multiplicand in columns format
• Now, using multiplicative identity, any number multiplied by 1 is the number, for multiplying 65 and 1
• Calculate the number of digits of the multiplier and put zeroes one less than the calculated number of digits
• Further, multiplying the tens digit of the multiplier with the multiplicand
• Adding the two obtained partial products i.e. 65 + 1950 = 2015
Hence, the required result of the multiplication of 65 × 31 is 2015.
Showing the Multiplication of Two Digit Number Whose One Digit is 1
## Practice Problems
Some practice problems based on the concept, any number multiplied by 1 is the number itself are given below. These should be solved by the children on their own for a better understanding of the concepts.
Q 1. 59 × 1
Ans: 59
Q 2. 84 × 1
Ans: 84
Q 3. 1 × 18
Ans: 18
Q 4. 34 × 1
Ans: 34
Q 5. 1 × 32
Ans: 32
## Summary
To wrap up here with the topic of number multiplication. The main motive of this article is to impart knowledge of multiplication by covering every topic including, what is multiplication, the steps involved in solving mathematics problems, whether any number multiplied by 1 is the number, etc. It is formed using the solved examples and images which makes learning interesting and exciting. Hoping the writing helped you in capturing the concept of multiplication and you enjoyed reading it. Feel comfortable to ask about your problems.
## FAQs on Multiplicative Identity Property (Number Multiplied by 1)
1. What is the general use of multiplication?
The most general use of multiplication is when we need to calculate a large number of things for a given small amount. Multiplication is nothing but a less time taking process of repetitive addition. So it takes less time to calculate the sum. The multiplication rule is also used to calculate the probability of occurrence of two events.
2. Can the multiplier and multiplicand be the same?
Yes, a multiplier and multiplicand can be the same in a multiplication problem. There is no such rule in maths that these both should be different. When a number is multiplied by its own, the result is always the square of the given number. For example 2 × 2 = 2² = 4, 13 × 13 = 13² = 169, etc.
3. What happens when one is multiplied by zero?
When 1 is multiplied by 0, the answer is always zero. However, we know that when any number is multiplied by 1 the product is the number itself, similarly we know that any number multiplied by zero results in zero. |
find the equation of the plane passing through the point (3,3,1) and perpendicular to the line joining (3,4,-1) and (2,-1,5).
Video Solution
Text Solution
Generated By DoubtnutGPT
To find the equation of the plane passing through the point (3, 3, 1) and perpendicular to the line joining the points (3, 4, -1) and (2, -1, 5), we can follow these steps:Step 1: Identify the Point and Normal VectorThe plane passes through the point A(3,3,1). We need to find a normal vector to the plane, which can be derived from the line joining the points B(3,4,−1) and C(2,−1,5).Step 2: Find the Direction Vector of the LineTo find the direction vector →AB from point B to point C:→AB=→C−→B=(2−3,−1−4,5−(−1))=(−1,−5,6)Thus, the direction vector →n (which is also the normal vector to the plane) is:→n=−1^i−5^j+6^kStep 3: Write the Equation of the PlaneThe general equation of a plane can be expressed in the form:→r⋅→n=→a⋅→nwhere →r is the position vector of any point on the plane, →n is the normal vector, and →a is the position vector of the given point A(3,3,1).Step 4: Calculate →a⋅→nThe position vector →a for point A is:→a=3^i+3^j+1^kNow, we calculate the dot product →a⋅→n:→a⋅→n=(3)(−1)+(3)(−5)+(1)(6)=−3−15+6=−12Step 5: Substitute into the Plane EquationNow substituting into the plane equation:→r⋅(−1^i−5^j+6^k)=−12This can be simplified to:→r⋅^i+5→r⋅^j−6→r⋅^k=12Step 6: Final Equation of the PlaneThe equation of the plane can be expressed as:−x−5y+6z=12SummaryThe equation of the plane passing through the point (3, 3, 1) and perpendicular to the line joining (3, 4, -1) and (2, -1, 5) is:−x−5y+6z=12
|
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Q:
# Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Use a 0.05 significance level to test for a difference between the measurements from the two arms. What do you conclude? Assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal.Right arm102101947979Left arm175169182146144
Accepted Solution
A:
Answer:The p value is lower than the significance level given $$\alpha=0.05$$, so then we can conclude that we can reject the null hypothesis that no difference between the two measures. So there is a significant difference between the measurements from the two arms.Step-by-step explanation:We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test. A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it. Let put some notation x=test value with right arm , y = test value with left arm x: 102, 101, 94, 79, 79 y: 175, 169, 182, 146, 144 The system of hypothesis for this case are: Null hypothesis: $$\mu_y- \mu_x = 0$$ Alternative hypothesis: $$\mu_y -\mu_x \neq 0$$ The first step is calculate the difference $$d_i=y_i-x_i$$ and we obtain this: d: 73, 68, 88, 67, 65 The second step is calculate the mean difference $$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{361}{5}=72.2$$ The third step would be calculate the standard deviation for the differences, and we got: $$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =9.311$$ The 4 step is calculate the statistic given by : $$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{72.2 -0}{\frac{9.311}{\sqrt{5}}}=17.339$$ The next step is calculate the degrees of freedom given by: $$df=n-1=5-1=4$$ Now we can calculate the p value, since we have a two tailed test the p value is given by: $$p_v =2*P(t_{(4)}>17.339) =6.49x10^{-5}$$ The p value is lower than the significance level given $$\alpha=0.05$$, so then we can conclude that we can reject the null hypothesis that no difference between the two measures. So there is a significant difference between the measurements from the two arms. |
# What Is 8/22 as a Decimal + Solution With Free Steps
The fraction 8/22 as a decimal is equal to 0.363.
Proper fractions are formed when we divide two numbers and we get a fractional result with a numerator less than the denominator. This fraction is always less than a whole (value 1)
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 8/22.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 8
Divisor = 22
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 8 $\div$ 22
This is when we go through the Long Division solution to our problem. Given is the Long Division process in Figure 1:
Figure 1
## 8/22 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 8 and 22, we can see how 8 is Smaller than 22, and to solve this division, we require that 8 be Bigger than 22.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 8, which after getting multiplied by 10 becomes 80.
We take this 80 and divide it by 22; this can be done as follows:
80 $\div$ 22 $\approx$ 3
Where:
22 x 3 = 66
This will lead to the generation of a Remainder equal to 80 – 66 = 14. Now this means we have to repeat the process by Converting the 14 into 140 and solving for that:
140 $\div$ 22 $\approx$ 6
Where:
22 x 6 = 132
This, therefore, produces another Remainder which is equal to 140 – 132 = 8. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 80.
80 $\div$ 22 $\approx$ 3
Where:
22 x 3 = 66
Finally, we have a Quotient generated after combining the three pieces of it as 0.363, with a Remainder equal to 14.
Images/mathematical drawings are created with GeoGebra. |
Question Video: Using Place Value to Write Numbers in Expanded Form | Nagwa Question Video: Using Place Value to Write Numbers in Expanded Form | Nagwa
# Question Video: Using Place Value to Write Numbers in Expanded Form Mathematics
Which of the following is equivalent to 94315? [A] 90000 + 4000 + 300 + 10 + 5 [B] 90000 + 4000 + 100 + 30 + 5 [C] 90000 + 3000 + 400 + 10 + 5 [D] 50000 + 1000 + 300 + 40 + 9.
02:06
### Video Transcript
To help us work out which is equivalent to our number, we can look at each of the digits and how much it’s worth. So our number has five ones. So if you look at that column, we can see that the first three choices also have five ones.
Now the bottom option has nine ones, which is incorrect. So we can eliminate this option. If we look at the 10s digit, which is one, we know that our number should have 10. So our first option has 10. Our second option has 30. And our third option has 10.
So let’s eliminate the incorrect response. Next we move along to the 100s digit, which is three worth 300. Our first choice does have 300. But the other remaining choice has 400. So we need to get rid of that.
Let’s just check the 1000s digit, which is four. So our number should have 4000. And that’s correct! Finally, we can check the 10000s digit, which is nine. So we should have 90000. And we do, so 90000 add 4000 add 300 add 10 add five is equivalent to 94315. |
## Tuesday, October 17, 2017
### Lesson 4-3: Using an Automatic Drawer (Day 43)
This is what Theoni Pappas writes on page 290 of her Magic of Mathematics:
"Determine how to cut this shape into three pieces with two straight cuts, so that the three pieces can be rearranged into a square."
This is the final page of the section "Some Mathematical Recreations," and our last mathematical recreation is called "The Square Transformation." Once again, we have a completely visual puzzle, and so I have no choice but to describe it here in words.
The figure fits in a rectangle of base 18 and height 27. But the upper-left corner -- a square of length 9 -- has been removed. So we are technically left with a hexagon, alternating between vertical and horizontal sides of length 27, 18, 18, 9, 9, 9.
As usual, I'll post the complete answer tomorrow. But let's look for some hints. We notice that the figure consists of five squares of length 9, and so its area is 405 square units. A square with the same area must have sides of length sqrt(405) or 9sqrt(5) units. So somehow our cuts must lead to sides of this irrational length. This strongly implies that the cuts are neither horizontal nor vertical, but are diagonal (or oblique). A cut across the diagonal of a 2 * 1 rectangle is sqrt(5) units, and so our cuts will ultimately have to be nine times this length. That's all I'll say about this until tomorrow!
Meanwhile, here are the solutions to last week's doublets puzzle:
(1) EYE-LYE-LIE-LID
(2) PITY-PITS-FITS-FINS-FIND-FOND-FOOD-GOOD
(3) TREE-FREE-FLEE-FLED-FEED-WEED-WELD-WOLD-WOOD
(4) OAT-RAT-ROT-ROE-RYE
You can see how this works. EYE to LID was straightforward -- just change the first letter, then the second letter, then the third letter. But OAT to RYE doesn't work as easily -- RAT is a word, but RYT is not. But at least RAT gives us a start -- and remember the hint last week about vowels, so at some point we needed a word with two vowels (ROE). The four-letter words were indirect -- it's interesting that in both cases we had intermediate words beginning with F.
We didn't cover much in the Pappas recreation section -- mainly because half of the section was blocked by the weekend and three straight non-blogging days. But I do wish to write about one of the skipped pages, 238.
In the past on the blog, I've mentioned that October is Hexaflexagon Month. The month of October was selected because it's the birthday of their inventor, Martin Gardner. Notice that Gardner has appeared in many of my recent posts since both Pappas and Hoffman keep mentioning the famous recreational mathematician in their books. But Gardner's actual birthday is the 21st -- a Saturday this year, hence it's yet another non-posting day blocked by the weekend.
Anyway, on page 238, Pappas writes about something called a "hexatetraflexagon." In this case, the prefix "hexa-" implies that the "toy," if you will, has six faces, but each face has only "tetra-," or four sides (a square), rather than the usual six.
Sarah Carter, one of the most famous math teacher bloggers, would write an annual post about hexaflexagons in her classroom. Even though Gardner's birthday is in October, it would usually be November or December by the time she writes the post. But she didn't write about hexaflexagons in 2016 at all. Instead, here's a link to her November 2015 post:
https://mathequalslove.blogspot.com/2015/11/hexaflexagon-day-2015.html
So I'll certainly be on the lookout for her 2017 hexaflexagon post the next two months! Meanwhile, the queen of the hexaflexagon is Vi Hart. Here's a link to one of her videos:
You may be asking, is there a video about hexatetraflexagons, since that is what actually appears on Pappas page 238? The answer is yes -- here's a video by Jill Britton:
That's good -- I won't have to type in the instructions from page 238, since you can just watch the video above.
OK, so let's get back to Paul Hoffman's book. On Friday, we read two chapters in order to cover the extra Chapter e. Today is Chapter 3 -- and naturally, if there's a Chapter e, there's also a chapter numbered for the other famous transcendental number, pi. So once again we have two chapters. Just like Friday, I'll probably just summarize the chapters and skip over large portions of them in posting about them on the blog.
Chapter 3 of Paul Hoffman's The Man Who Loved Only Numbers is "Einstein vs. Dostoyevsky." As usual, the chapter begins with a quote:
"My own greatest debt to Erdos arises from a conversation 30 years ago in the Hotel Parco del Principi in Rome...."
Wow, I usually like to post the complete opening quote, but this one's quite long (two paragraphs). I already told you that I'm skipping today, so let's skip to the end of the quote:
"Ever since then, I've realized that I'm infinitely rich: not just in the material sense that I have everything I need, but infinitely rich in spirit in having mathematics and having known Erdos."
-- Richard Guy
By the way, Richard Guy is a British mathematician. As of today's post, he is still alive -- last month he celebrated his 101st birthday! So if you want the full story of his meeting with Erdos, ask him!
Chapter 3 opens in 1959 when Erdos is allowed to enter the U.S. for a math conference -- in other words, he catches (Uncle) Sam in a good mood. On his return to Hungary, he meets a twelve-year old prodigy, Louis Posa, whose mother wants him to meet the famous mathematician.
And so Erdos asks him, "Prove that if you have n + 1 integers less than or equal to 2n there are always two of them which are relatively prime." Hoffman gives us the following example:
"As an example, choose n to be 5. Then the conjecture is that if you take any six integers from the set 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10, you can't avoid choosing two that are relatively prime."
Hoffman tells us Posa's response -- "The two are neighbors." As it turns out, this is yet another example of the Pigeonhole Principle that I mentioned in my last post. We divide the 2n natural numbers into n pairs -- {1, 2}, {3, 4}, {5, 6}, all the way up to {2n - 1, 2n}. Since there are n pairs and n + 1 numbers to choose, two of them must be in the same pair -- and so the GCD of the two of them must be 1.
Ultimately, Erdos homeschools the young "epsilon" Posa. When the youngster is 14, he gains the Erdos number of 1 as he and the famous mathematician write a paper together. Hoffman tells us that Posa enjoys algebra, but doesn't care for calculus or -- gasp -- geometry. (The gasp is because this is a geometry blog!) The chapter title comes from Posa, who announces that he'd rather be Dostoyevsky, the Russian novelist, than Einstein. This is at the age of 20, when he becomes an elementary teacher.
The young Posa is also distracted by girls, and so he asks Erdos why there are so few girl mathematicians around. Naturally, Erdos responds with a story about "slaves" and "bosses." He says, "Suppose the slave children would be brought up with the idea that if they are very clever, the bosses will not like them. Would there then be many boys who do mathematics?"
Here Erdos implies that the reason for the lack of female mathematicians is that guys say that they are turned off by smart women. This, of course, is the image that Danica McKellar tries to fight. I also mention this in my February 21st post, when some of the eighth grade girls in my class aren't eager to learn math. My support provider dispels the notion that guys are turned off by intelligent women, and then I bring out McKellar's book.
Later on in this chapter, Hoffman writes about another time when Erdos and his mother are allowed to enter the United States. They stay with Vazsonyi, a childhood friend and fellow mathematician, right here in Southern California -- Manhattan Beach, to be precise. Hoffman reminds us again that attended school too -- who tries to make her friendship with Erdos more than platonic anyway. Of course, the mathematician rejects her advances.
The only woman Erdos is ever close to is his mother. This is why he is very depressed when she dies of an ulcer in 1971. Hoffman closes the chapter by telling us how from that day on, Erdos spends 19 hours a day doing math, in order to cope with his mom's passing.
Chapter pi of Paul Hoffman's The Man Who Loved Only Numbers is "Dr. Worst Case" As usual, the chapter begins with a quote:
"Dear Ron, When Paul Erdos's mother died, someone told me that she was the last of his living relatives...."
Really, Hoffman -- the opening quote is an entire letter! Let's skip again to the end:
"And to you, who have invested more than any person in smoothing his way in this rough society, I express my sympathy and my thanks for all you have done. With my sincerest good wishes, Gordon Raisbeck."
The author of this letter is, of course, another mathematician acquaintance. But our focus is the person to whom the letter is addressed -- Ron Graham. Hoffman mentions Graham earlier in this book, but now Hoffman begins Chapter pi by writing about Graham's story.
Graham is born right here in California -- albeit Taft, in Northern California. (He is still alive -- his 82nd birthday is coming up on Halloween.) Hoffman tells us that Graham first becomes interested in math when he memorizes numbers on his paper route, and his fifth grade teacher helps him use this mental ability to calculate square and cube roots.
His seventh grade teacher poses a problem -- find the size of a population of rats if the rats die at a rate proportional to the size of the population -- and the older man uses this problem to motivate Graham to learn calculus independently. He ultimately graduates high school at 15 and is awarded a scholarship to attend the University of Chicago. (Hey, that sounds familiar, since of course I write about the U of Chicago text all the time.)
Three years later, Graham's father convinces him to transfer from "the dangerous, leftist" U of Chicago to "an all-American school" like UC Berkeley. (This is ironic from a 2017 perspective, considering the political images of those schools now. But recall that this was the 1950's.)
Hoffman tells us that Graham's Ph.D. dissertation is on unit fractions. These fractions go all the way to the ancient Egyptians, who wrote all fractions as the sum of distinct unit fractions. Graham's dissertation is all about how to convert fractions most efficiently. For example, he writes:
3/7 = 1/3 + 1/11 + 1/231
3/7 = 1/6 + 1/7 + 1/14 + 1/21
3/7 = 1/4 + 1/7 + 1/28
Which of these is the "best" -- the first one (greedy algorithm, where we have the largest possible fraction 1/3), the second one (smallest maximum denominator 21), or the third (fewest number of terms, 3, with a small maximum denominator)?
I could keep going on with more of Graham's discoveries in this chapter -- but I really do need to get to the textbook that comes from Graham's original college, the U of Chicago.
Lesson 4-3 of the U of Chicago text is called "Using an Automatic Drawer." The newer Third Edition of the U of Chicago text diverges wildly from my old Second Edition -- after Lesson 4-2, the lessons don't really line up again until Chapter 7. Many lessons from Chapters 4, 5, and 6 appear are placed in a different order between the Second and Third Editions. Indeed, Lesson 4-3 of the Third Edition is the same as Lesson 6-4 of the Second (on miniature golf and billiards), and in general the rest of the new Chapter 4 is the old Chapter 6 (on the other transformations -- translations and rotations). On the other hand, the old Lesson 4-3 doesn't appear in the new text at all (due to updates in technology).
This is what I wrote two years ago about today's lesson:
Today's scheduled lesson is another technology-based lesson. Just as I did with Lesson 2-3 three weeks ago, I'm supplementing this with an extra worksheet. It's also about graphing -- except this worksheet involves making reflections on graph paper.
The relationship between the coordinate plane transformations -- including reflections -- in Common Core Geometry is a bit complex. On one hand, many of the properties of the coordinate plane, such as the slopes of parallel and perpendicular lines, depend on dilations and similarity -- and we know that this is emphasized in the standards. This ultimately affects reflections on the plane -- suppose we have the coordinates of a point P and the equation of a line l, and we wish to find the coordinates of P', the reflection image of P. Now by the definition of reflection, line l is the perpendicular bisector of PP', which means that lines PP' and l have opposite reciprocal slopes. So just to perform the reflection, we need slopes and thus ultimately, dilations. And so we wouldn't be able to work on the coordinate plane until after the unit on dilations.
But on the other hand, reflections are easier for students to visualize -- and therefore understand -- if students can draw them on the coordinate plane. This is especially true for the simplest mirrors, namely the x- and y-axes. We don't need to know anything about slope in order to perform reflections over the coordinate axes. And indeed, there's a brief reference to such reflections over the axes on my Lesson 4-1 worksheet.
Yet this isn't nearly enough emphasis on the coordinate plane when we consider the Common Core exams such as PARCC and SBAC. Of the four questions on the PARCC Practice Exam that mention reflections, three of them take place on the coordinate plane. As usual for the blog, the PARCC exam takes priority over all other considerations. My duty on this blog is to make sure that students are prepared to do well on the Common Core exams.
The reflections that appear on the PARCC usually have one of the coordinate axes as a mirror, but we've also seen other horizontal and vertical mirrors, as well as y = x and y = -x as mirrors. It can be argued that one doesn't really need dilations or slope to reflect over horizontal or vertical mirrors, provided we take it for granted that any horizontal line is perpendicular to any vertical line and that we can easily find distance along a horizontal or vertical line.
Is it possible to prove that the reflection image of (xy) over y = x is (yx) without having previously to prove anything about dilations or slope? On one hand, it may seem that we could prove that the line y = x forms a 45-degree angle with either axis simply by showing, for example, that (0, 0), (x, 0), and (xx) are the vertices of an isosceles right triangle. Then the line y = -x also forms a 45-degree angle with the axes, and so the angle between y = x and y = -x must be 45 + 45, or 90, degrees. And so we can show that the lines y = x and y = -x are perpendicular, which is a start.
And all of this, of course, requires us to prove that the graph of y = x is even a line! (Interestingly enough, today I subbed in an art class where the students were learning the concept of line. Art defines the word line differently from geometry -- according to a video featuring several famous artists, a point is a dot, and a line is a dot that moves. A line in art can be any shape, even a circle. I noticed that one artist in the video was using software that looked very similar to the Geogebra program that I mention later in this post.)
But even after proving that the equation of y = x really is linear, we'd still need to find distance along the the oblique lines y = x and y = -x, and this seems to be impossible without having a Distance Formula, which comes from the Pythagorean Theorem, which in turn comes from similarity and dilations. So it indeed appears impossible to show that the reflection image of (xy) is (yx) before the similarity chapter.
And so I've decided to create a worksheet just with reflections over the coordinate axes. I've added on a "reflection square" from last year, which students can fold to see the reflections.
Now I like including technology sections, since these show to the students that geometry isn't just something done in the classroom, but is actually performed out in the real world. But the last time there was a technology chapter -- Lesson 2-3 -- I converted the BASIC programs given in the U of Chicago text into TI-BASIC programs for the graphing calculator. But this section will be more difficult, precisely because the TI-83 or TI-84 is not an automatic drawer. The TI was designed to graph functions and equations -- in other words, do algebra. It was not designed to measure distances, and especially not angles -- in other words, do geometry. So many of the tasks described in the text are not doable on the TI.
As it turns out, there does exist an online graphics program that performs both geometry and algebra -- appropriately enough, it's called Geogebra:
http://www.geogebra.org/cms/en/
I'm not familiar with Geogebra, since I've never downloaded it on used it in a classroom. But based on what I've heard about it, Geogebra can perform all of the tasks described in Lesson 4-3. Much of what I know about Geogebra I read on the blog of John Golden, a mathematics professor from Michigan who calls himself the "Math Hombre." Here's a link directly to the "Geogebra" tag on Golden's blog:
http://mathhombre.blogspot.com/search/label/Geogebra
One thing I learned about Geogebra is not only can it reflect figures over a line -- which is of course the topic for the current chapter -- but it can reflect figures over a circle as well! A circle reflection is not, however, one of the transformations required on Common Core. But I think that it's interesting to compare circle reflections to the Common Core transformations, just in case someone sees that option on Geogebra and wants to know what a circle reflection is.
As you might expect, a circle reflection maps points inside the reflecting circle to points outside the circle, and vice versa -- and just as with line reflections, the image of a point on the reflecting circle is the point itself. Preimage points close to the center of the reflecting circle have points that are far away from the center -- indeed, halving the distance from the preimage to the center ends up doubling the distance from the image to the center. This means that if the preimage is the center itself, its image must be infinitely far away. It's a special imaginary point called "the point at infinity."
A circle reflection is definitely not an isometry -- that is, the Reflection Postulate certainly doesn't hold for circle reflection. Part b of that postulate states that the image of a line is a line. But circle reflections don't preserve collinearity. As it turns out, though, the image of a "line-or-circle" is a "line-or-circle" -- if the preimage line passes through the center, then its image is itself, otherwise, the image ends up being a circle.
My favorite part is what happens when we find the composition of two circle reflections. As we will find out later in the U of Chicago text (and as I mentioned last year), the composition of two reflections in parallel lines is a translation. Well, the composition of two reflections in two concentric circles happens to be -- a dilation! And just as we can easily find the direction and distance of the translation -- its direction is perpendicular to the two reflecting lines, its distance is double that between the two lines -- we can find the center and scale factor of the dilation. The center of the dilation is the common center of the two reflecting circles, while the scale factor is the square of the ratio of the radius of the second reflecting circle to that of the first. (So the dilation is an enlargement if the second circle is larger than the first and a reduction if the second circle is smaller than the first.)
But let's return to the TI. For the sake of those teachers who have access to TI in the classroom, but not Geogebra, let me make Lesson 4-3 into a lesson fit for the TI-83 or TI-84. Here are some commands that will be helpful for drawing on the TI. (Before beginning the following, make sure that there are no functions turned on under Y=.)
First, we'll usually want to turn the axes off for this. So we press 2nd FORMAT (which is the ZOOM key) to choose AxesOff. If we press GRAPH, the screen should be blank. If it isn't, we press 2nd DRAW (which is the PRGM key) to choose ClrDraw. Many of the following commands can be found on this 2nd DRAW menu.
The command Line( draws a line -- segment that is. The arrow keys and ENTER are used to select the starting and ending points. We can also draw an individual point by moving to the right of the DRAW menu to the POINTS menu and choosing Pt-On(.
Now we're in the reflection chapter, so I want to bring this back to reflections. Unfortunately, the TI doesn't automatically reflect for us. So the students will have to reflect instead. One way of doing is to divide the class into partners, and give a calculator to each pair. Then one partner can draw the preimage triangle, and the other add the image onto the picture. Example 2 on the U of Chicago text may be awkward, though, since the reflecting line is oblique (that is, neither horizontal nor vertical), s one might want to try a horizontal or vertical reflecting line first before trying an oblique line.
Example 4 is especially nice. The first partner can draw triangle ABC first, then the second partner can reflect it to draw triangle ABD, and then the first partner takes the calculator back to draw both triangles CEF and DEF.
Interestingly enough, a question in the text that's very suitable for TI drawing is Question 22, in the Exploration (or Bonus) section of the Questions. Part a -- a spiral made up of straight line segments -- is extremely easy to draw on the TI. One can use the Line( command to draw each segment, or even use the Pen command (choice A, the final choice on the Draw menu). After selecting Pen, all the student has to do is press ENTER at the beginning of the spiral, then move with the arrow keys until reaching the end of the spiral, then pressing ENTER again.
Part b is more of a challenge, though. Since this picture contains circles, the Circle( command (choice 9 on the Draw menu) will come in handy. Notice that the endpoints of all the segments in the picture are either points on the circles or centers of the circle. Because the picture has reflectional symmetry, this is also a good picture for drawing from the command line. The necessary commands happen to be Line(X1X2Y1Y2) to draw a line segment from (X1X2) to (Y1Y2), and Circle(X,Y,R) to draw a circle with center (X,Y) and radius R. If a student uses this method, it will be a good idea to make the viewing window symmetrical and square by choosing ZSquare or ZDecimal from the ZOOM menu. (I personally prefer ZDecimal, since it makes the pixels correspond to integers and multiples of .1, which is easier and also makes the graphs more accurate.)
On my worksheet, I give some simple commands for TI drawing, then move on to the Exercises based on the Questions in the book. For simplicity, I decided to keep Questions 1-7, but they are reworded to so that they work in classrooms with Geogebra, TI, or no technology at all (where today's lesson would be simply a second day of Section 4-2).
First, Questions 1-2 ask about automatic drawers. Since technically TI is not an automatic drawer, I changed these to simply ask about graphing technology. In a classroom without technology, the students can be made aware of graphing technology without actually using it.
Questions 3-4 involve measuring with a ruler and drawing by hand. So these can be completed in any of the classrooms I described earlier.
Questions 5-7 ask to use an automatic drawer like Geogebra. Classes with TI or no technology can just do these problems by hand like Questions 3-4.
Then I include three review problems that can be completed in any classroom. Finally, I included Question 22 as a Bonus, since these can be completed on either Geogebra or TI. Since it's a bonus question, classes without technology can just ignore this one.
By the way in case you're wondering, Math Hombre's blog is still active. Although his most recent Geogebra post is dated 2015 (where he shows how to draw various grids on this software), he's written other material since. His last post is all about a college course he's teaching called "The Nature of Modern Mathematics." Notice that he surveyed his students, and when he asked them to name five milestones in the history of mathematics, the two most common answers (with seven responses each) are Geometry and Euclid. [2017 update -- his most recent post now has a similar survey. Euclid is still up there, joined now by Pythagoras.] |
# Video: GCSE Mathematics Foundation Tier Pack 3 • Paper 2 • Question 6
GCSE Mathematics Foundation Tier Pack 3 • Paper 2 • Question 6
04:25
### Video Transcript
An airline company is selling tickets to fly from London to New York. The cost of two adult tickets and two child tickets is 1620 pounds. The cost of three adult tickets is 1650 pounds. Calculate the cost for one adult ticket and the cost for one child ticket.
So first of all, before we actually try to solve the problem, what I’m gonna say is that actually the cost of one adult ticket is gonna be 𝐴 and the cost of one child ticket is going to be 𝐶. So we can use our 𝐴 and our 𝐶 to set up a couple of equations.
The first equation we’re gonna set up is two 𝐴 plus two 𝐶 equals 1620 and that’s because the cost of two adult tickets and the cost of two child tickets is 1620 pounds. And the other equation we can set up is that three 𝐴 equals 1650 and that’s because the cost of three adult tickets is 1650 pounds.
Well, now, we can actually use this equation to actually work out the value of an adult ticket because that’s the only variable we have in this equation. Well, in order to find out how much one adult ticket is going to cost, what we’re gonna do is actually divide each side of our equation by three. And that’s because we have three adult tickets. So we know the cost of it and we want to find one adult ticket. So therefore, we divided by three.
And when we do that, we’re gonna get 𝐴 is equal to 550. And that’s because three 𝐴 divided by three is just 𝐴. And 1650 divided by three is 550. And we could work that out using the bus stop method or short division.
So if we see how many threes go into 1650, first of all how many threes go into one is zero. Then, we carry the one. So then, we’ve got how many threes is going into 16 which is five, carry the one. So we’ve got how many threes are in 15 which is five with no remainder. So then, finally, how many threes is going to zero? There are none with no remainder. So therefore, that’s how we got our 550.
Okay, great, so now, we know the cost of an adult ticket. So what we’re gonna do now is actually substitute our value for 𝐴, so the cost of an adult ticket, into our original equation. So when we do that, we’re gonna have two multiplied by 550. And that’s because we said that 𝐴 is equal to 550 plus two 𝐶 is equal to 1620.
So therefore, we can say that 1100 plus two 𝐶 is equal to 1620. We can work that out using column multiplication. So we’d have 550 multiplied by two. So we do two multiplied by zero is just zero. Then, we’d have two multiplied by five which would give us 10. So we put a zero in the tens column and carry the one into the hundreds column. And then finally, we’d have two multiplied by the five which will come from the 500. Again, this gives us 10 but plus the one we carried gives us 11. So we put one in the hundreds column and one in the thousands column. So therefore, we’ve got the result 1100.
So now, what we need to do is we’re trying to find out the cost of a child’s ticket. So we want to find out what 𝐶 is. So in that case, we’re gonna subtract 1100 away from each side of our equation. So then, when we do that, we get two 𝐶 is equal to 520.
So then as that’s the amount for two 𝐶, what we want to do is find out one 𝐶. So to do that, we divide each side of the equation by two, which gives us 𝐶 is equal to 260. And that’s because two 𝐶 divided by two is just 𝐶 and 520 divided by two is 260. We can think of that as 52 divided by two is 26. And then we’ve got a zero cause that’s was 520. And therefore, we have a result of 260 or we could use the bus stop method as we showed previously.
So therefore, we can say that if an airline company is selling tickets to fly from London to New York and the cost of two adult tickets and two child tickets is 1620 pounds, the cost of three adult tickets is 1650 pounds, then the cost for one adult ticket is 550 pounds and the cost for one child ticket is 260 pounds. |
# Test: Calculations, Exponents And Basic Algebra
## 10 Questions MCQ Test Quantitative Reasoning for GMAT | Test: Calculations, Exponents And Basic Algebra
Description
Attempt Test: Calculations, Exponents And Basic Algebra | 10 questions in 15 minutes | Mock test for GMAT preparation | Free important questions MCQ to study Quantitative Reasoning for GMAT for GMAT Exam | Download free PDF with solutions
QUESTION: 1
### lies between:
Solution:
The first culprit in this expression is the radical in the denominator. Radicals in the denominator are dealt with by multiplying the fraction with an expression that is equal to 1 but contains a “canceling radical” in both the numerator and denominator.
For example, is simplified by multiplying by
In the case of a complex radical, such as we multiply by the conjugate, as follows:
Simplifying radicals in the denominator with conjugate radical expressions is very useful on challenging GMAT radical questions.
QUESTION: 2
### lies between:
Solution:
In order to rid the expression of square roots, let's first square the entire expression. We are allowed to do this as long as we remember to "unsquare" whatever solution we get at that end.
Notice that the new expression is of the form where
Recall that This is one of the GMAT's favorite expressions.
Returning to our expression:
Notice that x2 + y2 neatly simplifies to 48. This leaves only the 2xy expression left to simplify.
In order to simplify recall that
Thus,
Notice that the expression under the square root sign is of the form And recall that This is another one of the GMAT's favorite expressions. Returning to our expression:
Finally then:
But now we must remember to "unsquare" (or take the square root of) our answer:
QUESTION: 3
### A, B, C, D, E, F, G, and H are all integers, listed in order of increasing size. When these numbers are arranged on a number line, the distance between any two consecutive numbers is constant. If G and H are equal to 512 and 513, respectively, what is the value of A?
Solution:
The distance from G to H is 513 - 512.
The distance between and two consecutive points is constant, so the distance from A to G will be 6 times the distance from G to H or 6(513 – 512).
The value of A , therefore, will be equal to the value of G minus the distance from A to G :
QUESTION: 4
Solution:
Remember that when you multiply different bases raised to the SAME exponent, the product is simply the product of the bases raised to their common exponent.
QUESTION: 5
The three-digit positive integer x has the hundreds, tens, and units digits of a, b, and c, respectively. The three-digit positive integer y has the hundreds, tens, and units digits of k, l , and m, respectively. If (2a)(3b)(5c) = 12(2k)(3l)(5m), what is the value of x – y?
Solution:
First, let us simplify the exponential equation:
When the bases on both sides of an equation are equal and the bases are prime numbers, the exponents of the respective bases must also be equal: a = k + 2; b = l + 1; and c = m . Now recall that a , b , and c represent the hundreds, tens, and units digits of the three-digit integer x ; similarly, k , l , and m represent the hundreds, tens, and units digits of the three-digit integer y .
Therefore, the hundreds digit of x is 2 greater than the hundreds digit of y ; the tens digit of x is 1 greater than the tens digit of y ; finally, the units digit of x is equal to the units digit of y . Using this information, we can set up our subtraction problem and find the value of (x – y ):
QUESTION: 6
Solution:
A radical expression in a denominator is considered non-standard. To eliminate a radical in the denominator, we can multiply both the numerator and the denominator by the conjugate of that denominator.
QUESTION: 7
If 274x + 2 × 162-2x × 36x × 96 – 2x = 1, then what is the value of x?
Solution:
An effective strategy for problems involving exponents is to break the bases of all the exponents into prime factors. This technique will allow us to combine like terms:
274x + 2 × 162-2x × 36x × 96 – 2x = 1 (33)4x + 2 × (2 × 34)-2x × (22 × 32)x × (32)6 – 2x = 1
312x + 6 × 2-2x × 3-8x × 22x × 32x × 312 4x = 1
2-2x + 2x × 312x + 6 – 8x + 2x + 12 – 4x = 1
20 × 3 2x + 18 = 1
3 2x + 18 = 1
3 2x + 18 = 30
2x + 18 = 0
2x = -18
x = -9
QUESTION: 8
If (22x+1)(32y-1) = 8x27y, then x + y =
Solution:
Let's rewrite the right side of the equation in base 2 and base 3: (22x+1)(32y-1) = (23)x(33)y. This can be rewritten as: (22x+1)(32y-1) = 23x33y Since both bases on either side of the equation are prime, we can set the exponents of each respective base equal to one another: 2x + 1 = 3x , so x = 1 2y – 1 = 3y, so y = -1 Therefore, x + y = 1 + (-1) = 0.
QUESTION: 9
x2 + bx + 72 = 0 has two distinct integer roots; how many values are possible for 'b'?
Solution:
In quadratic equations of the form represents the sum of the roots of the quadratic equation and c/a
represents the product of the roots of the quadratic equation.
In the equation given a = 1, b = b and c = 72
So, the product of roots of the quadratic equation = 72/1 = 72
And the sum of roots of this quadratic equation= -b/1 = -b
We have been asked to find the number of values that 'b' can take.
If we list all possible combinations for the roots of the quadratic equation, we can find out the number of values the sum of the roots of the quadratic equation can take.
Consequently, we will be able to find the number of values that 'b' can take.
The question states that the roots are integers.
If the roots are r1 and r2, then r1 * r2 = 72, where both r1 and r2 are integers.
Possible combinations of integers whose product equal 72 are : (1, 72), (2, 36), (3, 24), (4, 18), (6, 12) and (8, 9) where both r1 and r2 are positive. 6 combinations.
For each of these combinations, both r1 and r2 could be negative and their product will still be 72.
i.e., r1 and r2 can take the following values too : (-1, -72), (-2, -36), (-3, -24), (-4, -18), (-6, -12) and (-8, -9). 6 combinations.
Therefore, 12 combinations are possible where the product of r1 and r2 is 72.
Hence, 'b' will take 12 possible values.
QUESTION: 10
If x > 0, how many integer values of (x, y) will satisfy the equation 5x + 4|y| = 55?
Solution:
5x + 4|y| = 55
The equation can be rewritten as 4|y| = 55 - 5x.
Because |y| is non-negative, 4|y| will be non-negative. Therefore, (55 - 5x) cannot take negative values.
Because x and y are integers, 4|y| will be a multiple of 4.
Therefore, (55 - 5x) will also be a multiple of 4.
55 is a multiple of 5. 5x is a multiple of 5 for integer x. So, 55 - 5x will always be a multiple of 5 for any integer value of x.
So, 55 - 5x will be a multiple of 4 and 5.
i.e., 55 - 5x will be a multiple of 20.
Integer values of x > 0 that will satisfy the condition that (55 - 5x) is a multiple of 20:
1. x = 3, 55 - 5x = 55 - 15 = 40.
2. x = 7, 55 - 5x = 55 - 35 = 20
3. x = 11, 55 - 5x = 55 - 55 = 0.
When x = 15, (55 - 5x) = (55 - 75) = -20. Because (55 - 5x) has to non-negative x = 15 or values greater than 15 are not possible.
So, x can take only 3 values viz., 3, 7, and 11.
We have 3 possible values for 55 - 5x. So, we will have these 3 values possible for 4|y|.
Possibility 1: 4|y| = 40 or |y| = 10. So, y = 10 or -10.
Possibility 2: 4|y| = 20 or |y| = 5. So, y = 5 or -5.
Possibility 3: 4|y| = 0 or |y| = 0. So, y = 0.
Number of values possible for y = 5.
The correct choice is (C) and the correct answer is 5.
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# How do you write the equation in point slope form given ( -2 , 5 ) , ( 4 , -3 )?
Jun 1, 2017
Answers: $y - 5 = - \frac{4}{3} \left(x + 2\right)$ or $y + 3 = - \frac{4}{3} \left(x - 4\right)$
#### Explanation:
Write point-slope form of a linear equation given points $\left(- 2 , 5\right)$ and $\left(4 , - 3\right)$.
Note that the general form of point-slope form is:
$y - {y}_{1} = m \left(x - {x}_{1}\right)$ where $m$ is the slope of the line and $\left({x}_{1} , {y}_{1}\right)$ is a point on the line.
First, we need to find the slope, m:
We know that $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ with points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$
So, we can plug in our given values:
$m = \frac{5 - \left(- 3\right)}{- 2 - 4}$
$m = \frac{8}{- 6}$
$m = - \frac{4}{3}$
Now, we choose one of the points to write the equation in point slope form, (I will write both possible equations in point-slope form):
For $\left(- 2 , 5\right)$
$y - 5 = - \frac{4}{3} \left(x + 2\right)$
For $\left(4 , - 3\right)$
$y + 3 = - \frac{4}{3} \left(x - 4\right)$ |
## How do we find the derivative of 1/(1+x) ?
I would recommend rewriting the expression first.
Sometimes if you change how something looks, it becomes easier to work with.
Now you can do the power rule and the chain rule.
## Tutoring Calculus, Series, Sequences, Improper Integrals and Limits
We focused on series and sequences and looked a bit at improper integrals.
All of these are connected to limits. If something diverges, which is easier to determine than convergence, you simply say it diverges. If a test is inconclusive, you try another test.
Geometric series might be the easiest to deal with. Sometimes a series doesn’t exactly look like a typical geometric series and needs to be manipulated a bit algebraically. If you have the ratio and the first term, there is a simple formula to determine the sum.
Many times for series, the variable with the highest exponent becomes more important.
An alternating series just needs to be decreasing to be convergent. To be absolutely convergent requires more tests.
The algebra before doing the calculus can be very important. Changing the look of an expression can make it easier to work with.
Knowing the trig functions is necessary for some of these problems.
U-substitution was a useful technique for some of the integration. Before calculus 3, it’s probably a good idea to review integration by parts.
e ~ 2.7
Remember the chain rule when taking the derivative of e^u (most of the time) unless taking the derivative with respect to u.
If you make a substitution in integration, you must change the bounds or label the bounds more carefully.
## Reviewing Some Ideas From Calculus I and leading up to it
Tutored calculus for a couple of hours the other day, just before the student started Calculus II. Here are some things we went over.
We talked about the purpose of a derivative and how you can optimize things early on and how integrals can find the areas under various curves and on into three dimensions, etc.
Went over how logarithms can be useful, what the natural number e is, and the behavior of the natural logarithm.
How algebra can simplify calculus before doing derivatives or integrals.
How to deal with fractional exponents.
How the order of approach for fractional exponents can simplify things.
Thinking about the behavior of functions for limits.
How exponents like 1/4 and 1/6 work.
How the ln function increases and decreases very slowly.
How to divide by fractions by multiplying by the reciprocal.
Using common denominators.
Going over integration by parts.
Doing some derivatives and keeping track of the sign.
How squared trig functions look.
Factoring out algebraic terms.
Use the product rule when differentiating a product.
Picking u and v for integration by parts won’t always work out, if it seems to not be working, you may want to choose something else.
Doing an integral is taking an anti-derivative and finding something that when you take the derivative of it will give you what you started with between the integral sign and the differential.
## Always take the chain rule for derivatives
Sometimes, if you take the derivative with respect to the variable in the expression (and there is only one variable), you can ‘not’ use it since, for example, dx/dx = 1. But in general, you use the chain rule for derivatives.
Reviewing for the final, we started with the earlier version of a derivative which involved limits.
The letter ‘a’ is used sometimes to denote a constant.
With multiple variables, the chain rules becomes even more important.
The general approach for a problem involving changing dimensions of a rectangle was to find an expression for a particular characteristic and then take the derivative. Certain rates were known in the problem.
## How small mistakes propagate and multiply in Calculus problems
We looked some at the previous test. But focused on material for the upcoming test.
One thing was the product rule, he had made a mistake before with it on another test.
Often there were small algebraic errors. The new material seems fairly comfortable though.
If you make a mistake early on it can continue to be in your work for the entirety of the problem and depending on the direction of where you go with the problem, it can get bigger. Let’s say you have four instead of two, then you square that number, suddenly the mistake got larger.
Calculus involves a lot of algebra, geometry, etc. Sometimes there can be many steps and mistakes add up and get bigger. Often times it is better to slow down a bit and be careful rather than hunt for mistakes later on.
## “What are some tips on how to solve Integration problems using U-Substitution?”
Question from Quora
There are basically two terms here, the 5x and the 1- x^2 in parentheses.
The general idea for u-substitution is to make u something that when you take the derivative of it can be substituted for what is left over (with a little manipulation).
So if you chose 5x as your u, then du would be 5dx which you could not easily substitute.
If what you choose seems to not be working, either you made a mistake or you should try a different option.
For this problem choosing u = 1 -x^2 seems to work better. You can probably even take the derivative of that in your head to figure out in advance if it will work well.
At that point it becomes much easier to solve.
## How to do a phasor calculation with radians using a principle of Eulers Formula
“How has this been calculated? I try to convert the 7.11 to radians and sum it up, then converting it back to degrees. However I do not get 1.61.”
-Quora question
My response:
Dividing the coefficients does not seem to be the problem you had.
If you divide the e^x terms you would normally subtract the jπ/4 from the -j7.11.
That would get you
~-j7.8954
It’s best to save this in memory to not lose accuracy.
From there we can think about Euler’s formula. Shown on the right side in green. Much like the trigonometric functions we learn much before doing anything with phasors, you can add or subtract 2π to the input and get the same result for periodic functions. You can also do that with e^ix.
For this problem we add j2π to the -j7.8954 to get -j1.61. The answer is an approximation.
## “Should I skip Pre-Calculus?”
Trigonometry, if it’s included, is very important for several things and could very well be included in precalculus. Check out the syllabus and book for what would be included.
I would also think about who teaches the class. If it’s a challenging class with a good teacher, you could learn a lot. If it’s easy, maybe you could learn it on your own.
I skipped Spanish 2 in high school. That would have been no problem if the Spanish 3 class was easy, but it wasn’t. So I worked very hard and got a lot of help to catch up.
What is the rush to take calculus?
Seems like you might want to take precalculus in the summer since it would likely be at an accelerated pace given the constrained time frame. Or you could maybe just go through it on your own. But just reading the book probably wouldn’t be enough. You would want to go through a lot of problems.
## “why does $x^{1/2} = +\sqrt{x}$ not $±\sqrt{x}$?”
I think you’re asking why it is not true that x^{1/2} = ±\sqrt{x}?
Since it seems logical that it would be similar to this, \sqrt{x^{2}} = ±x
Basically, I would say it’s because you can write -x^{1/2}
If that’s what you want to say, you put the negative sign in front of it. If you want it to be positive, you don’t write the negative sign.
If you square ±\sqrt{x}
In either case, you will get x.
## “What are the practical uses of calculus, other than calculating the area under curves?”
Among other things, one thing you learn early on in calculus that can be very powerful is Optimization.
If you know how to take a derivative, you can find where the first derivative is equal to zero which can often be a maximum or minimum.
Knowing exactly how to get the maximum or minimum of something can be very useful.
Let’s say you have a certain amount of material to build something and you want the maximum volume. Use optimization.
Let’s say you want to minimize the cost of doing something or making something. Use optimization. |
# How do you differentiate f(x)=1/e^sqrt(1-(3x+5)^2) using the chain rule.?
$\textcolor{b l u e}{f ' \left(x\right) = \frac{- \left(3 x + 5\right) \cdot \sqrt{1 - {\left(3 x + 5\right)}^{2}} \cdot {e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}}}{3 {x}^{2} + 10 x + 8}}$
#### Explanation:
The given equation is $f \left(x\right) = \frac{1}{e} ^ \left(\sqrt{1 - {\left(3 x + 5\right)}^{2}}\right)$
The solution:
$f \left(x\right) = \frac{1}{e} ^ \left(\sqrt{1 - {\left(3 x + 5\right)}^{2}}\right)$
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{1}{e} ^ \left(\sqrt{1 - {\left(3 x + 5\right)}^{2}}\right)\right) = \frac{d}{\mathrm{dx}} \left({e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}}\right)$
$f ' \left(x\right) = {e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}} \cdot \frac{d}{\mathrm{dx}} \left(- \sqrt{1 - {\left(3 x + 5\right)}^{2}}\right)$
$f ' \left(x\right) = {e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}} \cdot \left(- 1\right) \cdot \frac{d}{\mathrm{dx}} \left(\sqrt{1 - {\left(3 x + 5\right)}^{2}}\right)$
$f ' \left(x\right) =$
${e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}} \cdot \left(- \frac{1}{2 \sqrt{1 - {\left(3 x + 5\right)}^{2}}}\right) \frac{d}{\mathrm{dx}} \left(1 - {\left(3 x + 5\right)}^{2}\right)$
$f ' \left(x\right) =$
${e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}} \cdot \left(- \frac{1}{2 \sqrt{1 - {\left(3 x + 5\right)}^{2}}}\right) \left(0 - 2 {\left(3 x + 5\right)}^{1} \cdot 3\right)$
$f ' \left(x\right) =$
${e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}} \cdot \left(- \frac{1}{2 \sqrt{1 - {\left(3 x + 5\right)}^{2}}}\right) \left(- 6 \left(3 x + 5\right)\right)$
$f ' \left(x\right) = {e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}} \cdot \left(\frac{9 x + 15}{\sqrt{1 - {\left(3 x + 5\right)}^{2}}}\right)$
We simplify by rationalization
$f ' \left(x\right) = \frac{- \left(9 x + 15\right) \cdot \sqrt{1 - {\left(3 x + 5\right)}^{2}} \cdot {e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}}}{9 {x}^{2} + 30 x + 24}$
Reducing the numbers by canceling common numerical coefficients
$\textcolor{b l u e}{f ' \left(x\right) = \frac{- \left(3 x + 5\right) \cdot \sqrt{1 - {\left(3 x + 5\right)}^{2}} \cdot {e}^{- \sqrt{1 - {\left(3 x + 5\right)}^{2}}}}{3 {x}^{2} + 10 x + 8}}$
God bless....I hope the explanation is useful. |
# Math 10 – week 18 – elimination
This week in math 10 we learned how to solve equations with using elimination. Elimination in incredibly similar to using insertion.
The first step to solving something using elimination is to add or subtract. in this case I will use subtract.
7x+y=15
3x+y=3
I’m going to make all of the number in the second equation negative
7x+y=15
-3x-y=-3
Now to subtract. Note: when using elimination, make zero pares with x or y
4x=12
Now to divide by 4
X=3
We can then insert x back into the equation to find out y
7(3)+y=15
21+y=15
X=-6
my core competencies reflection:
Taking too long?
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# Math 10 – week 15 – graphing
This week in math 10, we learned how to tell what a graph will look like by looking at a linear equation. The equation is: y=mx+b
m = slope
b = y intercept
lets take a look at this picture and find the equation.
Remember the equation for slope is y/x or rise/run. So, after finding to nice points, the slope is 7/1, witch can be simplified to just 7. The y-intercept for this line -8, so the equation for this line is: Y=7x – 8.
Know, lets to the opposite, take an equation and make a graph from it: Y=1/3x + 4.
My recommendation would be to start form the y-intercept and then add the slope
Y=1/4x + 2
# Math 10 – week 17 – substation
This week in math 10 I learned how to solve equations with using substitution. The steps for substitute are as fallows:
1. Pick an equation to rearrange
2. Substitute equation
3. Solve equation
4. Substitute the number just found, back into the rearranged equation
5. Verify solution on both equations
The equations being solved for this example will be y-x=1 and 2x+3y=18.
The first step is to pick an equation to rearrange and rearrange it. The best option is always the equation with a coefficient of 1. So, the best option for rearranging is y-x=1.
y-x=1
y-x+x=1+x
y=1+x
The second step is to insert the rearranged equation and substitute into the second equation. We are going to put y=1+x into 2x+3y=18. We are going to want to put 1+x into the equation where the y is in the second equation because y is equal to 1+x.
y=1+x and 2x+3y=18 becomes 2x+3(1+x)=18
If you notice, the equation has become a simple 9th grade equation.
The next step is to solve the equation
2x+3(1+x)=18
2x+3+3x=18
5x+3=18
5x+3-3=18-3
5x=15
5x/5=15/5
x=3
Now to take the number we found (3) and insert it back into the rearranged equation (y=1+x).
y=1+4
y=4
Now to verify the numbers in both equation
y-x=1
4-3=1
2x+3y=18
2(3)+3(4)=18
6+12=18
# Math 10 – Week 16 – converting equations
This week in math 10, we learned about the different form of equations and how to change between them.
Here are the different equations:
Slope intercept form
y=mx+b
Slope Point form
y-y2=m(x-x2)
General form
0=mx+y+b
Slope intercept form to Slope Point form (y=mx+b to y-y2=m(x-x2))
We want to take the equation y=4x+8 and convert it to slope point form. This is one of the most simple conversion.
The first step is to fin a point to convert: (0,8) being put into the slope changes to (1,12)
Next, you would just inster the number into the equation: y-12=4(x-1)
Slope intercept form to General form (y=mx+b to 0=mx+y+b)
This conversion is also very simple. All you need to do to subtract y:
y=4x+8
-y
0=4x-y+8
Slope Point form to General form (y-y2=m(x-x2) to 0=mx+y+b)
This conversion is a little more complicated. The first step is to convert form slope point form to slope intercept form:
We need to find the y-intercept, you can do that by applying the slope to the point in the equation: y-12=4(x-1) turn into y=4x+8.
You then can apply the conversion for Slope intercept form to General form.
# Math 10 – week 13 – analysis
This week in math team we learned how to dissect graphs and connect it to a story. Here the graphs that we will be dissecting.
George and his family watched a movie together and the graph show the relationship between the amount of popcorn and time.
We can dissect George’s graph first. It starts on the y axis so we can assume he has a full bole of popcorn at the start of the movie. The line on the graph goes down really fast and ends up the x axis really early so we can deduce that he ate all of his pop corn really fast.
Know lets dissect George’s sister’s graph. Her graph starts at the same place as before, so she had a full bole of popcorn at that start of the movie. her graph is really curvy, so we can assume that she was eating the popcorn slowly at times and fast too.
Geroge’s dad’s graph is next. There is a strate line in the graph so he did not eat any popcorn for a while.
Gorege’s mom did not get her pop corn right at the star of the movie, I know this because it dose not start at the y-axis. There is a long strate line and we can
# Math 10 – week 12 – function notation
This week in math we learned how to write function notation. There are 3 different types, equation, mapping, and function notation.
1: equation
5x+4
You can input any number in this equation and there will be no overlapping outputs. This is the t-chart for the equation that can then be charted on a plane
x y -3 -11 -2 -6 -1 -1 0 4 1 9 2 14 3 19
2: mapping
This is the same equation but in map notation: f:(x)→5x+4
The f at the start is the name of the function (eg: g,h,k). the arrow means happed onto. So if I where to use function f to solve if x=3, it would look like this:
f:(3)→5(3)+4
f:(3)→15+4
f:(3)→19
3: function
This is what the same equation would look like in function notation: f(x)=5x+4
The function still has a name(f) and the equal sign(=) is mapped onto.
This is what it looks like to solve if x=3:
f(3)=5(3)+4
f(3)=15+4
f(3)=19
# Math 10 – week 11 – domain and Range
Domain and Range
This week in math 10, I learned how to identify a coordinate plain’s Domain and Range.
The horizontal line, also known as the x-axis, is used to find the domain, and the vertical line, also known as the y-axis, is used to find the range.
To understand how to find Domain and Range, I will simplify it to only one line. Know to explain the next few examples.
Example #1
To describe this line, you list out the places where the dots are: {2,3,4,5}. You can not answer the question like this: {2-5} because that would imply decimals.
Example #2
The that arrow means that all numbers smaller than -6 are possible solutions for the equation and the unshaded circle means that -6 is not a possible solution. The answer would be written like this: {-6˃x}
Example #3
The fact that there are 2 circles means that all the possible solutions have to bee in between those 2 numbers. The shaded circle means that that the number circled is a possible answer. The answer would be written like this: {-1≤x˂7}
Know to introduce Range in.
Example #4
Starting with the Domain. The answer would be any real number because there is no base point and the arrows go both ways. For the range, the answer would be 8 because the height dose not vary.
D {xR}
R {8}
# Math 10 – Week 10 – Factoring Ugly trinomials
This week in math 10, we learned how to factor ugly trinomials. What makes a trinomial ugly? The coefficients are big numbers. For this blog post, I will solve this polynomial: 2x2+17x+35
First I must multiply and leading coefficient (2) and the constant together (35), to get the total (2*35=70).
Next, I will list out all of the factors of 70 and see with set can add to 17:
70
1*70
2*35
5*14
7*10
7+10=17 so I can now put all of these number into the box to do the reverse box method.
2x2 7x 10x 35
Note: It dose not matter where you but the xs.
Now I can fill not the box. To find the outside find what is similar between the number in the row or column. 2x2 and 7x have x common. 10x and 35 have 5 in common. 2x2 and 10x have 2x in common. 7x and 35 have 7 in common
2x 7 x 2x2 7x 5 10x 35
Now I can take the out side number and put them into an equation.
(x+5)(2x+7)
# Week 7 – Math 10 – Factoring Polynomials
This week in math 10 we learned how to Factoring Polynomials. There are several ways to factor polynomials.
Common factors: To make an equation easier to understand, you can find a common factor in the numbers. The example I will be using is: $2x^2$ + 12x + 10
all number in this equation have a common factor of 2, so we have to divide all the hole equation. and this is the result: 2($x^4$ + 6x + 5)
Trinomials: $x^2$ -11 + 28
In order to factor the example above, we need to find 2 numbers that add or subtract to -11 and multiply to 28. To Start, I will create a list of factors for 28: 1,2,4,7,14,28
7+4=11 so -7-4=-11 and that means our simplified equation is:
(x-7)(x-4)
Binomials: $x^2$ – 25
(x-5)(x+5)
# Week 6 – Math 10 – Multiplying polynomials
This week we learned how to multiply polynomials in several different ways. In order to multiply polynomials, you need to understand the distributive property and the multiplication exponent law, but I will explain them when they come up. I this I will explain how to multiply a trinomial (3 terms) by a binomial (2 terms) in 2 ways. The example I will be solving is (5x2-3x+6)(4x+5)
Way 1: algebraically
First I can start with expanding. Each term in the first bracket will multiply into terms in the second, because of the distributive property. This is what the equation would look expanded:
5x2(4x)+5x2(5)-3x(4x)-3x(5)+5(4x)+5(5)
Know I can do the individual multiplication:
20x3+25x2-12x2-15x+20x+20
The reason 5x2(4x)=20x3 is according to the multiplication exponent law. There in an invisible exponent of 1 on the 4x.
Before I combine like terms I like to organize my equation to make things easier for myself. I organize the equation by like terms then I combine. This equation is already organized so I can skip that step.
Like terms combined: 20x3+13x2+5x+20
Way 2: visually
(5x2-3x+6)(4x+5)
You start with making a box with lines separating it to the number of terms
Then you fill in the polynomials to their respective sides. put each term on a line like this
4x 5 5x2 -3x 6
Then I would do the multiplication in the individual squares
4x 5 5x2 20x3 25x2 -3x -13x2 -15x 6 24x 30
Know I can right out the hole equation and combine like terms.
Written out: 20×3+25×2-12×2-15x+20x+20
Like terms combined: 20×3+13×2+5x+20 |
Subsets and Splits
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