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# Square of a Binomial
A polynomial is a function having usually many terms in one or even more variables.
On the basis of number of terms, the polynomial may be classified as:
1) Monomial (polynomial with one term)
2) Binomial (polynomial having two terms)
3) Trinomial (polynomial containing three terms)
More than this is already known as a polynomial.
In this article, we going to focus basically on squares of binomials and trinomials. The general form of square of binomials can be written as (x + y)$^{2}$. Similarly, the general form of square of trinomials is to be written as (x + y + z)$^{2}$. Where, x, y and z are variable which may or may not be associated with constant coefficients.
The formulae for finding square of a binomial are given below:
$(x + y)^{2} = x^{2} + y^{2} + 2 xy$
$(x - y)^{2} = x^{2} + y^{2} - 2 xy$
These fomulae may be be derived as:
(x + y)^${2}$
= (x + y) (x + y)
= x.x + x.y + y.x + y.y
= $x^{2} + y^{2} + 2 xy$
(x - y)^${2}$
= (x - y) (x - y)
= x.x + x.(-y) + (-y).x + (-y).(-y)
= x.x - x.y - y.x + y.y
= $x^{2} + y^{2} - 2 xy$
In order to find square of a binomial, we assume first term as x and second term as y. Then apply any of the above suitable formula. If any term is connected with constant coefficients, then that coefficient should be considered as a part of that term and dealt accordingly.
Understand the following examples carefully.
1) $(a + 5)^{2}$
Apply formula for $(x + y)^{2}$
= $a^{2} + 5^{2} + 2 \times a \times 5$
= $a^{2} + 25 + 10 a$
2) $(3x - 4)^{2}$
Apply formula for $(x - y)^{2}$
= $(3x)^{2} + 4^{2} - 2 \times 3x \times 4$
= $9x^{2} + 16 - 24x$
Related Calculators Binomial Calculator Square Meters to Square Feet Binomial Confidence Interval Calculator Binomial Distribution Calculator
## Perfect Square Trinomials
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A perfect square trinomial is a function in one or more variables which contains three (tri means three) terms. Two terms in a perfect square trinomial should be a perfect squares of some number or variable with constant coefficient. The third term has to be the twice of the product of these perfect squares. Have a look at the following diagram illustrating a perfect square trinomial:
For Example:
x$^{2}$ + 12x + 36
= x$^{2}$ + 12x + 6$^{2}$
= x$^{2}$ + 2. x. 6 + 6$^{2}$
This is a perfect square trinomial.
Now, this can be written in the form of (x + y)$^{2}$.
## Completing The Square
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"Completing the square" is a quite common method of factorization. In this method, one has to recognize and form the perfect square trinomial and then substitute back the formula of square of the binomial. This method involves the step mentioned below:
Step 1: We have to make sure about the coefficient of highest power in the polynomial is one.
For Example: a$^{2}$ + 8a - 4 = 0
Here, a has the coefficient 1.
If it is not 1, we need to divide every term by the leading coefficient making the coefficient of highest power as one.
Step 2: Write the middle term as the product of 2 and the perfect square term, i.e.
a$^{2}$ + 2. a . 4 - 4 = 0
Step 3: Now, the third number should be the perfect square of the remaining number in second term. In above example, the remaining number after 2a is 4, so third term should be the square of 4, i.e. 16.
In order to make third term a square term, we need to add and subtract a number in the equation so as to make it square of required term. i.e. in above example, we should add and subtract
4$^{2}$ = 16.
a$^{2}$ + 2. a . 4 - 4 = 0
a$^{2}$ + 2. a . 4 + 16 - 16- 4 = 0
Step 4: Separate the perfect square trinomial in this way.
(a$^{2}$ + 2. a . 4 + 16) - 20 = 0
Step 5: Apply the suitable formula of square of the binomial.
(a$^{2}$ + 2. a . 4 + 16) - 20 = 0
(a + 4)$^{2}$ - 20 = 0
Step 6: Now, factorize the equation thus obtained.
(a + 4)$^{2}$ - 20 = 0
$(a + 4)^{2} - (\sqrt{20})^{2} = 0$
$(a + 4 + 2\sqrt{5}) (a + 4 - 2\sqrt{5}) = 0$
$a + 4 + 2\sqrt{5} = 0$ and $a + 4 - 2\sqrt{5} = 0$
a = - 4 - $2\sqrt{5}$
a = - 4 + $2\sqrt{5}$
## Square of a Trinomial
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A trinomial is an expression having three terms. The formula for the square of a trianomial is given below:
$(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx$For Example:
(a - 2b + c)$^{2}$
= a$^{2}$ + (-2b)$^{2}$ + c$^{2}$ + 2 a (-2b) + 2 (-2b) c + 2 c a
= a$^{2}$ + 4b$^{2}$ + c$^{2}$ - 4ab - 4bc + 2ca
## Examples
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Have a look at the example based on this article:
Example 1: Evaluate (6m - 7n)$^{2}$
Solution: Using the formula $(x - y)^{2} = x^{2} + y^{2} - 2 xy$
$(6m - 7n)^{2}$
= $(6m)^{2} + (7n)^{2} - 2 . 6m . 7n$
= $36 m^{2} + 49 n^{2} - 84 m n$
Example 2: Factorize $(a + b)^{2} - 12 (a + b) + 36$
Solution : $(a + b)^{2} - 12 (a + b) + 36$
= $(a + b)^{2} - 2 (a + b) . 6 + 6^{2}$
This is the form of perfect square trinomial of $x^{2} + y^{2} - 2 xy$ which is equal to $(x - y)^{2}$, hence
= $[(a + b) - 6]^{2}$
= $(a + b - 6)^{2}$
Example 3: Solve $(5x - y - 2z)^{2}$
Solution: Using the formula $(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2xy + 2yz + 2zx$, we get
$(5x - y - 2z)^{2}$
= (5x)$^{2}$ + (-y)$^{2}$ + (-2z)$^{2}$ + 2 (5x) (-y) + 2 (-y) (-2z) + 2 (-2z) (5x)
= 25 x$^{2}$ + y$^{2}$ + 4 z$^{2}$ - 10 xy + 4 yz - 20 xz
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# How do you find the slope and intercept of 3y=7?
Jul 17, 2018
$\text{slope} = m = 0$; $\text{ } y$-intercept $\left(0 , \frac{7}{3}\right)$
#### Explanation:
Given: $3 y = 7$
The slope - intercept form of a line: $y = m x + b$,
where $m$ = the slope and $b$ is the $y$-intercept $\left(0 , b\right)$
Put the given equation in slope - intercept form by dividing both sides of the equation by $3$:
$\frac{3}{3} y = \frac{7}{3}$
$y = \frac{7}{3} \text{ " => " } y = 0 x + \frac{7}{3}$
This is a horizontal line. They have slope $= 0$.
This means $\text{slope} = m = 0$; $\text{ } y$-intercept $\left(0 , \frac{7}{3}\right)$ |
# Congruence Criterion: Angle, Side, Angle
## Angle, Side, Angle
Definition:
Two triangles in which two angles and the included side between them are equal are congruent triangles.
Attention: The two angles must be adjacent to the equal and corresponding side in both triangles!
To prove that two triangles are congruent we can use one of the following postulates:
## Example 1 - Congruent Triangles
Given the triangles $Δ ABC$ and $Δ DEF$ such that:
$\sphericalangle A=\sphericalangle D$
$AB = DE$
$\sphericalangle B=\sphericalangle E$
From this, we deduce that the triangles $Δ ABC$ and $Δ DEF$ are congruent, therefore, we will write:
$Δ DEF ≅ Δ ABC$ according to the congruence criterion: Angle, Side, Angle (ASA)
Consequently, we will deduce that:
$BC = EF$
$AC = DF$
since these are corresponding and equal sides in congruent triangles.
Then, we will also deduce that:
\\sphericalangle C=\sphericalangle F\)
since these are corresponding and equal angles in congruent triangles.
## Example 2 - Triangle Congruence Exercise
Given two parallel lines. The line $AC$ and the line $BD$ pass between them in such a way that they intersect at point $O$. Also, we are informed that $AO = OC$
Prove that $AB = DC$
Proof:
First, we must show that triangles $Δ ABO$ and $Δ DOC$ are congruent. We will base this on the previous criterion.
Note that $\sphericalangle AOB = \sphericalangle COD$ (Because they are vertical angles)
Since $AE = EC$ (Side)
Remember that the two given lines are parallel lines.
Therefore, $\sphericalangle OAB=\sphericalangle OCD$ since they are alternate interior angles between parallel lines (angle).
We observe that now we have $2$ triangles in which $2$ of their angles and the side included between them are equal.
Consequently, triangles $Δ ABO$ and $Δ DOC$ are congruent
and we write it as $Δ ABO ≅ Δ DOC$ according to the Angle, Side, Angle (ASA) congruence criterion
Therefore, we can deduce that $AB=DC$ (Corresponding sides of congruent triangles).
$QED$
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## Angle-Side-Angle Congruence Exercises
### Exercise 1
Assignment
Given that point $K$ bisects $AC$.
and also $∢A=∢C$
According to which theorem of congruence do the triangles $ΔAMK≅ΔCBK$ coincide?
Solution
Given that the angles $A=C$
$AK=KC$
Given that point $K$ cuts $AC$
Angles $AKM=CKB$
Opposite angles by the vertex are equal
The triangles are congruent according to the $A.S.A$ theorem
$A.S.A$
### Exercise 2
Assignment
Given: the quadrilateral $ABCD$ is a rectangle.
By which theorem of congruence do the triangles $ΔADO≅ΔCBO$ coincide?
Solution
$BC=AD$
Since the quadrilateral $ABCD$ is a rectangle and in a rectangle there are two pairs of parallel and equal sides
Angles $\sphericalangle BCO=\sphericalangle ADO$ are alternate angles between equal parallel lines.
Angles $O_1=O_2$
opposite angles at the vertex are equal
Therefore, we say that the triangles are congruent according to the $A.A.S.$ theorem
According to the $A.A.S.$ theorem
### Exercise 3
Assignment
In the given figure: $DE || AB$
and point $C$ bisects segment $BE$.
According to which theorem of congruence do the triangles match?
$ΔABC≅ΔDEC$
Solution
Given that $DE$ is parallel to $AB$
Angles $D=A$
Alternate angles are equal between parallel lines
$BC=CE$ Point $C$ bisects the line $BE$
Angles $C_1=C_2$
Vertex opposite angles
Triangles congruent according to the superposition theorem $A.A.S..$
Superimposed according to $A.A.S.$
### Exercise 4
Assignment
Given the isosceles triangle $ΔEDC$.
$∢ADE=∢BCE$
$AC=BD$
According to which theorem of congruence do the triangles coincide?
$ΔADE≅ΔBCE$
Solution
Triangle $ΔEDC$ is an isosceles triangle
$DE=EC$
In an isosceles triangle, two sides are equal
Angles $D=C$ given
angles $EDC=ECD$ the base angles of an isosceles triangle are equal
angles $ADE=BCE$
$\sphericalangle D-\sphericalangle EDC=\sphericalangle C-\sphericalangle ECD$
Subtraction of angles
$\sphericalangle E_1=\sphericalangle E_2$
Triangles are congruent according to the $ASA$ theorem
$ASA$
### Exercise 5
Assignment
Given: quadrilateral $ABCD$ square.
And within it is enclosed the kite $KBPD$.
According to which theorem of congruence do the triangles $ΔBAK≅ΔBCP$ coincide?
Solution
$ABCD$ is a square
$AB=BC$
In a square all sides are equal
Given that $KBPD$ is a kite
$BK=BP$
In a kite two pairs of adjacent sides are equal
$\sphericalangle C=\sphericalangle A$
Equal angles in a square are $90°$ degrees
$\sphericalangle BPC=\sphericalangle BKA$
The angles of the kite are equal $\sphericalangle K=\sphericalangle P$
therefore $180° -\sphericalangle K=180° -\sphericalangle P$
$\sphericalangle ABK=\sphericalangle PBC$
if two angles are equal then the third is also equal
The triangles are congruent according to $S.A.S$
$S.A.S$
## Review Questions
### What is a criterion in geometry?
In mathematics, a criterion is a guideline that allows us to determine certain characteristics depending on the case or topic being studied. In the case of geometry, it allows us to judge certain characteristics for given figures.
### What is the triangle congruence criterion?
There are $4$ triangle congruence criteria, which help us determine when two triangles are congruent, that is, they help us determine when two triangles have the same dimensions and corresponding angles, thus having the same shape and side measurements regardless of the orientation of these triangles.
### What is the ASA criterion?
We say that two triangles are congruent with the ASA criterion when two of their angles and a non-included side are congruent.
### What is the angle-side-angle criterion?
This criterion tells us that two triangles are congruent when two angles and the side between them are congruent.
### How do you determine the criterion of a triangle?
This depends on the corresponding angles and sides of both triangles.
#### For example, given two triangles:
If we have three corresponding sides of two triangles that are congruent, then we are talking about the SSS criterion.
If it is found that the two angles and the included side between the corresponding angles are congruent, we refer to the ASA criterion.
When observing that two of their angles and the non-included side are congruent respectively, then we are discussing the AAS criterion.
And finally, when two pairs of corresponding sides and the included angle between these sides are congruent, it will be the SAS criterion.
Based on this, we can determine which congruence criterion we are referring to and deduce whether or not they are congruent triangles.
Related Subjects |
RS Aggarwal Solutions Chapter 10 Quadratic Equation Exercise 10D Class 10 Maths
Chapter Name RS Aggarwal Chapter 10 Quadratic Equation Book Name RS Aggarwal Mathematics for Class 10 Other Exercises Exercise 10AExercise 10BExercise 10CExercise 10EExercise 10F Related Study NCERT Solutions for Class 10 Maths
Exercise 10D Solutions
1. Find the nature of the roots of the following quadratic equations:
(i) 2x2 – 8x + 5 = 0
(ii) 3x2 - 2√6x + 2 = 0
(iii) 5x2 – 4x + 1 = 0
(iv) 5x(x – 2) + 6 = 0
(iv) 5x(x – 2) + 6 = 0
(v) 12x2 + 4√15 + 5 = 0
(vi) x2 – x + 2 = 0
Solution
(i) The given equation is 2x2 – 8x + 5 = 0
This is of the form ax2 + bx + c = 0, where a = 2, b = - 8 and c = 5.
∴ Discriminant, D = b2 – 4ac = (-8)2 – 4 × 2 × 5 = 64 – 40
= 24 > 0
Hence, the given equation has real and unequal roots.
(ii) The given equation is 3x2 - 2√6x + 2 = 0
This is of the form ax2 + bx + c = 0, where a = 3, b = -2√6 and c = 2
∴ Discriminant, D = b2 – 4ac
= (-2√6)2 – 4 × 3 × 2
= 24 – 24
= 0
Hence, the given equation has real and equal roots.
(iii) The given equation is 5x2 – 4x + 51 = 0.
This is of the form ax2 + bx + c = 0, where a = 5, b = - 4 and c = 1.
∴ Discriminant,
D = b2 – 4ac
= (-4)2 – 4 × 5 × 1’
= 16 – 20
= - 4 < 0
Hence, the given equation has no real roots.
(iv) The given equation is
5x(x – 2) + 6 = 0
⇒ 5x2 – 10x + 6 = 0
This is of the form ax2 + bx + c = 0, where a = 5, b = - 10 and c = 6.
∴ Discriminant. D = b2 – 4ac
= (-10)2 – 4 × 5 × 6
= 100 – 120
= -20 < 0
Hence, the given equation has no real roots.
(v)
(vi) The given equation is x2 – 2x + 2 = 0
This is of the form ax2 + bx + c = 0, where a = 1, b = - 1 and c = 2.
∴ Discriminant, D = b2 – 4ac = {(-1)2 – 4 × 1 × 2}
= 1 – 8
= -7 < 0
Hence, the given equation has no real roots.
2. If a and b are distinct real numbers, show that the quadratic equations 2(a2 + b2)x2 + 2(a + b)x + 1 = 0. has no real roots.
Solution
The given equation is 2(a2 + b2)x2 + 2(a + b)x + 1 = 0.
∴ D = [2(a + b)]2 – 4 × 2(a2 + b2) × 1
= 4(a2 + 2ab + b2) – 8(a2 + b2)
= 4a2 + 8ab + 4b2 – 8a2 – 8b2
= -4a2 + 8ab – 4b2
= -4(a2 – 2ab + b2)
= -4(a – b)2 < 0
Hence, the given equation has no real roots.
3. Show that the roots of the equation x2 + px – q2 = 0 are real for all real values of p and q.
Solution
x2 + px – q2 = 0
Here,
a = 1, b = p and c = - q2
Discriminant D is given by:
D = (b2 – 4ac)
= p2 - 4 × 1 × (-q)2
= (p2 + 4q2) > 0
D > 0 for all real values of p and q.
Thus, the roots of the equation are real.
4. For what values of k are the roots of the quadratic equation 3x2 + 2kx + 27 = 0 real and equal ?
Solution
3x2 + 2kx + 27 = 0
Here,
a = 3, b = 2k and c = 27
It is given that the roots of the equation are real and equal; therefore, we have:
D = 0
⇒ (2k)2 – 4 × 3 × 27 = 0
⇒ 4k2 – 324 = 0
⇒ 4k2 = 324
⇒ k2 = 81
⇒ k2 = ± 9
∴ k = 9 or k = -9
5. For what value of k are the roots of the quadratic equation kx(x - 2√5) + 10 = 0 real and equal.
Solution
The given equation is
kx(x - 2√5) + 10 = 0
⇒ kx2 - 2√5kx + 10 = 0
This is of the form ax2 + bx + c = 0, where a = k, b = -2√5k and c = 10.
∴ D = b2 – 4ac
= (-2√5k)2 – 4 × k × 10
= 20k2 – 40k
The given equation will have real and equal roots if D = 0.
∴ 20k2 – 40k = 0
⇒ 20k(k – 2) = 0
⇒ k = 0 or k – 2 = 0
⇒ k = 0 or k = 2
But, for k = 0, we get 10 = 0, which is not true
Hence, 2 is the required value of k.
6. For what values of p are the roots of the equation 4x2 + px + 3 = 0. real and equal?
Solution
The given equation is 4x2 + px + 3 = 0
This is of the form ax2 + bx + c = 0, where a = 4, b = p and c = 3.
∴ D = b2 – 4ac
= p2 – 4 × 4 × 3
= p2 – 48
The given equation will have real and equal roots if D = 0.
∴ p – 48 = 0
⇒ p2 = 48
Hence, 4√3 and -4√3 are the required values of p.
7. Find the nonzero value of k for which the roots of the quadratic equation 9x2 – 3kx + k = 0.
Solution
The given equation is 9x2 – 3kx + k = 0.
This is of the form ax2 + bx + c = 0, where a = 9, b = - 3k and c = k.
∴ D = b2 – 4ac
= (-3k)2 – 4 × 9 × k
= 9k2 – 36k
The given equation will have real and equal roots is D = 0.
∴ 9k2 – 36k = 0
⇒ 9k(k – 4) = 0
⇒ k = 0 or k – 4 = 0
⇒ k = 0 or k = 4
But, k ≠ 0 (Given)
Hence, the required values of k is 4.
8Find the values of k for which the quadratic equation (3k + 1)x2 + 2(k + 1)x + 1 = 0. has real and equal roots.
Solution
The given equation is (3k + 1)x2 + 2(k + 1)x + 1 = 0.
This is of the form ax2 + bx + c = 0, where a = 3k + 1, b = 2(k + 1) and c = 1.
∴ D = b2 – 4ac
= [2(k + 1)]2 – 4 × (3k + 1) × 1
= 4(k2 + 2k + 1) – 4(3k +1)
= 4k2 + 8k + 4 – 12k – 4
= 4k2 – 4k
The given equation will have real and equal roots if D = 0.
∴ 4k2 – 4k = 0
⇒ 4k(k – 1) = 0
⇒ k = 0 or k – 1 = 0
⇒ k = 0 or k = 1
Hence, 0 and 1 are the required values of k.
9. Find the value of p for which the quadratic equation (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0. has real and equal roots.
Solution
The given equation is (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0
This is of the form ax2 + bx + c = 0, where a = 2p + 1, b = -(7p + 2) and c = 7p – 3
∴ D = b2 – 4ac
= -[-(7p + 2)]2 – 4 × (2p + 1) × (7p – 3)
= (49p2 + 28p + 4) - 4(14p2 + p – 3)
= 49p2 + 28p + 4 – 56p2 – 4p + 12
= -7p2 + 24p + 16
The given equation will have real and equal roots if D = 0.
∴ -7p2 + 24p + 16 = 0
⇒ 7p2 – 24p – 16 = 0
⇒ 7p2 – 28p + 4p – 16 = 0
⇒ 7p(p – 4) + 4(p – 4) = 0
⇒ (p – 4)(7p + 4) = 0
⇒ p – 4 = 0 or 7p + 4 = 0
⇒ p = 4 or p = -(4/7)
Hence, 4 and –(4/7) are the required values of p.
10. Find the values of p for which the quadratic equation (p + 1)x2 – 6(p + 1)x + 3(x + 9) = 0., p ≠ - 1has equal roots. hence, find the roots of the equation.
Solution
The given equation is (p + 1)x2 – 6(p + 1)x + 3(p + 9) = 0
This is of the form ax2 + bx + c = 0, where a = p + 1, b = -6(p + 1) and c = 3(p + 9).
∴ D = b2 – 4ac
= [-6(p+ 1)]2 – 4 × (p + 1) × 3(p + 9)
= 12(p + 1)[3(p + 1) – (p + 9)]
= 12(p + 1)(2p – 6)
The given equation will have real and equal roots if D = 0.
∴ 12(p + 1)(2p – 6) = 0
⇒ p + 1 = 0 or 2p – 6 = 0
⇒ p = -1 or p = 3
But, p ≠ -1 (Given)
Thus, the value of p is 3
Putting p = 3, the given equation becomes 4x2 – 24x + 36 = 0
4x2 – 24x + 36 = 0
⇒ 4(x2 – 6x + 9) = 0
⇒ (x – 3)2 = 0
⇒ x – 3 = 0
⇒ x = 3
Hence, 3 is the repeated root of this equation.
11. If – 5 is a root of the quadratic equation 2x2 + px – 15 = 0. and the quadratic equation p(x2 + x) + k = 0 has equal
Solution
It is given that -5 is a root of the quadratic equation 2x2 + px - 15 = 0.
∴ 2(-5)2 + p × (-5) – 15 = 0
⇒ - 5p + 35 = 0
⇒ p = 7
The roots of the equation px2 + px + k = 0 = 0 are equal.
∴ D = 0
⇒ p2 – 4pk = 0
⇒ (7)2 – 4 × 7 × k = 0
⇒ 49 – 28k = 0
⇒ k = 49/28 = 7/4
Thus, the value of k is 7/4.
12. If 3 is a root of the quadratic equation x2 – x + k = 0., find the value of p so that the roots of the equation x2 + 2kx + (k2 + 2k + p) = 0 are equal.
Solution
It is given that 3 is a root of the quadratic equation x2 – x + k = 0
∴ (3)2 – 3 + k = 0
⇒ k + 6 = 0
⇒ k = - 6
The roots of the equation x2 + 2kx + (k2 + 2k + p) = 0 are equal.
∴ D = 0
⇒ (2k)2 – 4 × 1 × (k2 + 2k + p) = 0
⇒ 4k2 – 4k2 – 8k – 4p = 0
⇒ - 8k – 4p = 0
⇒ p = 8k/-4 = - 2k
⇒ p = 8k/-4 = -2k
⇒ p = -2 × (-6) = 12
Hence, the value of p is 12.
13. If -4 is a root of the equation x2 + 2x + 4p = 0. find the value of k for the which the quadratic equation x2 + px(1 + 3k) + 7(3 + 2k) = 0 has real roots.
Solution
It is given that -4 is a root of the quadratic equation x2 + 2x + 4op = 0
∴ (-4)2 + 2 × (-4) + 4p = 0
⇒ 16 – 8 + 4p = 0
⇒ 4p + 8 = 0
⇒ p = - 2
The equation x2 + px(1 + 3k) + 7(3 + 2k) = 0 has real roots.
∴ D = 0
⇒ [p(1 + 3k)]2 – 4 × 1 × 7(3 + 2k) = 0
⇒ [-2(1 + 3k)]2 – 28(3 + 2k) = 0
⇒ 4(1 + 6k + 9k2) – 28(3 + 2k) = 0
⇒ 4(1 + 6k + 9k2 – 21 – 14k) = 0
⇒ 9k2 – 8k – 20 = 0
⇒ 9k2 – 18k + 10k – 20 = 0
⇒ 9k(k – 2) + 10(k – 2) = 0
⇒ (k – 2)(9k + 10) = 0
⇒ k – 2 = 0 or 9k + 10 = 0
⇒ k = 2 or k = -(10/9)
Hence, the required value of k is 2 or –(10/9).
14. If the quadratic equation (1 + 4m2)x2/ + 2mcx + (c2 – a2) = 0 has equal roots, prove that c2 = a2 (1 + m2).
Solution
Given:
(1 + m2)x2 + 2mcx + (c2 – a2) = 0
Here,
a = (1 + m2), b = 2mc and c = (c2 – a2)
It is given that the roots of the equation are equal; therefore, we have:
D = 0
⇒ (b2 – 4ac) = 0
⇒ (2mc)2 – 4 × (1 + m2) × (c2 – a2) = 0
⇒ 4m2c2 – 4(c2 – a2 + m2c2 - m2a2) = 0
⇒ 4m2c2 – 4c2 + 4a2 – 4m2c2 + 4m2a2 = 0
⇒ - 4c2 + 4a2 + 4m2a2 =0
⇒ a2 + m2a2 = c2
⇒ a2(1 + m2) = c2
⇒ c2 = a2(1 + m2)
Hence proved.
15. If the roots of the quadratic equation (c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0 are real and equal, show that either a = 0 or (a3 + b3 + c3 = 3abc)
Solution
Given:
(c2 – ab)x2 – 2(a2 – bc)x + (b2 – ac) = 0
Here,
a = (c2 – ab), b = - 2(a2 – bc), c = (b2 – ac)
It is given that the roots of the equation are real and equal; therefore, we have:
D = 0
⇒ (b2 – 4ac) = 0
⇒ {-2(a2 – bc)}2 – 4 × (c2 – ab) × (b2 – ac) = 0
⇒ 4(a4 – 2a2bc + b2c2) – 4(b2c2 – ac3 – ab3 + a2bc) = 0
⇒ a4 – 2a2bc + b2c2 – b2c2 + ac3 + ab3 – a2bc = 0
⇒ a4 – 3a2bc + ac3 + ab3 = 0
⇒ a(a3 – 3abc + c3 + b3) = 0
Now,
a = 0 or a3 – 3abc + c3 + b3 = 0
a = 0 or a3 + b3 + c3 = 3abc
16. Find the value of p for which the quadratic equation 2x2 + px + 8 = 0 has real roots.
Solution
Given:
2x2 + px + 8 = 0
Here,
a = 2, b = p and c = 8
Discriminant D is given by
D = (b2 – 4ac)
= p2 – 4 × 2 × 8
= (p2 – 64)
If D ≥ 0, the roots of the equation will be real
⇒ (p2 – 64) ≥ 0
⇒ (p + 8)(p – 8) ≥ 0
⇒ p ≥ 8 and p ≤ - 8
Thus, the roots of the equation are real for p ≥ 8 and p ≤ -8.
17. Find the value of a for which the equation (α – 12)x2 + 2(α – 12)x + 2 = 0 has equal roots.
Solution
(α – 12)x2 + 2(α – 12x + 2 = 0
Here,
a = (α – 12), b = 2(α – 12) and c = 2
It is given that the roots of the equation are equal; therefore, we have
D = 0
⇒ (b2 – 4ac) = 0
⇒ {2(α – 12)}2 – 4 × (α – 12) × 2 = 0
⇒ 4(α2 - 24α + 144) - 8α + 96 = 0
⇒ 4α2 - 96α + 576 - 8α + 96 = 0
⇒ 4α2 - 104α + 672 = 0
⇒ α2 - 26α + 168 = 0
⇒ α2 - 14α - 12α + 168 = 0
⇒ α(α – 14) – 12(α – 14) = 0
⇒ (α – 14)(α – 12) = 0
∴ α = 14 or α = 12
If the value of α is 12, the given equation becomes non-quadratic.
Therefore, the value of α will be 14 for the equation to have equal roots.
18. Find the value of k for which the roots of 9x2 + 8kx + 16 = 0 are real and equal
Solution
Given:
9x2 + 8kx + 16 = 0
Here,
a = 9, b = 8k and c = 16
It is given that the roots of the equation are real and equal; therefore, we have:
D = 0
⇒ (b2 – 4ac) = 0
⇒ (8k)2 – 4 × 9 × 16 = 0
⇒ 64k2 – 576 = 0
⇒ 64k2 = 576
⇒ k2 = 9
⇒ k = ± 3
∴ k = 3 or k = -3
19. Find the values of k for which the given quadratic equation has real and distinct roots:
(i) kx2 + 6x + 1 = 0.
(ii) x2 – kx + 9 = 0
(iii) 9x2 + 3kx + 4 = 0
(iv) 5x2 – kx + 1 = 0
Solution
(i) The given equation is kx2 + 6x + 1 = 0.
∴ D = 62 – 4 × k × 1 = 36 – 4k
The given equation has real and distinct roots if D > 0.
∴ 36 – 4k > 0
⇒ 4k < 36
⇒ k < 9
(ii) The given equation is x2 – kx + 9 = 0
∴ D = (-k)2 – 4 × 1 × 9
= k2 – 36
The given equation has real and distinct roots if D > 0
∴ k2 – 36 > 0
⇒ (k – 6)(k + 6) > 0
⇒ k < -6 or k > 6
(iii) the given equation is 9x2 + 3kx + 4 = 0
∴ D = (3k)2 – 4 × 9 × 4
= 9k2 – 144
The given equation has real and distinct roots if D > 0.
∴ 9k2 – 144 > 0
⇒ 9(k2 – 16) > 0
⇒ (k – 4)(k + 4) > 0
⇒ k < - 4 or k > 4
(iv) The given equation is 5x2 – kx + 1 = 0.
∴ D = (-k)2 – 4 × 5 × 1
= k2 – 20
The given equation has real and distinct roots if D > 0.
∴ k2 – 20 > 0
⇒ k2 - (2√5)2 > 0
⇒ (k - 2√5)(k + 2√5) > 0
⇒ k = -2√5 or k > 2√5
20. If a and b are real and a ≠ b then show that the roots of the equation
(a – b)x2 + 5(a + b)x – 2(a – b) = 0 are equal and unequal.
Solution
The given equation is (a – b)x2 + 5(a + b)x – 2(a – b) = 0.
∴ D = [5(a + b)]2 – 4 × (a – b) × [-2(a – b)]
= 25(a + b)2 + 8(a – b)2
Since a and b are real and a ≠ b, so (a – b)2 > 0 and (a + b)2 > 0
∴ 8(a – b)2 > 0 ...(1) (Product of two positive numbers is always positive)
Also, 25(a + b)2 > 0 ...(2) (Product of two positive numbers is always positive)
Adding (1) and (2), we get
25(a + b)2 + 8(a – b)2 > 0 (Sum of two positive numbers is always positive)
⇒ D > 0
Hence, the roots of the given equation are real and unequal.
21. If the roots of the equation (a2 + b2)x2 – 2(ac + bd)x = (c2 + d2) = 0 are equal, prove that a/b = c/d.
Solution
It is given that the roots of the equation (a2 + b2)x2 - 2(ac + bd)x + (c2 + d2) = 0 are equal.
∴ D = 0
⇒ [-2(ac + bd)]2 – 4(a2 + b2)(c2 + d2) = 0
⇒ 4(a2c2 + b2d2 + 2abcd) – 4(a2c2 + a2d2 + b2c2 + b2d2) = 0
⇒ 4(a2c2 + b2d2 + 2abcd – a2c2 – a2d2 – b2c2 – b2d2) = 0
⇒ (-a2d2 + 2abcd – b2c2) = 0
⇒ -(a2d2 – 2abcd + b2c2) = 0
⇒ (ad – bc)2 = 0
⇒ ad – bc = 0
⇒ a/b = c/d
Hence proved.
22. If the roots of the equations ax2 + 2bx + c = 0 and bax2 - 2√acx + b = 0 are simultaneously real then prove that b2 = ac
Solution
It is given that the roots of the equation ax2 + 2bx + c = 0 are real.
∴ D1 = (2b)2 – 4 × a × c ≥ 0
⇒ 4(b2 – ac) ≥ 0
⇒ b2 – ac ≥ 0 ....(1)
⇒ 4(ac – b2) ≥ 0
⇒ - 4(b2 – ac) ≥ 0
⇒ b2 – ac ≥ 0 ...(2)
The roots of he given equations are simultaneously real if (1) and (2) holds true together.
this is possible if
b2 – ac = 0
⇒ b2 = ac |
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 2 (P2-9709/02) | Year 2005 | Oct-Nov | (P2-9709/02) | Q#4
Hits: 62
Question
The equation of the curve is .
i. Show that
ii. Find the equation of the tangent to the curve at the point (2, 4), giving your answer in the form ax+by=c.
Solution
i.
We are given that;
Therefore;
Rule for differentiation of is:
If and are functions of , and if , then;
If , then;
Rule for differentiation of is:
To find from an implicit equation, differentiate each term with respect to , using the chain rule to differentiate any function as .
ii.
We are required to find the equation of tangent to the curve at the point (2,4).
To find the equation of the line either we need coordinates of the two points on the line (Two-Point form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope form of Equation of Line).
We already have coordinates of the point on the curve (and tangent) as (2,4).
Next we need to find slope of tangent in order to write its equation.
The slope of a curve at a particular point is equal to the slope of the tangent to the curve at the same point;
Therefore, if we can find slope of the curve C at point then we can find slope of the tangent to the curve at this point.
Gradient (slope) of the curve at the particular point is the derivative of equation of the curve at that particular point.
Gradient (slope) of the curve at a particular point can be found by substituting x- coordinates of that point in the expression for gradient of the curve;
We are given equation of the curve implicitly;
We have found in (i) that;
Therefore;
Hence;
With coordinates of a point on the tangent and its slope in hand, we can write equation of the tangent.
Point-Slope form of the equation of the line is; |
# How do you find the product of (x+2)(x+2)?
Aug 6, 2016
${x}^{2} + 4 x + 4$
#### Explanation:
We must ensure that each term in the 2nd bracket is multiplied by each term in the 1st bracket.
This can be achieved as follows.
$\left(\textcolor{red}{x + 2}\right) \left(x + 2\right) = \textcolor{red}{x} \left(x + 2\right) + \textcolor{red}{2} \left(x + 2\right)$
distribute each pair of brackets.
$\textcolor{red}{x} \left(x + 2\right) + \textcolor{red}{2} \left(x + 2\right) = {x}^{2} + 2 x + 2 x + 4 = {x}^{2} + 4 x + 4$ |
# 7.1: Polar Necessities
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Trigonometry, Chapter 6, Lesson 4.
ID: 12558
Time Required: 15 minutes
## Activity Overview
Students will explore what is necessary to understand the calculus of polar equations. Students will graphically and algebraically find the slope of the tangent line at a point on a polar graph. Finding the area of a region of a polar curve will be determined using the area formula.
Topic: Polar Equations
• Find the slope of a polar equation at a particular point.
• Find the area of polar equation.
Teacher Preparation and Notes
• Make sure each students' calculator is in RADIANS (RAD) and POLAR (POL) in the MODE menu.
Associated Materials
## Plotting Coordinates & Exploring Polar Graphs
Students begin the activity by plotting points on a polar graph. This should be a refresher of polar coordinates for most students. Students practice using the calculator to graph a polar equation.
Discussion Questions
• What do you think it means to have a negative angle, like \begin{align*}\left(- \frac{\pi}{3}, 3 \right)\end{align*}?
• What about if r was negative? For example, move to \begin{align*}\left(\frac{\pi}{2}, -6 \right)\end{align*}.
## Solutions
1. See image below.
2. If \begin{align*}r(\theta) = \cos(\theta), r \left(\frac{\pi}{3} \right) = 0.5\end{align*}.
3. a heart or cardioid
4. A circle is in the form \begin{align*}r = a\end{align*}, where \begin{align*}a\end{align*} is a constant.
A polar rose with even petals is in the form \begin{align*}r(\theta) = a \cdot \sin(n \theta)\end{align*}, where \begin{align*}n\end{align*} is even.
A polar rose with odd petals is in the form \begin{align*}r(\theta) = a \cdot \sin(n \theta)\end{align*}, where \begin{align*}n\end{align*} is odd.
A limaçon with an inner loop comes form \begin{align*}r(\theta) = b + a \cdot \cos(\theta)\end{align*}, where \begin{align*}b < a\end{align*}.
### Notes/Highlights Having trouble? Report an issue.
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# 2018 AMC 10A Problems/Problem 12
The following problem is from both the 2018 AMC 12A #10 and 2018 AMC 10A #12, so both problems redirect to this page.
## Problem
How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?
$$x+3y=3$$
$$\big||x|-|y|\big|=1$$
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$
## Solutions
### Solution 1
We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this: $[asy] draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); [/asy]$ Now, it becomes clear that there are $\boxed{\textbf{(C) } 3}$ intersection points.
### Solution 2
$x+3y=3$ can be rewritten to $x=3-3y$. Substituting $3-3y$ for $x$ in the second equation will give $||3-3y|-y|=1$. Splitting this question into casework for the ranges of $y$ will give us the total number of solutions.
$\textbf{Case 1:}$ $y>1$: $3-3y$ will be negative so $|3-3y| = 3y-3.$ $|3y-3-y| = |2y-3| = 1$
Subcase 1: $y>\frac{3}{2}$
$2y-3$ is positive so $2y-3 = 1$ and $y = 2$ and $x = 3-3(2) = -3$
Subcase 2: $1
$2y-3$ is negative so $|2y-3| = 3-2y = 1$. $2y = 2$ and so there are no solutions ($y$ can't equal to $1$)
$\textbf{Case 2:}$ $y = 1$: It is fairly clear that $x = 0.$
$\textbf{Case 3:}$ $y<1$: $3-3y$ will be positive so $|3-3y-y| = |3-4y| = 1$
Subcase 1: $y>\frac{4}{3}$
$3-4y$ will be negative so $4y-3 = 1$ $\rightarrow$ $4y = 4$. There are no solutions (again, $y$ can't equal to $1$)
Subcase 2: $y<\frac{4}{3}$
$3-4y$ will be positive so $3-4y = 1$ $\rightarrow$ $4y = 2$. $y = \frac{1}{2}$ and $x = \frac{3}{2}$. Thus, the solutions are: $(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)$, and the answer is $\boxed{\textbf{(C) } 3}$.
### Solution 3
Note that $||x| - |y||$ can take on either of four values: $x + y$, $x - y$, $-x + y$, $-x -y$. Solving the equations (by elimination, either adding the two equations or subtracting), we obtain the three solutions: $(0, 1)$, $(-3,2)$, $(1.5, 0.5)$ so the answer is $\boxed{\textbf{(C) } 3}$. One of those equations overlap into $(0, 1)$ so there's only 3 solutions.
~trumpeter, ccx09 ~minor edit, XxHalo711
### Solution 4
Just as in solution $2$, we derive the equation $||3-3y|-|y||=1$. If we remove the absolute values, the equation collapses into four different possible values. $3-2y$, $3-4y$, $2y-3$, and $4y-3$, each equal to either $1$ or $-1$. Remember that if $P-Q=a$, then $Q-P=-a$. Because we have already taken $1$ and $-1$ into account, we can eliminate one of the conjugates of each pair, namely $3-2y$ and $2y-3$, and $3-4y$ and $4y-3$. Find the values of $y$ when $3-2y=1$, $3-2y=-1$, $3-4y=1$ and $3-4y=-1$. We see that $3-2y=1$ and $3-4y=-1$ give us the same value for $y$, so the answer is $\boxed{\textbf{(C) } 3}$
~Zeric Hang
### Solution 5
Just as in solution $2$, we derive the equation $x=3-3y$. Squaring both sides in the second equation gives $x^2+y^2-2|xy|=1$. Putting $x=3-3y$ and doing a little calculation gives $10y^2-18y+9-2|3y-3y^2|=1$. From here we know that $3y-3y^2$ is either positive or negative.
When positive, we get $2y^2-3y+1=0$ and then, $y=1/2$ or $y=1$. When negative, we get $y^2-3y+2=0$ and then, $y=2$ or $y=1$. Clearly, there are $3$ different pairs of values and that gives us $\boxed{\textbf{(C) } 3}$
~OlutosinNGA
### Solution 6(Desperate)
Since the absolute value is the square root of the square, we get that the first equation is quartic(degree $4$) and the other is linear. Subtract to get $\boxed{\textbf{(C) } 3}$.
~integralarefun
~savannahsolver |
# Dividing Integers
Integers are divided in the same fashion as whole numbers, except that certain rules should be applied.
Word of Caution. Remember, that we can't divide by 0.
Another interesting property is that color(red)(0/a=0) for any number a. For example, 0/5=0.
If you divide integers with different signs, i.e. one is positive and another is negative, then divide numbers ignoring minus and place minus in front of result.
Example 1. Find 86/(-2).
Ignore signs: 86/2=43. Since numbers have different signs then place minus in front of result: -43.
So, 86/(-2)=-43 .
Next example.
Example 2 . Find (-72)/3.
Ignore signs: 72/3=24. Since numbers have different signs then place minus in front of the result: -24.
So, (-72)/3=-24 .
• If you divide two positive numbers, you're actually dividing whole numbers.
• If you divide two negative numbers, multiply numbers ignoring minuses, i.e. color(green)((-a)/(-b)=a/b).
Example 3. Find 48/3.
48/3=16.
Another example.
Example 4. Find (-75)/(-5).
Ignore signs, because we divide numbers with same signs:
(-75)/(-5)=75/5=15.
So, (-75)/(-15)=5 .
Now, it's your turn. Take pen and paper and solve following problems.
Exercise 1. Find 12/(-3).
Next exercise.
Exercise 2. Find (-60)/4.
Next exercise.
Exercise 3. Find 90/5.
Exercise 4. Find (-1632)/(-24).
Exercise 5. Find 0/(-7). |
## Relations Between Roots Solution8
The page relations between roots solution8 is containing solution of some practice questions from the worksheet relationship between roots and coefficients.
(11) If one root of the equation 2 x² - a x + 64 = 0 is twice the other, then find the value of "a"
Solution:
Roots of any quadratic equation are α and β
here one root is twice the other
α = 2 β
by comparing the given equation with general form of quadratic equation we get a = 2 b = -a and c = 64
Sum of the roots α + β = -b/a
= -(-a)/2
= a/5
Product of roots α β = c/a
= 64/2
= 32
α + β = a/5
2 β + β = a/5
3 β = a/5
α β = 32
2 β (β) = 32
2 β² = 32
β² = 32/2
β² = 16
β= √16
β= √4 x 4
β= 4
3 β= a/5
3(4) = a/5
12 (5) = a
a = 60
(12) If α and β are the roots of 5 x² - p x + 1 = 0 and α - β = 1, then find p.
Solution:
Roots of any quadratic equation are α and β
by comparing the given equation with general form of quadratic equation we get a = 5 b = -p and c = 1
Sum of the roots α + β = -b/a
= -(-p)/5
= p/5
Product of roots α β = c/a
= 1/5
α - β = 1
α - β = √(α + β)² - 4 α β
(p/5)² - 4 (1/5) = 1
p²/25 - 4/5 = 1
(p² - 20)/25 = 1
(p² - 20) = 25
p² = 25 + 20
p² = 45
p = 3√5 |
# Help? Algebra 2 Math - Solve X,Y,Z
• MHB
In summary, the sum of three numbers is 95, with the second number being 5 more than the first, and the third number being 3 times the second. The numbers are x=15, y=20, and z=60.
The sum of three numbers is 95. The second number is 5 more than the first. The third number is 3 times the second. What are the numbers?
Begin by turning the word problem into a system of equations:
Let $$x + y + z = 95$$, $$y = x + 5$$, and $$z = 3y$$.
You can now use Elimination, Substitution, or Matrices to solve. I will use substitution by taking the 2nd and 3rd equations and getting the 1st equation in terms of $$y$$.
$$y = x+ 5$$ subtract $$5$$ from both sides.
$$x = y - 5$$
Substitute $$x = y - 5$$ and $$z = 3y$$ into $$x + y + z = 95$$:
$$( y -5 ) + y + (3y) = 95$$ combine like terms
$$5y - 5 = 95$$ add $$5$$ to both sides
$$5y = 100$$ divide by $$5$$ on both sides
$$y = 20$$
Substitute $$y = 20$$ into $$x = y - 5$$:
$$x = (20) - 5$$ simplify
$$x = 15$$
Substitute $$y = 20$$ into $$z = 3y$$:
$$z = 3(20)$$ simplify
$$z = 60$$
ANSWER: $$x = 15$$, $$y = 20$$, and $$z = 60$$
Beer soaked comment follows.
SquareOne said:
Begin by turning the word problem into a system of equations:
Let $$x + y + z = 95$$, $$y = x + 5$$, and $$z = 3y$$.
You can now use Elimination, Substitution, or Matrices to solve. I will use substitution by taking the 2nd and 3rd equations and getting the 1st equation in terms of $$y$$.
$$y = x+ 5$$ subtract $$5$$ from both sides.
$$x = y - 5$$
Substitute $$x = y - 5$$ and $$z = 3y$$ into $$x + y + z = 95$$:
$$( y -5 ) + y + (3y) = 95$$ combine like terms
$$5y - 5 = 95$$ add $$5$$ to both sides
$$5y = 100$$ divide by $$5$$ on both sides
$$y = 20$$
Substitute $$y = 20$$ into $$x = y - 5$$:
$$x = (20) - 5$$ simplify
$$x = 15$$
Substitute $$y = 20$$ into $$z = 3y$$:
$$z = 3(20)$$ simplify
$$z = 60$$
ANSWER: $$x = 15$$, $$y = 20$$, and $$z = 60$$
Prepare thyself for more questions.
Spoon feeding can be very addictive.
jonah said:
Prepare thyself for more questions.
Spoon feeding can be very addictive.
Spoon feeding can be very "additive." (Dance)
-Dan |
# Probability Solved Sums
There are 15 boys and 10 girls in the class. If three students are selected at random, what is the probability that 1 girl and 2 boys are selected?
A. 1/40
B. 1/ 2
C. 21/46
D. 7/ 41
E. None of these
The correct option is: C
Solution:
Total number of ways of selecting 3 students from 25 students = 25C3
Number of ways of selecting 1 girl and 2 boys = selecting 2 boys from 15 boys and 1 girl from 10 girls
⇒ Number of ways in which this can be done = 15C2 × 10C1
⇒ Required probability = (15C2 × 10C1)/ (25C3)
The formula for combinations is nCk = n! / (k! * (n – k)!), where “!” denotes factorial.
Let’s calculate the combinations step by step:
1. Calculate 15C2: 15C2 = 15! / (2! * (15 – 2)!) = 15! / (2! * 13!) = (15 * 14) / 2 = 210
2. Calculate 10C1: 10C1 = 10! / (1! * (10 – 1)!) = 10! / (1! * 9!) = 10
3. Calculate 25C3: 25C3 = 25! / (3! * (25 – 3)!) = 25! / (3! * 22!) = (25 * 24 * 23) / (3 * 2) = 2300
Now, let’s put it all together and solve the expression:
(15C2 × 10C1) / (25C3) = (210 × 10) / 2300 = 2100 / 2300 ≈ 0.91304348
Therefore, (15C2 × 10C1) / (25C3) is approximately 0.913.
Two friends Harish and Kalyan appeared for an exam. Let A be the event that Harish is selected and B is the event that Kalyan is selected. The probability of A is 2/5 and that of B is 3/7. Find the probability that both of them are selected.
A. 35/36
B. 5/35
C. 5/12
D. 6/35
E. None of these
The correct option is : D
Solution:
Given, A be the event that Harish is selected and
B is the event that Kalyan is selected.
P(A)= 2/5
P(B)=3/7
Let C be the event that both are selected.
P(C)=P(A)×P(B) as A and B are independent events:
P(C) = 2/5*3/7
P(C) =6/35
The probability that both of them are selected is 6/35
A card is drawn from a well-shuffled pack of 52 cards. What is the probability of getting a queen or club card?
A. 17/52
B. 15/52
C. 4/13
D. 3/13
E. None of these
The correct option is : C
Solution:
The probability of getting a queen card = 4/52
The probability of getting a club card = 13/52
The club card contains already a queen card, therefore the required probability is,
4/52 + 13/52 – 1/52 = 16/52 = 4/13
16 people shake hands with one another at a party. How many shaking hands took place?
A. 124
B. 120
C. 165
D. 150
E. None of these
The correct option is: B
Solution:
Total possible ways = 16C2
= 120
2 dice are thrown simultaneously. What is the probability that the sum of the numbers on the faces is divisible by either 3 or 5?
A. 7/36
B. 19/36
C. 9/36
D. 2/7
E. None of these
1. The correct option is: B
Solution:
Clearly n(s) = 6*6 = 36
Let E be the event that the sum of the numbers on the 2 faces is divisible by either 3 or 5. Then
E = {(1,2), (1,4), (1,5), (2,1), (2,3), (2,4), (3,2), (3,3), (3,6), (4,1), (4,2), (4,5), (4,6), (5,1), (5,4), (5,5), (6,3), (6,4), (6,6)}
n(E) = 19
Hence P(E) = n(E) / n(s)
= 19/ 36
Daniel speaks the truth in 2/5 cases and Sherin lies in 3/7 cases. What is the percentage of cases in which both Daniel and Sherin contradict each other in stating a fact?
A. 72.6%
B. 51.4%
C. 62.3%
D. 47.5%
E. None of these
The correct option is: B
Solution:
Daniel and Sherin will contradict each other when one speaks the truth and the other speaks lies.
The probability of Daniel speaking the truth and Sherin lies
=2/5*3/7
=6/35
The probability of Sherin speaking the truth and Daniel lies
=4/7*3/5
=12/35
The two probabilities are mutually exclusive.
Hence, probabilities that Daniel and Sherin contradict each other:
=6/35 +12/35
=18/35
=18/35*100
=51.4%
The names of 5 students from Section A, 6 students from Section B and 7 students from Section C were selected. The age of all 18 students was different. Again, one name was selected from them and it was found that it was of section B. What was the probability that it was the youngest student of the section B?
A. 1/18
B. 1/15
C. 1/6
D. 1/12
E. None of these
The correct option is: C
Solution:
The total number of students = 18
When 1 name was selected from 18 names, the probability that he was of section B
= 6 = 1
18 3
But from the question, there are 6 students from section B and the age of all 6 are different therefore, the probability of selecting one i.e. youngest student from 6 students will be 1/6
There are a total of 18 balls in a bag. Out of them 6 are red in colour, 4 are green in colour and 8 are blue in colour. If Vishal picks three balls randomly from the bag, then what will be the probability that all three balls are not of the same colour?
A. 95/102
B. 19/23
C. 21/26
D. 46/51
E. None of these
The correct option is : D
Solution:
Number of ways in which the person can pick three balls out of 18 balls
= 18C3 = 816
Number of ways of picking 3 balls of same colour = 6C3+ 4C3 + 8C3 = (20 + 4 + 56) = 80
Probability of picking three balls of the same color
= 80 = 5
816 51
Required probability = 1 – the probability of picking three balls of the same colour
= 1 – 5 = 46
Bag A contains 3 green and 7 blue balls. While bag B contains 10 green and 5 blue balls. If one ball is drawn from each bag, what is the probability that both are green?
A. 29/30
B. 1/5
C. 1/3
D. 1/30
E. None of these
1. The correct option is: B
Solution:
The required probability = 3C1/10C1 × 10C1/15C1
= 3/10 × 10/15 = 1/5
Ram and Shyam are playing chess together. Ram knows the two rows in which he has to put all the pieces in but he doesn’t know how to place them. What is the probability that he puts all the pieces in the right place?
A. 8!/16!
B. 8!/(2*15!)
C. 8!/15!
D. (2*8!)/16!
E. None of these
The correct option is: B
Solution:
Total boxes = 16
Total pieces = 16
Similar pieces = 8 pawns, 2 bishops, 2 rooks, 2 knights
Total ways of arranging these 16 pieces in 16 boxes
= 16! = 16!
(8! 2! 2! 2!) (8 × 8!)
Ways of correct arrangement = 1
Probability of correct arrangement = 1
(16! / (8 × 8!)
= (8 × 8!) = 8!
16! (2 × 15!)
To solve the expression 16! / (2 × 15!), we can simplify it step by step:
1. Calculate 15!: 15! = 15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
2. Multiply 2 by 15!: 2 × 15! = 2 × (15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)
3. Simplify 16! / (2 × 15!) = 16! / (2 × (15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1))
Now, let’s look at the factorial expressions:
• 16! = 16 × 15! (because 16! = 16 × 15!),
• 15! = 15 × 14! (because 15! = 15 × 14!),
• 14! = 14 × 13! (because 14! = 14 × 13!), and so on until 2! = 2 × 1 (because 2! = 2 × 1).
We can now rewrite the expression as:
16! / (2 × 15!) = (16 × 15!) / (2 × (15 × 14 × 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1))
Cancel out the common factors:
16! / (2 × 15!) = (16 × 15!) / (2 × 15!) = 16 / 2 = 8
So, 16! / (2 × 15!) equals 8.
MBA & BBA ADMISSION | MBA TUITION | BBA TUITION | ASSIGNMENT MCQ | PROJECT | DIGITAL MARKETING | EXECUTIVE COACHING | OTHER STUDY MATERIALS | STUDY MATERIALS
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# Write an equation in standard form for the line described perpendicular
You might use these, communicate as an example: Type an observation or a fiction. Find the Tenacity I had an introduction containing 45 problems, and I am drawn of 4, can you learned and see if the answers are able.
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Write the standard form of the beginning of each line given the slipe and y-intercept.
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Students graph two strategies and should make that parallel lines have the same basic. And this is the other of the line in point writing form if you wanna put it in life intercept form. Example: Find the equation of the line that is perpendicular to 2x = y - 5 and that passes through the point (4, -3).
Write the equation in standard form. Solution: Step 1: Find the slope of the equation given.
I want a line that is perpendicular to it, so I need to find the "negative reciprocal" of the other line's slope. For example, the line described at the top of this page is given via the equation 2x + y = 4.
It could also be described by y = - 2 x + 4, for example. This last form, y = - 2 x + 4, is called the slope-intercept form of the equation. Find the equation, in standard form of the line perpendicular to 2x-3y=-5 and passing through (3,-2) Write the equation in standard form with all integer coefficient.
Hi Kristy. I can show you how it's done with a similar problem, then you can follow those steps in solving your question. Perpendicular Line Calculator Find the equation of the perpendicular line step-by-step. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.
If you know the slope (m) any y-intercept (b) of a line, this page will show you how to find the equation of the line. The slope-intercept form is, where is the slope and is the y-intercept. Using the slope-intercept form, the slope is.
The equation of a perpendicular line to must have a slope that is the negative reciprocal of the original slope.
Write an equation in standard form for the line described perpendicular
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Physics
# 8.1Linear Momentum, Force, and Impulse
Physics8.1 Linear Momentum, Force, and Impulse
## Section Learning Objectives
By the end of this section, you will be able to do the following:
• Describe momentum, what can change momentum, impulse, and the impulse-momentum theorem
• Describe Newton’s second law in terms of momentum
• Solve problems using the impulse-momentum theorem
## Teacher Support
### Teacher Support
The learning objectives in this section will help your students master the following standards:
• (6) Science concepts. The student knows that changes occur within a physical system and applies the laws of conservation of energy and momentum. The student is expected to:
• (C) calculate the mechanical energy of, power generated within, impulse applied to, and momentum of a physical system.
## Section Key Terms
change in momentum impulse impulse–momentum theorem linear momentum
## Teacher Support
### Teacher Support
[BL][OL] Review inertia and Newton’s laws of motion.
[AL] Start a discussion about movement and collision. Using the example of football players, point out that both the mass and the velocity of an object are important considerations in determining the impact of collisions. The direction as well as the magnitude of velocity is very important.
## Momentum, Impulse, and the Impulse-Momentum Theorem
Linear momentum is the product of a system’s mass and its velocity. In equation form, linear momentum p is
$p=mv . p=mv .$
You can see from the equation that momentum is directly proportional to the object’s mass (m) and velocity (v). Therefore, the greater an object’s mass or the greater its velocity, the greater its momentum. A large, fast-moving object has greater momentum than a smaller, slower object.
Momentum is a vector and has the same direction as velocity v. Since mass is a scalar, when velocity is in a negative direction (i.e., opposite the direction of motion), the momentum will also be in a negative direction; and when velocity is in a positive direction, momentum will likewise be in a positive direction. The SI unit for momentum is kg m/s.
Momentum is so important for understanding motion that it was called the quantity of motion by physicists such as Newton. Force influences momentum, and we can rearrange Newton’s second law of motion to show the relationship between force and momentum.
Recall our study of Newton’s second law of motion (Fnet = ma). Newton actually stated his second law of motion in terms of momentum: The net external force equals the change in momentum of a system divided by the time over which it changes. The change in momentum is the difference between the final and initial values of momentum.
In equation form, this law is
$F net = Δp Δt , F net = Δp Δt ,$
where Fnet is the net external force, $Δp Δp$ is the change in momentum, and $Δt Δt$ is the change in time.
We can solve for $Δp Δp$ by rearranging the equation
$F net = Δp Δt F net = Δp Δt$
to be
$Δp= F net Δt . Δp= F net Δt .$
$F net Δt F net Δt$ is known as impulse and this equation is known as the impulse-momentum theorem. From the equation, we see that the impulse equals the average net external force multiplied by the time this force acts. It is equal to the change in momentum. The effect of a force on an object depends on how long it acts, as well as the strength of the force. Impulse is a useful concept because it quantifies the effect of a force. A very large force acting for a short time can have a great effect on the momentum of an object, such as the force of a racket hitting a tennis ball. A small force could cause the same change in momentum, but it would have to act for a much longer time.
## Teacher Support
### Teacher Support
[OL][AL] Explain that a large, fast-moving object has greater momentum than a smaller, slower object. This quality is called momentum.
[BL][OL] Review the equation of Newton’s second law of motion. Point out the two different equations for the law.
## Newton’s Second Law in Terms of Momentum
When Newton’s second law is expressed in terms of momentum, it can be used for solving problems where mass varies, since $Δp=Δ(mv) Δp=Δ(mv)$ . In the more traditional form of the law that you are used to working with, mass is assumed to be constant. In fact, this traditional form is a special case of the law, where mass is constant. $F net =ma F net =ma$ is actually derived from the equation:
$F net = Δp Δt F net = Δp Δt$
For the sake of understanding the relationship between Newton’s second law in its two forms, let’s recreate the derivation of $F net =ma F net =ma$ from
$F net = Δp Δt F net = Δp Δt$
by substituting the definitions of acceleration and momentum.
The change in momentum $Δp Δp$ is given by
$Δp=Δ(mv) . Δp=Δ(mv) .$
If the mass of the system is constant, then
$Δ(mv)=mΔv . Δ(mv)=mΔv .$
By substituting $mΔv mΔv$ for $Δp Δp$, Newton’s second law of motion becomes
$F net = Δp Δt = mΔv Δt F net = Δp Δt = mΔv Δt$
for a constant mass.
Because
$Δv Δt =a , Δv Δt =a ,$
we can substitute to get the familiar equation
$F net =ma F net =ma$
when the mass of the system is constant.
## Teacher Support
### Teacher Support
[BL][OL][AL] Show the two different forms of Newton’s second law and how one can be derived from the other.
## Tips For Success
We just showed how $F net =ma F net =ma$ applies only when the mass of the system is constant. An example of when this formula would not apply would be a moving rocket that burns enough fuel to significantly change the mass of the rocket. In this case, you can use Newton’s second law expressed in terms of momentum to account for the changing mass without having to know anything about the interaction force by the fuel on the rocket.
## Snap Lab
### Hand Movement and Impulse
In this activity you will experiment with different types of hand motions to gain an intuitive understanding of the relationship between force, time, and impulse.
• one ball
• one tub filled with water
Procedure:
1. Try catching a ball while giving with the ball, pulling your hands toward your body.
2. Next, try catching a ball while keeping your hands still.
3. Hit water in a tub with your full palm. Your full palm represents a swimmer doing a belly flop.
4. After the water has settled, hit the water again by diving your hand with your fingers first into the water. Your diving hand represents a swimmer doing a dive.
5. Explain what happens in each case and why.
What are some other examples of motions that impulse affects?
1. a football player colliding with another, or a car moving at a constant velocity
2. a car moving at a constant velocity, or an object moving in the projectile motion
3. a car moving at a constant velocity, or a racket hitting a ball
4. a football player colliding with another, or a racket hitting a ball
## Teacher Support
### Teacher Support
[OL][AL] Discuss the impact one feels when one falls or jumps. List the factors that affect this impact.
## Teacher Support
### Teacher Support
Talk about the different strategies to be used while solving problems. Make sure that students know the assumptions made in each equation regarding certain quantities being constant or some quantities being negligible.
## Worked Example
### Calculating Momentum: A Football Player and a Football
(a) Calculate the momentum of a 110 kg football player running at 8 m/s. (b) Compare the player’s momentum with the momentum of a 0.410 kg football thrown hard at a speed of 25 m/s.
## Strategy
No information is given about the direction of the football player or the football, so we can calculate only the magnitude of the momentum, p. (A symbol in italics represents magnitude.) In both parts of this example, the magnitude of momentum can be calculated directly from the definition of momentum:
$p=mv p=mv$
Discussion
Although the ball has greater velocity, the player has a much greater mass. Therefore, the momentum of the player is about 86 times greater than the momentum of the football.
## Worked Example
### Calculating Force: Venus Williams’ Racquet
During the 2007 French Open, Venus Williams (Figure 8.3) hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What was the average force exerted on the 0.057 kg tennis ball by Williams’ racquet? Assume that the ball’s speed just after impact was 58 m/s, the horizontal velocity before impact is negligible, and that the ball remained in contact with the racquet for 5 ms (milliseconds).
Figure 8.3 Venus Williams playing in the 2013 US Open (Edwin Martinez, Flickr)
## Strategy
Recall that Newton’s second law stated in terms of momentum is
$F net = Δp Δt . F net = Δp Δt .$
As noted above, when mass is constant, the change in momentum is given by
$Δp=mΔv=m( v f − v i ) , Δp=mΔv=m( v f − v i ) ,$
where vf is the final velocity and vi is the initial velocity. In this example, the velocity just after impact and the change in time are given, so after we solve for $Δp Δp$, we can use $F net = Δp Δt F net = Δp Δt$ to find the force.
Discussion
This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact. This problem could also be solved by first finding the acceleration and then using Fnet = ma, but we would have had to do one more step. In this case, using momentum was a shortcut.
## Practice Problems
1 .
A 5 kg bowling ball is rolled with a velocity of 10 m/s. What is its momentum?
1. $0.5\,\text{kg} \cdot \text{m/s}$
2. $2\,\text{kg} \cdot \text{m/s}$
3. $15\,\text{kg} \cdot \text{m/s}$
4. $50\,\text{kg} \cdot \text{m/s}$
2 .
(credit: modification of work from Pinterest)
A 155-g baseball is incoming at a velocity of 25 m/s. The batter hits the ball as shown in the image. The outgoing baseball has a velocity of 20 m/s at the angle shown.
What is the magnitudde of the impulse acting on the ball during the hit?
1. 2.68 kg⋅m/s.
2. 5.42 kg⋅m/s.
3. 6.05 kg⋅m/s.
4. 8.11 kg⋅m/s.
## Check Your Understanding
3.
What is linear momentum?
1. the sum of a system’s mass and its velocity
2. the ratio of a system’s mass to its velocity
3. the product of a system’s mass and its velocity
4. the product of a system’s moment of inertia and its velocity
4.
If an object’s mass is constant, what is its momentum proportional to?
1. Its velocity
2. Its weight
3. Its displacement
4. Its moment of inertia
5.
What is the equation for Newton’s second law of motion, in terms of mass, velocity, and time, when the mass of the system is constant?
1. $F net = Δv ΔmΔt F net = Δv ΔmΔt$
2. $F net = mΔt Δv F net = mΔt Δv$
3. $F net = mΔv Δt F net = mΔv Δt$
4. $F net = ΔmΔv Δt F net = ΔmΔv Δt$
6.
Give an example of a system whose mass is not constant.
1. A spinning top
2. A baseball flying through the air
3. A rocket launched from Earth
4. A block sliding on a frictionless inclined plane
## Teacher Support
### Teacher Support
Use the Check Your Understanding questions to assess whether students master the learning objectives of this section. If students are struggling with a specific objective, the assessment will help identify which objective is causing the problem and direct students to the relevant content.
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# Amazing Math Tricks to Astound, Confound (and Practice Mental Math)
Want to be a mind reader? Have x-ray vision? No problem - you’ll just need a little bit of math to get you going.
These math tricks are great to play on your little ones, and with a bit of practice (which will come in handy for strengthening their math muscle), they’ll be able to wow their friends too.
You probably know that the numbers opposite each other on a dice add up to 7 - you can use this knowledge in a trick to make your kids think you have x-ray vision, and also make them practice their 7 times table without them even realising it. First, get some dice...raid the Monopoly box if you have to!
Roll the dice and amaze your child by announcing the hidden face down numbers.
How?
Well, given that you know that the opposite faces always add up to 7, then you know that if the top side is a 6, the bottom is a 1, and if the top is a 3, that the bottom is a 4 and so on.
Once you’ve convinced your kids that you’ve got super powers, you can let them in on the secret and they can try out the trick themselves. A bit of practice at this is a great way of reinforcing number bonds to 7 for younger children.
The next stage in this trick is to do it with a stack of dice.
• Roll three dice and pile them on top of one another in a stack.
• Tell your audience that you’re going to use your x-ray vision to see the hidden sides of the dice in the stack, and add them all up.
• Because the opposite ends always add up to seven then this means that the top and bottom numbers on your stack of three will always add up to 21. (NB this works no matter how many dice you have - just multiply 7 by the number of dice.)
• Let’s say the number at the top of the stack is 4. All you have to do is subtract this from 21
• This means that the remaining top and bottom faces of the dice in the stack that are hidden will add up to 17.
• Reveal the hidden dice sides one by one, adding as you go to reach the number you’ve predicted.
This trick uses the 7 times table and lots of single digit addition, which is great for working the math muscle!
Always three
• Now ask them to double that number.
• Subtract 3.
• Divide by 2.
• And finally, subtract the original number.
This trick is a slow burner and a real party piece. You’ll need a good thick book and some paper and a pencil for this one. Go to page 108 of the book and find the ninth word – write this word onto some paper and place it in an envelope.
Explain to your audience that you have written a word on a piece of paper which has been placed inside a sealed envelope. The following trick will reveal the word. Hand the pen and paper to an audience member and ask them to do the following:
• Write down any three digit number.
• Directly underneath it, write the number with the digits reversed, eg 321 becomes 123
• Now take the smaller number away from the larger number eg 321-123 = 198
• Now reverse the digits of this number (891)
• Then add these last two numbers together, eg 198 + 891 = 1089
• Hand them the book, and ask them to go to the page indicated by the first three numbers (108)
• Once they’re on the page, tell them to go to the word indicated by the last number (9)
• When they read out the word, ask them to open the envelope and show the exact word to the rest of the amazed audience.
After a bit of practice, get your child to try it at a family gathering and everyone will love it - give it a miss at the 9 year old’s birthday party though as it doesn’t have quite the same wow factor as pulling a rabbit out of a hat.
About Komodo – Komodo is a fun and effective way to boost K-5 math skills. Designed for 5 to 11 year olds to use in the home, Komodo uses a little and often approach to learning math (15 minutes, three to five times per week) that fits into the busy family routine. Komodo helps users develop fluency and confidence in math – without keeping them at the screen for long.
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### How Wordle Can Help with Math Anxiety
Wordle, which you might think is simply a word game, is actually a mathematical puzzle. Yes, it may use words and letters instead of numbers, and a wide vocabulary doesn't hurt, but the logic, strategy, probability and elimination processes we use to solve it are actually math skills.
### Baking Math
Baking is great for putting maths into practice, and showing children how they can use maths skills in real life is a great motivator to keep them interested and keen to learn. |
# Kunihiko Chikaya's Inequality
### Solution 1
Under the constraints, the inequality is equivalent to
$\displaystyle \small{\frac{(a^{10}-b^{10})(b^{10}-c^{10})(a^{10}-c^{10})}{(a^{9}+b^{9})(b^{9}+c^{9})(c^{9}+a^{9})}\ge\frac{125}{3}[(a-c)^3-(a-b)^3-(b-c)^3]}.$
But
$\displaystyle (a-c)^3-(a-b)^3-(b-c)^3=3(a-b)(b-c)(a-c).$
Hence, suffice it to show that
$\displaystyle \frac{(a^{10}-b^{10})(b^{10}-c^{10})(a^{10}-c^{10})}{(a^{9}+b^{9})(b^{9}+c^{9})(c^{9}+a^{9})}\ge 125(a-b)(b-c)(a-c).$
By the AM-GM inequality, employed repeatedly, or by Muirhead's inequality, employed just once,
$\displaystyle \frac{u^{10}-v^{10}}{u-v}\le 5(u^9+v^9).$
Substituting $a,b,c$ in pairs and taking the product proves the required inequality.
### Solution 2
The $RHS$ simplifies to $125(a-b)(b-c)(c-a)$. If any two variables are equal, the inequality is trivially satisfied. Thus, we assume that no two variables are equal.
The constraint implies that both $LHS$ and $RHS$ are non-positive. Thus, we can negate both sides and flip the direction of the inequality. The inequality can then be written as
\begin{align} &&\left[5(a-b)(a^9+b^9)\right]\left[5(b-c)(b^9+c^9)\right]\left[5(a-c)(a^9+c^9)\right] \\ &&\geq (a^{10}-b^{10}) (b^{10}-c^{10}) (a^{10}-c^{10}). \end{align}
If we prove that for $x>y$,
$5(x-y)(x^9+y^9)\geq (x^{10}-y^{10}),$
we are done. Dropping the common positive factor $(x-y)$ the inequality becomes
\displaystyle \begin{align} &5(x^9+y^9)-\sum_{i=0}^{9}x^iy^{9-i}\geq 0 \\ &[(x^9+y^9)-(x^8y+y^8x)]+[(x^9+y^9)-(x^7y^2+y^7x^2)]+ \\ &[(x^9+y^9)-(x^6y^3+y^6x^3)]+ [(x^9+y^9)-(x^5y^4+y^5x^4)]\geq 0. \end{align}
Each box bracket is $\geq 0$ from Muirhead and the inequality follows.
### Solution 3
The $RHS$ simplifies to $125(a-b)(b-c)(c-a)$. If any two variables are equal, the inequality is trivially satisfied. Thus, we assume that no two variables are equal.
The constraint implies that both $LHS$ and $RHS$ are non-positive. Thus, we can negate both sides and flip the direction of the inequality. The inequality can then be written as
\begin{align} &&\left[5(a-b)(a^9+b^9)\right]\left[5(b-c)(b^9+c^9)\right]\left[5(a-c)(a^9+c^9)\right] \\ &&\geq (a^{10}-b^{10}) (b^{10}-c^{10}) (a^{10}-c^{10}). \end{align}
If we prove that for $x>y$,
$5(x-y)(x^9+y^9)\geq (x^{10}-y^{10}),$
we are done. Dropping the common positive factor $(x-y)$ the inequality becomes
$\displaystyle 5(x^9+y^9)-\sum_{i=0}^{9}x^iy^{9-i}\geq 0.$
and follows from
\displaystyle \begin{align} \sum_{i=0}^{9}x^iy^{9-i} &= \sum_{i=0}^{9}\left\{[x^9]^i[y^9]^{9-i}\right\}^{1/9} \leq\sum_{i=0}^{9} \frac{i\cdot x^9+(9-i)\cdot y^9}{9}~\text{(AM-GM)} \\ &= x^9\left(\frac{\displaystyle \sum_{i=0}^{9} i}{9}\right) + y^9\left(\frac{\displaystyle \sum_{i=0}^{9} (9-i)}{9}\right) \\ &= (x^9+y^9)\left(\frac{\displaystyle \sum_{i=0}^{9} i}{9}\right)= 5(x^9+y^9). \end{align}
### Solution 4
We can rewrite the rhs:
$rhs=\frac{125}{3} (-3 a^2 b+3 a^2 c+3 a b^2-3 a c^2-3 b^2 c+3 b c^2)$,
so:
$rhs=125 (a-b) (c-a) (b-c)$.
Since $\displaystyle \sum _{i=0}^{n-1} a^{n-i-1}b^i =\frac{a^n-b^n}{a-b}$ and with $n=10$,
$\displaystyle \sum _{i=0}^{n-1} a^{n-i-1}b^i\bigg|_{n=10} =\sum_{sym}a^5 b^4+\sum_{sym}a^6 b^3+\sum_{sym}a^7 b^2+\sum_{sym}a^8 b^1+\sum_{sym}a^9 b^0$
where $\displaystyle \sum_{sym}a^{p_1}b^{p_2}$ is the symmetric sum over all permutations $a^{p_1} b^{p_2}$.
So we need to prove that $a^{10}-b^{10}\geq 5 (a^9+b^9)(a-b)$ (cycl.)
$\displaystyle 5 \sum_{sym}a^9 b^0=5 (a^9 + b^9)\geq \sum_{sym}a^5 b^4+\sum_{sym}a^6 b^3+\sum_{sym}a^7 b^2+\sum_{sym}a^8 b^1+\sum_{sym}a^9 b^0$
We note for the nonpositive case $(c^{10}-a^{10})$, that it cancels out because $c-a$ is also nonpositive.
### Solution 5
Multiplying both sides by $\displaystyle \underset {cycl} \sum a$, we write the initial equation as:
$4 \mathfrak{M}_{a,b,c,d}^{\{1,1,1,0\}}-8 \mathfrak{M}_{a,b,c,d}^{\{2,1,0,0\}}+4 \mathfrak{M}_{a,b,c,d}^{\{3,0,0,0\}} \geq0$
where $\mathfrak{M}_{a,b,c,d}^{\sigma=\{\sigma_1,\sigma_2,\sigma_3,\sigma_4\}}$ is the mean across all permutations $\sigma$. Less elegantly:
$\small{ (a b c+a b d+a c d+b c d)\\ -8 \left(\frac{a^2 b}{12}+\frac{a^2 c}{12}+\frac{a^2 d}{12}+\frac{a b^2}{12}+\frac{a c^2}{12}+\frac{a d^2}{12}+\frac{b^2 c}{12}+\frac{b^2 d}{12}+\frac{b c^2}{12}+\frac{b d^2}{12}+\frac{c^2 d}{12}+\frac{c d^2}{12}\right)\\ +\left(a^3+b^3+c^3+d^3\right) }$
We have from a generalization of Schur for 4 variables:
$(a-b) (a-c) (a-d) a^t+(b-a) (b-c) (b-d) b^t\\ +(c-a) (c-b) (c-d) c^t+(d-a) (d-b) (d-c) d^t\geq 0$
Which we can write the inequality using the $\mathfrak{M}$ notation as (for $t=0$):
$8 \mathfrak{M}_{a,b,c,d}^{\{1,1,1,0\}}-12 \mathfrak{M}_{a,b,c,d}^{\{2,1,0,0\}}+4 \mathfrak{M}_{a,b,c,d}^{\{3,0,0,0\}}\geq0$
and by Miurhead's inequality, since $\{2,1,0,0\}$ majorizes $\{1,1,1,0\}$, we have:
$\mathfrak{M}_{a,b,c,d}^{\{2,1,0,0\}} \geq \mathfrak{M}_{a,b,c,d}^{\{1,1,1,0\}}$
which allows the final proof.
### Solution 5'
Multiplying both sides by $\displaystyle \underset {cycl} \sum a$, we write the initial equation as:
$4 \mathfrak{M}_{a,b,c,d}^{\{1,1,1,0\}}-8 \mathfrak{M}_{a,b,c,d}^{\{2,1,0,0\}}+4 \mathfrak{M}_{a,b,c,d}^{\{3,0,0,0\}} \geq0$
where $\mathfrak{M}_{a,b,c,d}^{\sigma=\{\sigma_1,\sigma_2,\sigma_3,\sigma_4\}}$ is the mean across all permutations $\sigma$. Less elegantly:
$\small{ (a b c+a b d+a c d+b c d)\\ -8 \left(\frac{a^2 b}{12}+\frac{a^2 c}{12}+\frac{a^2 d}{12}+\frac{a b^2}{12}+\frac{a c^2}{12}+\frac{a d^2}{12}+\frac{b^2 c}{12}+\frac{b^2 d}{12}+\frac{b c^2}{12}+\frac{b d^2}{12}+\frac{c^2 d}{12}+\frac{c d^2}{12}\right)\\ +\left(a^3+b^3+c^3+d^3\right) }$
Using Shur's inequality:
$S_{x,y,z}^t=x^t (x-y) (x-z)+y^t (y-x) (y-z)+z^t (z-x) (z-y) \geq 0$
for $t=1$, and cycling:
$\displaystyle \underset {cycl} \sum S_{a,b,c}^t \geq 0$
which can be written as:
$\displaystyle 4 \mathfrak{M}_{a,b,c,d}^{\{1,1,1,0\}}-8 \mathfrak{M}_{a,b,c,d}^{\{2,1,0,0\}}+4 \mathfrak{M}_{a,b,c,d}^{\{3,0,0,0\}} \geq 0.$
### Acknowledgment
Leo Giugiuc has kindly communicated to me the above problem by Kunihiko Chikaya, along with a solution of his. Solutions 2 and 3 are by Amit Itagi that make the references to Muirhead's and the AM-GM inequalities more explicit. Solutions 4, 5 and 5' are by N. N. Taleb |
# CLASS-6PIE CHART - PROBLEM & SOLUTION
PIE CHART - PROBLEM & SOLUTION -
Example.1) Observe the following pie chart that represents the money spent by Ana at the funfair. The indicated color shows the amount spent on each category. The total value of the data is 20 and the amount spent on each category is interpreted as follows:
· Ice Cream - 4
· Toffees - 4
· Popcorn - 2
· Rides - 10
Ans.) To convert this into pie chart percentage, we apply the formula:-
(Frequency ÷ Total Frequency) × 100
Let us convert the above data into a percentage.
Amount spent on rides: (10/20) × 100 = 50%
Amount spent on toffees: (4/20) × 100 = 20%
Amount spent on popcorn: (2/20) × 100 = 10%
Amount spent on ice-cream: (4/20) × 100 = 20% (Ans.)
Example.2) Observe the following pie chart that recommends a low-carb diet on a day.
Ans.) We measure the angles of each slice. We get that Protein measures 180°, Carb measures 108°, and Fats measures 72°.
To find the percentage, we divide each angle by 360 and multiply it by 100.
Protein = (180/360) × 100 = 50%
Carb = (108/360) × 100 = 30%
Fats = (72/360) × 100 = 20%
Example.3) The pie chart shown below shows the percentages of types of transportation used by 500 students to come to school. With this given information, answer the following questions:
a) How many students come to school by bicycle?
b) How many students do not walk to school?
c) How many students come to school by bus and car?
Ans.)
a) The students who come by bicycle = 25%; (25/100) × 500 = 25 × 5 = 125.
b) The students who do not walk to school- We need to add the values of all the remaining means, i.e., bus + car + bicycle = 26 + 32 + 25 = 83.
Hence, (83/100) × 500 = 83 × 5 = 415 students do not walk to school.
c) The students who come by bus and car [(32 + 26)/100] × 500 = 58 × 5 = 290.
Example.4) The following chart shows the various activities done by Diana in a week.
a) Calculate the central angle subtended at sleeping.
b) Find the portion of time spent by Diana at school.
c) Find the central angle subtended in playing.
Ans.)
a) Time spent in sleeping = 34%; (34/100) × 360 = 122.4°. Therefore, the central angle subtended at sleeping = 122.4°.
b) Time spent at school = 25%; 25/100 = 1/4. Therefore, she spends 1/4th of her time in school.
c) Time spent on playing = 8%; (8/100) × 360 = 28.8°. Therefore, the central angle subtended at playing = 28.8°.
Example.5) The pie chart shows the favorite subjects of students in a class. Using the information given in the pie chart, find the percentage of students who chose English.
Ans.) Let's first determine the percentage of students who chose English by looking at the pie chart.
We know that 144° + 36° + 72° + 108°= 360°
The percentage of students who chose English:- (72/360) × 100 = 20 Therefore, the percentage of students who chose English = 20% (Ans.)
Example.6) A pie chart is divided into 3 parts with the angles measuring as x, 4x, and 5x respectively. Find the value of x in degrees.
Ans.)
We know, the sum of all angles in a pie chart would give 360º as result.
x + 4x + 5x = 360º
⇒ 10 x = 360º
⇒ x = 360º/10
⇒ x = 36º
Therefore, the value of x is 36º. |
#### Perform division of polynomials
• Find the quotient and the remainder of $$(3x^2 – 6x + 2) ÷ (x – 1)$$
• $$\begin{array}{c}\hspace{3cm}{\rm{ }}3x{\rm{ }} - 3\\x - 1\left){\vphantom{1\begin{array}{l}3{x^2} - 6x + 2\\\underline {3{x^2} - 3x{\rm{ }}} \\{\rm{ }} - 3x + 2\\{\rm{ }}\underline { - 3x + 3} \\{\rm{ }} - 1\end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}3{x^2} - 6x + 2\\\underline {3{x^2} - 3x{\rm{ }}} \\{\rm{ }} - 3x + 2\\{\rm{ }}\underline { - 3x + 3} \\{\rm{ }} - 1\end{array}}}\end{array}$$
∴ Quotient $$= \underline{\underline {3x - 3}}$$
Remainder $$= \underline{\underline { - 1}}$$
#### Understand the remainder theorem
• For any real value function f(x) divided by $$ax-b$$, the remainder, r, is given by $$f(\frac{b}{a})$$
• Consider $$f(x) = 2{x^3} + 6{x^2} + 2x - 7$$.
(a) Find the values of f(1) and f(–1).
(b) Hence, find the remainder of f(x) ÷ (x + 1).
• $$\begin{array}{1}a)\qquad f(1) \\= 2{(1)^3} + 6{(1)^2} + 2(1) - 7\\= 2 + 6 + 2 - 7\\= \underline{\underline 3} \end{array}$$
$$\begin{array}{1}f( - 1) \\ = 2{( - 1)^3} + 6{( - 1)^2} + 2( - 1) - 7\\ = - 2 + 6 - 2 - 7\\ = \underline{\underline { - 5}} \end{array}$$
(b) By the remainder theorem, we have
remainder = f(–1)
= –5
#### Understand the factor theorem
• For any real value function f(x), $$ax-b$$ is a linear factor if $$f(\frac{b}{a})=0$$
• Let $$h(x) = 6{x^3} - 4{x^2} - 9x + 6$$.
Determine whether each of the following is a factor of h(x) by using the factor theorem.
(a) $$x - 1$$
(b) $$2x + 3$$
• $$\begin{array}{1}a)\qquad h(1)\\ = 6{(1)^3} - 4{(1)^2} - 9(1) + 6\\ = - 1\\ \ne 0\end{array}$$
$$\therefore x -1$$ is not a factor of h(x).
$$\begin{array}{1}b)\qquad h( - \frac{3}{2}) \\= 6{( - \frac{3}{2})^3} - 4{( - \frac{3}{2})^2} - 9( - \frac{3}{2}) + 6\\ = - \frac{{39}}{4}\\ \ne 0\end{array}$$
$$\therefore 2x+ 3$$ is not a factor of h(x). |
# What are the first and second derivatives of f(x)=(ln x)^(1/2)?
Apr 15, 2016
$f ' \left(x\right) = \frac{1}{2 x \sqrt{\ln x}}$
and
$f ' ' \left(x\right) = - \frac{1}{2 {x}^{2}} \left[\frac{1}{2 \sqrt{{\ln}^{3} x}} + \frac{1}{\sqrt{\ln x}}\right]$
#### Explanation:
To get the first derivative, we'll use the chain rule :
$f ' \left(x\right) = \frac{d f \left(x\right)}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} {\left(\ln x\right)}^{\frac{1}{2}} = \frac{1}{2} {\left(\ln x\right)}^{- \frac{1}{2}} \cdot \frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{2} {\left(\ln x\right)}^{- \frac{1}{2}} \cdot \frac{1}{x}$
the second derivative again uses first the product rule and then the chain rule:
$f ' ' \left(x\right) = \frac{d f ' \left(x\right)}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \frac{1}{2} {\left(\ln x\right)}^{- \frac{1}{2}} \cdot \frac{1}{x} = - \frac{1}{4} {\left(\ln x\right)}^{- \frac{3}{2}} \frac{d}{\mathrm{dx}} \left(\ln x\right) \cdot \frac{1}{x} + \frac{1}{2} {\left(\ln x\right)}^{- \frac{1}{2}} \left(- \frac{1}{x} ^ 2\right) = - \frac{1}{4} {\left(\ln x\right)}^{- \frac{3}{2}} \frac{1}{x} ^ 2 - \frac{1}{2} {\left(\ln x\right)}^{- \frac{1}{2}} \frac{1}{x} ^ 2$
We can also choose to write these two answers in simpler forms:
$f ' \left(x\right) = \frac{1}{2 x \sqrt{\ln x}}$
and
$f ' ' \left(x\right) = - \frac{1}{2 {x}^{2}} \left[\frac{1}{2 \sqrt{{\ln}^{3} x}} + \frac{1}{\sqrt{\ln x}}\right]$ |
Did you understand that 100 is the the smallest 3 number number? Multiples the 100 space the products when an essence is multiply by 100. Multiples the 100 room the easiest to learn. In this mini-lesson, we will certainly calculate the multiples the 100, and interesting facts around these multiples.
You are watching: What are the multiples of 100
First 5 multiples the 100: 100, 200, 300, 400 and also 500Prime administrate of 100: 100 = 22 × 52
Let us explore a little bit an ext about the multiples of 100 and also its properties of it.
1 What space the Multiples that 100? 2 First 20 Multiples that 100 3 Important Notes 4 FAQs ~ above Multiples that 100
## What room the Multiples of 100?
Multiples of 100 space the numbers that deserve to be split by 100 without any remainder. To create a perform of multiples of 100,
We an initial multiply 100 through 1 to gain the an initial multiple of 100 i m sorry is 100,Then us multiply 100 through 2 to get the 2nd multiple the 100 i m sorry is 200,Then we multiply 100 through 3 to get the 3rd multiple the 100 which is 300, and also so on.
## List of an initial 20 Multiples of 10
The first 10 multiples the 100 are derived by the outcomes of the times table 100.
100 × 1 100 100 × 2 200 100 × 3 300 100× 4 400 100 × 5 500 100 × 6 600 100 × 7 700 100 × 8 800 100 × 9 900 100 × 10 1000
The following 10 multiples of 100 as much as 20 time are:
100 × 11 1100 100 × 12 1200 100 × 13 1300 100× 14 1400 100 × 15 1500 100 × 16 1600 100 × 17 1700 100 × 18 1800 100 × 19 1900 100 × 20 2000
To understand the ide of finding multiples, let united state take a few more examples.
Important Notes:
When you main point a number by 100 you will get the number attached through two zeroes together the result. Eg: 100 × 2 = 200.If a number is ending with 00 then it is always divisible by 100.
## Multiples of 100 fixed Examples
Example 1: Aaron was given a list of numbers: 50, 200, 500, 750, 600, 1000, 1020. Have the right to you assist him uncover out multiples of 100?
Solution: We recognize that multipes of 100 end with dual zero. The numbers through zero a unit and tens place are 200, 500, 600, 1000. Hence, multipes the 100 room 200, 500, 600, and also 1000.
Example 2: Fabian needs to solve the following puzzle the was provided by a friend. Uncover the numbers which fit all these statements:
It is a typical multiple that 100 and also 80It is lie in between 1 and 1000
Solution: Common multiples the 100 and 80 space 400, 800, 1200 and also so on.The multiple of 100 and 80 lie between 1 and 1000 is 400 and 800. Hence, answer is 400 and also 800.
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## Interactive Questions
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## FAQs top top Multiples of 100
### How countless multiples walk 100 have?
The variety of multiples of any number is infinite. Eg: 100, 200, 300, 400, 500, 600, etc.
### What is the the smallest multiple of 100?
The the smallest multiple of 100 is 100 itself.
### How do you find multiples the 100?
By Multiplying any number with 100 us can discover multiples of 100.
See more: What Kind Of Drug Test Does Costco Do, Does Costco Drug Test In 2021
### What is the Least usual Multiple of 100 and also 150?
The least usual multiples of two numbers a and b can be calculated utilizing the formula: (a × b)/GCF(a, b). Here GCF(a,b) is the greatest usual factors the a and also b. GCF(100, 150) = 50LCM(100, 150) = (100 × 150)/GCF(100, 150) = 15000/50 = 300
### What are the very first five multiples the 100?
The very first five multiples that 100 are 100, 200, 300, 400 and 500. |
what is the definition of domain and range of a function
what is the definition of domain and range of a function
Free online calculator to find the domain and range of a function.It also shows plots of the function and illustrates the domain and range on a number line to enhance your mathematical intuition. These wont be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. Small sets of points are generally the simplest sorts of relations, so your book starts with those. We will now attempt to formalize our definition of function by providing three textbook definitions of the concept of function.3. What happens if the domain and range have the same number of elements? The domain of a function is the set of all its allowable inputs.Often, it is much easier to get the range from a graph of the function (which is the topic of a future section). In this exercise, you are only asked to find the range for very simple functions. Definition of a Composite Function.What are the domain and range of k h? Solution. Substitute the expression for h(x) into the expression for k. We are familiar with Domain of a Function and Range of a Function. But what does it mean? before diving deeper in to the topic, let us understand what is a function?Definition 2: The set of all possible values which qualify as inputs to a function, is known as the domain of the function. Definition Domain: The domain of a function is the set of all possible input values (usually x), which allows the function formula to work.While only a few types have limited domains, you will frequenty see functions with unusual ranges.
For finding domain and range of a function which consists of absolute value, one should use above properties.According to this definition a>0 is an enough condition for defining the exponential function if u(x) is a real valued function. What is the domain, codomain and range of R? Is this relation a function?Definition 6 A function which has either R or one of its subsets as its range is called a real valued function. Define the function of range and domain? Let the function be f(x) 1/(x-1) The domain is all allowable values for which the function can be defined. Here, except 1, any number would give the function a meaningful value. The plot of function the function definition is here: httpsAlso what would be the Domain Range of this function? Please also provide a method for finding it without plotting it.2. What is the range and domain of the following function. 1. Functions assign outputs to inputs. The domain of a function is the set of all possible inputs for the function.If youre behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked.
Learn what the domain and range mean, and how to determine the domain and range of a given function.When working with functions, we frequently come across two terms: domain range. What is a domain? Function: Identifying the Domain from an Equation Example 3: State the domain of Recall from the definition that each member of the domain must correspond to exactly one member (a real number) of the range. In math, the domain and range are algebraic values on the coordinate plane. Discover the definition of domains and ranges in math with tips from a What Is the Domain and Range of a Function?Finding the Domain and Range: Given a Function. In order to find the domain of a function, if it isnt stated to begin with, we need to look at the function definition to determine what values are not allowed. DEFINITION 2. Set Form of the Denition of a Function.If the domain and range of a function are innite sets, then the rule of correspondence cannot be displayed in a table, and it is not possible to actually list all the ordered pairs belonging to the function. Definition of a Function.You can use the vertical line test to test whether a relation is a function. Example. What are the domain and range of the relation graphed at the right? Graph the linear function What is the domain and the range of f?Mappings and functions C3 core mathematics key concepts: definition of a function domain range inverse function. The range of a function is the complete set of all possible resulting values of the dependent variable (y, usually), after we have substituted the domain. In plain English, the definition means In general, the definitions hold when the domain and codomain of a function are linearly ordered. Onto (Surjective) Functions. A function f:AB is onto or surjective or a surjection iff its range is equal to its codomain (bB, aA: f(a)b). More Articles. Definition of Table of Values. How to Find the Domain of a Function Defined by an Equation. How to Write Functions in Math.
How to Find the Range of a Square Root Function. Domain and Range. The domain of a function is the set of all possible input values. If the domain is not specified, we usually take it to be the set of all real numbers for which the function is defined. In order to figure out the domain of a function, youll have to look at the values thats possible (or that were allowed to use) for the independent variable.State the domain, range, and whether it is a function. Math 102 3.1 "Concept of a Function". Objectives: Dene function, domain, and range. Use function notation and evaluate functions. Find the domain and range of a function. Example: a simple function like f(x) x2 can have the domain (what goes in) of just the counting numbers 1,2,3, and the range will then be the set 1,4,9The Codomain is the set of values that could possibly come out. The Codomain is actually part of the definition of the function. What is the rule f ?Domain and range: As suggested by the figure, a function moves from X to Y via some operation (rule) f : We choose an element x in X, perform the operation (apply the rule), and get an element y in Y. Indeed, by the definition, we get exactly one element y in Y. [If you take an Update: Wow- to the person being rude to me and accusing me of not studying or paying attention to my course- Im double checking that my understanding of the definition is correct.For example, y1/x, the domain of this function is (-inf, 0) (0, inf) because 1/x isDomain is x and range is y. Definitions The domain of a relation is the set of all first coordinates.c. What are the domain and range? d. Is this relation a function? Certain functions have defined domains and range.For example, if you have the function f(x)1/(x2 9), you can exclude any values of x (the domain) that make the denominator equal to zero (because division by zero is not defined). (1 is paired with 2, and 2 is paired with 2.) This does not violate the definition of a function, since the first components (x-values) are different and each is paired with only one second componentThis is what makes functions so important in applications. OBJECTIVE 3 Find the domain and range. What is the domain and range of a relation? Show Answer.On the other hand, relation 2 has TWO distinct y values a and c for the same x value of 5 . Therefore, relation 2 does not satisfy the definition of a mathematical function. However, if one element in the domain maps to two elements in the range, then we no longer have a function.Therefore, our definition of a surjection says that all elements x in the domain map to all elements y in the codomain at least once. The range of f is the set of all values that the function takes when x takes the values given in the domain.Domain : (-, ). Since this function is a quadratic function, its graph will be a parabola which will open upwards since the leading coefficient is positive. Definition of a Composite Function.What are the domain and range of k h? Solution. Substitute the expression for h(x) into the expression for k. In these notes we will cover various aspects of functions. We will look at the denition of a function, the domain and range of a function, and what we mean by specifying the domain of a function. 1.1 What is a function? In taking both domain and range into account, a function is any mathematical formula that produces one and only one result for each input.A relation is a definition where one item in the definitions domain maps to more than one item in the definitions range. We use the terms domain and range Before we tackle the idea of the domain and range of a function, we first need to spend a few minutes talking about how to picture what any particular function does. One way is to simply plug a few numbers into the function and see what output it gives you. The set of possible y-values is called the range. If you want to know how to find the domain of a function in a variety of situations, just follow these steps.Learn the definition of the domain. Definition. As mentioned earlier, there are chances that the output of a function may not exist in some cases for a given input or a set of inputs.Example 1: What is the domain and range of f(x) -x2 6x 4. Solution: The given function is a quadratic function, a type of polynomial functions. 1 Graphing Piecewise Functions. 2 Example. Graph the piecewise (0,5) defined function What is the domain of the function?Picture of a Function Definition of Function Illustration There is a domain, a range, and a rule. Here we will discuss about domain, co-domain and range of function. Let : A B (f be function from A to B), then. Set A is known as the domain of the function f. The range of the function: All possible output values. What is Domain and Range ?definition of domain and range of a function. Functions: Domain and Range. We apologize but this resource is not available to you.The review should include a precise definition of a function: a rule that assigns to each input exactly one output. 9. (Philosophy) philosophy range of significance (esp in the phrase domain of definition).11. (Computer Science) computing a group of computers, functioning and administered as a unit, that are identified by sharing the same domain name on the internet. 5.1 - Introduction to functionsThe argument and value of a functionIdentifying the domain and range of a functionIf it could, that inverse would be one-to-many and this would violate the definition of a function. A step by step tutorial, with detailed solutions, on how to find the domain and range of real valued functions is presented. First the definitions of these two concepts are presented. range and domain is determined by hit and trial method. you have to put different values of a variable in the function and check wheather the value of the function is defined or not. Function Notation and Linear Functions. With the definition of a function comes special notation. If we consider each x-value to be the input that produces exactly one output, then we can use the notation.89. Is a vertical line a function? What are the domain and range of a vertical line? In mathematics, and more specifically in naive set theory, the domain of definition (or simply the domain) of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or value for each member of the domain. |
# 1.2 Prime factorization
Page 1 / 1
This module is from Elementary Algebra by Denny Burzynski and Wade Ellis, Jr. This chapter contains many examples of arithmetic techniques that are used directly or indirectly in algebra. Since the chapter is intended as a review, the problem-solving techniques are presented without being developed. Therefore, no work space is provided, nor does the chapter contain all of the pedagogical features of the text. As a review, this chapter can be assigned at the discretion of the instructor and can also be a valuable reference tool for the student.
## Overview
• Prime And Composite Numbers
• The Fundamental Principle Of Arithmetic
• The Prime Factorization Of A Whole Number
## Prime and composite numbers
Notice that the only factors of 7 are 1 and 7 itself, and that the only factors of 23 are 1 and 23 itself.
## Prime number
A whole number greater than 1 whose only whole number factors are itself and 1 is called a prime number.
The first seven prime numbers are
2, 3, 5, 7, 11, 13, and 17
The number 1 is not considered to be a prime number, and the number 2 is the first and only even prime number.
Many numbers have factors other than themselves and 1. For example, the factors of 28 are 1, 2, 4, 7, 14, and 28 (since each of these whole numbers and only these whole numbers divide into 28 without a remainder).
## Composite numbers
A whole number that is composed of factors other than itself and 1 is called a composite number. Composite numbers are not prime numbers.
Some composite numbers are 4, 6, 8, 10, 12, and 15.
## The fundamental principle of arithmetic
Prime numbers are very important in the study of mathematics. We will use them soon in our study of fractions. We will now, however, be introduced to an important mathematical principle.
## The fundamental principle of arithmetic
Except for the order of the factors, every whole number, other than 1, can be factored in one and only one way as a product of prime numbers.
## Prime factorization
When a number is factored so that all its factors are prime numbers, the factorization is called the prime factorization of the number.
## Sample set a
Find the prime factorization of 10.
$10=2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5$
Both 2 and 5 are prime numbers. Thus, $2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5$ is the prime factorization of 10.
Find the prime factorization of 60.
$\begin{array}{lllll}60\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}30\hfill & \hfill & \text{30\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}prime}.\text{\hspace{0.17em}}30=2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}15\hfill \\ \hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}15\hfill & \hfill & 15\text{\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}prime}.\text{\hspace{0.17em}}15=3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill \\ \hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill & \hfill & \text{We'll\hspace{0.17em}use\hspace{0.17em}exponents}\text{.\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2={2}^{2}\text{\hspace{0.17em}}\hfill \\ \hfill & =\hfill & {2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill & \hfill & \hfill \end{array}$
The numbers 2, 3, and 5 are all primes. Thus, $2^{2}·3·5$ is the prime factorization of 60.
Find the prime factorization of 11.
11 is a prime number. Prime factorization applies only to composite numbers.
## The prime factorization of a whole number
The following method provides a way of finding the prime factorization of a whole number. The examples that follow will use the method and make it more clear.
1. Divide the number repeatedly by the smallest prime number that will divide into the number without a remainder.
2. When the prime number used in step 1 no longer divides into the given number without a remainder, repeat the process with the next largest prime number.
3. Continue this process until the quotient is 1.
4. The prime factorization of the given number is the product of all these prime divisors.
## Sample set b
Find the prime factorization of 60.
Since 60 is an even number, it is divisible by 2. We will repeatedly divide by 2 until we no longer can (when we start getting a remainder). We shall divide in the following way.
$\begin{array}{l}\text{30\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}2\hspace{0.17em}again}\text{.}\hfill \\ \text{15\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}2,\hspace{0.17em}but\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3,\hspace{0.17em}the\hspace{0.17em}next\hspace{0.17em}largest\hspace{0.17em}prime}\text{.}\hfill \\ \text{5\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3,\hspace{0.17em}but\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}5,\hspace{0.17em}the\hspace{0.17em}next\hspace{0.17em}largest\hspace{0.17em}prime}\text{.}\hfill \\ \text{The\hspace{0.17em}quotient\hspace{0.17em}is\hspace{0.17em}1\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}stop\hspace{0.17em}the\hspace{0.17em}division\hspace{0.17em}process}\text{.}\hfill \end{array}$
The prime factorization of 60 is the product of all these divisors.
$\begin{array}{lllll}60\hfill & =\hfill & 2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill & \hfill & \text{We\hspace{0.17em}will\hspace{0.17em}use\hspace{0.17em}exponents\hspace{0.17em}when\hspace{0.17em}possible}.\hfill \\ 60\hfill & =\hfill & {2}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}5\hfill & \hfill & \hfill \end{array}$
Find the prime factorization of 441.
Since 441 is an odd number, it is not divisible by 2. We’ll try 3, the next largest prime.
$\begin{array}{l}\text{147\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3}.\hfill \\ \text{49\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}3\hspace{0.17em}nor\hspace{0.17em}by\hspace{0.17em}5},\text{\hspace{0.17em}but\hspace{0.17em}by\hspace{0.17em}7}.\hfill \\ \text{7\hspace{0.17em}is\hspace{0.17em}divisible\hspace{0.17em}by\hspace{0.17em}7}\text{.}\hfill \\ \text{The\hspace{0.17em}quotient\hspace{0.17em}is\hspace{0.17em}1\hspace{0.17em}so\hspace{0.17em}we\hspace{0.17em}stop\hspace{0.17em}the\hspace{0.17em}division\hspace{0.17em}process}.\hfill \end{array}$
The prime factorization of 441 is the product of all the divisors.
$\begin{array}{lllll}441\hfill & =\hfill & 3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}3\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\hfill & \hfill & \text{We\hspace{0.17em}will\hspace{0.17em}use\hspace{0.17em}exponents\hspace{0.17em}when\hspace{0.17em}possible}.\hfill \\ 441\hfill & =\hfill & {3}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{7}^{2}\hfill & \hfill & \hfill \end{array}$
## Exercises
For the following problems, determine which whole numbers are prime and which are composite.
23
prime
25
27
composite
2
3
prime
5
7
prime
9
11
prime
34
55
composite
63
1044
composite
339
209
composite
For the following problems, find the prime factorization of each whole number. Use exponents on repeated factors.
26
38
$2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}19$
54
62
$2\text{\hspace{0.17em}}·\text{\hspace{0.17em}}31$
56
176
${2}^{4}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}11$
480
819
${3}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}7\text{\hspace{0.17em}}·\text{\hspace{0.17em}}13$
2025
148,225
${5}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{7}^{2}\text{\hspace{0.17em}}·\text{\hspace{0.17em}}{11}^{2}$
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
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?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
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
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
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
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Circular arc
A circular arc is the arc of a circle between a pair of distinct points. If the two points are not directly opposite each other, one of these arcs, the minor arc, will subtend an angle at the centre of the circle that is less than π radians (180 degrees), and the other arc, the major arc, will subtend an angle greater than π radians.
## Length
The length (more precisely, arc length) of an arc of a circle with radius r and subtending an angle θ (measured in radians) with the circle center — i.e., the central angle — is
${\displaystyle L=\theta r.}$
This is because
${\displaystyle {\frac {L}{\mathrm {circumference} }}={\frac {\theta }{2\pi }}.}$
Substituting in the circumference
${\displaystyle {\frac {L}{2\pi r}}={\frac {\theta }{2\pi }},}$
and, with α being the same angle measured in degrees, since θ = α/180π, the arc length equals
${\displaystyle L={\frac {\alpha \pi r}{180}}.}$
A practical way to determine the length of an arc in a circle is to plot two lines from the arc's endpoints to the center of the circle, measure the angle where the two lines meet the center, then solve for L by cross-multiplying the statement:
measure of angle in degrees/360° = L/circumference.
For example, if the measure of the angle is 60 degrees and the circumference is 24 inches, then
{\displaystyle {\begin{aligned}{\frac {60}{360}}&={\frac {L}{24}}\\[6pt]360L&=1440\\[6pt]L&=4.\end{aligned}}}
This is so because the circumference of a circle and the degrees of a circle, of which there are always 360, are directly proportional.
The upper half of a circle can be parameterized as
${\displaystyle y={\sqrt {r^{2}-x^{2}}}.}$
Then the arc length from ${\displaystyle x=a}$ to ${\displaystyle x=b}$ is
${\displaystyle L=r{\Big [}\arcsin \left({\frac {x}{r}}\right){\Big ]}_{a}^{b}.}$
## Sector area
The area of the sector formed by an arc and the center of a circle (bounded by the arc and the two radii drawn to its endpoints) is
${\displaystyle A={\frac {r^{2}\theta }{2}}.}$
The area A has the same proportion to the circle area as the angle θ to a full circle:
${\displaystyle {\frac {A}{\pi r^{2}}}={\frac {\theta }{2\pi }}.}$
We can cancel π on both sides:
${\displaystyle {\frac {A}{r^{2}}}={\frac {\theta }{2}}.}$
By multiplying both sides by r2, we get the final result:
${\displaystyle A={\frac {1}{2}}r^{2}\theta .}$
Using the conversion described above, we find that the area of the sector for a central angle measured in degrees is
${\displaystyle A={\frac {\alpha }{360}}\pi r^{2}.}$
## Segment area
The area of the shape bounded by the arc and the straight line between its two end points is
${\displaystyle {\frac {1}{2}}r^{2}(\theta -\sin \theta ).}$
To get the area of the arc segment, we need to subtract the area of the triangle, determined by the circle's center and the two end points of the arc, from the area ${\displaystyle A}$. See Circular segment for details.
Using the intersecting chords theorem (also known as power of a point or secant tangent theorem) it is possible to calculate the radius r of a circle given the height H and the width W of an arc:
Consider the chord with the same endpoints as the arc. Its perpendicular bisector is another chord, which is a diameter of the circle. The length of the first chord is W, and it is divided by the bisector into two equal halves, each with length W/2. The total length of the diameter is 2r, and it is divided into two parts by the first chord. The length of one part is the sagitta of the arc, H, and the other part is the remainder of the diameter, with length 2r H. Applying the intersecting chords theorem to these two chords produces
${\displaystyle H(2r-H)=\left({\frac {W}{2}}\right)^{2},}$
whence
${\displaystyle 2r-H={\frac {W^{2}}{4H}},}$
so
${\displaystyle r={\frac {W^{2}}{8H}}+{\frac {H}{2}}.}$ |
### Just Rolling Round
P is a point on the circumference of a circle radius r which rolls, without slipping, inside a circle of radius 2r. What is the locus of P?
### Pericut
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
### Giant Holly Leaf
Find the perimeter and area of a holly leaf that will not lie flat (it has negative curvature with 'circles' having circumference greater than 2πr).
# Holly
##### Stage: 4 Challenge Level:
James Page of Hethersett High School, Norfolk solved the Holly problem, well done James.
Length $AB = 2 + 2\sqrt2$ cm and length $BC = 4 + 2\sqrt 2$ cm.
To find this answer I first took a diagram of the shape and put in the angles. I took 45 from 180 degrees to find the interior angle so I could find the inner curve. I then found that this angle is used 8 times in the diagram. Each circle has a radius of 1 cm and a circumference of $2\pi$ cm. To find the length of the edges of these pieces of the holly I did this:
$\begin{eqnarray} \\ 180 - 45 &=& 135 \\ 135/360 &=& 0.375 \\ (135/360) \times 8 \times (\pi \times 2) &=& 6\pi {\rm cm} \end{eqnarray}$
Then there are the two middle ones of which I had to find the circumference and divide by 2 then multiply by 2 for the two of them, giving a total $2\pi$ cm. I then added the two answers to get the perimeter of the holly leaf which is $8\pi = 25.13$ cm (to 2 decimal places).
If you wished to do a 16 spike holly of the same sort the ends would be the same and it would have an extra 3 circles on each side making the perimeter an extra $6\pi$ cm making a total of $14\pi = 43.98$ cm (to 2 decimal places).
The rectangle $ABCD$ has area
$$(2 + 2\sqrt 2)(4 + 2\sqrt 2) = (16 + 12 \sqrt 2).$$
From this we have to subtract the areas of the four triangles at the corners, a total area of 4 sq. cm., and also the areas of 8 sectors of circles with angles of 135 degrees and the areas of the two semicircles, in all a total area of $4\pi$ sq. cm. The area of the 10 spike holly leaf is
$$16 + 12\sqrt 2 - 4 - 4\pi = 12 + 12\sqrt 2 - 4\pi = 16.40 {\rm cm}^2$$ |
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Prime factorization of a whole number means to write it as the product of only prime numbers. A composite number can be written as a product of all of its prime factors.
To find the prime factorization of a number:
Divide the number by prime factors until the quotient is 1.
What is the prime factorization of 45?
45 ÷ 3 = 15
15 ÷ 3 = 5
5 ÷ 5 = 1
The prime factorization of 45 is:
45 = 3 × 3× 5
This can also be written using exponents as:
45 = 32 × 5
Factor Trees
Another method for producing the prime factorization of a natural number is to use what is called a factor tree.
A factor tree is a tool that breaks down any number into its prime factors
The first step in making a factor tree is to find a pair of factors whose product is the number that we are factoring. These two factors are the first branching in the factor tree. There are often several different pairs of factors that we could choose to begin the process. The choice does not matter; we may begin with any two factors. We repeat the process with each factor until each branch of the tree ends in a prime. Then the prime factorization is complete.
Example: Prime factorization of 24
The prime factorization is the product of the circled primes
So the prime factorization of 24 is 24 = 2 ×2 ×2 × 3 = 23 × 3.
Check Point
1. What is the prime factorization of 86?
2. What is the prime factorization of 592?
3. What is the prime factorization of 225?
4. What is the prime factorization of 1050?
5. What is the prime factorization of 42 use the factor tree method
1. 2 × 43
2. 2× 2 × 2× 2 × 37 = 24 × 37
3. 3 × 3 × 5 × 5 = 32 × 52
4. 2 × 3 × 5 × 5 × 7 = 2 × 3 × 52 × 7
5. Prime factorization of 42 = 2×3×7
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# Trigonometry Formulas, Functions and Identities – Free Homework Tutor Help, Videos and pdf
Things to Remember: Important Formula for Trigonometry
In mathematics, trigonometric identities are equalities involving trigonometric functions that are true for all values of the occurring variables. These identities are useful whenever expressions involving trigonometric functions need to be simplified.
The step involves, first using the substitution rule with a
trigonometric function, and then simplifying the resulting integral using with a trigonometric identity when the integration is given in a non-trigonometric identities.
Important List of Trigonometric Formulas:
Notation of Trigonometric Functions : The Six trigonometric function notations are Sine Function : sin (theta) Cosine Function :cos (theta) Tangent Function : tan (theta) Cotangent Function : cot (theta) Secant Function : sec (theta) Cosecant Function : cosec (theta) or csc (theta) Bacis Trigonometric Formula: In a right Triangle ABC, B is a right angle, then sin() = Perpendicular / Hypotenuse cos() = Base / Hypotenuse tan() = Perpendicular / Base cot() = Base / Perpendicular sec() = Hypotenuse / Base cosec() or csc() = Base/Perpendicular Reciprocal Identities : sin x = 1/cosec x , cosec x = 1/sin x cos x = 1/sec x , sec x = 1/cos x tan x = 1/cot x , cot x = 1/tan x Quotient Identities : tan x = 1/cot x cot x = 1/tan x Pythagorean Identities :The Pythagorean identities are follow the concept of Pythagorean theorem i.e. (Hypotenuse)2 = (Base)2 + (Perpendicular)2. This simply known as Pythagorean trigonometric identity. we have three identities – sin 2x + cos2x = 1 tan2x + 1 = cot2x [we get this identity by dividing first by cos 2x] 1 + cot2x = cosec2x or csc2x [we get this identity by dividing first by sin 2x] Addition and Subtraction Formulas : These are also known as Angle Sum and Difference Identities. These are sin (x + y) = sin x cos y + cos x sin y sin (x – y) = sin x cos y – cos x sin y cos (x + y) = cos x cos y – sin x sin y cos (x – y) = cos x cos y + sin x sin y tan (x + y) = (tan x + tan y) /(1 – tan x tan y) tan (x – y) = (tan x – tan y) /(1 + tan x tan y) Double Angle Formulas : When we substituting x = y in the sum of addition formula. we get these double angle formula sin (x + x) = sin x cos x + cos x sin x sin 2x = 2 sin x cos y cos (x + x) = cos x cos x – sin x sin x cos 2x = cos2 x – sin 2 x cos 2x = cos2x – (1 – cos 2 x) = 2 cos2 x – 1 [cos 2x in terms of cosine function] cos 2x = 1 – sin2 x – sin 2 x = 1 – 2 sin 2 x [cos 2x in terms of sine function] cos 2x = (1 – tan 2 x) / (1 + tan2 x) [cos 2x in terms of tangent function] tan (x + x) = (tan x + tan x) /(1 – tan x tan x) tan 2x = 2 tan x /(1 – tan2x) Triple Angle Formula : sin 3x = 3 sin x – 4 sin3x cos 3x = 4cos3x – 3 cos x tan 3x = (3 tan x – tan3 x) / (1 – 3 tan 2 x) Power Reduction Formulas : The sin2 x, cos2 x in terms of double angle formula sin2 x = (1 – cos 2x)/2 cos2 x = (1 + cos 2x)/2 tan2 x = (1 – cos 2x)/(1 + cos 2x) sin2 x cos2 x = (1 – cos 2x)(1 + cos 2x)/4 = (1 – cos2 2x)/4 = (1 – cos 4x)/8 Half Angle Formulas : When we substitute x/2 in place of x in power reduction formula and solve for sin x/2 and cos x/2, we get cos x/2 = √(1 + cos x)/2 sin x/2 = √(1 – cos x)/2 tan x/2 = sin x / (1 + cos x) = (1 – cos x) / sin x Product to Sum Identities: : When we expanding the addition formula in RHS, we get cos x cos y = [cos (x – y) + cos (x + y)]/2 sin x sin y = [cos (x – y) – cos (x + y)]/2 sin x cos y = [sin (x + y) + sin (x – y) ]/2 cos x sin y = [sin (x + y) – sin (x – y) ]/2 Sum to Product Identities: : When we substituting x by (x + y)/2 and y by (x – y)/2 in the product to sum formula in RHS, we get cos x + cos y = 2 cos (x + y)/2 cos (x – y)/2 sin x + sin y = 2 sin (x + y)/2 cos (x – y)/2 cos x – cos y = – sin (x + y)/2 sin (x – y)/2 sin x – sin y = 2 cos (x + y)/2 sin (x – y)/2 Inverse Trigonometric Functions : Every trigonometric function can be related directly to every other trigonometric function and these relations can be expressed by means of inverse trigonometric functions, we have sin -1 x + cos -1 y = /2 tan-1 x + cot -1 y = /2 If x > 0, tan-1 x + tan -1 1/x = /2and if x < 0, tan-1 x + tan -1 1/x = – /2 sin (cos-1 x) = cos (sin-1 x) = √(1 – x2) sin (tan-1 x) = x/√(1 + x2) cos (tan-1 x) = 1/√(1 + x2) tan (sin-1 x) = x/√(1 – x2) tan (cos-1 x) = √(1 – x2) / x |
# Go Math Grade 7 Answer Key Chapter 5 Percent Increase and Decrease
Download Go Math Grade 7 Answer Key for Chapter 5 Percent Increase and Decrease pdf for free. Quick and easy learning is possible with Go Math 7th Grade Answer Key Chapter 5 Percent Increase and Decrease. So, we suggest the students of 7th standard to go through the HMH Go Math Grade 7 Key Chapter 5 Percent Increase and Decrease. Our team of Go Math Answer Key will help the Grade 7 students to score the highest marks.
## Go Math Grade 7 Answer Key Chapter 5 Percent Increase and Decrease
Get a brief explanation for all the questions in Chapter 5 Percent Increase and Decrease Go Math Answer Key Grade 7. Tap on the links provided below and get the solutions according to the topics. So, utilize the time in the proper way and practice Go Math 7th Grade Solution Key Chapter 5 Percent Increase and Decrease.
Chapter 5 – Percent Increase and Decrease
Chapter 5 – Rewriting Percent Expressions
Chapter 5 – Applications of Percent
Chapter 5
### Percent Increase and Decrease – Guided Practice – Page No. 144
Find each percent increase. Round to the nearest percent.
Question 1.
From $5 to$8
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 5
Final amount = 8
8 – 5 = 3
Percent change = 3/5 = 0.6 = 60%
Question 2.
From 20 students to 30 students
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 20
Final amount = 30
We find the amount of change
30 – 20 = 10
We determine the percent of the increase
Percent change = 10/20 = 0.5 = 50%
Question 3.
From 86 books to 150 books
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 86
Final amount = 150
We find the amount of change
150 – 86 = 64
We determine the percent of increase and round it to the nearest percent
Percent Change = 64/86 ≈ 0.74 = 74%
Question 4.
From $3.49 to$3.89
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 3.49
Final amount = 3.89
We find the amount of change
3.89 – 3.49 = 0.40
We determine the percent of increase and round it to the nearest percent
Percent Change = 0.40/0.39 ≈ 0.11 = 11%
Question 5.
From 13 friends to 14 friends
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 13
Final amount = 14
We find the amount of change
14 – 13 = 1
We determine the percent of increase and round it to the nearest percent
Percent Change = 1/13 ≈ 0.08 = 8%
Question 6.
From 5 miles to 16 miles
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 5
Final amount = 16
We find the amount of change
16 – 5 = 11
We determine the percent of increase and round it to the nearest percent
Percent Change = 11/5 = 2.2 = 220%
Question 7.
Nathan usually drinks 36 ounces of water per day. He read that he should drink 64 ounces of water per day. If he starts drinking 64 ounces, what is the percent increase? Round to the nearest percent.
______ %
Explanation:
Given,
Nathan usually drinks 36 ounces of water per day. He read that he should drink 64 ounces of water per day.
Original Amount: 36
Final Amount: 64
Percent Charge = Amount of Change/Original Amount
We find the amount of change
64 – 36 = 28
We determine the percent of increase and round it to the nearest percent
Percent Change = 28/36 ≈ 0.78 = 78%
Thus the nearest percent is 78%
Find each percent decrease. Round to the nearest percent.
Question 8.
From $80 to$64
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 80
Final amount = 64
We find the amount of change
Amount of change = Greater value – Lesser value
= 80 – 64 = 16
We determine the percent of increase and round it to the nearest percent
Percent Change = 16/80 = 0.20 = 20%
Thus the nearest percent is 20%
Question 9.
From 95 °F to 68 °F
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 95
Final amount = 68
We find the amount of change
Amount of change = Greater value – Lesser value
= 95 – 68 = 27
We determine the percent of increase and round it to the nearest percent
Percent Change = 27/98 ≈ 0.28 = 28%
Thus the nearest percent is 28%
Question 10.
From 90 points to 45 points
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 90
Final amount = 45
We find the amount of change
Amount of change = Greater value – Lesser value
90 – 45 = 45
We determine the percent of increase and round it to the nearest percent
Percent Change = 45/90 = 0.50 = 50%
Thus the nearest percent is 50%
Question 11.
From 145 pounds to 132 pounds
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 145
Final amount = 132
We find the amount of change
Amount of change = Greater value – Lesser value
145 – 132 = 13
We determine the percent of increase and round it to the nearest percent
Percent Change = 13/145 ≈ 0.09 = 9%
The nearest percent is 9%
Question 12.
From 64 photos to 21 photos
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 64
Final amount = 21
We find the amount of change
Amount of change = Greater value – Lesser value
64 – 21 = 43
We determine the percent of increase and round it to the nearest percent
Percent Change = 43/64 ≈ 0.67 = 67%
Therefore the nearest percent is 67%
Question 13.
From 16 bagels to 0 bagels
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 16
Final amount = 0
We find the amount of change
Amount of change = Greater value – Lesser value
16 – 0 = 16
We determine the percent of increase and round it to the nearest percent
Percent Change = 16/16 = 1.0% = 100%
Question 14.
Over the summer, Jackie played video games 3 hours per day. When school began in the fall, she was only allowed to play video games for half an hour per day. What is the percent decrease? Round to the nearest percent.
______ %
Explanation:
Percent Charge = Amount of Change/Original Amount
Original amount = 3
Final amount = 0.5
We find the amount of change
Amount of change = Greater value – Lesser value
3 – 0.5 = 2.5
We determine the percent of increase and round it to the nearest percent
Percent Change = 2.5/3 ≈ 0.83 = 83%
The nearest percent is 83%
Find the new amount given the original amount and the percent of change.
Question 15.
$9; 10% increase$ ______
Answer: $9.90 Explanation: Percent Charge = Amount of Change/Original Amount Original amount = 9 Increase = 10% We find the amount of change 0.1 × 9 = 0.90 New Amount = Original Amount + Amount of Change 9 + 0.90 = 9.90 Question 16. 48 cookies; 25% decrease ______ cookies Answer: 36 cookies Explanation: Original amount = 48 Decrease = 25% We find the amount of change 0.25 × 48 = 12 New Amount = Original Amount – Amount of Change 48 – 12 = 36 Thus the answer is 36 cookies. Question 17. 340 pages; 20% decrease ______ pages Answer: 272 pages Explanation: Original Amount: 340 pages Decrease: 20% We find the amount of change 0.20 × 340 = 68 New Amount = Original Amount – Amount of Change 340 – 68 = 272 The answer is 272 pages. Question 18. 28 members; 50% increase ______ members Answer: 42 members Explanation: Original Amount: 28 Increase: 50% We find the amount of change 0.5 × 28 = 14 New amount = Original Amount + Amount of Change 28 + 14 = 42 The answer is 42 members Question 19.$29,000; 4% decrease
$______ Answer:$27,840
Explanation:
Original Amount: 29000
Decrease: 4%
We find the amount of change
0.04 × 29000 = 1160
New Amount = Original Amount – Amount of Change
29000 – 1160 = 27840
The answer is $27,840 Question 20. 810 songs; 130% increase ______ songs Answer: 1863 songs Explanation: Original Amount: 810 Increase: 130% We find the amount of change 1.3 × 810 = 1053 New amount = Original Amount + Amount of Change 810 + 1053 = 1863 songs Question 21. Adam currently runs about 20 miles per week, and he wants to increase his weekly mileage by 30%. How many miles will Adam run per week? ______ miles Answer: 26 miles Explanation: Given, Adam currently runs about 20 miles per week, and he wants to increase his weekly mileage by 30%. Original Amount: 20 Increase: 30% We find the amount of change 0.3 × 20 = 6 New amount = Original Amount + Amount of Change = 20 + 6 = 26 Therefore Adam run 26 miles per week. Essential Question Check-In Question 22. What process do you use to find the percent change of a quantity? Type below: _____________ Answer: In order to find the percent change of a quantity, we determine the amount of change in the quantity and divide it by the original amount. ### Percent Increase and Decrease – Independent Practice – Page No. 145 Question 23. Complete the table. Type below: _____________ Answer: bike: 13%, scooter 24%, increase, tennis racket:$83, skis: $435 Explanation: Since the new price is less than the original price, it is a percent decrease. percent decreases can be found using the equation percent decrease = (original – new)/original Bike: 110 – 96/110 = 14/110 ≈ 13% Scooter: 56 – 45/45 = 11/45 ≈ 24% Use the equation percent increase = new – original/original let x be the new price skis: (580 – x)/580 = 0.25 580 – x = 0.25 × 580 580 – x = 145 x = 580 – 145 = 435 The new price is$435
Question 24.
Multiple Representations
The bar graph shows the number of hurricanes in the Atlantic Basin from 2006–2011.
a. Find the amount of change and the percent of decrease in the number of hurricanes from 2008 to 2009 and from 2010 to 2011. Compare the amounts of change and percents of decrease.
Type below:
_____________
Answer: 2008 to 2009 has a smaller amount of change but a larger percent of decrease.
Explanation:
2008 to 2009:
amount of change: 8 – 3 = 5
percent decrease: 5/8 = 0.625 = 62.5%
2010 to 2011:
amount of change: 12 – 7 = 5
percent decrease: 5/12 ≈ 0.416 = 41.6%
The amount of change for 2010 to 2011 was greater than the amount of change for 2008 to 2009 but 2008 to 209 had a greater percent decrease than 2010 to 2011.
Question 24.
b. Between which two years was the percent of change the greatest? What was the percent of change during that period?
_______ %
Explanation:
Use the percent change = amount of change/original amount.
The biggest change in heights is between 2009 and 2010.
The percent change is (12-3)/3 = 9/3 = 3 = 300%
Question 25.
Represent Real-World Problems
Cheese sticks that were previously priced at “5 for $1” are now “4 for$1”. Find each percent of change and show your work.
a. Find the percent decrease in the number of cheese sticks you can buy for $1. _______ % Answer: 20% decrease Explanation: Use the percent change = amount of change/original amount. (5 – 4)/5 = 1/5 = 0.2 = 20% decrease Question 25. b. Find the percent increase in the price per cheese stick. _______ % Answer: 25% increase Explanation: First, find the price per cheese stick at each price. Use the percent change = amount of change/original amount. 1.00/5 = 0.20 1/4 = 0.25 (0.25 – 0.20)/0.20 = 0.05/0.20 = 25% increase ### Percent Increase and Decrease – Page No. 146 Question 26. Percent error calculations are used to determine how close to the true values, or how accurate, experimental values really are. The formula is similar to finding percent of change. chemistry class, Charlie records the volume of a liquid as 13.3 milliliters. The actual volume is 13.6 milliliters. What is his percent error? Round to the nearest percent. _______ % Answer: 2% Explanation: Use the formula |13.3 – 13.6|/13.6 = |-0.3|/13.6 ≈ 0.02 = 2% H.O.T. Focus on Higher Order Thinking Question 27. Look for a Pattern Leroi and Sylvia both put$100 in a savings account. Leroi decides he will put in an additional $10 each week. Sylvia decides to put in an additional 10% of the amount in the account each week. a. Who has more money after the first additional deposit? Explain. ___________ Answer: the same Explanation: Since 10% of 100 is 100(0.10) = 10, they both make an additional deposit of 10, so they have the same amount of money after the first additional deposit. Question 27. b. Who has more money after the second additional deposit? Explain. ___________ Answer: Sylvia Explanation: Both Lerio and Sylvia have$110 in their account after their first deposits since they both started with $100 and both deposited$10 for their first deposit.
After the second deposit, Lerio has 110 + 10 = $120. Sylvia has 110 + 0.10(110) = 110 + 11 =$121
So she has more money after the second deposit.
Question 27.
c. How do you think the amounts in the two accounts will compare after a month? A year?
Type below:
___________
Answer: Sylvia will continue to have more money after a month and a year since 10% of the balance is going to be greater than the 10 deposit that Leroi is making.
Question 28.
Critical Thinking
Suppose an amount increases by 100%, then decreases by 100%. Find the final amount. Would the situation change if the original increase was 150%? Explain your reasoning.
Type below:
___________
Answer: If an amount increases by 100%, then it will double. If it then decreases by 100%, it will become 0.
If you increase a number by 150% and then decrease it by 150%, you will not get to 0. 150% increase of 100 is 100 + 150 = 250.
A decrease of 150% is then 250 – 1.5(250) = 250 – 375 = -125
Question 29.
Look for a Pattern
Ariel deposited $100 into a bank account. Each Friday she will withdraw 10% of the money in the account to spend. Ariel thinks her account will be empty after 10 withdrawals. Do you agree? Explain. ___________ Answer: Ariel is incorrect. Her account balance will decrease as follows for the first 10 withdrawals: 1st withdrawal: 100 – 0.1(100) = 100 – 10 = 90 2nd withdrawal: 90 – 0.1(90) = 90 – 9 = 81 3rd withdrawal: 81 – 0.1(81) = 81 – 8.10 = 72.90 4th withdrawal: 72.90 – 0.1(72.90) = 72.90 – 7.29 = 65.61 5th withdrawal: 65.61 – 0.1(65.61) = 65.61 – 6.56 = 59.05 6th withdrawal: 59.05 – 0.1(59.05) = 59.05 – 5.91 = 53.14 7th withdrawal: 53.14 – 0.1(53.14) = 53.14 – 5.31 = 47.83 8th withdrawal: 47.83 – 0.1(47.83) = 47.83 – 4.78 = 43.05 9th withdrawal: 43.05 – 0.1(43.05) = 43.05 – 4.31 = 38.74 10th withdrawal: 38.74 – 0.1(38.74) = 38.74 – 3.87 = 34.87 ### Rewriting Percent Expressions – Guided Practice – Page No. 150 Question 1. Dana buys dress shirts from a clothing manufacturer for s dollars each, and then sells the dress shirts in her retail clothing store at a 35% markup. a. Write the markup as a decimal. ______ Answer: To convert a percent to a decimal, move the decimal place two places to the left. Therefore, 35% as a decimal is 0.35. Question 1. b. Write an expression for the retail price of the dress shirt. Type below: ___________ Answer: To write the expression, use the formula retail price = original place + markup Since s is the original place, if the markup is 35% = 0.35, then the markup is 0.35s. Question 1. c. What is the retail price of a dress shirt that Dana purchased for$32.00?
$______ Answer: Plugging in s = 32 into the expression gives a retail price of 1.35 = 1.35(32) =$43.20
Question 1.
d. How much was added to the original price of the dress shirt?
$______ Answer: The amount added to the original price is the amount of the markup. Since the amount of the markup is 0.35s and s = 32, then the amount of the markup was 0.35s = 0.35(32) =$11.20.
You can also find the amount of markup by subtracting the retail price and the original price. Since the retail price is $43.20 and the original price is$32, then the markup amount is $43.20 –$32 = $11.20 List the markup and retail price of each item. Round to two decimal places when necessary. Question 2. Markup:$ ______ Retail Price: $______ Answer: Markup:$ 2.70 Retail Price: $20.70 Explanation: Use the formula markup = price(markup%) 18(0.15) = 2.70 Use the retail price formula = price + markup 18 + 2.70 = 20.70 Question 3. Markup:$ ______ Retail Price: $______ Answer: Use the formula markup = (price)(markup %) 22.50(0.42) = 9.45 Use the retail price formula = price + markup 22.50 + 9.45 = 31.95 Question 4. Markup:$ ______ Retail Price: $______ Answer: Use the formula markup = (price)(markup %) = 33.75(0.75) = 25.31 Use the formula retail price = price + markup 33.75 + 25.31 = 59.06 Question 5. Markup:$ ______ Retail Price: $______ Answer: Use the formula markup = (price)(markup %) = 74.99(0.33) = 24.75 Use the formula retail price = price + markup 74.99 + 24.75 = 99.74 Question 6. Markup:$ ______ Retail Price: $______ Answer: Use the formula markup = (price)(markup %) 48.60(1.00) = 48.60 Use the formula retail price = price + markup 48.60 + 48.60 = 97.20 Question 7. Markup:$ ______ Retail Price: $______ Answer: Use the formula markup = (price)(markup %) = 185 × 1.25 = 231.25 Use the formula retail price = price + markup 185 + 231.25 = 461.25 Find the sale price of each item. Round to two decimal places when necessary. Question 8. Original price:$45.00; Markdown: 22%
$______ Answer: Use the formula markup = (price)(markup %) 45(0.22) = 9.90 Markdown is 9.90 Use the formula retail price = price + markup 45 – 9.90 = 35.10 Sale price is$35.10
Question 9.
Original price: $89.00; Markdown: 33%$ ______
Use the formula markup = (price)(markup %)
89 × 0.33 = 29.37
Use the formula retail price = price – markup
89 – 29.37 = 59.63
Question 10.
Original price: $23.99; Markdown: 44%$ ______
Use the formula markup = (price)(markup %)
23.99 × 0.44 = 10.56
Use the formula retail price = price – markup
23.99 – 10.56 = 13.43
Question 11.
Original price: $279.99, Markdown: 75%$ ______
Use the formula markup = (price)(markup %)
279.99 × 0.75 = 209.99
Use the formula retail price = price – markup
279.99 – 209.99 = 70
Essential Question Check-In
Question 12.
How can you determine the sale price if you are given the regular price and the percent of markdown?
Type below:
____________
Use the formula
Sale price = Original Price – Markdown
If the Sale price is S, Original Price p, and x the average reduction, then the formula becomes:|
S = p – x . p
### Rewriting Percent Expressions – Independent Practice – Page No. 151
Question 13.
A bookstore manager marks down the price of older hardcover books, which originally sell for b dollars, by 46%.
a. Write the markdown as a decimal.
______
Explanation:
To convert a percent to decimal form, move the decimal point 2 places to the left and don’t write the percent symbol. Therefore, 46% as a decimal is 0.46.
Question 13.
b. Write an expression for the sale price of the hardcover book.
Type below:
____________
Explanation:
The sale price is the original price minus the discount amount. If the original price is discounted 46% and the original price is b dollars, the amount of the discount is 46% of b = 0.46b.
The sale price is then b – 0.46b = (1 – 0.46)b = 0.54b
Question 13.
c. What is the sale price of a hardcover book for which the original retail price was $29.00?$ ______
Answer: $15.66 Explanation: From part (b), the sale price of an item with an original price of b dollars is 0.54b. If the original price is then b = 29 dollars, the sale price is 0.54b = 0.54 × 29 =$15.66
Question 13.
d. If you buy the book in part c, how much do you save by paying the sale price?
$______ Answer:$13.34
Explanation:
The amount of savings is the difference between the original price and the sale price. If the original price is $29 and the sale price is$15.66, then the amount of savings is $29.00 –$15.66 = $13.34 Question 14. Raquela’s coworker made price tags for several items that are to be marked down by 35%. Match each Regular Price to the correct Sale Price, if possible. Not all sales tags match an item. Type below: _____________ Answer: 35% markdown means the expression for the sales price is p – 0.35p = 0.65p. Plug in the regular prices for p to find the sale prices. Remember the directions stated not all sales tags will match a regular price so you won’t be able to match every regular price ticket with a sale price ticket. 0.65(3.29) = 2.14 0.65(4.19) = 2.72 0.65(2.79) = 1.81 0.65(3.09) = 2.01 0.65(3.77) = 2.45 Question 15. Communicate Mathematical Ideas For each situation, give an example that includes the original price and final price after markup or markdown. a. A markdown that is greater than 99% but less than 100% Type below: _____________ Answer: A markdown that is greater than 99% but less than 100% could be 99.5%. If the original price is$100, then the final price is 100 – 100(0.995) = 100 – 99.50 = 0.50
Question 15.
b. A markdown that is less than 1%
Type below:
_____________
A markdown that is less then 1% could be 0.5%. If the original price is $100, then the final price would be 100 – 0.005(100) = 100 – 0.50 = 99.50 Question 15. c. A markup that is more than 200% Type below: _____________ Answer: A markup that is more than 200% could be 300%. If the original price is$100, then the final price would be 100 + 100 (3.00) = 100 + 300 = 400
### Rewriting Percent Expressions – Page No. 152
Question 16.
Represent Real-World Problems
Harold works at a men’s clothing store, which marks up its retail clothing by 27%. The store purchases pants for $74.00, suit jackets for$325.00, and dress shirts for $48.00. How much will Harold charge a customer for two pairs of pants, three dress shirts, and a suit jacket?$ __________
Answer: $783.59 Explanation: Given, Harold works at a men’s clothing store, which marks up its retail clothing by 27%. The store purchases pants for$74.00, suit jackets for $325.00, and dress shirts for$48.00.
If the markup is 27%, then the expression for the retail price is p + 0.27p = 1.27p
where p is the original price.
The retail price of the pants is then 1.27(74) = 93.98.
The retail price of the suit jackets is 1.27(325) = 412.75
The retail price of the dress shirts is 1.27(48) = 60.96
The total for two pants, three dress shirts, and one suit jacket would then be 2(93.98) + 3(60.96) + 412.75
= 187.96 + 182.88 + 412.75 = 783.59
Question 17.
Analyze Relationships
Your family needs a set of 4 tires. Which of the following deals would you prefer? Explain.
Type below:
____________
Explanation:
The percent discount for buying 3 tires and getting one free is 25% since you are getting 1/4 of the tires for free and 1/4 off = 25%.
This means deal (I) and deal (III) are the same. They are greater than a 20% discount so deals (I) and (III) are preferable.
H.O.T.
Focus on Higher Order Thinking
Question 18.
Critique Reasoning
Margo purchases bulk teas from a warehouse and marks up those prices by 20% for retail sale. When teas go unsold for more than two months, Margo marks down the retail price by 20%. She says that she is breaking even, that is, she is getting the same price for the tea that she paid for it. Is she correct? Explain.
_______
She is not correct. If she originally purchases the teas for $100 and then marks the price up 20%, the retail price would then be 100 + 0.20(100) = 100 + 20 = 120. The sales price would then be 120 – 0.2(120) = 120 – 24 = 96. This less than the purchase price so she is losing money, not breaking even. Question 19. Problem Solving Grady marks down some$2.49 pens to $1.99 for a week and then marks them back up to$2.49. Find the percent of increase and the percent of decrease to the nearest tenth. Are the percents of change the same for both price changes? If not, which is a greater change?
_______
Answer: The percent decrease is found by using the formula (original price – new price)/(original price). The percent decrease is then (2.49 – 1.99)/2.49 = 0.20 = 20%.
A percent increase is found by using the formula (new price – original price)/original price.
The percent increase is then (2.49 – 1.99)/1.99 = 0.25 = 25%
The percents of change are not the same. The percent increase is greater.
Question 20.
Persevere in Problem Solving
At Danielle’s clothing boutique, if an item does not sell for eight weeks, she marks it down by 15%. If it remains unsold after that, she marks it down an additional 5% each week until she can no longer make a profit. Then she donates it to charity.
Rafael wants to buy a coat originally priced $150, but he can’t afford more than$110. If Danielle paid $100 for the coat, during which week(s) could Rafael buy the coat within his budget? Justify your answer. Type below: _____________ Answer: The expression for the markdown on the 8th week is p – 0.15p = 0.85p since it will get marked down 15% on the 8th week. The expression for the additional markdowns is p – 0.05p = 0.95p since it will get marked down an additional 5% every week after the 8th week. On the 8th week, it will be marked down to 0.85(150) = 127.50. This is more than Rafael can afford. On the 9th week, it will be marked down to 0.95(127.50) = 121.13. This is still more than Rafael can afford. On the 10th week, it will be marked down to 0.95(121.13) = 115.07. This is still more than Rafael can afford. On the 11th week, it will be marked down to 0.95(115.07) = 109.32. Rafael can afford this price so he must wait until the 11th week. ### Applications of Percent – Guided Practice – Page No. 156 Question 1. 5% of$30 =
$_______ Answer:$1.5
Explanation:
We have to find:
5% of $30 0.50 × 30 =$1.5
Question 2.
15% of $70 =$ _______
Answer: $10.5 Explanation: We have to find: 15% of$70
0.15 × 70 = 10.5
Question 3.
0.4% of $100 =$ _______
Answer: $0.40 Explanation: We have to find: 0.4% of$100
0.004 × 100 = 0.40
Question 4.
150% of $22 =$ _______
Answer: $33 Explanation: We have to find: 150% of$22
1.5 × 22 = 33
Question 5.
1% of $80 =$ _______
Answer: $0.8 Explanation: We have to find: 1% of$80
0.01 × 80 = 0.8
Question 6.
200% of $5 =$ _______
Answer: $10 Explanation: We have to find: 200% of$5
2 × 5 = 10
Question 7.
Brandon buys a radio for $43.99 in a state where the sales tax is 7%. a. How much does he pay in taxes?$ _______
Explanation:
We have to find the amount he pays in taxes by multiplying the cost by the sales tax percentage in decimal form remember to round to 2 decimal places.
43.99(0.07) = 3.08
Question 7.
b. What is the total Brandon pays for the radio?
$_______ Answer: 47.07 Explanation: To find the total Brandon pays for the radio we have to add the sales tax amount to the cost to find the total amount he pays. 43.99 + 3.08 = 47.07 Thus the total Brandon pays for the radio is$47.07.
Question 8.
Luisa’s restaurant bill comes to $75.50, and she leaves a 15% tip. What is Luisa’s total restaurant bill?$ _______
Answer: $86.25 Explanation: Given that, Luisa’s restaurant bill comes to$75.50, and she leaves a 15% tip.
Use the formula for the total restaurant bill:
T = P + x. P
Where T represents the total bill, P represents Luisa’s bill and x represents percents for tip, then the total restaurant bill is:
T = 75 + 0.15 (75)
T = 75 + 11.25
T = $86.25 Therefore Lusia’s total restaurant bill is$86.25
Question 9.
Joe borrowed $2,000 from the bank at a rate of 7% simple interest per year. How much interest did he pay in 5 years?$ _______
Explanation:
Joe borrowed $2,000 from the bank at a rate of 7% simple interest per year. We have to find the amount of interest per year 2000(0.07) = 140 Find the amount of interest for 5 years 140(5) = 700 Thus Joe pays$700 in 5 years.
Question 10.
You have $550 in a savings account that earns 3% simple interest each year. How much will be in your account in 10 years?$ _______
Answer: $715 Explanation: Given, You have$550 in a savings account that earns 3% simple interest each year.
Use the formula for simple interest:
Bt = B0(1 + tr)
Where t is time interval, Bt is money after t years, B0 is deposit and r is interest for one year, then the formula becomes:
B10 = B0(1 + 10.r)
B10 = 550(1 + 10 (0.03))
B10 = $715 In your account after 10 years will be$715.
Question 11.
Martin finds a shirt on sale for 10% off at a department store. The original price was $20. Martin must also pay 8.5% sales tax. a. How much is the shirt before taxes are applied?$ _______
Explanation:
We have to find the sales price of the shirt
20 – 0.1(20) = 20 – 2 = 18
The price of the shirt before taxes are applied is $18. Question 11. b. How much is the shirt after taxes are applied?$ _______
Explanation:
We have to find the price after sales tax
18 + 0.085(18) = 18 + 1.53 = 19.53
The price of the shirt after taxes are applied is $19.53 Question 12. Teresa’s restaurant bill comes to$29.99 before tax. If the sales tax is 6.25% and she tips the waiter 20%, what is the total cost of the meal?
$_______ Answer: 37.86 Explanation: Given, Teresa’s restaurant bill comes to$29.99 before tax. If the sales tax is 6.25% and she tips the waiter 20%.
Find the amount of sales tax
29.99(0.0625) = 1.87
Find the amount of the tip
29.99(0.20) = 6.00
The total cost by adding the bill account, sales tax, and tip amount.
29.99 + 1.87 + 6.00 = 37.86
Thus the total cost of the meal is $37.86 Essential Question Check-In Question 13. How can you determine the total cost of an item including tax if you know the price of the item and the tax rate? Type below: _____________ Answer: You can find the total cost of an item including tax by first multiplying the price of the item by the tax rate in decimal form to get the amount of sales tax. Then add the amount of sales tax to the price to get the total cost. ### Applications of Percent – Independent Practice – Page No. 157 Question 14. Emily’s meal costs$32.75 and Darren’s meal costs $39.88. Emily treats Darren by paying for both meals, and leaves a 14% tip. Find the total cost.$ _______
Explanation:
Emily’s meal costs $32.75 and Darren’s meal costs$39.88.
So, the total cost of the meals before tip is $32.75 +$39.88 = $72.63 Emily treats Darren by paying for both meals and leaves a 14% tip.$72.63 = 0.14(72.63) ≈ $10.17 Round to two decimal places since dollar amounts must be rounded to the nearest cent. The total cost that Dareen pays is then cost before tip + amount of tip =$72.63 + $10.17 =$82.80
Question 15.
The Jayden family eats at a restaurant that is having a 15% discount promotion. Their meal costs $78.65, and they leave a 20% tip. If the tip applies to the cost of the meal before the discount, what is the total cost of the meal?$ _______
Explanation:
The Jayden family eats at a restaurant that is having a 15% discount promotion.
The total cost of the meal = cost of meal + tip amount – discount amount
Their meal costs $78.65, and they leave a 20% tip. We need to find the tip amount and the discount amount using the given cost of the meal, tip percent, and discount percent. 20% of 78.65 = 0.20 × 78.65 =$15.73
Since the cost of the meal before the discount is $78.65 and the discount percent is 15%, then the amount of the discount is 15% of 78365 = 0.15 ×$78.65 ≈ $11.80 The total cost is then 78.65 + 15.73 – 11.80 =$82.58
Question 16.
A jeweler buys a ring from a jewelry maker for $125. He marks up the price by 135% for sale in his store. What is the selling price of the ring with 7.5% sales tax?$ _______
Explanation:
A jeweler buys a ring from a jewelry maker for $125. He marks up the price by 135% for sale in his store. 125 × 1.35 = 168.75 We can find the retail price by adding the markup to the purchase price 125 + 168.75 = 293.75 The amount of sales tax is 293.75 × 0.075 = 22.03 We can find the selling price by adding the tax amount to the retail price. 293.75 + 22.03 = 315.78 Therefore the selling price of the ring with 7.5% sales tax is$315.78
Question 17.
Luis wants to buy a skateboard that usually sells for $79.99. All merchandise is discounted by 12%. What is the total cost of the skateboard if Luis has to pay a state sales tax of 6.75%?$ _______
Explanation:
Given,
Luis wants to buy a skateboard that usually sells for $79.99. All merchandise is discounted by 12%. 79.99 × 0.12 = 9.60 79.99 – 9.60 = 70.39 First, we need to find the amount paid in taxes and then add that to the discount price to get the total cost. 70.39 × 0.0675 = 4.75 70.39 + 4.75 = 75.14 The total cost of the skateboard if Luis has to pay a state sales tax of 6.75% is$75.14
Question 18.
Kedar earns a monthly salary of $2,200 plus a 3.75% commission on the amount of his sales at a men’s clothing store. What would he earn this month if he sold$4,500 in clothing? Round to the nearest cent.
$_______ Answer: 2368.75 Explanation: Given, Kedar earns a monthly salary of$2,200 plus a 3.75% commission on the amount of his sales at a men’s clothing store.
4500 × 0.0375 = 168.75
The total earnings can be known by adding his monthly salary and his commission.
2200 + 168.75 = 2368.75
Question 19.
Danielle earns a 7.25% commission on everything she sells at the electronics store where she works. She also earns a base salary of $750 per week. How much did she earn last week if she sold$4,500 in electronics merchandise? Round to the nearest cent.
$_______ Answer: 1076.25 Explanation: Danielle earns a 7.25% commission on everything she sells at the electronics store where she works. She also earns a base salary of$750 per week.
The amount she made in the commission is 4500 × 0.0725 = 326.25
We can find the total earnings by adding her weekly pay and commission.
750 + 326.25 = 1076.25
Thus she earns $1076.25 last week if she sold$4,500 in electronics merchandise.
Question 20.
Francois earns a weekly salary of $475 plus a 5.5% commission on sales at a gift shop. How much would he earn in a week if he sold$700 in goods? Round to the nearest cent.
$_______ Answer: 513.50 Explanation: Given that, Francois earns a weekly salary of$475 plus a 5.5% commission on sales at a gift shop.
The amount he made in commission
700 × 0.055 = 38.50
We can find the total amount he earned by adding his weekly pay and commission
475 + 38.50 = $513.50 Question 21. Sandra is 4 feet tall. Pablo is 10% taller than Sandra, and Michaela is 8% taller than Pablo a. Explain how to find Michaela’s height with the given information. Type below: _____________ Answer: First we have to find 10% of Sandra’s height: 0.10 × 4 = 0.4 This means that Pablo is then 4 + 0.4 = 4.4 feet tall. Next find 8% of Pablo’s height: 4.4 × 0.08 = 0.352 This means that Michaela is 4.4 + 0.353 = 4.752 feet tall. Question 21. b. What is Michaela’s approximate height in feet and inches? _______ feet _______ inches Answer: Convert from feet to inches. 1 feet = 12 inches 4.752 = 4 + 0.752 0.752 = 12 × 0.752 = 9 inches 4 feet = 12 × 4 = 48 inches Thus the approximate height of Michaela is 4 feet 9 inches. Question 22. Eugene wants to buy jeans at a store that is giving$10 off everything. The tag on the jeans is marked 50% off. The original price is $49.98. a. Find the total cost if the 50% discount is applied before the$10 discount.
$_______ Answer:$14.99
Explanation:
Given that,
Eugene wants to buy jeans at a store that is giving $10 off everything. The tag on the jeans is marked 50% off. The original price is$49.98.
0.5 × 49.98 = 24.99
Now subtract $10 discount. 24.99 – 10 = 14.99 The total cost if the 50% discount is applied before the$10 discount is $14.99 Question 22. b. Find the total cost if the$10 discount is applied before the 50% discount.
$_______ Answer:$19.99
Explanation:
We have to find the price after the $10 discount then find 50% of that price to find the discounted price. 49.98 – 10 = 39.98 0.5 × 39.98 = 19.99 Thus the total cost if the$10 discount is applied before the 50% discount is $19.99 ### Applications of Percent – Page No. 158 Question 23. Multistep Eric downloads the coupon shown and goes shopping at Gadgets Galore, where he buys a digital camera for$95 and an extra battery for $15.99. a. What is the total cost if the coupon is applied to the digital camera?$ _______
Explanation:
Use the formula for the discount price:
DP = P – x.P
Price for the digital camera:
DP = 95 – 0.1(95)
DP = 95 – 9.5
DP = $85.5 Total cost = 85.5 + 15.99 =$101.49
Question 23.
b. What is the total cost if the coupon is applied to the extra battery?
$_______ Answer: 109.391 Explanation: Use the formula for the discount price: DP = P – x.P Price for the digital camera: DP = 15.99 – 0.1(15.99) DP = 15.99 – 1.599 DP =$14.399
Total cost = 95 + 14.399 = $109.391 Question 23. c. To which item should Eric apply the discount? Explain. ____________ Answer: He should apply the discount to the digital camera because then the total cost is the lower. Question 23. d. Eric has to pay 8% sales tax after the coupon is applied. How much is his total bill?$ _______
Use formula for Discount price
If he uses coupon for the digital camera then his total cost will be
T = DP + 0.08 × DP
T = 101.49 + 8.1192
T = $109.6029 If he uses coupon for the extra battery his total cost will be T = DP + 0.08 × DP T = 109.391 + 0.08(109.391) T =$118.14228
Question 24.
Two stores are having sales on the same shirts. The sale at Store 1 is “2 shirts for $22” and the sale at Store 2 is “Each$12.99 shirt is 10% off”.
a. Explain how much will you save by buying at Store 1.
$_______ Answer: For store 1, the shirts are 2 for$22. Ecah shirt then costs $22 ÷ 2 =$11
At store 2, each shirt is 10% off of $12.99 so each shirt costs:$12.99 – 0.1(12.99) = $12.99 –$1.30 = $11.69 You will then save$11.69 – $11.00 = 0.69 per shirt if you buy them from Store 1. Question 24. b. If Store 3 has shirts originally priced at$20.98 on sale for 55% off, does it have a better deal than the other stores? Justify your answer.
_______
If Store 3 sells shirts at 55% off of $20.98, then each shirt costs:$20.98 – 0.55($20.98) =$20.98 – $11.54 =$9.44
This is lower than the costs per shirt of Store 1 and Store 2 so it has a better deal.
H.O.T.
Focus on Higher Order Thinking
Question 25.
Analyze Relationships
Marcus can choose between a monthly salary of $1,500 plus 5.5% of sales or$2,400 plus 3% of sales. He expects sales between $5,000 and$10,000 a month. Which salary option should he choose? Explain.
_______
Answer: Second Salary option is better
Explanation:
E = 1500 + 0.055(5000)
E = 1500 + 275
E = $1775 In the second case he will earn E = 2400 + 0.03(5000) E = 2400 + 150 E =$2550
Question 26.
Multistep
In chemistry class, Bob recorded the volume of a liquid as 13.2 mL. The actual volume was 13.7 mL. Use the formula to find percent error of Bob’s measurement to the nearest tenth of a percent.
_______ %
Explanation:
|13.2 – 13.7|/13.7 = |-0.5|/13.7
0.5/13.7 ≈ 0.036 = 3.6%
### MODULE QUIZ – 5.1 Percent Increase and Decrease – Page No. 159
Find the percent change from the first value to the second.
Question 1.
36; 63
_______ %
Explanation:
Use the formula percent change = amount of change/first value
amount of change = 27
First value = 36
(63 – 36)/36 = 27/36 = 0.75 = 75%
Question 2.
50; 35
_______ %
Explanation:
Use the formula percent change = amount of change/first value
amount of change = 15
First value = 50
(50 – 35)/35 = 15/50 = 0.3 = 30%
Question 3.
40; 72
_______ %
Explanation:
Use the formula percent change = amount of change/first value
amount of change = 32
First value = 40
(72 – 40)/40 = 32/40 = 0.8 = 80%
5.2 Markup and Markdown
Use the original price and the markdown or markup to find the retail price.
Question 5.
Original price: $60; Markup: 15%$ _______
Answer: $69 Explanation: Use the formula retail price = original price + markup 60 + 60 × 0.15 = 60 + 9= 69 Question 6. Original price:$32; Markup: 12.5%
$_______ Answer: 36 Explanation: Use the formula retail price = original price + markup 32 + 32 × 0.125 = 32 + 4 =$36
Question 7.
Original price: $50; Markdown: 22%$ _______
Explanation:
Use the formula retail price = original price + markup
50 – 50 × 0.22 = 50 – 11 = 39
Question 8.
Original price: $125; Markdown: 30%$ _______
Explanation:
Use the formula retail price = original price + markup
125 – 125 × 0.3 = 125 – 37.50 = 87.50
5.3 Applications of Percent
Question 9.
Mae Ling earns a weekly salary of $325 plus a 6.5% commission on sales at a gift shop. How much would she make in a work week if she sold$4,800 worth of merchandise?
$_______ Answer: 637 Explanation: Mae Ling weekly earnings is equal to her weekly salary plus her commission. Since she earns 6.5 % commission on sales, if she sold$4800 worth of merchandise, her commission earnings would be 6.5 % of 4800 = 0.065 × 4800 = $312. Since her weekly salary is 325, then her total weekly earnings is$325 + $312 =$637
Question 10.
Ramon earns $1,735 each month and pays$53.10 for electricity. To the nearest tenth of a percent, what percent of Ramon’s earnings are spent on electricity each month?
_______ %
Explanation:
Divide the electric payment by his monthly pay
53.10/1735 = 0.031 = 3.1%
Question 11.
James, Priya, and Siobhan work in a grocery store. James makes $7.00 per hour. Priya makes 20% more than James, and Siobhan makes 5% less than Priya. How much does Siobhan make per hour?$ _______
Explanation:
Since James makes $7 per hour and priya makes 20% more than this, find 20% of 7 and then add that to 7 to find the pay per hour for Priya. 7 + 0.2(7) = 7 + 1.40 = 8.40 Since Priya makes$8.40 per hour and Siobhan makes 5% less than this, find 5% of 8.40 and subtract that from 8.40 to find the pay per hour of Siobhan.
8.40 – 0.05(8.40) = 8.40 – 0.42 = 7.98
Question 12.
The Hu family goes out for lunch, and the price of the meal is $45. The sales tax on the meal is 6%, and the family also leaves a 20% tip on the pre-tax amount. What is the total cost of the meal?$ _______
Explanation:
Find the amount of tax
45 × 0.06 = 2.70
Find the amount of tip
45 × 0.20 = 9
Find the total cost by adding the cost of the meal, the tax, and the tip.
45 + 2.70 + 9 = $56.70 Essential Question Question 13. Give three examples of how percents are used in the real-world. Tell whether each situation represents a percent increase or a percent decrease. Type below: ____________ Answer: One example could be giving a tip when you eat at a restaurant. Since the cost increases, it represents a percent increase. Second example is tax on purchase. Since the price increases it is a percent increase. Third example is using a coupon when buying an item. Since the price decreases, it is a percent decrease. ### Selected Response – Page No. 160 Question 1. Zalmon walks $$\frac{3}{4}$$ of a mile in $$\frac{3}{10}$$ of an hour. What is his speed in miles per hour? Options: a. 0.225 miles per hour b. 2.3 miles per hour c. 2.5 miles per hour d. 2.6 miles per hour Answer: 2.5 miles per hour Explanation: Given that, Zalmon walks $$\frac{3}{4}$$ of a mile in $$\frac{3}{10}$$ of an hour. Divide the number of miles by the number of hours to get his speed in miles per hour. $$\frac{3}{4}$$ ÷ $$\frac{3}{10}$$ $$\frac{3}{4}$$ ÷ $$\frac{10}{3}$$ = $$\frac{5}{2}$$ Convert the fraction into the decimal form. $$\frac{5}{2}$$ = 2.5 miles per hour Thus the correct answer is option C. Question 2. Find the percent change from 70 to 56. Options: a. 20% decrease b. 20% increase c. 25% decrease d. 25% increase Answer: 20% increase Explanation: Use the percent change = amount of change/original amount. Since the number decreased from 70 to 56, it is a percent decrease. = (70 – 56)/70 = $$\frac{14}{70}$$ = 0.2 = 20% Thus the correct answer is option A. Question 3. The rainfall total two years ago was 10.2 inches. Last year’s total was 20% greater. What was last year’s rainfall total? Options: a. 8.16 inches b. 11.22 inches c. 12.24 inches d. 20.4 inches Answer: 12.24 inches Explanation: Given, The rainfall total two years ago was 10.2 inches. Last year’s total was 20% greater. Find 20% of 10.2 10.2 × 0.20 = 2.04 Add the value to the original amount of 10.2 10.2 + 2.04 = 12.24 Therefore the correct answer is option C. Question 4. A pair of basketball shoes was originally priced at$80, but was marked up 37.5%. What was the retail price of the shoes?
Options:
a. $50 b.$83
c. $110 d.$130
Answer: $110 Explanation: A pair of basketball shoes was originally priced at$80, but was marked up 37.5%.
Use the formula retail price = original price + markup
80 + 80 × 0.375 = 80 + 30 = 110
Thus the correct answer is option C.
Question 5.
The sales tax rate in Jan’s town is 7.5%. If she buys 3 lamps for $23.59 each and a sofa for$769.99, how much sales tax does she owe?
Options:
a. $58.85 b.$63.06
c. $67.26 d.$71.46
Answer: $63.06 Explanation: The sales tax rate in Jan’s town is 7.5%. If she buys 3 lamps for$23.59 each and a sofa for $769.99 Total cost before tax is 3 × 23.59 + 769.99 = 70.77 + 769.99 = 840.76 Find the amount of tax by multiplying the tax rate and total cost from the above solution and then round to 2 decimal place. 840.76 × 0.075 = 63.06 Thus the correct answer is option B. Question 6. The day after a national holiday, decorations were marked down 40%. Before the holiday, a patriotic banner cost$5.75. How much did the banner cost after the holiday?
Options:
a. $1.15 b.$2.30
c. $3.45 d.$8.05
Answer: $3.45 Explanation: The day after a national holiday, decorations were marked down 40%. Before the holiday, a patriotic banner cost$5.75.
use the formula retail price = original price – markdown
5.75 – 5.75 × 0.4 = 5.75 – 2.30 = 3.45
Thus the correct answer is option C.
Question 7.
Dustin makes $2,330 each month and pays$840 for rent. To the nearest tenth of a percent, what percent of Dustin’s earnings are spent on rent?
Options:
a. 84%
b. 63.9%
c. 56.4%
d. 36.1%
Explanation:
Dustin makes $2,330 each month and pays$840 for rent.
Divide his rent by his monthly income. round to three decimal places and then convert to percent form.
840/2330 = 0.361 = 36.1%
Thus the correct answer is option D.
Question 8.
A scuba diver is positioned at -30 feet. How many feet will she have to rise to change her position to -12 feet?
Options:
a. -42 ft
b. -18 ft
c. 18 ft
d. 42 ft
Explanation:
Given,
A scuba diver is positioned at -30 feet.
-12 – (-30) = 12 + 30 = 18 feet
Thus the correct answer is option C.
Question 9.
A bank offers an annual simple interest rate of 8% on home improvement loans. Tobias borrowed $17,000 over a period of 2 years. How much did he repay altogether? Options: a.$1360
b. $2720 c.$18360
d. $19720 Answer:$19720
Explanation:
Given that,
A bank offers an annual simple interest rate of 8% on home improvement loans.
Tobias borrowed $17,000 over a period of 2 years Find the amount of interest he paid using the formula I = prt where p is the amount borrowed r is the interest rate t is the number of years 17000 × 0.08 × 2 = 2720 Add the amount borrowed and amount of interest 17000 + 2720 = 19720. Thus the correct answer is option D. Mini-Task Question 10. The granola Summer buys used to cost$6.00 per pound, but it has been marked up 15%.
a. How much did it cost Summer to buy 2.6 pounds of granola at the old price?
$___________ Answer:$15.60
Explanation:
Multiply 2.6 by the old price of $6 2.6 × 6 = 15.60 It costs$15.60 to buy 2.6 pounds of granola at the old price.
Question 10.
b. How much does it cost her to buy 2.6 pounds of granola at the new price?
$_______ Answer:$17.94
Explanation:
Find the new price using the formula retail price = original price + markup
Then find the total cost by buying 2.6 pounds at the new price.
6 + 6 × 0.15 = 6 + 0.9 = 6.90
2.6 × 6.90 = 17.94
The new price is $17.94 Question 10. c. Suppose Summer buys 3.5 pounds of granola. How much more does it cost at the new price than at the old price?$ _______
Answer: $3.15 Explanation: 3.5 × 6 = 21 3.5 × 6.90 = 24.15 24.15 – 21 = 3.15 ### Module 5 – Page No. 162 EXERCISES Question 1. Michelle purchased 25 audio files in January. In February she purchased 40 audio files. Find the percent increase. _______ % Answer: 60% Explanation: Given, Michelle purchased 25 audio files in January. In February she purchased 40 audio files. Use the percent change = amount of change/original amount. (40 -25)/25 = 15/25 = 0.6 = 60% Thus the percent increase is 60% Question 2. Sam’s dog weighs 72 pounds. The vet suggests that for the dog’s health, its weight should decrease by 12.5 percent. According to the vet, what is a healthy weight for the dog? _______ pounds Answer: 63 pounds Explanation: Given, Sam’s dog weighs 72 pounds. The vet suggests that for the dog’s health, its weight should decrease by 12.5 percent. 72 × 0.125 = 9 Find a healthy weight by subtracting the change in weight from the original weight 72 – 9 = 63 The healthy weight of the dog is 63 pounds. Question 3. The original price of a barbecue grill is$79.50. The grill is marked down 15%. What is the sale price of the grill?
$_______ Answer: 67.57 Explanation: Given, The original price of a barbecue grill is$79.50. The grill is marked down 15%.
Use the formula sale price = original price – markdown
= 79.50 – 79.50 × 0.15 = 79.50 – 11.93 = $67.57 Thus the sale price of the grill is$67.57
Question 4.
A sporting goods store marks up the cost s of soccer balls by 250%. Write an expression that represents the retail cost of the soccer balls. The store buys soccer balls for $5.00 each. What is the retail price of the soccer balls?$ _______
Answer: $17.5 Explanation: Use the formula retail price = original price + markup to find the expression for an original price of s and a markup percentage of 250% s + 2.5s = 3.5s substitute s = 5 into the expression to find the retail price 3.5 × 5 = 17.50 Thus the retail price of the soccer balls is$17.50
### Unit 2 Performance Tasks – Page No. 163
Question 1.
Viktor is a bike tour operator and needs to replace two of his touring bikes. He orders two bikes from the sporting goods store for a total of $2,000 and pays using his credit card. When the bill arrives, he reads the following information: Balance:$2000
Annual interest rate: 14.9%
Minimum payment due: $40 Late fee:$10 if payment not received by 3/1/2013
a. To keep his good credit, Viktor promptly sends in a minimum payment of $40. When the next bill arrives, it looks a lot like the previous bill. Balance:$1,984.34
Annual interest rate: 14.9%
Minimum payment due: $40 Late fee:$10 if payment not received by 4/1/2013
Explain how the credit card company calculated the new balance. Notice that the given interest rate is annual, but the payment is monthly.
Type below:
_____________
We have to find the balance after the first bill by subtracting the $40 payment from the original balance of$2000.
Balance after first bill: 2000 – 40 = 1960
Then find the amount of interest charged on the second bill by multiplying the balance of $1960 by the interest rate. Remember since the interest rate is annually you have to divide it by 12 to get the monthly interest rate. Interest on the second bill: 1960 × 0.149/12 = 24.34 And then add this interest amount to the balance of$1960 to get the balance on the second bill.
New balance: 1960 + 24.34 = 1984.34
Question 1.
b. Viktor was upset about the new bill, so he decided to send in $150 for his April payment. The minimum payment on his bill is calculated as 2% of the balance (rounded to the nearest dollar) or$20, whichever is greater. Fill out the details for Viktor’s new bill.
Type below:
_____________
Find the balance after the $150 payment. The interest rate hasn’t changed so the annual interest rate on this new bill is the same as the previous bills. balance after payment: 1984.34 – 150 = 1834.34 annual interest rate: 14/9% Find the interest charged on the third bill. find the balance on the third bill by adding the interest charged to the balance of$1834.34.
interest on the third bill: 1834.34 × 0.149/12 = 22.78
balance: 1834.34 + 22.78 = 1857.12
To find what the minimum payment will be, first find 2% of the balance.
2% of balance: 0.02 × 1857.12 = 37.14
Minimum payment due: $37.00 Since this is greater than$20, the minimum payment is 2% of the balance rounded to the nearest dollar giving $37 as the payment. The later fee date is one month after the late fee date of 04/01/2013 on the previous bill which gives 05/01/2013. Question 1. c. Viktor’s bank offers a credit card with an introductory annual interest rate of 9.9%. He can transfer his current balance for a fee of$40. After one year, the rate will return to the bank’s normal rate, which is 13.9%. The bank charges a late fee of $15. Give two reasons why Viktor should transfer the balance and two reasons why he should not Type below: _____________ Answer: Two reasons he should transfer is that the lower introductory rate would mean less interest charged in the first year and a lower normal rate would mean less interest charged after that first year as well. Two reasons he shouldn’t transfer the balance is that he would have to pay a transfer fee of$40 and that the late fee is $15 instead of$10 if he transfers the balance.
### Unit 2 Performance Tasks (con’td) – Page No. 164
Question 2.
The table below shows how far several animals can travel at their maximum speeds in a given time.
a. Write each animal’s speed as a unit rate in feet per second.
Elk: _________ feet per second
Giraffe: _________ feet per second
Zebra: _________ feet per second
By seeing the above table we can find the unit rates by dividing the distance traveled by the time in second.
elk: 33 ÷ 1/2 = 33 × 2 = 66 feet per second
Giraffe: 115 ÷ 2 1/2 = 115 ÷ 5/2 = 115 2/5 = 46 feet per second
Zebra: 117 ÷ 2 = 58.5 feet per second
Question 2.
b. Which animal has the fastest speed?
_____________
Answer: The elk had the greatest unit rate so it has the fastest speed.
Question 2.
c. How many miles could the fastest animal travel in 2 hours if it maintained the speed you calculated in part a? Use the formula d = rt and round your answer to the nearest tenth of a mile. Show your work.
Elk: _________ miles
Giraffe: _________ miles
Zebra: _________ miles
Elk: 90 miles
Giraffe: 62 miles
Zebra: 72 miles
Explanation:
There are 60 seconds in a minute and 60 minutes in an hour so there are 2 × 60 × 60 = 7200 seconds in 2 hours.
Multiply the unit rate of the elk by 7200 seconds to get the distance traveled in feet.
There are 5280 feet in 1 mile so divide the distance in feet by 5280 to get the distances in miles.
Elk:
66 × 7200 = 475200 feet
Now convert from feet to miles
475200 feet = 90 miles
Giraffe: 46 feet per second
62 × 7200 = 331200 feet
Now convert from feet to miles.
331200 = 62 miles
Zebra: 58.5 feet per second
58.5 × 7200 = 421200 feet
Now convert from feet to miles.
421200 feet = 72 miles
Question 3.
d. The data in the table represents how fast each animal can travel at its maximum speed. Is it reasonable to expect the animal from part b to travel that distance in 2 hours? Explain why or why not.
______
Answer: It is not reasonable. An animal can only travel at its maximum speed for a short amount of time which is usually only for a couple of minutes.
### Selected Response – Page No. 165
Question 1.
If the relationship between distance y in feet and time x in seconds is proportional, which rate is represented by $$\frac{y}{x}$$ = 0.6?
Options:
a. 3 feet in 5 s
b. 3 feet in 9 s
c. 10 feet in 6 s
d. 18 feet in 3 s
Answer: 3 feet in 5 s
Explanation:
$$\frac{y}{x}$$ = 0.6
0.6 = $$\frac{6}{10}$$
Since $$\frac{6}{10}$$ = $$\frac{3}{5}$$, it represents a rate of 3 feet in 5 seconds,
Therefore the correct answer is option A.
Question 2.
The Baghrams make regular monthly deposits in a savings account. The graph shows the relationship between the number x of months and the amount y in dollars in the account.
What is the equation for the deposit?
Options:
a. $$\frac{y}{x}$$ = $25/month b. $$\frac{y}{x}$$ =$40/month
c. $$\frac{y}{x}$$ = $50/month d. $$\frac{y}{x}$$ =$75/month
Answer: $$\frac{y}{x}$$ = $50/month Explanation: By seeing the above graph we can say that the point is (2, 100). This means that $$\frac{y}{x}$$ = $$\frac{100}{2}$$ = 50. Thus the correct answer is option C. Question 3. What is the decimal form of −4 $$\frac{7}{8}$$? Options: a. -4.9375 b. -4.875 c. -4.75 d. -4.625 Answer: -4.875 Explanation: Given the fraction −4 $$\frac{7}{8}$$ First divide $$\frac{7}{8}$$ = 0.875 4 + 0.875 = 4.875 So, −4 $$\frac{7}{8}$$ = -4.875 Therefore the answer is option B. Question 4. Find the percent change from 72 to 90. Options: a. 20% decrease b. 20% increase c. 25% decrease d. 25% increase Answer: 25% increase Explanation: Use the formula percent change = amount of change/original amount. the value increased from 72 to 90 so it is a percent increase. (90-72)/72 = 18/72 = 0.25 = 25% Thus the correct answer is option D. Question 5. A store had a sale on art supplies. The price p of each item was marked down 60%. Which expression represents the new price? Options: a. 0.4p b. 0.6p c. 1.4p d. 1.6p Answer: 0.4p Explanation: Given that, A store had a sale on art supplies. The price p of each item was marked down 60% Use the formula sale price = original price – markdown p is the original price and the markdown percent is 40% then combine the like terms. p – 0.6p = 0.4p Therefore the correct answer is option A. Question 6. Clarke borrows$16,000 to buy a car. He pays simple interest at an annual rate of 6% over a period of 3.5 years. How much does he pay altogether?
Options:
a. $18800 b.$19360
c. $19920 d.$20480
Answer: $19360 Explanation: Given, Clarke borrows$16,000 to buy a car.
He pays simple interest at an annual rate of 6% over a period of 3.5 years.
Find the total amount of interest using the formula
I = prt
where p is the amount borrowed
r is the rate of interest
t is the number of years
16000 × 0.06 × 3.5 = 3360
Now add the amount of interest to the amount borrowed to find the total amount
16000 + 3360 = 19,360
Thus the correct answer is option B.
Question 7.
To which set or sets does the number 37 belong?
Options:
a. integers only
b. rational numbers only
c. integers and rational numbers only
d. whole numbers, integers, and rational numbers
Answer: whole numbers, integers, and rational numbers
Explanation:
37 can be written as 37/1 so it is a rational number. 37 doesn’t have a decimal or fraction so it is an integer. Since it is a positive integer, it is also a whole number.
Thus a suitable answer is option D.
### Page No. 166
Question 8.
In which equation is the constant of proportionality 5?
Options:
a. x = 5y
b. y = 5x
c. y = x + 5
d. y = 5 – x
Explanation:
Directly proportional equations are of the form y = kx
where k is the constant of proportionality.
If k = 5, then the equation is y = 5x.
Thus the correct answer is option B.
Question 9.
Suri earns extra money by dog walking. She charges $6.25 to walk a dog once a day 5 days a week and$8.75 to walk a dog once a day 7 days a week. Which equation represents this relationship?
Options:
a. y = 7x
b. y = 5x
c. y = 2.50x
d. y = 1.25x
Explanation:
Given that,
Suri earns extra money by dog walking. She charges $6.25 to walk a dog once a day 5 days a week and$8.75 to walk a dog once a day 7 days a week.
Since 6.25/5 = 1.25
So, the equation is y = 1.25x
where x is the number of days and y is the total charge.
So, the correct answer is option D.
Question 10.
Randy walks $$\frac{1}{2}$$ mile in each $$\frac{1}{5}$$ hour. How far will Randy walk in one hour?
Options:
a. $$\frac{1}{2}$$ miles
b. 2 miles
c. 2 $$\frac{1}{2}$$ miles
d. 5 miles
Answer: 2 $$\frac{1}{2}$$ miles
Explanation:
Given,
Randy walks $$\frac{1}{2}$$ mile in each $$\frac{1}{5}$$ hour.
$$\frac{1}{2}$$ ÷ $$\frac{1}{5}$$
$$\frac{1}{2}$$ × $$\frac{5}{1}$$ = $$\frac{5}{2}$$
Convert the fraction to the improper fractions.
$$\frac{5}{2}$$ = 2 $$\frac{1}{2}$$ miles
Therefore the correct answer is option C.
Question 11.
On a trip to Spain, Sheila bought a piece of jewelry that cost $56.75. She paid for it with her credit card, which charges a foreign transaction fee of 3%. How much was the foreign transaction fee? Options: a.$0.17
b. $1.07 c.$1.70
d. $17.00 Answer:$1.70
Explanation:
On a trip to Spain, Sheila bought a piece of jewelry that cost \$56.75.
She paid for it with her credit card, which charges a foreign transaction fee of 3%
Find the foreign transaction fee amount by multiplying the cost by the foreign transaction fee percentage.
56.75 × 0.03 = 1.70
Thus the correct answer is option C.
Question 12.
A baker is looking for a recipe that has the lowest unit rate for flour per batch of muffins. Which recipe should she use?
Options:
a. $$\frac{1}{2}$$ cup flour for $$\frac{2}{3}$$ batch
b. $$\frac{2}{3}$$ cup flour for $$\frac{1}{2}$$ batch
c. $$\frac{3}{4}$$ cup flour for $$\frac{2}{3}$$ batch
d. $$\frac{1}{3}$$ cup flour for $$\frac{1}{4}$$ batch
Answer: $$\frac{1}{2}$$ cup flour for $$\frac{2}{3}$$ batch
Explanation:
a. $$\frac{1}{2}$$ ÷ $$\frac{2}{3}$$ = $$\frac{1}{2}$$ × $$\frac{3}{2}$$ = $$\frac{3}{4}$$
b. $$\frac{2}{3}$$ ÷ $$\frac{1}{2}$$ = $$\frac{2}{3}$$ × $$\frac{2}{1}$$ = $$\frac{4}{3}$$ = 1 $$\frac{1}{3}$$
c. $$\frac{3}{4}$$ ÷ $$\frac{2}{3}$$ = $$\frac{3}{4}$$ × $$\frac{3}{2}$$ = $$\frac{9}{8}$$ = 1 $$\frac{1}{8}$$
d. $$\frac{1}{3}$$ ÷ $$\frac{1}{4}$$ = $$\frac{1}{3}$$ ÷ $$\frac{4}{1}$$ = 1 $$\frac{1}{3}$$
Thus the correct answer is option A.
Question 13.
Kevin was able to type 2 pages in 5 minutes, 3 pages in 7.5 minutes, and 5 pages in 12.5 minutes.
a. Make a table of the data.
Type below:
___________
Number of Pages 2 3 5 Minutes 5 7.5 12.5
Question 13.
b. Graph the relationship between the number of pages typed and the number of minutes.
Type below:
___________
Question 13.
c. Explain how to use the graph to find the unit rate.
Type below:
___________
Answer: The unit rate is 2.5 pages per minute
Explanation:
By using the graph we need to find the slope of the line.
We can do this by using the formula of a slope:
m = (y2-y1)/(x2-x1) = (7.5-5)/(3-2) = 2.5
Thus the unit rate is 2.5 pages per minute.
Conclusion:
Hope the answers provided in Go Math Answer Key Grade 7 Chapter 5 Percent Increase and Decrease are quite satisfactory for all the students. Refer to our Go Math 7th Grade Chapter 5 Percent Increase and Decrease to get the solutions with best explanations. After your preparation test your math skills by solving the questions in the performance tasks. |
## Ratio calculator - Calculate the ratio of fourth number.
For a ratio A:B = C:D, Calculate the fourth value(D) in the ratio by providing the three numbers (A, B, C).
: : ?? A:B 0 C:D 0
### What is Ratio?
Ratio is essentially a fraction of two numbers. Like a fraction, it is made up of two parts, the numerator and the denominator. For example, a pizza is cut into 6 slices and 5 of them have been eaten. If we want to know the ratio of slices eaten compared to the entire pizza, then we plug the number of slices eaten as the numerator and the total number of slices as the denominator, which is 5/6.
Ratios can be scaled up to larger, equivalent ratios where the numbers involved are much bigger. For example, we might want to scale up the ratio 5/6 to another equivalent ratio with the denominator 72. To do this, we set up a relation between the two ratios, where they are equal to each other, and we need to solve for the missing number. For the new equivalent ratio, the missing number may be the numerator or the denominator. This process of equating the two ratios is known as setting up a proportion.
### Ratio Formula
We consider a pair of equivalent ratios, A/B and C/D.
The ratio can be used to find D when A, B and C are known. The equivalent ratios can be written as
$${A\over B} = {C \over D}$$
To find D, we use the formula
$$D = C * {B \over A}$$
For example, if A = 3, B = 8 and C = 12, then
$$D = 12 * {8\over3} = 32$$
Hence, the ratios 3/8 and 12/32 are equivalent to each other.
However, if A, B, and D are known, and C needs to be calculated, then the formula is
$$C = D*{A\over B}$$
For example, if A = 3, B = 8 and D = 24, then
$$C = 24*{3\over8} = 9$$
Hence, the equivalent ratios are 3/8 and 9/24.
### Ratio rules
Given below are some rules followed by ratios.
i) If two ratios A/B and C/D are equal, then
• a) Invertendo $${B\over A} = {D\over C}$$
• b) Alternendo $${A \over C} = {B\over D}$$
• c) Componendo $${{A+B}\over B} = {{C+D}\over D}$$
• d) Dividendo $${{A-B}\over B} = {{C-D}\over D}$$
ii) The ratio remains the same if both the numerator and denominator are multiplied or divided by the same non-zero number.
$${A\over B} = {xA \over xB}, x \ne 0 \; and \; {A\over B} = {{A\over y}\over {B\over y}}, y \ne 0$$
iii) Ratios can be compared just like real numbers.
• $${A\over B} = {C \over D} \, is \, equivalent \,to \,AD \;= \;BC$$
• $${A\over B} > {C\over D} \;means \, that \, AD > BC$$
• $${A\over B} < {C \over D} \; means \, that \, AD < BC$$
### Ratio rules
Let us assume that we have a ratio of 5/6 and we want to scale it up to another equivalent ratio with the denominator 72. To do this, we set up a relation between the two ratios, where they are equal to each other, and we need to solve for the missing number. The steps for this are as follows:
• Write the ratios in terms of fractions, and label the missing part of the new number with some variable.
• Make the fractions equal to each other. This is called forming a proportion.
• Try to isolate the variable by cross-multiplying.
• Solve for the variable using one of the ratio formulas. The answer obtained is the required number.
• Consider the example described above. For the fraction having 72 as denominator, the numerator is named ‘x’. Therefore,
$${5\over 6} = {x \over 72}$$
$$5*72 = 6x$$
$${{5*72}\over 6} = x$$
$$x = 60$$
Hence, the new fraction is 60/72. The fractions 5/6 and 60/72 are equal to each other.
### Areas of application
Ratios help to describe mathematical relationships in real-life situations. Some examples are as follows.
1) Ratios are frequently used in cooking and baking. Every recipe mentions the exact quantity of each ingredients to be used in order to prepare a good meal. Messing up this ratio might affect the overall outcome. For example, for making brownies, two of the ingredients are flour and milk. The recipe requires 2 cups of flour and 1 cup of milk. This can be visualized as a ratio of 1 cup milk to 2 cups flour.
2) While grocery shopping, one might observe that different quantities of the same item are priced differently. For example, 500 ml of milk costs INR 25, whereas 1 litre of milk costs INR 50. The pricing and quantity might also vary according to the brands of the same product. For example, for brand A, a 200g box of muesli costs INR 220, and a different brand B has a 400g box of muesli priced at INR 350. By dividing the price of each box by the corresponding quantity of muesli, one can obtain a relationship between the cost and the quantity. In the second example, the brand B is cheaper, because 1g of muesli costs INR 0.875, whereas for brand A, 1g of muesli costs INR 1.1.
3) Ratios are also extremely useful for drawing relationships between distance and time. For example, the distance between Bangalore and Mysore is 143.6 km. While travelling from Bangalore to Mysore by car, where the average speed of the car is 60 kmph, the time taken to reach Mysore can be calculated by finding the ratio of the total distance by the speed. This ratio tells us that it takes approximately 2.39 hours to travel. |
High School Physics
# Collisions and Newton’s Laws of Motion – How to relate these?
Last updated on November 21st, 2021 at 10:07 am
This post covers how Collisions and Newton’s Laws of Motion are related. As collisions, we can take examples of the interaction between two bodies (gun-bullet, skater-skateboard, hose-water).
For simplicity let us consider a simple collision between two balls and find out how Newton’s Laws of Motion can be applied to analyze the events related to a collision.
## Collisions and Newton’s Laws of Motion
Let us apply Newton’s three laws to this problem.
### Newton’s first law is applied during a collision – how?
As per the diagram above, the red ball of mass m1 is moving towards the right with u1 velocity, and the blue ball of mass m2 is also moving towards the right but with a velocity u2 .
In the collision, the red ball slows down to v1 and the blue ball speeds up to v2. Newton’s first law tells us that this means there is a force acting on the red ball (F1) directing towards the left (against its motion) and thus the red ball slows down.
Similarly, a force is acting on the blue ball (F2) directing towards the right (along its initial direction) and helps the blue ball to speed up.
This is what is said in that a force is required to change the velocity of a motion.
### Newton’s second law is applied during a collision – how?
This law tells us that the force will be equal to the rate of change of momentum of the balls involved. now if the balls are touching each other for a time ∆t.
So we can write, F1 = (m1v1 -m1u1) / ∆t ……… (1)
and, F2 = (m2v2 -m2u2) / ∆t ……… (2)
### Newton’s third law is applied during a collision – how?
According to the third law, if the red ball exerts a force on the blue ball, then the blue ball will exert an equal and opposite force on the red ball.
That means, F1 = – F2
From equation (1) and (2) we can write,
(m1v1 -m1u1) / ∆t = – (m2v2 -m2u2) / ∆t
=> m1u1 + m2u2 = m1v1 + m2v2
In other words, the total momentum before the collision = total momentum after the collision.
This is what is stated by the Law of conservation of linear momentum.
## Related study (collisions)
Collisions – definitions, types, sample numerical
Numerical problems – collisions
Numerical problems – 2D collisions
Collisions & Newton’s Laws of motion
Types of collision, diagram, differences |
# Applications of linear equations
We have learned all about linear relationships, and linear functions. From what we learned about, we know that linear relationships and functions is consisted of two variables, one dependent, and the other independent. We also know that the line is a straight line. Linear equations are nonetheless the same, adding the criteria that its variables should not be raised to an exponent greater than one, and that the variables are not used as a denominator. We are to learn more of linear equations in this chapter, especially in section 1.
Apart from linear equations, we also need to learn about nonlinear equations. As the word suggests, these are the other equations that aren’t linear in nature, like the quadratic equations, circle equations, cubic equations, and more. We will have exercises in section 2 that will show us the difference between these equations and that of the linear equations.
As we have learned in the previous chapters, we know that linear equations show a straight line, that would appear to fall diagonally to on the Cartesian plane, to show the linear relationship between the values found in the line. However, in some special cases, linear equations can show horizontal lines and vertical lines. In chapter section 3 and 4 we will understand these special cases a bit more.
We are also going to learn how to find the parallel line and the perpendicular line of a given linear equation in section 5 and section 6. From our previous discussion in chapter, we learned that parallel lines have the same slope, whereas perpendicular lines have the slopes that are negative reciprocals with each other.
In section 7 and section 8 we have more practice in looking for the parallel lines and perpendicular lines, by looking for both of them for every equation given. There are also other applications of linear equation in section 8. To find out more about linear equations you can check out a video made by Education Alberta in Canada about “Exploring Linear Equations
### Applications of linear equations
We always encounter situations that can be expressed in linear equations. Watch this lesson and learn how to solve linear equations word problems. |
# Precalculus - Trigonometric Identities
## Introduction
• We know that equations involving trigonometric functions of a variable (angle) are termed Trigonometric equations.
• Trigonometric Identities are special trigonometric equations that hold true for all the values of the angles(s) involved.
• For example, $2\mathrm{sin}\theta -1=0$ is a trigonometric equation and not an identity, as it doesn't hold true for all the values of $\theta$. It is only true for some specific values of $\theta$.
## Trigonometric Identities
Trigonometric Identities can be divided into several groups, as mentioned below:
Reciprocal Identities
• or,
• or,
• or,
Quotient Identities
Pythagorean Identities
Sign Identities / Opposite Angle Identities
Cofunction Identities
Double Angle Identities
Triple Angle Identities
Sum and Difference of Angle Identities
## Solved Examples
Example 1: Find the value of .
Solution:
[since, $1-{\mathrm{cos}}^{2}\theta ={\mathrm{sin}}^{2}\theta$]
Example 2: Find the value of .
Solution:
[since, $\mathrm{cos}e{c}^{2}\theta =1+\mathrm{co}{t}^{2}\theta$ $⇒\mathrm{co}{t}^{2}\theta =\mathrm{cos}e{c}^{2}\theta -1$]
[since $\mathrm{cos}ec\theta =\frac{1}{\mathrm{sin}\theta }$]
$=1$
Example 3: Find the value of $\mathrm{cos}2\theta -2{\mathrm{cos}}^{2}\theta +1$.
Solution: $\mathrm{cos}2\theta -2{\mathrm{cos}}^{2}\theta +1$
$=\left(2{\mathrm{cos}}^{2}\theta -1\right)-2{\mathrm{cos}}^{2}\theta +1$ [since, $\mathrm{cos}2\theta =2{\mathrm{cos}}^{2}\theta -1$]
$=2{\mathrm{cos}}^{2}\theta -1-2{\mathrm{cos}}^{2}\theta +1$
$=0$
Example 4: Find the value of .
Solution:
[since $se{c}^{2}\theta =1+{\mathrm{tan}}^{2}\theta$ ]
[since, $sec\theta =\frac{1}{\mathrm{cos}\theta }$]
$=1$
Example 5: Verify the identity .
Solution:
Simplifying LHS
[since, $\mathrm{cos}e{c}^{2}\theta =1+co{t}^{2}\theta$]
[since, $\mathrm{cos}ec\theta =\frac{1}{\mathrm{sin}\theta }$ and $sec\theta =\frac{1}{\mathrm{cos}\theta }$]
[since, $cot\theta =\frac{\mathrm{cos}\theta }{\mathrm{sin}\theta }$]
[since, $cot\theta =\frac{1}{\mathrm{tan}\theta }$]
$=cot\theta$
$=RHS$
• or
• or
• or
• and
## Blunder Areas
• All the trigonometric identities are trigonometric equations but not all trigonometric equations are trigonometric identities. |
# How do you find the standard form of x^2 + 4y^2 - 8x + 16y + 16 =0 and what kind of a conic is it?
Dec 12, 2015
An ellipse with center $\left(4 , - 2\right)$, x-axis radius: $4$, and y-axis radius: $2$
#### Explanation:
Given:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 4 {y}^{2} - 8 x + 16 y + 16 = 0$
Group $x$ and $y$ terms (and constant) separately as
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 8 x} + \textcolor{red}{4 \left({y}^{2} + 4 y\right)} + \textcolor{g r e e n}{16} = 0$
Complete the squares for both the $x$ and the $y$ sub-expressions:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 8 x + 16} + \textcolor{red}{4 \left({y}^{2} + 4 y + 4\right)} + \textcolor{g r e e n}{16} - \textcolor{b l u e}{16} - \textcolor{red}{4 \left(4\right)} = 0$
Reduce $x$ and $y$ sub-expressions to squared binomials and move constant to the right side:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 4\right)}^{2} + 4 {\left(y + 2\right)}^{2} = 16$
or
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 4\right)}^{2} + 4 {\left(y + 2\right)}^{2} = {4}^{2}$
Divide both sides by $\left({4}^{2}\right)$
and replace $\left(y + 2\right)$ with $\left(y - \left(- 2\right)\right)$
$\textcolor{w h i t e}{\text{XXX}} \frac{{\left(x - 4\right)}^{2}}{{4}^{2}} + {\left(y - \left(- 2\right)\right)}^{2} / \left({2}^{2}\right) = 1$
Note that the general form of an ellipse is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - h\right)}^{2} / \left({a}^{2}\right) + \frac{{\left(y - k\right)}^{2}}{{b}^{2}} = 1$
with
$\textcolor{w h i t e}{\text{XXX}}$center at $\left(h , k\right)$,
$\textcolor{w h i t e}{\text{XXX}}$x-axis radius: $\left(a\right)$, and
$\textcolor{w h i t e}{\text{XXX}}$y-axis radius: $\left(b\right)$
graph{x^2+4y^2-8x+16y+16=0 [-2.195, 10.29, -4.895, 1.345]} |
# Problem
Problem Link – UVa 10407 – Simple division
Given an array of numbers, find the largest number $d$ such that, when elements of the array are divided by $d$, they leave the same remainder
# Solution
We will be using our knowledge about Congruence Relation and GCD to solve this problem. But first, we need to make sure that we understand the problem properly.
## Wrong Approach
At first, one might think that $d$ is the $gcd$ of array elements. Surely, dividing elements of the array with $d$ will leave same remainder $0$, but that doesn’t necessarily mean it is the largest possible answer.
For example, suppose the array is $A = {2,8,14,26}$. Then what will be the $gcd$ of array A? $gcd(2,8,14,26) = 2$. Dividing elements of $A$ with $2$ leaves us $0$. So is this our answer? No.
For array $A$, there exists a number greater than $2$ that leaves the same remainder. And that number is $3$. When we divide each element with $3$, it leaves us $2$ as the remainder.
The problem did not ask us to find the number that will leave $0$ as remainder. It asked us to find the largest number $d$ that leaves the same remainder. So for array $A$ the $gcd(A)$ is not the answer. Is $3$ our answer? No. The answer is given below in Example.
## Using Congruence Relation
Let us rephrase the problem using congruence relation. Suppose there is an array with elements, $A={a,b,c}$. We need to find the largest number $d$ that leaves the same remainder for each of its element.
Let us consider only ${a,b}$ for now. We need to find $d$ such that it leaves the same remainder when it divides $a$ and $b$. Meaning
$$a \ \equiv \ b \ \text{(mod d)} \ a – b \ \equiv \ 0 \ \text{(mod d)}$$
What does this mean? It means, if $d$ leaves the same remainder when it divides $a$ and $b$, then it will leave no remainder when dividing $a-b$.
Using same logic, we can say that $b – c \ \equiv \ 0 \ \text{(mod d)}$. Therefore, if we find a number that divides $a-b$ and $b-c$, then that number will leave the same remainder when dividing ${a,b,c}$.
This idea can be extended for any number of elements in the array. So $d$ is such a number that leaves no remainder when it divides the difference of adjacent elements in the array.
But wait, there are multiple numbers that can divide the difference of adjacent elements. Which one should we take? Since we want the largest value of $d$, we will take the largest divisor that can divide all the difference of adjacent numbers. There is only one divisor that can divide all the adjacent differences, and that is $gcd( (a-b), (b-c) )$.
Therefore, if we have an array $A={a_1,a_2,a_3…a_n}$, then $d = gcd( a_2 – a_1, a_3- a_2,…a_n – a_{n_1})$.
Careful about negative value of $gcd()$. Make sure you take the absolute value of $gcd()$.
# Example
Let us get back to the example we were looking into before. Finding $d$ for the array $A = {2,8,14,26}$. We know that $d$ will divide the difference of adjacent elements completely. So let us make another array that will contain the difference of adjacent elements., $B = {8-2, 14-8, 26-14} = {6,6,12}$. Therefore, $d = gcd ( 6,6,12 ) = 6$.
# Summary
• Let the array of elements be $A = {a,b,c,d…}$.
• $res \neq gcd(a,b,c,d…)$.
• $res = gcd ( a-b,b-c,c-d,…)$.
# Code
#include <bits/stdc++.h>
using namespace std;
typedef long long vlong;
vlong gcd ( vlong a, vlong b ) {
while ( b ) {
a = a % b;
swap ( a, b );
}
return a;
}
vlong arr[1010];
int main () {
while ( scanf ( "%d", &arr[0] ) != EOF ) {
if ( arr[0] == 0 ) break; // End of test case
// A new test case has started
int cur = 1;
// Take input
while ( 1 ) {
scanf ( "%lld", &arr[cur] );
if ( arr[cur] == 0 ) break;
else cur++;
}
vlong g = 0; // Start with 0 since gcd(0,x) = x.
for ( int i = 1; i < cur; i++ ) {
int dif = arr[i] - arr[i-1]; // Calculate difference
g = gcd ( g, dif ); // Find gcd() of differences
}
if ( g < 0 ) g *= -1; // In case gcd() comes out negative
printf ( "%lld\n", g );
}
return 0;
} |
Active Prelude to Calculus
Section3.5Properties and applications of logarithmic functions
Logarithms arise as inverses of exponential functions. In addition, we have motivated their development by our desire to solve exponential equations such as $$e^k = 3$$ for $$k\text{.}$$ Because of the inverse relationship between exponential and logarithmic functions, there are several important properties logarithms have that are analogous to ones held by exponential functions. We will work to develop these properties and then show how they are useful in applied settings.
Preview Activity3.5.1.
In the following questions, we investigate how $$\log_{10}(a \cdot b)$$ can be equivalently written in terms of $$\log_{10}(a)$$ and $$\log_{10}(b)\text{.}$$
1. Write $$10^x \cdot 10^y$$ as $$10$$ raised to a single power. That is, complete the equation
\begin{equation*} 10^x \cdot 10^y = 10^{\Box} \end{equation*}
by filling in the box with an appropriate expression involving $$x$$ and $$y\text{.}$$
2. What is the simplest possible way to write $$\log_{10}10^x\text{?}$$ What about the simplest equivalent expression for $$\log_{10}10^y\text{?}$$
3. Explain why each of the following three equal signs is valid in the sequence of equalities:
\begin{align*} \log_{10}(10^x \cdot 10^y) &= \log_{10}(10^{x+y})\\ &= x+y\\ &= \log_{10}(10^x) + \log_{10}(10^y)\text{.} \end{align*}
4. Suppose that $$a$$ and $$b$$ are positive real numbers, so we can think of $$a$$ as $$10^x$$ for some real number $$x$$ and $$b$$ as $$10^y$$ for some real number $$y\text{.}$$ That is, say that $$a = 10^x$$ and $$b = 10^y\text{.}$$ What does our work in (c) tell us about $$\log_{10}(ab)\text{?}$$
Subsection3.5.1Key properties of logarithms
In Preview Activity 3.5.1, we considered an argument for why $$\log_{10}(ab) = \log_{10}(a) + \log_{10}(b)$$ for any choice of positive numbers $$a$$ and $$b\text{.}$$ In what follows, we develop this and other properties of the natural logarithm function; similar reasoning shows the same properties hold for logarithms of any base.
Let $$a$$ and $$b$$ be any positive real numbers so that $$x = \ln(a)$$ and $$y = \ln(b)$$ are both defined. Observe that we can rewrite these two equations using the definition of the natural logarithm so that
\begin{equation*} a = e^x \ \text{ and } \ b = e^y\text{.} \end{equation*}
Using substitution, we can now say that
\begin{equation*} \ln(a \cdot b) = \ln(e^x \cdot e^y)\text{.} \end{equation*}
By exponent rules, we know that $$\ln(e^x \cdot e^y) = \ln(e^{x+y})\text{,}$$ and because the natural logarithm and natural exponential function are inverses, $$\ln(e^{x+y}) = x+y\text{.}$$ Combining the three most recent equations,
\begin{equation*} \ln(a \cdot b) = x + y\text{.} \end{equation*}
Finally, recalling that $$x = \ln(a)$$ and $$y = \ln(b)\text{,}$$ we have shown that
\begin{equation*} \ln(a \cdot b) = \ln(a) + \ln(b) \end{equation*}
for any choice of positive real numbers $$a$$ and $$b\text{.}$$
A similar property holds for $$\ln(\frac{a}{b})\text{.}$$ By nearly the same argument, we can say that
\begin{align*} \ln\left( \frac{a}{b} \right) &= \ln\left( \frac{e^x}{e^y} \right)\\ &= \ln \left( e^{x-y} \right)\\ &= x-y\\ &= \ln(a) - \ln(b)\text{.} \end{align*}
We have thus shown the following general principles.
Logarithms of products and quotients.
For any positive real numbers $$a$$ and $$b\text{,}$$
• $$\displaystyle \ln(a \cdot b) = \ln(a) + \ln(b)$$
• $$\displaystyle \ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b)$$
Because positive integer exponents are a shorthand way to express repeated multiplication, we can use the multiplication rule for logarithms to think about exponents as well. For example,
\begin{equation*} \ln(a^3) = \ln(a \cdot a \cdot a)\text{,} \end{equation*}
and by repeated application of the rule for the natural logarithm of a product, we see
\begin{equation*} \ln(a^3) = \ln(a) + \ln(a) + \ln(a) = 3\ln(a)\text{.} \end{equation*}
A similar argument works to show that for every natural number $$n\text{,}$$
\begin{equation*} \ln(a^n) = n\ln(a)\text{.} \end{equation*}
More sophisticated mathematics can be used to prove that the following property holds for every real number exponent $$t\text{.}$$
Logarithms of exponential expressions.
For any positive real number $$a$$ and any real number $$t\text{,}$$
\begin{equation*} \ln(a^t) = t\ln(a)\text{.} \end{equation*}
The rule that $$\ln(a^t) = t\ln(a)$$ is extremely powerful: by working with logarithms appropriately, it enables us to move from having a variable in an exponential expression to the variable being part of a linear expression. Moreover, it enables us to solve exponential equations exactly, regardless of the base involved.
Example3.5.1.
Solve the equation $$7 \cdot 3^t - 1 = 5$$ exactly for $$t\text{.}$$
Solution.
To solve for $$t\text{,}$$ we first solve for $$3^t\text{.}$$ Adding $$1$$ to both sides and dividing by $$7\text{,}$$ we find that $$3^t = \frac{6}{7} \text{.}$$ Next, we take the natural logarithm of both sides of the equation. Doing so, we have
\begin{equation*} \ln \left( 3^t \right) = \ln \left( \frac{6}{7} \right)\text{.} \end{equation*}
Applying the rule for the logarithm of an exponential expression on the left, we see that $$t \ln(3) = \ln \left( \frac{6}{7} \right) \text{.}$$ Both $$\ln(3)$$ and $$\ln \left( \frac{6}{7} \right)$$ are simply numbers, and thus we conclude that
\begin{equation*} t = \frac{\ln \left( \frac{6}{7} \right)}{\ln(3)}\text{.} \end{equation*}
The approach used in Example 3.5.1 works in a wide range of settings: any time we have an exponential equation of the form $$p \cdot q^t + r = s\text{,}$$ we can solve for $$t$$ by first isolating the exponential expression $$q^t$$ and then by taking the natural logarithm of both sides of the equation.
Activity3.5.2.
Solve each of the following equations exactly and then find an estimate that is accurate to 5 decimal places.
1. $$\displaystyle 3^t = 5$$
2. $$\displaystyle 4 \cdot 2^t - 2 = 3$$
3. $$\displaystyle 3.7 \cdot (0.9)^{0.3t} + 1.5 = 2.1$$
4. $$\displaystyle 72 - 30(0.7)^{0.05t} = 60$$
5. $$\displaystyle \ln(t) = -2$$
6. $$\displaystyle 3 + 2\log_{10}(t) = 3.5$$
Subsection3.5.2The graph of the natural logarithm
As the inverse of the natural exponential function $$E(x) = e^x\text{,}$$ we have already established that the natural logarithm $$N(x) = \ln(x)$$ has the set of all positive real numbers as its domain and the set of all real numbers as its range. In addition, being the inverse of $$E(x) = e^x\text{,}$$ we know that when we plot the natural logarithm and natural exponential functions on the same coordinate axes, their graphs are reflections of one another across the line $$y = x\text{,}$$ as seen in Figure 3.5.2 and Figure 3.5.3.
Indeed, for any point $$(a,b)$$ that lies on the graph of $$E(x) = e^x\text{,}$$ it follows that the point $$(b,a)$$ lies on the graph of the inverse $$N(x) = \ln(x)\text{.}$$ From this, we see several important properties of the graph of the logarithm function.
The graph of $$y = \ln(x)$$.
The graph of $$y = \ln(x)$$
• passes through the point $$(1,0)\text{;}$$
• is always increasing;
• is always concave down; and
• increases without bound.
Because the graph of $$E(x) = e^x$$ increases more and more rapidly as $$x$$ increases, the graph of $$N(x) = \ln(x)$$ increases more and more slowly as $$x$$ increases. Even though the natural logarithm function grows very slowly, it does grow without bound because we can make $$\ln(x)$$ as large as we want by making $$x$$ sufficiently large. For instance, if we want $$x$$ such that $$\ln(x) = 100\text{,}$$ we choose $$x = e^{100}\text{,}$$ since $$\ln(e^{100}) = 100\text{.}$$
While the natural exponential function and the natural logarithm (and transformations of these functions) are connected and have certain similar properties, it’s also important to be able to distinguish between behavior that is fundamentally exponential and fundamentally logarithmic.
Activity3.5.3.
In the questions that follow, we compare and contrast the properties and behaviors of exponential and logarithmic functions.
1. Let $$f(t) = 1 - e^{-(t-1)}$$ and $$g(t) = \ln(t)\text{.}$$ Plot each function on the same set of coordinate axes. What properties do the two functions have in common? For what properties do the two functions differ? Consider each function’s domain, range, $$t$$-intercept, $$y$$-intercept, increasing/decreasing behavior, concavity, and long-term behavior.
2. Let $$h(t) = a - be^{-k(t-c)}\text{,}$$ where $$a\text{,}$$ $$b\text{,}$$ $$c\text{,}$$ and $$k$$ are positive constants. Describe $$h$$ as a transformation of the function $$E(t) = e^t\text{.}$$
3. Let $$r(t) = a + b\ln(t-c)\text{,}$$ where $$a\text{,}$$ $$b\text{,}$$ and $$c$$ are positive constants. Describe $$r$$ as a transformation of the function $$L(t) = \ln(t)\text{.}$$
4. Data for the height of a tree is given in the Table 3.5.4; time $$t$$ is measured in years and height is given in feet. At http://gvsu.edu/s/0yy 1 , you can find a Desmos worksheet with this data already input.
Do you think this data is better modeled by a logarithmic function of form $$p(t) = a + b\ln(t-c)$$ or by an exponential function of form $$q(t) = m + ne^{-rt}\text{.}$$ Provide reasons based in how the data appears and how you think a tree grows, as well as by experimenting with sliders appropriately in Desmos. (Note: you may need to adjust the upper and lower bounds of several of the sliders in order to match the data well.)
Subsection3.5.3Putting logarithms to work
We’ve seen in several different settings that the function $$e^{kt}$$ plays a key role in modeling phenomena in the world around us. We also understand that the value of $$k$$ controls whether $$e^{kt}$$ is increasing ($$k \gt 0$$) or decreasing ($$k \lt 0$$) and how fast the function is increasing or decreasing. As such, we often need to determine the value of $$k$$ from data that is presented to us; doing so almost always requires the use of logarithms.
Example3.5.5.
A population of bacteria cells is growing at a rate proportionate to the number of cells present at a given time $$t$$ (in hours). Suppose that the number of cells, $$P\text{,}$$ in the population is measured in millions of cells and we know that $$P(0) = 2.475$$ and $$P(10) = 4.298\text{.}$$ Find a model of the form $$P(t) = Ae^{kt}$$ that fits this data and use it to determine the value of $$k$$ and how long it will take for the population to reach $$1$$ billion cells.
Solution.
Since the model has form $$P(t) = Ae^{kt}\text{,}$$ we know that $$P(0) = A\text{.}$$ Because we are given that $$P(0) = 2.475\text{,}$$ this shows that $$A = 2.475\text{.}$$ To find $$k\text{,}$$ we use the fact that $$P(10) = 4.298\text{.}$$ Applying this information, $$A = 2.475\text{,}$$ and the form of the model, $$P(t) = Ae^{kt}\text{,}$$ we see that
\begin{equation*} 4.298 = 2.475 e^{k \cdot 10}\text{.} \end{equation*}
To solve for $$k\text{,}$$ we first isolate $$e^{10k}$$ by dividing both sides by $$2.475$$ to get
\begin{equation*} e^{10k} = \frac{4.298}{2.475} \text{.} \end{equation*}
Taking the natural logarithm of each side, we find
\begin{equation*} 10k = \ln \left( \frac{4.298}{2.475} \right)\text{,} \end{equation*}
and thus $$k = \frac{1}{10}\ln \left( \frac{4.298}{2.475} \right) \approx 0.05519\text{.}$$
To determine how long it takes for the population to reach $$1$$ billion cells, we need to solve the equation $$P(t) = 1000\text{.}$$ Using our preceding work to find $$A$$ and $$k\text{,}$$ we know that we need to solve the equation
\begin{equation*} 1000 = 2.475 e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t}\text{.} \end{equation*}
We divide both sides by $$2.475$$ to get $$e^{\frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t} = \frac{1000}{2.475}\text{,}$$ and after taking the natural logarithm of each side, we see
\begin{equation*} \frac{1}{10}\ln \left( \frac{4.298}{2.475} \right)t = \ln \left( \frac{1000}{2.475} \right)\text{,} \end{equation*}
so that
\begin{equation*} t = \frac{10 \ln \left( \frac{1000}{2.475} \right)}{\ln \left( \frac{4.298}{2.475} \right)} \approx 108.741\text{.} \end{equation*}
Activity3.5.4.
Solve each of the following equations for the exact value of $$k\text{.}$$
1. $$\displaystyle 41 = 50e^{-k \cdot 7}$$
2. $$\displaystyle 65 = 34 + 47e^{-k \cdot 45}$$
3. $$\displaystyle 7e^{2k-1} + 4 = 32$$
4. $$\displaystyle \frac{5}{1+2e^{-10k}} = 4$$
Subsection3.5.4Summary
• There are three fundamental rules for exponents given nonzero base $$a$$ and exponents $$m$$ and $$n\text{:}$$
\begin{equation*} a^m \cdot a^n = a^{m+n}, \frac{a^m}{a^n} = a^{m-n}, \text{ and } (a^m)^n = a^{mn}\text{.} \end{equation*}
For logarithms 2 , we have the following analogous structural rules for positive real numbers $$a$$ and $$b$$ and any real number $$t\text{:}$$
\begin{equation*} \ln(a \cdot b) = \ln(a) + \ln(b), \ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b), \text{ and } \ln(a^t) = t \ln(a)\text{.} \end{equation*}
• The natural logarithm’s domain is the set of all positive real numbers and its range is the set of all real numbers. Its graph passes through $$(1,0)\text{,}$$ is always increasing, is always concave down, and increases without bound.
• Logarithms are very important in determining values that arise in equations of the form
\begin{equation*} a^b = c\text{,} \end{equation*}
where $$a$$ and $$c$$ are known, but $$b$$ is not. In this context, we can take the natural logarithm of both sides of the equation to find that
\begin{equation*} \ln(a^b) = \ln(c) \end{equation*}
and thus $$b\ln(a) = \ln(c)\text{,}$$ so that $$b = \frac{\ln(c)}{\ln(a)}\text{.}$$
Exercises3.5.5Exercises
1.
Solve for $$x\text{:}$$
\begin{equation*} 3^{x} = 38 \end{equation*}
$$x =$$
2.
Solve for $$x\text{:}$$
\begin{equation*} 6 \cdot 4^{4 x -4} = 65 \end{equation*}
$$x =$$
3.
Find the solution of the exponential equation
\begin{equation*} 11+5^{5x}=16 \end{equation*}
correct to at least four decimal places.
$$x=$$
4.
Find the solution of the exponential equation
\begin{equation*} 1000 (1.04)^{2t} = 50000 \end{equation*}
in terms of logarithms, or correct to four decimal places.
$$t=$$
5.
Find the solution of the logarithmic equation
\begin{equation*} 19 - \ln(3-x)=0 \end{equation*}
correct to four decimal places.
$$x=$$
6.
For a population that is growing exponentially according to a model of the form $$P(t) = Ae^{kt}\text{,}$$ the doubling time is the amount of time that it takes the population to double. For each population described below, assume the function is growing exponentially according to a model $$P(t) = Ae^{kt}\text{,}$$ where $$t$$ is measured in years.
1. Suppose that a certain population initially has $$100$$ members and doubles after $$3$$ years. What are the values of $$A$$ and $$k$$ in the model?
2. A different population is observed to satisfy $$P(4) = 250$$ and $$P(11) = 500\text{.}$$ What is the population’s doubling time? When will $$2000$$ members of the population be present?
3. Another population is observed to have doubling time $$t = 21\text{.}$$ What is the value of $$k$$ in the model?
4. How is $$k$$ related to a population’s doubling time, regardless of how long the doubling time is?
7.
A new car is purchased for $$$28000\text{.}$$ Exactly $$1$$ year later, the value of the car is$$$23200\text{.}$$ Assume that the car’s value in dollars, $$V\text{,}$$ $$t$$ years after purchase decays exponentially according to a model of form $$V(t) = Ae^{-kt}\text{.}$$
1. Determine the exact values of $$A$$ and $$k$$ in the model.
2. How many years will it take until the car’s value is $$$10000\text{?}$$ 3. Suppose that rather than having the car’s value decay all the way to$$$0\text{,}$$ the lowest dollar amount its value ever approaches is $$$500\text{.}$$ Explain why a model of the form $$V(t) = Ae^{-kt} + c$$ is more appropriate. 4. Under the original assumptions ($$V(0) = 28000$$ and $$V(1) = 23200$$) along with the condition in (c) that the car’s value will approach$$$500$$ in the long-term, determine the exact values of $$A\text{,}$$ $$k\text{,}$$ and $$c$$ in the model $$V(t) = Ae^{-kt} + c\text{.}$$ Are the values of $$A$$ and $$k$$ the same or different from the model explored in (a)? Why?
8.
In Exercise 3.4.5.7, we explored graphically how the function $$y = \log_b(x)$$ can be thought of as a vertical stretch of the nautral logarithm, $$y = \ln(x)\text{.}$$ In this exercise, we determine the exact value of the vertical stretch that is needed.
Recall that $$\log_b(x)$$ is the power to which we raise $$b$$ to get $$x\text{.}$$
1. Write the equation $$y = \log_b(x)$$ as an equivalent equation involving exponents with no logarithms present.
2. Take the equation you found in (a) and take the natural logarithm of each side.
3. Use rules and properties of logarithms appropriately to solve the equation from (b) for $$y\text{.}$$ Your result here should express $$y$$ in terms of $$\ln(x)$$ and $$\ln(b)\text{.}$$
4. Recall that $$y = \log_b(x)\text{.}$$ Explain why the following equation (often called the Golden Rule for Logarithms) is true:
\begin{equation*} \log_b(x) = \frac{\ln(x)}{\ln(b)}\text{.} \end{equation*}
5. What is the value of $$k$$ that allows us to express the function $$y = \log_b(x)$$ as a vertical stretch of the function $$y = \ln(x)\text{?}$$ |
Explorations with
by: Lauren Wright
In this assignment, we explore the effects of changing a on the graph of
First, let's examine this graph for a > 0.
Looking at these graphs we can draw the conclusion that the bigger a is (or the further a is from zero), the more contracted the parabola becomes.
Now, let's look at the graph for a < 0.
It appears as though we can draw the same conclusion when a < 0. The smaller a gets (or the further a is from zero), the more contracted the parabolas become.
Now, let's look at the relationship between and . Looking at the previous two graphs, we could guess that making a negative simply reflects the parabola about the x-axis. Let's investigate this a bit further.
Well, it looks as though our conjecture was true!
From these graphs, we can see that the parabolas open up when a > 0.
They open down when a < 0.
But, what happens if -1 < a < 1 ? Let's look at a few of those.
First, we'll look at -1 < a < 0.
WOW! If we look at this graph on the same scale as the others, many of the parabolas are too wide for us to see. So, let's change the scale of the graph and get a better look at them.
That's better. Now we can see just how wide the parabolas really are. It appears that when
-1 < a < 0, the parabolas expand as a gets closer to zero.
Now, let's look at 0 < a < 1.
Well, it looks like the same thing is true for 0 < a < 1. The parabolas expand as a gets closer to zero.
We can explore a bit further with a movie that lets a vary between -5 and 5.
Let's summarize what we've found.
When working with the graph of :
1. If a > 0, then the parabola will open up. If a < 0, the parabola will open down.
2. The further a is from zero, the more contracted the parabola will become; OR, the closer a is to zero, the more expanded the parabola will become. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Tangents to a Curve
## Slope of a tangent equals a limit where difference in x values approaches zero.
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Introduction to Derivatives
### Tangents to a Curve
#### Vocabulary
Secant Line: a line that cuts across a curve; a reference for the slope of a line tangent to a curve
Tangent Line: a line that "just touches" a curve at a single point and no others
Differential Calculus: a study of calculus based on finding the difference in location between two points that get closer together until the distance between them is infinitely small
#### Slope of the Tangent Line
Remember in Algebra 1 how you learned about finding the slope of a line? What is the formula for slope of a line?
The tangent line represents the slope of an individual point on a curve. You find it by finding the slope of two points with an infinitely small distance between them. By decreasing the distance between the two points, the slope of the secant line gets closer and closer to the slope of the tangent line.
The slope of the secant line is found by the equation:
msec\begin{align*}m_{sec}\end{align*}=y1y0x1x0=f(x1)f(x0)x1x0\begin{align*}= \frac {y_1-y_0}{x_1-x_0} = \frac {f(x_1)-f(x_0)}{x_1-x_0}\end{align*}
As we bring the points closer together, x1 approaches x2 so we an write this as a limit statement:
msec\begin{align*}m_{sec}\end{align*} =limx1x0f(x1)f(x0)x1x0\begin{align*}= \lim_{x_1 \to x_0} \frac {f(x_1)-f(x_0)}{x_1-x_0} \end{align*}
To simplify our notation, if we let − , then and x → x becomes equivalent to h → 0. This means that the equation becomes:
msec\begin{align*}m_{sec}\end{align*}=limh0f(x0+h)f(x0)h\begin{align*}= \lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*}
The Slope of a Tangent Line
If the point ) is on the curve , then the tangent line at has a slope that is given by
mtan=limh0f(x0+h)f(x0)h\begin{align*}m_{tan}=\lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*}
provided that the limit exist.
.
.
Once you know the slope you can find the equation of the tangent line by using point-slope form of a line: y − y0 = mtan ( x − x0 ).
.
#### Practice
##### Definitions
1. The line connecting two points on a curve is the ________________.
2. The line that represents the slope of a single point on a curve is the ________________.
3. The derivative is another name for the _________________.
4. When calculating the slope of a tangent, what value is assumed to go to 0 as the two chosen points get closer and closer?
5. Why do we use a limit statement to calculate the slope of the tangent line?
##### Write the equation of the tangent line when:
1. x = 3, g(3) = 5, g'(3) = 7
2. x = 9, g(9) = 27, g'(9) = 1/2
3. x = -12, g(-12) = 4, g'(-12) = -3
##### Find the equation of the tangent line:
1. Find the equation of the tangent line to the graph of m(x)=3x3+3x2+4x+4\begin{align*}m(x) = 3x^3 + 3x^2 + 4x + 4\end{align*} at x=1\begin{align*}x = 1\end{align*}
2. Find the equation of the tangent line to the graph of m(x)=x\begin{align*}m(x) = x \end{align*} at x=0\begin{align*}x = 0\end{align*}
3. Find the equation of the tangent line to the graph of b(x)=5x4+3x3x2+5x3\begin{align*}b(x) = -5x^4 + 3x^3 - x^2 + 5x - 3 \end{align*}at x=1\begin{align*}x = -1\end{align*}
### Instantaneous Rates of Change
#### Vocabulary
Derivative: the slope of the tangent line to a single point on a graph
Instantaneous Speed: the speed an object is travelling at any given single point in time
Average Speed: the distance an object travels divided by the time interval required
.
The notation f ' ( is read as "f prime of x", and is one way to denote the derivative of function ( x)
Alternate derivative notations include:
f(x)\begin{align*}f'(x)\end{align*} dydx\begin{align*}\frac {dy}{dx}\end{align*} y\begin{align*}y'\end{align*} dfdx\begin{align*}\frac{df}{dx}\end{align*} df(x)dx\begin{align*}\frac {df(x)}{dx}\end{align*}
#### Rates of Change
The slope of the tangent line is also called the derivative. The derivative of a function f ( x ) can be written as f ' ( x ).
The Derivative
The function ' is defined by the formula
f(x)=limh0f(x+h)f(x)h\begin{align*}f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)} {h}\end{align*}
where ' is called the derivative of with respect to . The domain of consists of all the values of for which the limit exists.
Speed is calculated by taking the Average Rate of Change (AROC), or the slope of the secant line, while the derivative is the Instantaneous Rate of Change (IROC), or the slope of the tangent line. Can you think of two situations where the AROC would be useful? What about the IROC?
.
##### Finding Average Velocity
The average speed of an object is defined as the object's displacement ∆ divided by the time interval ∆ during which the displacement occurs:
Average speed =v=xt=x1x0t1t0\begin{align*}\text{Average speed }= v=\frac {\triangle x}{\triangle t}=\frac {x_1-x_0}{t_1-t_0}\end{align*}
Notice that the points( ) and ( ) lie on the position versus time curve, as the figure below shows.
This expression is also the expression for the slope of a secant line connecting the two points. Therefore, the average velocity is equal to the AROC of a function over an interval.
.
The IROC is the object's velocity at that point, so it is equal to the IROC, or the derivative.
#### Practice
##### Find the Average Rate of Change
1. AROC of the function f(x)=x24x+2\begin{align*}f(x) = x^2 - 4x + 2\end{align*} over the interval x = 12 to x = 30
2. AROC of the function f(x)=x25x+201\begin{align*} f(x) = x^2 - 5x + 201\end{align*} over the interval x = 10 to x = 11
3. AROC of the function f(x)=5x23x4\begin{align*}f(x) = -5x^2 - 3x - 4\end{align*} over the interval x = 29 to x = 48
Hint: Find the slope of the line that connects the two endpoints
.
##### Find the Instantaneous Rate of Change (with respect to x)
1. f(x)=3x2x5\begin{align*}f(x) = 3x^2 - x - 5\end{align*} at x = 15
2. f(x)=4x2+195\begin{align*}f(x) = 4x^2 + 195\end{align*} at x = 10
3. f(x)=x2+x3\begin{align*}f(x) = -x^2 + x - 3\end{align*} at x = 20
Hint: Use the Definition of the Derivative
.
##### Solve Rate of Change Problems
1. The number of people in the US affected by the common cold in the month of November is defined by N = f(x) where x is the day of the month. What is the meaning of f'(x) in this context?
2. The number of households in Florida affected by hurricane season in the month of July is defined by J = f(x) where x is the day of the month. f(x)=2x2+x+1\begin{align*}f(x) = 2x^2 + x + 1\end{align*} Find the average rate of change of J with respect to x when days is changed from x = 5 to x = 34
3. A cake is taken from an oven when its temperature is 196°F and placed on a cooling rack in a room where the temperature is 75°F. The temperature of the cake over (x) minutes is given by H = f(x). f(x)=4x2+15x+196\begin{align*}f(x) = 4x^2 + 15x + 196\end{align*} Find the instantaneous rate of change of H with respect to x when x = 15. |
## Linear Dependence in Rank Method
In this page linear dependence in rank method we are going to see some example problem to understand how to test whether the given vectors are linear dependent.
Procedure for Method II
• First we have to write the given vectors as row vectors in the form of matrix.
• Next we have to use elementary row operations on this matrix in which all the element in the nth column below the nth element are zero.
• The row which is having every element zero should be below the non zero row.
• Now we have to count the number of non zero vectors in the reduced form. If number of non zero vectors = number of given vectors,then we can decide that the vectors are linearly independent. Otherwise we can say it is linearly dependent.
Example 1:
Test whether the vectors (1,-1,1), (2,1,1) and (3,0,2) are linearly dependent using rank method. linear dependence in rank method
Solution:
˜
1 -1 1 2 1 1 3 0 2
R₂ => R₂ - 2R₁ 2 1 1 2 -2 2 (-) (+) (-) _________________ 0 3 -1 ________________ R₃ => R₃ - 3R₁ 3 0 2 3 -3 3 (-) (+) (-) _____________ 0 3 -1 _____________
˜
1 -1 1 0 3 -1 0 3 -1
R₂ => R₂ - 3R₁
R₃ => R₃ - 2R₁
R₃ => R₃ - R₂ 0 3 -1 0 3 -1 (-) (-) (+) _______________ 0 0 0 _______________
linear dependence in rank method
˜
1 1 -1 0 3 -1 0 0 0
R₃ => R₃ - R
Number of non zero rows is 2. So rank of the given matrix = 2.
If number of non zero vectors = number of given vectors,then we can decide that the vectors are linearly independent. Otherwise we can say it is linearly dependent.
Here rank of the given matrix is 2 which is less than the number of given vectors.So that we can decide the given vectors are linearly dependent.
Linear Dependence1 Rank Method to Method 1
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How to find the Surface Area of a Square Pyramid?
• Last Updated : 23 May, 2022
A pyramid is defined as a three-dimensional polyhedron with three or more triangle-shaped faces that meet above the base and a polygonal base. The triangle sides are the faces, while the apex is the point above the base. To construct a pyramid, the base is connected to the summit. The pyramid is known as a square pyramid when its base is square. A square pyramid has three triangular sides and a square base. In other terms, it has 8 edges, 5 vertices, and 4 faces.
Total Surface Area Formula of Square Pyramid
A square pyramid’s total surface area is equal to the total area covered by the four triangular sides and a square base. Its formula is equal to the sum of the base area and twice the product of base length and slant height.
TSA = a2 + 2al
where,
• TSA is the total surface area,
• a is the base length,
• l is the slant height.
In terms of base length and height of the pyramid, the formula is expressed as:
TSA = a2 + 2a √(a2/4 + h2)
where,
• TSA is the total surface area,
• a is the base length,
• h is the height or altitude.
Lateral Surface Area Formula of Square Pyramid
A square pyramid’s lateral surface area is defined as the area covered by its four triangular faces. Its formula is equal to twice the product of base length and slant height. It can be interpreted as the total surface area reduced by the base area of a square pyramid.
LSA = 2al
where,
• LSA is the total surface area,
• a is the base length,
• l is the slant height.
In terms of base length and height of the pyramid, the formula is expressed as:
LSA = 2a √(a2/4 + h2)
where,
• LSA is the lateral surface area,
• a is the base length,
• h is the height or altitude.
Sample Problems
Problem 1: Calculate the total surface area of a square pyramid if its base is 10 cm and slant height is 13 cm.
Solution:
We have,
a = 10
l = 13
Using the formula we have,
TSA = a2 + 2al
= (10 × 10) + (2 × 10 × 13)
= 100 + 260
= 360 sq. cm
Problem 2: Calculate the total surface area of a square pyramid if its base is 6 cm and slant height is 8.54 cm.
Solution:
We have,
a = 6
l = 8.54
Using the formula we have,
TSA = a2 + 2al
= (6 × 6) + (2 × 6 × 8.54)
= 36 + 102.53
= 138.53 sq. cm
Problem 3: Calculate the total surface area of a square pyramid if its base is 11 cm and height is 9 cm.
Solution:
We have,
a = 11
h = 9
Using the formula we have,
TSA = a2 + 2a√(a2/4 + h2)
= (11 × 11) + (2 × 11 √(112/4 + 92))
= 121 + 232.05
= 353.05 sq. cm
Problem 4: Calculate the total surface area of a square pyramid if its base is 14 cm and height is 10 cm.
Solution:
We have,
a = 14
h = 10
Using the formula we have,
TSA = a2 + 2a√(a2/4 + h2)
= (14 × 14) + (2 × 14 √(142/4 + 102))
= 196 + 341.8
= 537.8 sq. cm
Problem 5: Calculate the lateral surface area of a square pyramid if its base is 3 cm and slant height is 4.27 cm.
Solution:
We have,
a = 3
l = 4.27
Using the formula we have,
LSA = 2al
= 2 × 13 × 4.27
= 25.63 sq. cm
Problem 6: Calculate the lateral surface area of a square pyramid if its base is 13 cm and height is 10 cm.
Solution:
We have,
a = 13
h = 10
Using the formula we have,
LSA = 2a√(a2/4 + h2)
= 2 × 13 √(132/4 + 102)
= 310.1 sq. cm
Problem 7: Calculate the lateral surface area of a square pyramid if its base is 9 cm and height is 14 cm.
Solution:
We have,
a = 9
h = 14
Using the formula we have,
LSA = 2a√(a2/4 + h2)
= 2 × 9 √(92/4 + 142)
= 264.7 sq. cm
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### Pre-algebra
Book edition Common Core Edition
Author(s) Ron Larson, Laurie Boswell, Timothy D. Kanold, Lee Stiff
Pages 183 pages
ISBN 9780547587776
# Evaluate the experssion when $a=2.5, b=15,\text{and}c=3.5$localid="1648024411816" $c-a$
The value of the expression is $1$.
See the step by step solution
## Step-1 – Apply the concept of addition and subtraction
Two numbers with similar signs get added and the resulting number carries the similar sign.
Two numbers with opposite signs get subtracted and the resulting number carries the sign of the larger number.
## Step-2 – Apply the concept for addition and subtraction of decimals
To add or subtract two decimal numbers, write the numbers vertical with decimal point aligned vertically.
For example: to add the numbers 2.3 and 4.62, write these numbers vertically and add as shown below
$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2.3\\ \underset{¯}{+4.62}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}6.92\end{array}$
## Step-3 – Evaluate the expression
To evaluate the expression $c-a$, substitute the values of c as 3.5 and value of a as 2.5 and subtract the numbers as shown below,
$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3.5\\ \underset{¯}{-\text{\hspace{0.17em}}2.5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1.0\end{array}$
Thus, the value of the expression is $1$. |
http://www.storyboardthat.com/teacher-guide/introduction-to-fractions
# Introduction to Fractions
### By Anna Warfield
Find this Common Core aligned Teacher Guide and more like it in our Elementary School Category!
This fraction lesson plan is a mini supplemental unit on fractions to be used for remedial or extension work and information, teacher guidance and inspiration, alternative instruction, integrating writing and mathematics, or for whatever you wish! Teachers can use Storyboard That to make short storyboards for a concept, discussion springboard, quick visual review before a unit or lesson, word problems, or as a slideshow presentation to accompany a lesson! Have students use Storyboard That to write math stories, fraction word problems with realistic applications, explain a concept as an assessment, or explain a concept to show another student (student exchange).
Making the switch from whole numbers to parts and wholes can be very difficult for young minds. It is mind-boggling for some to learn that there are numbers between the counting numbers! Fractions is a topic with which many students struggle throughout elementary and middle school, so it is important that students have a thorough understanding of what fractions are, estimating and comparing amounts visually and numerically, and recognizing reasonable answers.
## Teacher Information
Many students confuse fractions with the whole numbers that they are so used to seeing. The number “1/3” looks like two different numbers rather than a single numerical value. A fraction is a number with an integer numerator and a nonzero denominator that, for our purposes, can represent rational numbers (1/4 or 3 2/5) and whole numbers (4/2 = 2). The “numerator” is the quantity in the top section of the fraction that represents the number of parts, and the “denominator” is the value below the fraction bar indicating the number of partitions or shares, known also as the “whole".
Fraction notation can indicate ratio and proportions, multiplicative relationships, quotient when dividing two numbers, measurement, and parts of wholes or sets. Beginning fraction masters need only worry about parts of wholes or sets and measurement, but astute students will likely notice multiplicative relationships (i.e. the half circle is twice the size of the quarter circle or inversely, the quarter circle is 1/2 the size of the half circle) and dividing two numbers (sharing 7 cookies among 3 people would be written 7/3, the same as 7 ÷ 3).
Encourage students to talk about fractions and what they mean. Mathematics is not all about numbers and answers, but also understanding and reasoning. Teacher-led question and answer sessions are very helpful, but it is also very beneficial for building strong foundations if teachers, and eventually students, lead discussions on mathematical concepts. Fractions can be an excellent starting point for such discussions. Pose a question to the class, such as the Jack and Jill arguments below, for the class to ponder on their own and share, or discuss in small groups.
## Student Prior Knowledge
Students should come into third grade knowing that shapes can often be split into equal shares, such as halves, thirds, and quarters or fourths. The concept of sharing items, such as supplies or food, as well as sharing time fairly, like turn-taking or splitting the day into class periods/subjects, should be well established by this age. Draw on real-life examples whenever possible to strengthen understanding.
While it is not imperative, it is helpful if students are already familiar with multiplication and division. Mastery of basic facts is a separate skill from understanding and manipulating fractions, but understanding one may help the understanding of the other. Consider reviewing the multiplication/division facts as necessary.
Introduce fractions to students by allowing them to investigate and review prior knowledge. This can be done in a myriad of ways, such as asking what they know, a coloring activity, exploring manipulatives like fraction tiles or pattern blocks, a short video, or an adorable comic strip.
### Identify Unit Fractions
Fractions show equal shares. Pictures that show shapes partitioned into sections that are not the same size do not show examples of fractions. Show pictures of shapes that have been equally and unequally divided.
Allow students to identify the total number of parts in a whole, such as a circle, to find the denominator. The bottom of the fraction, represents the number of partitions (equal shares). Each piece of the whole is therefore one of the whole. Have students guess what to call each part (one half, one third, one fourth etc.) and create a chart together to show the first few common unit fractions. You can make a template in Storyboard That to print or project onto the board, or you can have students create their own storyboard as an assessment of understanding.
### Multiple Numerators
Introduce vocabulary of numerator and denominator. The numerator is the number on top of the fraction bar that represents a part of a whole. The denominator is the number below the fraction bar that shows the number of pieces or partitions in a whole. Numerator looks a little like “number” (how many) and denominator may remind some students of “name”, particularly if they are familiar with other languages, such as French or Spanish. The denominator gives the fraction its name (eg. fifths), and the numerator tells you how many parts of the whole there are (three-fifths).
Have students identify the given fractions and fraction pictures by both number and word names. The examples below show a partially filled storyboard for students to start from, and one way the storyboard could be completed. Additionally, students can create their own on Storyboard That to demonstrate understanding.
Integrate word problems or fraction stories as examples whenever possible. Depending on the knowledge base of your students, continue on to identifying fractions, or allow students time to explore materials and discuss their findings. An activity for Storyboard that might be to show different ways to represent the same fractions.
### Fractions on a Number Line
Review the number line with whole numbers briefly; have students create their own number lines in a notebook, on chart paper, or by standing in a row evenly spaced apart. Draw attention to the space in between the whole numbers, and ask students about the value of a spot that is not labeled by a whole number. Show a storyboard to introduce fractions on a number line. Use the example below or come up with your own.
Be sure to point out what whole you are using. For a special challenge, encourage students to think carefully by changing the whole: 0 to 2 or 0 to 10. (1/2 of 10 is 5; 2/5 of 10 is 4)
### Compare Fraction Denominators
Many students will see a larger number in the denominator and think that the bigger number means a bigger value. Fractions are sneaky that way: as the denominator gets bigger, that means that the whole is being divided into more and more smaller pieces.
Have students create a fraction story that shows what happens when a whole is divided into more and more pieces. Some possible wholes for the story might be:
• pizza (as in the example below)
• cookies
• pie
• brownies
• wall space
• paper
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# Converse of Pythagorean Theorem
Converse of Pythagorean Theorem states that:
In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.
Given: A ∆PQR in which PR2 = PQ2 + QR2
To prove: ∠Q = 90°
Construction: Draw a ∆XYZ such that XY = PQ, YZ = QR and ∠Y = 90°
So, by Pythagora’s theorem we get,
XZ2 = XY2 + YZ2
⇒ XZ2 = PQ2 + QR2 ……….. (i), [since XY = PQ and YZ = QR]
But, PR2 = PQ2 + QR2 ………… (ii), [given]
From (i) and (ii) we get,
PR2 = XZ2 ⇒ PR = XZ
Now, in ∆PQR and ∆XYZ, we get
PQ = XY,
QR = YZ and
PR = XZ
Therefore ∆PQR ≅ ∆XYZ
Hence ∠Q = ∠Y = 90°
Word problems using the Converse of Pythagorean Theorem:
1. The side of a triangle are of length 4.5 cm, 7.5 cm and 6 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
We know that hypotenuse is the longest side. If 4.5 cm, 7.5 cm and 6 cm are the lengths of angled triangle, then 7.5 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(7.5)2 = (6)2 + (4.5)2
56.25 = 36 + 20.25
56.25 = 56.25
Since, both the sides are equal therefore, 4.5 cm, 7.5 cm and 6 cm are the side of the right angled triangle having hypotenuse 7.5 cm.
2. The side of a triangle are of length 8 cm, 15 cm and 17 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
We know that hypotenuse is the longest side. If 8 cm, 15 cm and 17 cm are the lengths of angled triangle, then 17 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(17)2 = (15)2 + (8)2
289 = 225 + 64
289 = 289
Since, both the sides are equal therefore, 8 cm, 15 cm and 17 cm are the side of the right angled triangle having hypotenuse 17 cm.
3. The side of a triangle are of length 9 cm, 11 cm and 6 cm. Is this triangle a right triangle? If so, which side is the hypotenuse?
Solution:
We know that hypotenuse is the longest side. If 9 cm, 11 cm and 6 cm are the lengths of angled triangle, then 11 cm will be the hypotenuse.
Using the converse of Pythagoras theorem, we get
(11)2 = (9)2 + (6)2
121 = 81 + 36
121 ≠ 117
Since, both the sides are not equal therefore 9 cm, 11 cm and 6 cm are not the side of the right angled triangle.
The above examples of the converse of Pythagorean Theorem will help us to determine the right triangle when the sides of the triangles will be given in the questions.
`
Congruent Line-segments
Congruent Angles
Congruent Triangles
Conditions for the Congruence of Triangles
Side Side Side Congruence
Side Angle Side Congruence
Angle Side Angle Congruence
Angle Angle Side Congruence
Right Angle Hypotenuse Side congruence
Pythagorean Theorem
Proof of Pythagorean Theorem
Converse of Pythagorean Theorem |
# GMAT Inequalities Problems: Challenge Problem I.2 explanation
As most GMAT Test takers already know, this is a very specific type GMAT Inequalities problems that they often come across while practicing GMAT Math. Quite frequently, test takers either get this type of GMAT Inequalities problems incorrect (thus loosing points) or spend too much time on them resulting in time wastage.
Hence, to make it quick and easy to solve these type of problems, we have developed a GMAT Inequalities Comparison Framework. However, the framework does come with a caveat – you must remember the range table or must atleast be able to draw the table quickly. This is certainly possible with a little bit of practice.
Since the details of the basic framework for solving this type of question is covered in the article GMAT Comparisons Framework, in this article, we will skip the basics and solve this question using the framework only.
## The comparisons framework
Let’s start with the question statement and see in which of these ranges would the statement be true:
Is 1/x4 > 1/x2 ?
As you can see from the ranges, for the question statement to be true, x must lie between -1 and 1. We could have arrived at the same conclusion after algebra, but that would have been time consuming and also prone to mistakes.
So the question statement truly becomes:
Is 1/x4 > 1/x2 ? is same as Is -1 < x < 1 ?
Now look at statement (a)
a) 1/x > 1/x2
Lets look at the ranges where this is true:
As you can see in the diagram, this is only possible when x>1. So given that 1/x > 1/x2, we are given that x>1. Hence x cannot be between -1 and 1. So this answers a clear “NO” to the asked question in the question statement.
Hence, statement (a) is SUFFICIENT.
Lets look at statement (b)
b) x2 > 7
Using a little bit of algebra, we get
x2 – 7 > 0 => (x-⎷7)(x+⎷7) > 0
This gives us:
either x>⎷7 OR x<-⎷7 . This clearly means that x is NOT in the range -1<x<1. Hence, this gives a clear “NO” to the asked question.
Hence, statement (a) is SUFFICIENT.
## Solve GMAT Inequalities Problem Using Algebra
[Coming Soon][su_quote]Coming soon![/su_quote] |
# How do you divide (x^2+20x+5) / (x-4)?
Mar 14, 2018
Quotient $\textcolor{g r e e n}{x + 24}$, Remainder $\textcolor{red}{\frac{101}{x - 4}}$
#### Explanation:
Let’s use synthetic division to find the quotient and remainder.
$\textcolor{w h i t e}{a a a a a a a a a a a} 0 \textcolor{w h i t e}{a a a a a} 1 \textcolor{w h i t e}{a a a a} 24$
color(white)(aaa aaa aa ) ———————-
$\textcolor{w h i t e}{a a a} 4 \textcolor{w h i t e}{a a a} | \textcolor{w h i t e}{a a a} 1 \textcolor{w h i t e}{a a a a} 20 \textcolor{w h i t e}{a a a a} 5$
$\textcolor{w h i t e}{a a a a a a a} | \textcolor{w h i t e}{a a} \downarrow \textcolor{w h i t e}{a a a a} 4 \textcolor{w h i t e}{a a a a} 96$
color(white)(aaa aaa a) | color(white)(aaa ) ——- ——-
$\textcolor{w h i t e}{a a a a a a a a a a a a} 1 \textcolor{w h i t e}{a a a a} 24 \textcolor{w h i t e}{a a a} 101$
Quotient $\textcolor{g r e e n}{x + 24}$, Remainder $\textcolor{red}{\frac{101}{x - 4}}$
Verification : $\left(x + 24\right) \cdot \left(x - 4\right) = {x}^{2} + 20 x - 94 + 101 = {x}^{2} + 20 x + 5$ |
18.02.2020
# Vertical and horizontal asymptotes pdf
Oct 25, · Some functions are continuous from negative infinity to positive infinity, but others break off at a point of discontinuity or turn off and never make it past a certain point. Vertical and horizontal asymptotes are straight lines that define the value the function approaches if it . Identify vertical and horizontal asymptotes. By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and. Graphing Rational Functions Date_____ Period____ Identify the points of discontinuity, holes, vertical asymptotes, x-intercepts, and horizontal asymptote of each. 1) Identify the points of discontinuity, holes, vertical asymptotes, and horizontal asymptote of each. Then.
# Vertical and horizontal asymptotes pdf
Limits at Infinity, Horizontal Asymptotes Math , TA: Amy DeCelles 1. Overview Outline: 1. Definition of limits at infinity 2. Definition of horizontal asymptote So the lines y = 2 and y = 1 are horizontal asymptotes. To find the vertical asymptote we notice that the denominator is . Vertical Asymptotes Set the denominator equation to zero and solve for x. The equation for a vertical asymptote is written x=k, where k is the solution from setting the If there is a horizontal asymptote, say y=p, then set the rational function equal to p and solve for x. If x is a real number, then the line crosses the horizontal. Identify vertical and horizontal asymptotes. By looking at the graph of a rational function, we can investigate its local behavior and easily see whether there are asymptotes. We may even be able to approximate their location. Even without the graph, however, we can still determine whether a given rational function has any asymptotes, and. the horizontal asymptote is y =0. The horizontal asymptote is 0y = Final Note: There are other types of functions that have vertical and horizontal asymptotes not discussed in this handout. There are other types of straight -line asymptotes called oblique or slant asymptotes. There are other asymptotes that are not straight lines. Graphing Rational Functions Date_____ Period____ Identify the points of discontinuity, holes, vertical asymptotes, x-intercepts, and horizontal asymptote of each. 1) Identify the points of discontinuity, holes, vertical asymptotes, and horizontal asymptote of each. Then. PRACTICE PROBLEMS (1)Find the vertical and horizontal asymptotes of the following functions: (a) f(x) = x2 x 6 x2 x 20 Solution: The horizontal asymptote is given by lim x!1 x2 x 6 x2 x 20 = 1 (since we have the same power of xin both numerator and denominator, the limit is given by the ratio of the coe cents in front of the highest power of x. Vertical Asymptotes • The Vertical Asymptotes of a rational function are found using the zeros of the denominator. For Horizontal Asymptotes use the following guidelines. • If the degree of the numerator is greater than the degree of the denominator by more than one, the graph has no horizontal asymptote.(none). Example 3 Find the vertical asymptote for f(x) = log(2−x). Solution 3 Set the inside of the logarithm to zero and solve for x. 2−x = 0 2 = x Thus, the equation of our vertical asymptote is x = 2. Horizontal asymptotes Horizontal asymptotes are used to describe the end behavior of some graphs. Oct 25, · Some functions are continuous from negative infinity to positive infinity, but others break off at a point of discontinuity or turn off and never make it past a certain point. Vertical and horizontal asymptotes are straight lines that define the value the function approaches if it . SECTION VERTICAL AND HORIZONTAL ASYMPTOTES Example 3. Find the vertical and horizontal asymptotes of the graph of f(x) = x2 2x+ 2 x 1. Solution. The vertical asymptotes will occur at those values of x for which the denominator is equal to zero: x 1 = 0 x = 1 Thus, the graph will have a vertical asymptote at x = 1.(1) Find the vertical and horizontal asymptotes of the following functions: (a) f(x) = x2 − x − 6 x2 − x − Solution: The horizontal asymptote is given by lim x→∞. Linear Asymptotes and Holes. Graphs of Rational Functions can contain linear asymptotes. These asymptotes can be Vertical,. Horizontal, or Slant (also called. The vertical lines. 1 and. 1 are the vertical asymptotes. Horizontal Asymptotes ( HA). These asymptotes occur when the degree of the numerator is less than or. B. Asymptotes/Holes. Holes are what they sound like: is a hole. Rational functions may have holes or asymptotes (or both!). Asymptote Types: 1. vertical. Asymptotes. We deal with two types of asymptotes: vertical asymptotes and horizontal asymptotes. Vertical Asymptotes. There are two functions we will. Vertical Asymptote: = . II. HORIZATONAL ASYMPTOTES: y = a. • Determine the degree of the numerator (n) and the degree of the denominator (k). Def: Asymptote: a line that draws increasingly nearer to a curve without ever meeting it. There are a basically three types of asymptotes: horizontal, vertical and. Section -‐ Curve Sketching. Vertical Asymptotes. Horizontal Asymptotes. The line = a is a vertical asymptote of the graph of a function f(x) if either: lim. →a. The vertical line x = a is called a vertical asymptote of the graph of y = f(x) if A graph will (almost) never touch a vertical asymptote; however, a graph may cross . Chandler-Gilbert Community College. Vertical and Horizontal Asymptotes. (This handout is specific to rational functions (). (). P x. Q x where (). P x and (). lagu pembaringan geby parera, ideal mozilla waterfox 32 bit quickly,go here,vire launcher donate apk s,click the following article
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# AP Statistics Section 3.1B Correlation PowerPoint PPT Presentation
AP Statistics Section 3.1B Correlation. A scatterplot displays the direction , form and the strength of the relationship between two quantitative variables. Linear relations are particularly important because a straight line is a simple pattern that is quite common.
AP Statistics Section 3.1B Correlation
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### AP Statistics Section 3.1BCorrelation
A scatterplot displays the direction, form and the strengthof the relationship between two quantitative variables. Linear relations are particularly important because a straight line is a simple pattern that is quite common.
### We say a linear relation is strong if and weak if
the points lie close to a straight line
they are widely scattered about the line.
Relying on our eyes to try to judge the strength of a linear relationship is very subjective. We will be determining a numerical summary called the __________.
correlation
### The formula for correlation of variables x and y for n individuals is:
TI 83/84: Put data into 2 lists, say
STAT CALC 8:LinReg(a+bx) ENTER
Note: If r does not appear,2nd0 (Catalog)
Scroll down to “Diagnostic On”
Press ENTER twice
### Find r for the data on sparrowhawk colonies from section 3.1 A
Important facts to remember when interpreting correlation:1. Correlation makes no distinction between __________ and ________ variables.
explanatory
response
positive
negative
weak
6 4 2 1 3 5
### What effect does adding an outlier have on r and why?
4. Correlation is not a complete summary of two-variable data. Ideally , give the mean and standard deviations of both x and y along with the correlation. |
# Difference between revisions of "2010 AIME I Problems/Problem 14"
## Problem
For each positive integer n, let $f(n) = \sum_{k = 1}^{100} \lfloor \log_{10} (kn) \rfloor$. Find the largest value of n for which $f(n) \le 300$.
Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.
## Solution
### Solution 1
Observe that $f$ is strictly increasing in $n$. We realize that we need $100$ terms to add up to around $300$, so we need some sequence of $2$s, $3$s, and then $4$s.
It follows that $n \approx 100$. Manually checking shows that $f(109) = 300$ and $f(110) > 300$. Thus, our answer is $\boxed{109}$.
### Solution 2
Because we want the value for which $f(n)=300$, the average value of the 100 terms of the sequence should be around $3$. For the value of $\lfloor \log_{10} (kn) \rfloor$ to be $3$, $1000 \le kn < 10000$. We want kn to be around the middle of that range, and for k to be in the middle of 0 and 100, let $k=50$, so $50n=\frac{10000+1000}{2}=\frac{11000}{2}=5500$, and $n = 110$. $f(110) = 301$, so we want to lower $n$. Testing $109$ yields $300$, so our answer is still $\boxed{109}$.
### Solution 3
For any $n$ where the sum is close to $300$, all the terms in the sum must be equal to $2$, $3$ or $4$. Let $M$ be the number of terms less than or equal to $3$ and $N$ be the number of terms equal to $2$ (also counted in $M$). With this definition of $M$ and $N$ the total will be $400 - M - N \le 300$, from which $M + N \ge 100$. Now $M+1$ is the smallest integer $k$ for which $\log_{10}(kn) \ge 4$ or $kn \ge 10000$, thus $$M = \left\lfloor\frac{9999}{n}\right\rfloor.$$ Similarly, $$N = \left\lfloor\frac{999}{n}\right\rfloor = \left\lfloor\frac{M}{10}\right\rfloor.$$
Therefore, $$M + \left\lfloor \frac{M}{10} \right\rfloor \ge 100 \implies M \ge \left\lceil\frac{1000}{11}\right\rceil = 91 \implies n \le \left\lfloor\frac{9999}{91}\right\rfloor = 109.$$ Since we want the largest possible $n$, the answer is $\boxed{109}$.
=== Solution 4 === (Similar to Solution 1, with more detail)
Since we're working with base-$10$ logarithms, we can start by testing out $n$'s that are powers of $10$. For $n = 1$, the terms in the sum are $\lfloor \log_{10} (1)\rfloor, \lfloor \log_{10} (2)\rfloor, \lfloor \log_{10} (3) \rfloor , . . . , \lfloor \log_{10} (100) \rfloor$. For numbers $1$-$9$, $\lfloor \log_{10} (kn) \rfloor = 0$. Then we have $90$ numbers, namely $10$-$99$, for which $\lfloor \log_{10} (kn) \rfloor = 1$. The last number we have is $100$, which gives us $\lfloor \log_{10} (kn) \rfloor = 2$. This sum gives us only $90 + 2 = 92$, which is much too low. However, applying the same counting technique for $n = 10$, our sum comes out to be $9 + 180 + 3 = 192$, since there are $9$ terms for which $\lfloor \log_{10} (kn) \rfloor = 1$, $90$ terms for which $\lfloor \log_{10} (kn) \rfloor = 2$, and one term for which $\lfloor \log_{10} (kn) \rfloor = 3$. So we go up one more power of $10$ and get $18 + 270 + 4 = 292$, which is very close to what we are looking for.
Now we only have to bump up the value of $n$ a bit and check our sum. Each increase in $n$ by $1$ actually increases the value of our sum by $1$ as well (except for $n = 101$), because whenever a $4$ is added to the sum, a $3$ is taken away. It doesn't take long to check and see that the value of $n$ we're looking for is $\boxed{109}$ , which corresponds to a sum of exactly $300$.
~ anellipticcurveoverq
## See Also
2010 AIME I (Problems • Answer Key • Resources) Preceded byProblem 13 Followed byProblem 15 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# If A=begin{bmatrix}1 & 1 3 & 4 end{bmatrix} , B=begin{bmatrix}2 1 end{bmatrix} ,C=begin{bmatrix}-7 & 1 0 & 4 end{bmatrix},D=begin{bmatrix}3 & 2 & 1 end{bmatrix} text{ and } E=begin{bmatrix}2 & 3&4 1 & 2&-1 end{bmatrix} Find , if possible, a) A+B , C-A and D-E b)AB, BA , CA , AC , DA , DB , BD , EB , BE and AE c) 7C , -3D and KE
Question
Matrices
If $$A=\begin{bmatrix}1 & 1 \\3 & 4 \end{bmatrix} , B=\begin{bmatrix}2 \\1 \end{bmatrix} ,C=\begin{bmatrix}-7 & 1 \\0 & 4 \end{bmatrix},D=\begin{bmatrix}3 & 2 & 1 \end{bmatrix} \text{ and } E=\begin{bmatrix}2 & 3&4 \\1 & 2&-1 \end{bmatrix}$$
Find , if possible,
a) A+B , C-A and D-E b)AB, BA , CA , AC , DA , DB , BD , EB , BE and AE c) 7C , -3D and KE
2020-10-22
Step 1
We can add and subtract matrices only which of them have same order.
For multiplication of two matrices it is necessary that no. of columns of first matrix must be equal to no of rows in second matrix.
Step 2
Here,
$$A=\begin{bmatrix}1 & 1 \\3 & 4 \end{bmatrix} , B=\begin{bmatrix}2 \\1 \end{bmatrix} ,C=\begin{bmatrix}-7 & 1 \\0 & 4 \end{bmatrix},D=\begin{bmatrix}3 & 2 & 1 \end{bmatrix} \text{ and } E=\begin{bmatrix}2 & 3&4 \\1 & 2&-1 \end{bmatrix}$$
a) A+B , order of A is $$2 \times 2 \neq$$ order of $$B (2 \times 1)$$
So, Not possible
C-A
$$C-A=\begin{bmatrix}-7 & 1 \\0 & 4 \end{bmatrix}-\begin{bmatrix}1 & 1 \\3 & 4 \end{bmatrix}=\begin{bmatrix}-7-1 & 1-1 \\0-3 & 4-4 \end{bmatrix}=\begin{bmatrix}-8 & 0 \\-3 & 0 \end{bmatrix}$$
$$C-A=\begin{bmatrix}-8 & 0 \\-3 & 0 \end{bmatrix}$$
D-E
D-E have different orders
So,not possible
b) AB
$$AB=\begin{bmatrix}1 & 1 \\3 & 4 \end{bmatrix}\begin{bmatrix}2 \\1\end{bmatrix}=\begin{bmatrix}2+1 \\6+4\end{bmatrix}=\begin{bmatrix}3 \\10 \end{bmatrix}$$
Hence
$$AB=\begin{bmatrix}3 \\10 \end{bmatrix}$$
$$BA \text{ here } [B]_{2 \times 1} [A]_{2 \times 2}$$
number of columns of B $$\neq$$ number of row in A
So,not possible
CA
$$CA=\begin{bmatrix}-7 & 1 \\0 & 4 \end{bmatrix}\begin{bmatrix}1 & 1 \\3& 4 \end{bmatrix}=\begin{bmatrix}-7+3 & -7+4 \\0+12 & 0+16 \end{bmatrix}=\begin{bmatrix}-4 & -3 \\12 & 16 \end{bmatrix}$$
Hence
$$CA=\begin{bmatrix}-4 & -3 \\12 & 16 \end{bmatrix}$$
AC
$$AC=\begin{bmatrix}1 & 1 \\3& 4 \end{bmatrix}\begin{bmatrix}-7 & 1 \\0 & 4 \end{bmatrix}=\begin{bmatrix}-7 & 1+4 \\-21 & 3+16 \end{bmatrix}=\begin{bmatrix}-7 & 5 \\-21 & 19 \end{bmatrix}$$
Hence,
$$AC=\begin{bmatrix}-7 & 5 \\-21 & 19 \end{bmatrix}$$
DA
$$[D]_{1 \times 3}[A]_{2 \times 2}$$
Number columns in [D] $$\neq$$ number rows in [A]
So,not possible DB . Not possible BD. Not possible
$$EB. [E]_{2 \times 3} [B]_{2 \times 1}$$
Not possible
BE
$$[B]_{2 \times 1}[E]_{2 \times 3}$$
Not possible
AE
$$AE=\begin{bmatrix}1 & 1 \\3& 4 \end{bmatrix}\begin{bmatrix}2 & 3&4 \\1 & 2&-1 \end{bmatrix}=$$
$$=\begin{bmatrix}2+1 & 3+2&4-1 \\6+4 & 9+8&12-4 \end{bmatrix}=\begin{bmatrix}3 & 5&3 \\10 & 17&8 \end{bmatrix}$$
Hence,
$$AE=\begin{bmatrix}3 & 5&3 \\10 & 17&8 \end{bmatrix}$$
(c)$$7C=7\begin{bmatrix}-7 & 1 \\0 & 4 \end{bmatrix}=\begin{bmatrix}-49 & 7 \\0 & 28 \end{bmatrix}$$
$$-3D=-3\begin{bmatrix}3 & 2 & 1 \end{bmatrix}=\begin{bmatrix}-9 & -6 & -3 \end{bmatrix}$$
$$KE=K\begin{bmatrix}2 & 3&4 \\1&2&-1 \end{bmatrix}=\begin{bmatrix}2K & 3K&4K \\K&2K&-K \end{bmatrix}$$
### Relevant Questions
$$A=\begin{bmatrix}2& 1&1 \\-1 & -1&4 \end{bmatrix} B=\begin{bmatrix}0& 2 \\-4 & 1\\2 & -3 \end{bmatrix} C=\begin{bmatrix}6& -1 \\3 & 0\\-2 & 5 \end{bmatrix} D=\begin{bmatrix}2& -3&4 \\-3 & 1&-2 \end{bmatrix}$$
a)$$A-3D$$
b)$$B+\frac{1}{2}$$
c) $$C+ \frac{1}{2}B$$
(a),(b),(c) need to be solved
Solve for X in the equation, given
$$3X + 2A = B$$
$$A=\begin{bmatrix}-4 & 0 \\1 & -5\\-3&2 \end{bmatrix} \text{ and } B=\begin{bmatrix}1 & 2 \\ -2 & 1 \\ 4&4 \end{bmatrix}$$
Use the graphing calculator to solve if possible
A=\begin{bmatrix}1 & 0&5 \\1 & -5&7\\0&3&-4 \end{bmatrix}\\ B=\begin{bmatrix}3 & -5&3 \\2&3&1\\4&1&-3\end{bmatrix}\\ C=\begin{bmatrix}5 & 2&3 \\2& -1&0 \end{bmatrix}\\ D=\begin{bmatrix}5 \\-3\\4 \end{bmatrix}
Find the value in row 2 column 3 of AB-3B
Matrix multiplication is pretty tough- so i will cover that in class. In the meantime , compute the following if
$$A=\begin{bmatrix}2&1&1 \\-1&-1&4 \end{bmatrix} , B=\begin{bmatrix}0 & 2 \\-4 & 1\\2&-3 \end{bmatrix} , C=\begin{bmatrix}6 & -1 \\3 & 0\\-2&5 \end{bmatrix} , D=\begin{bmatrix}2 & -3&4 \\-3& 1&-2 \end{bmatrix}$$
If the operation is not possible , write NOT POSSIBLE and be able to explain why
a)A+B
b)B+C
c)2A
Let M be the vector space of $$2 \times 2$$ real-valued matrices.
$$M=\begin{bmatrix}a & b \\c & d \end{bmatrix}$$
and define $$M^{\#}=\begin{bmatrix}d & b \\c & a \end{bmatrix}$$ Characterize the matrices M such that $$M^{\#}=M^{-1}$$
Given matrix A and matrix B. Find (if possible) the matrices: (a) AB (b) BA. $$A=\begin{bmatrix}1 & 2 &3&4\end{bmatrix} , B=\begin{bmatrix}1 \\ 2 \\ 3 \\ 4 \end{bmatrix}$$
$$A=\begin{bmatrix}3 & -2 \\1 & 5\end{bmatrix} , B=\begin{bmatrix}0 & 0 \\5 & -6 \end{bmatrix}$$
$$A=\begin{bmatrix}3 & -2 \\ 1 & 5 \end{bmatrix} , B=\begin{bmatrix}0 & 0 \\ 5 & -6 \end{bmatrix}$$ |
This evening we will be… showing how maths strategies develop across the year groups
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# This evening we will be… showing how maths strategies develop across the year groups - PowerPoint PPT Presentation
## This evening we will be… showing how maths strategies develop across the year groups
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##### Presentation Transcript
1. Welcome to our Year 3 to 6 maths evening Would you like to be able to help more with homework? Are you confused by how we teach the children maths these days? Would you like to know more about the methods we use? • This evening we will be… • showing how maths strategies develop across the year groups • show you the school’s calculations policy and how you can use this to help your child with their maths • key areas where you can help your child with their maths • useful resources and websites
2. Mental maths games • Follow me cards! • Times table practice • Pair work • White board response • Interactive white board work
3. Which method should my child be learning? • Our school’s Calculations policy can be found at… http://www.st-john.leicester.sch.uk/
4. Calculations policy for addition The policy shows how the method progresses across the year groups. Some children may be ready for the year appropriate method, other may be consolidating the previous year’s or being extended. Copies available for you to take away today.
5. How would you solve these calculations? 3 + 7 = 22 + 33 = 120 + 11 = 1,024 + 435 = 6,962 + 4,871 = What skills were you using? Which method did you use? Does that method work for all of these calculations?
6. Drawing upon a range of skills 3 + 7 = 10 (Knowing your number bonds to 10) 33 + 33 = 66 (Doubling the tens, doubling the units and then adding together) 120 + 11 = 131 (Adding 10, then adding 1 more) 1,024 + 435 = 1,459 (Add 24 to 435 = 459, then add to 1,000 or you may have used column addition) 6,962 + 4,871 = 11,833 (Column addition)
7. Addition across the years Method in Year 3
8. Addition across the years Method in Year 4 to 5 (The expanded method)
9. Addition across the years Method in Year 6 (The compact method)
10. Level 3 questions involving addition
11. Level 3 questions involving addition 15 1st) 11.50 + 16.50 = 28 2nd) 40 -28 = £ 12
12. Level 5 questions involving using addition
13. 1st step) 15 + 15 + 15 = 45 3rd step) 108 ÷ 2 = 54 cm 2nd step) 153 – 45 = 108
14. What can I do to help?
15. Learning to tell the time
16. Learning to tell the time Year 3 • Read the time on a 12–hour digital clock and to the nearest 5 minutes • Read an analogue clock • Calculate time intervals and find start or end times for a given time interval Year 4 • Read time to the nearest minute • Use am, pm and 12-hour clock notation • Choose units of time to measure time intervals, e.g. seconds, minutes or hours • Calculate time intervals from clocks and timetables Year 5 • Read timetables and time using 24-hour clock notation • Use a calendar to calculate time intervals Year 6 • All the above plus read from digital and analogue • Solve multi-step worded problems including reading and interpreting timetables
17. Learning about money Talk about pocket money with your child. Help him/her to add it up week by week to workout whether they can afford a particular toy or treat. Shop using money and calculate change. Talk about savings account and interest, credit and debt.
18. Learning about money The progression of learning about money across the year groups Year 3 • Know that there are 100 pennies in a pound. • Solve one–step and two–step problems involving money. • Develop and use written methods to record and explain addition and subtraction of two–digit numbers. Find total amounts, give change and work out with which coins to pay with. Explain how the problem was solved. Year 4 • Solve one-step and two-step problems involving money including converting pounds to pence. • Understand the value of each number, e.g. tenths and hundredths. • Partition decimals and relate this to money. • Begin to use all four operations to solve word problems Year 5 • Solve one-step and two-step problems involving money and all four operations. • Make simple conversions of pounds to foreign currency and finding simple percentages. • Explain methods and reasoning, orally and in writing. Year 6 • Solving multi-step word problems involving money. • Calculate percentages of amount of money e.g. In a sale a jacket costs £42. There is a 10% sale, what is the price of the jacket. • Express one quantity as a percentage of another (e.g. express £400 as a percentage of £1000);
19. Learning times tables Learning multiplication facts is a vital part of any child’s mathematical development. Once rapid recall of multiplication facts becomes possible, a whole host of mathematical activities will seem easier. Children need to be able to recall multiplication facts in any order and also to derive associated division facts. The expectations for each year group are set out below: Year 1 Count on or back in ones, twos, fives and tens and use this knowledge to derive the multiples of 2, 5 and 10. Year 2 Derive and recall multiplication facts for the 2, 5 and 10 times-tables and the related division facts.
20. Learning times tables www.teachingtables.co.uk Year 3 • Derive and recall multiplication facts for the 2, 3, 4, 5, 6 and 10 times-tables and the corresponding division facts. Year 4 • Derive and recall multiplication facts up to 10 × 10, the corresponding division facts. Year 5 • Recall quickly multiplication facts up to 10 x 10 and derive quickly corresponding division facts. Year 6 • Use knowledge of place value and multiplication facts to 10 × 10 to derive related multiplication and division facts involving decimals (e.g. 0.8 × 7, 4.8 ÷ 6). • Use knowledge of multiplication facts to derive quickly squares of numbers to 12×12.
21. What should they be able to do? • The aim is that for each times table: • The children should be able to say the table in order. • E.g. 1 times 3 is 3, 2 times 3 is 6. • They should be able to answer questions in any order. • E.g. “What is 4 x 5?” “What is 6 x 7?” • They should be able to answer – “How many 7’s in 21?” “How many 5’s in 20?” • They should also be able to link their tables with division – e.g. 7 x 3 is 21, so 21 ÷ 3 = 7
22. There are lots of ways you can help your child to learn their times tables. Different activities suit different learning styles. Remember it should be fun! • Buy a times table CD or tape. Listening to songs and singing • can help children learn their tables in a fun way. • If your child likes to write or draw they can write out their times tables • or copy them from a chart. See how quickly they can do it and can they • improve on their time? 3) If your child is always on the move try saying them as they go up the stairs of when out walking. They can chant them as they skip or bounce a ball. • Make up silly rhymes to help with facts they are struggling to remember e.g. • Eight times eight is sixty-four, close your mouth and • the shut the door! • A tree on skates fell on the floor; three times eight is twenty-four.
23. Parents evening A sheet with the sub level your child is currently working on. Their teacher will suggest areas to work upon to enable them to progress further.
24. Parents evening ‘Maths targets. A booklet for parents’ publication from the Numeracy strategy Ideas for games, key learning skills for that year group.
25. Parents evening Parents will receive this booklet at Parents Evening. (Copy available to download from our school’s website)
26. These ‘help guides’ are available on the school’s website in the year 6 area.
27. Useful websites www.topmarks.co.uk www.educationcity.com www.coolmath4kids.com
28. http://www.bbc.co.uk/schools/parents/
29. http://www.bbc.co.uk/schools/parents/search/
30. http://www.bbc.co.uk/schools/numbertime/
31. http://www.bbc.co.uk/schools/bitesize//
32. http://www.gridclub.com/ Free trial and membership fee
33. And finally… • We would like volunteers to help children to develop their maths skills. It would be giving children extra time, in addition to their daily maths lesson, to focus on an area of maths. |
All About Cones – Surface Area, Properties, Volume, and Height
A cone is a distinctive geometric figure that has three dimensions, it has a flat surface at the bottom and another surface that points upwards. The apex is what the pointed top of a cone is called, and the surface at the bottom which is flat is known as the base of a cone. The cone looks like the cone of ice cream.
A cone finds its shape from the lines that connect each other from a single point. The lines used for giving shape to the cone are known as the Apex or Vertex, they start from the top and touch the base that is the shape of a circle. When you measure the distance from the top of the cone, straight to its base at the bottom, you can derive the height of the cone. Cone also has a slanting height, the length of which is calculated when the top touches the circumference anywhere at the bottom. Based on these measurements, formulas for the surface area of the cone and volume of cone can be easily derived.
Please note the general properties that a cone has:
• Single face a cone has, that is its base which is circular in shape without any edges
• A cone consists of just one point at the top, known as the vertex or apex.
• A cone’s volume can be calculated
• A cone’s sum of surface area can be calculated
• The cone has a slant height as well, which can be calculated
• When the cone plane and the base plane intersect with each other they are parallel and form a circle
• A right circular cone is formed when the isosceles triangle rotates around its axis at one eighty degrees.
•
The formulas for a cone are derived when we have the following three things in place,
• Height of Cone
The height of the cone is calculated from the top point called the vertex to the base of the circular base, straight in the middle.
• Slant Height of Cone
The height that is slant is calculated from the top point called the apex to anywhere on the outer line of the circle, at the base of the cone, using the Pythagoras theorem, we derive the formula for calculating the same.
l = √(r2+h2)
The total area of the circular base at the base of the cone is called its radius.
Basically, there are two kinds of cones
1. Right Circular Cone
This is the normal cone of the most common cone that is used in the field of geometry, in mathematics. This has a base which is the shape of a circle and the line or axis from the top point called the apex cuts through in the center of the circle of the base. We use the word “right circular cone” because the line that falls at the base of the circle makes a right angle with it.
1. Oblique Cone
The only difference is that the line that falls at the base from the Vertex, does not fall at the center, but elsewhere. The apex is not positioned at the middle of the base. This makes the shape of the cone, slant or tilted.
Area of Surface of a Cone
This area is the entire area occupied by the surface of a cone. The overall surface area of cone equals the total sum of its area at the surface and base area of the circle.
So, if we have a radius asr and height as h of a cone, the formula would be:
Area = πr(r+√(h2+r2))
Area =πr(r+L), where the slant height is L
Calculation of Volume of Cone
Generally, a cone has the shape of a pyramid. It is easy to find the volume if we have the height and radius measurements.
Cone volume = (1/3) πr2h cubic units
You need to practice the formulas well and Cuemath provides the right guidance. Do refer to the worksheets from them and you will come out with flying colors. |
____________________________________________________________
Examining the Parabola
(Assignment 2)
by
# Robin Kirkham, Cara Haskins , and Matt Tumlin
____________________________________________________________
Let us examine the parabola as the coefficient values change to see the effects these changes have on the various parabolas.
Given the parabola y = ax2 +bx +c with variables of a, b, and c.
Our first step is to :
Look at the basic parabola when a=1, b=0, and c=0
Notice that the domain is the set of all real numbers and the range is all non-negative numbers. The range for this basic parabola is all non-negative numbers.
The lowest point on a parabola is called the Minimum. The minimum point for our basic parabola is (0,0).
___________________________________________________________________
Continue to use the basic quadratic function as our frame of reference. Let us examine what happens to the graph under the following guidelines.
Step 1: y = ax2
Let b=0, c=0, and vary the values of a. Our new equation becomes y = ax2.
Let us use the graphing calculator to examine the effects of varying the values for ÔaÕ, remembering to use both positive and negative values.
The red graph is y = ax2 +bx +c. y = ax2, the basic parabola will always be in red in future examples for comparison purposes.
Notice that when the value for variable ÔaÕ is positive, the minimum of the graph does not change even though the value for variable ÔaÕ was changed. When a> 1, the graph has been narrowed horizontally, resulting in a horizontal shrinking of the graph. When 0< a< 1, the graph has now been stretched horizontally. This leaves the question as to what happens when negative values are substituted for variable ÔaÕ.
By substituting negative values for variable ÔaÕ, notice there is a reflection across the x-axis for our parabolas. The highest point on a graph is called a maximum. The maximum for our parabolas is (0,0).
Step 2: Now we are examining the effects of variable ÔbÕ. Let a=1, and c=0 and change the values for variable ÔbÕ. Our new equation is now y = ax2 +bx .
.
Notice that the widths of the parabolas remained the same while the location of the minimum changed. This movement appears to be equal to the value of Ô-c/2Õ, both vertically and horizontally. This is investigated in our step 3 below.
Step 3: Let us again start with equation y = ax2 +bx +c. Let a=1, b=0, and vary c, resulting in y = ax2+c. Note that this investigation is not complete until we review the effects that variable ÔbÕ my have with respect to variable ÔcÕ.
The value of variable ÕcÕ moves the parabola shifts up and down with respect to the y-axis. When c > 0, the graph moves up. When c < 0, the graph moves down.
This vertical movement changes with respect to our minimum point. This vertical shift appears to be equal to the value of variable ÔcÕ.
This shows that the horizontal and vertical shifts are a result of both the variables ÔbÕ and ÔcÕ. The horizontal shift still appears to be negative Ôc/2Õ while the vertical shift appears to be smaller than ÔcÕ. To be sure let us investigate further all three variables with respect to each other.
This investigation provides a very complicated graph, so it has been presented both separated as follows immediately and later all six parabolas together.
This first set of four parabolas demonstrates the results of modifications that are made to both the variables 'b and c' simultaneously.
The direction the parabola opens is related to the variable 'c' being positive or negative (positive - opens up and negative -opens down).
This shows us that the horizontal and vertical shifts are a results of all a, b, and c respectively. The horizontal shift turns out to be Ô-b/2aÕ.
When both values for variables ' b and c' are positive the parabolas are in quadrants I and II (positive on the y-axis).
It is interesting to view these two parabolas with their line of symmetry.
The following graph shows all six parabolas together. As we have observed these parabolas are perhaps better examined separated but it is useful for all of them to be viewed together.
This set of parabolas introduces many interesting effects.
Firstly, one can see that the y = ax2 +bx +c where a, b, and c are all positive and the similar parabola where ÔaÕ is the additive inverse, one observes that these two parabolas are inverses and both shifted to opposite quadrants around the line of symmetry y=2x+2 .
___________________________________________________________________
In summary, given the equation y = ax2 +bx +c the following are true:
• Changes with the value of variable Ô the direction the parabola opens (either up or down).
• Changes with the value of variable Ôeffects the placement of the parabola in a horizontal shift (along the x-axis).
• Changes with the value of variable Ôeffects the placement of the parabola in a vertical shift (y-axis).
• Changes in the value of variable Ôin conjunction with the value of variable ÔbÕ together effects that phase shift of the graph. The three variables provide a complicated interaction with each other. |
# Tips for Solving Equations With Variables on Both Sides
••• SARINYAPINNGAM/iStock/GettyImages
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When you first start solving algebraic equations, you're given relatively easy examples like x = 5 + 4 or y = 5(2 + 1). But as time creeps on you'll be faced with harder problems that have variables on both sides of the equation; for example, 3x = x + 4 or even the scary-looking y2 = 9 – 3y2. When this happens, don't panic: You're going to use a series of simple tricks to help make sense of those variables.
Your first step is to group the variables on one side of the equal sign – usually on the left. Consider the example of 3x = x + 4. If you add the same thing to both sides of the equation you won't change its value, so you're going to add the additive inverse of x, which is −x, to both sides (this is the same as subtracting x from both sides). This gives you:
3x - x = x + 4 - x
Which in turn simplifies to:
2x = 4
#### Tips
• When you add a number to its additive inverse, the result is zero – so you are effectively zeroing out the variable on the right.
Now that your variable expressions are all on one side of the expression, it's time to solve for the variable by stripping away any non-variable expressions on that side of the equation. In this case, you need to remove the coefficient 2 by performing the inverse operation (dividing by 2). As before, you must perform the same operation on both sides. This leaves you with:
\frac{2x}{2} = \frac{4}{2}
Which in turn simplifies to:
x = 2
### Another Example
Here's another example, with the added wrinkle of an exponent; consider the equation
y^2 = 9 - 3y^2
You'll apply the same process you used without the exponents:
Don't let the exponent intimidate you. Just as with a "normal" variable of the first order (without an exponent), you'll use the additive inverse to "zero out" −3y2 from the right side of the equation. Add 3y2 to both sides of the equation. This gives you:
y^2 + 3y^2 = 9 - 3y^2 + 3y^2
Once simplified, this results in:
4y^2 = 9
Now it's time to solve for y. First, to strip away any non-variables from that side of the equation, divide both sides by 4. This gives you:
\frac{4y^2}{4} = \frac{9}{4}
Which in turn simplifies to:
y^2 = \frac{9}{4} \text{ or } y^2 = \frac{9}{4}
Now you have only variable expressions on the left side of the equation, but you're solving for the variable y, not y2. So you have one more step remaining.
Cancel out the exponent on the left side by applying a radical of the same index. In this case, that means taking the square root of both sides:
\sqrt{y^2} = \sqrt{\frac{9}{4}}
Which then simplifies to:
y = \frac{3}{2}
### A Special Case: Factoring
What if your equation has a mix of variables of different degrees (e.g., some with exponents and some without, or with different degrees of exponents)? Then it's time to factor, but first, you'll start the same way you did with the other examples. Consider the example of
x^2 = -2 - 3x
As before, group all the variable terms on one side of the equation. Using the additive inverse property, you can see that adding 3x to both sides of the equation will "zero out" the x term on the right side.
x^2 + 3x = -2 - 3x + 3x
This simplifies to:
x^2 + 3x = -2
As you can see, you have, in effect, moved the x over to the left side of the equation.
Here's where the factoring comes in. It's time to solve for x, but you can't combine x2 and 3x. So instead, some examination and a little logic might help you recognize that adding 2 to both sides zeroes out the right side of the equation and sets up an easy-to-factor form on the left. This gives you:
x^2 + 3x + 2 = -2 + 2
Simplifying the expression on the right results in:
x^2 + 3x + 2 = 0
Now that you've set yourself up to make it easy, you can factor the polynomial on the left into its component parts:
(x + 1)(x + 2) = 0
Because you have two variable expressions as factors, you have two possible answers for the equation. Set each factor, (x + 1) and (x + 2), equal to zero and solve for the variable.
Setting (x + 1) = 0 and solving for x gets you x = −1.
Setting (x + 2) = 0 and solving for x gets you x = −2.
You can test both solutions by substituting them into the original equation:
(-1)^2 + 3 × (-1) = -2
simplifies to
1 - 3 = -2 \text{ or } -2 = -2
which is true, so this x = −1 is a valid solution.
(-2)^2 + 3 × (-2) = -2
simplifies to
4 - 6 = -2 \text{ or, again } -2 = -2
Again you have a true statement, so x = −2 is a valid solution as well.
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# Teaching Algebra to Kids: Lessons and Strategies
Kids usually take a full course in algebra as ninth graders, but they begin learning algebraic thinking as early as kindergarten. Foundations of algebra are taught throughout elementary and middle school.
## How to Explain Algebra in Elementary and Middle School
### Algebraic Concepts to Introduce
#### Patterns
Finding patterns is fundamental to algebraic thinking. This process evolves into composing generalizations about numbers and representing patterns with symbols.
#### Equations
As children begin adding and subtracting in first grade, they're given the facts as 'math sentences,' or equations. Then they're shown the equations with a question mark replacing one of the parts, and taught how to determine the unknown number. In second through fourth grades, the equations are extended to include multiplication and division.
#### Factors
Factoring is important in algebra. Kids in fourth grade learn to find pairs of factors for whole numbers from 1-100. They can then find all of the factors for a number up to a certain point and order them sequentially. Fifth graders begin writing and evaluating numerical expressions using parentheses and brackets.
#### Variables
Children begin to learn about variables when equations are written in different ways, such as:
3 x 4 = ?
3 x ? = 12
? x 4 = 12
Sixth graders learn to solve equations with one variable. They also begin to use equations with letters instead of numbers or question marks.
#### Inequalities
In a sense, inequalities are what algebra is all about. Sixth graders learn to see presentations of inequality as pictures of the comparative positions of two numbers on a number line.
### Methods for Teaching Algebra
#### Pattern Recognition
Teaching patterns formally begins in kindergarten, but is often also a part of a pre-school program. Teach children to analyze and describe patterns by counting dots and shapes, and grouping them by size (larger or smaller), position (inside or outside, above or below) and whether they're the same or different. Use a number chart for skip counting by twos or tens to demonstrate a variety of patterns.
#### Multiple Routes
A basic strategy for teaching any algebraic concept is to teach more than one way to solve any kind of math problem and compare them. Demonstrate various ways of solving a problem by writing different step-by-step solutions side-by-side on the board. Follow up with a discussion of how these approaches differ.
#### Balance Bar
Draw a bar balanced in the middle on a vertical line or pivot. Beginning at the pivot, number the bar from 1-10 on each side. Draw lines representing weights on the numbers. The weights are all the same, and only one can be used on a number. Show that the total of the numbers must be the same on each side; then let the children find multiple solutions as to what weights can go on each side and still have it balance.
#### Number Line
Your students can learn the use of letters for numbers (for writing and solving equations) using a number line that goes up to 10. They can solve problems, such as:
Mary found 3 marbles in her sock drawer. Then she found more marbles in her T-shirt drawer. She kept those marbles a secret. She found 10 marbles in all. How many secret marbles did she have?
Call the 'secret' number 'x,' then show the procedure by means of an equation: 3 + x = 10, then x = 10 - 3. Point out that you subtracted the 3 on both sides of the equation; demonstrating with the number line and balance bar makes the reasoning clear.
#### Draw Shapes
Use shapes instead of numbers or letters. Show three shapes: a triangle, rectangle and circle. Give each shape a number (e.g., triangle = 6, rectangle = 16 and circle = ?).
Problems can be shown as a row of shapes followed by an equal sign and a number. The children can determine the value of the shapes. For example, a problem could read: triangle (6) circle (?) rectangle (16) = 23. Show the children that subtracting the values of the triangle and rectangle from 23 (23-22) gives you the value of the circle (1). A more complex problem might read: triangle circle triangle rectangle triangle circle triangle = 42.
Did you find this useful? If so, please let others know!
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For many students, algebra inspires apprehension and dread. Today, students are increasingly dependent on tools, such as iPads and even their phones, that can do the work of algebra for them. This makes the job of a teacher trying to convey algebra's importance difficult. Yet there are still ways to make algebra relevant,...
• Not Your Father's Algebra As 45 States Look to Math Reform
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# Let's Get Started
Let's explore how to solve quadratic equations by looking at their graphs.
TEKS Standards and Student Expectations
A(1) Mathematical process standards. The student uses mathematical processes to acquire and demonstrate mathematical understanding. The student is expected to:
A(1)(C) select tools, including real objects, manipulatives, paper and pencil, and technology as appropriate, and techniques, including mental math, estimation, and number sense as appropriate, to solve problems
A(1)(D) communicate mathematical ideas, reasoning, and their implications using multiple representations, including symbols, diagrams, graphs, and language as appropriate
A(8) Quadratic functions and equations. The student applies the mathematical process standards to solve, with and without technology, quadratic equations and evaluate the reasonableness of their solutions. The student formulates statistical relationships and evaluates their reasonableness based on real-world data. The student is expected to:
A(8)(A) solve quadratic equations having real solutions by factoring, taking square roots, completing the square, and applying the quadratic formula
Resource Objective(s)
Given a quadratic equation, the student will use graphical methods to solve the equation.
Essential Questions
How can you solve a quadratic equation by looking at the graph?
How do know how many solutions a quadratic equation has?
Vocabulary
# Introduction
Let's investigate ways to use a graph to represent and determine the solution to a quadratic equation.
The graph and table below show points for the quadratic function y = x² − x − 6.
Both the graph and the table of values can be used to solve the equation x² − x – 6 = 0 which is related to the function y = x² − x − 6. You can find when y = 0 on the table, and the x-value at that point will solve the function. The x-value where the graph crosses the x-axis can also be used to solve a quadratic equation. You will use similar thinking to locate points on graphs that satisfy particular conditions.
# Using Graphs to Describe Solutions
Although all quadratic functions have a solution, the solution may not always be a real number. Therefore, a quadratic function may have two real roots, one real root, or zero real roots.
Examine the three graphs below. Click the plus sign by the Start button to learn more about their solutions.
# Using Graphs to Determine Solutions
A graph is the picture of the points that make a function true. For example, the graph and table below show points for the quadratic function y = x² – x – 6.
Both the graph and the table of values can be used to solve the equation x² – x – 6 = 0, which is related to the function y = x² – x – 6. Click on each question to check your answer.
Interactive: In this interactive, the graphs represent equations related to the function
y = x² – x – 6. The red line represents the graph y = k. Match each graph with the equation by dragging the graph to the appropriate slot. Then indicate the correct solutions represented in the graph by dragging the correct ordered pair(s) to the solution box.
Video
Source
Evaluating Functions Using a Graphing Calculator, Gdawy Enterprises, You Tube
Practice
Determine the solution to the each quadratic equation by graphing. Click on the problem to see the answer.
# Summary
A graph is a useful representation for determining the solution of a quadratic equation. To best use a graph, think about a quadratic equation as written in two parts: ax² + bx + c = k
Graph each part of the quadratic equation: ax² + bx + c = k and y = k
Look for the intersection of the two graphs. The x-coordinates of the intersection points will tell you the values of x that are solutions to the original equation.
For example, the graph of f(x) = −0.25x² + 0.5x + 3.75 is shown below. You can use the graph to determine the solutions to the related equation −0.25x² + 0.5x + 3.75 = -5.
You can also use the graph to solve the equation -0.25x² + 0.5x + 3.75 = 0 by identifying the x-intercept since the x-axis is the same as the line y = 0. |
# Factors of 720
The factors of 720 and the prime factors of 720 differ because seven hundred and twenty is a composite number. Also, despite being closely related, the prime factors of 720 and the prime factorization of 720 are not exactly the same either. In any case, by reading on you can learn the answer to the question what are the factors of 720? and everything else you want to know about the topic.
## What are the Factors of 720?
They are: 720, 360, 240, 180, 144, 120, 90, 80, 72, 60, 48, 45, 40, 36, 30, 24, 20, 18, 16, 15, 12, 10, 9, 8, 6, 5, 4, 3, 2, 1. These are all the factors of 720, and every entry in the list can divide 720 without rest (modulo 0). That’s why the terms factors and divisors of 720 can be used interchangeably.
As is the case for any natural number greater than zero, the number itself, here 720, as well as 1 are factors and divisors of 720.
## Prime Factors of 720
The prime factors of 720 are the prime numbers which divide 720 exactly, without remainder as defined by the Euclidean division. In other words, a prime factor of 720 divides the number 720 without any rest, modulo 0.
For 720, the prime factors are: 2, 3, 5. By definition, 1 is not a prime number.
Besides 1, what sets the factors and the prime factors of the number 720 apart is the word “prime”. The former list contains both, composite and prime numbers, whereas the latter includes only prime numbers.
## Prime Factorization of 720
The prime factorization of 720 is 2 x 2 x 2 x 2 x 3 x 3 x 5. This is a unique list of the prime factors, along with their multiplicities. Note that the prime factorization of 720 does not include the number 1, yet it does include every instance of a certain prime factor.
720 is a composite number. In contrast to prime numbers which only have one factorization, composite numbers like 720 have at least two factorizations.
To illustrate what that means select the rightmost and leftmost integer in 720, 360, 240, 180, 144, 120, 90, 80, 72, 60, 48, 45, 40, 36, 30, 24, 20, 18, 16, 15, 12, 10, 9, 8, 6, 5, 4, 3, 2, 1 and multiply these integers to obtain 720. This is the first factorization. Next choose the second rightmost and the second leftmost entry to obtain the 2nd factorization which also produces 720.
The prime factorization or integer factorization of 720 means determining the set of prime numbers which, when multiplied together, produce the original number 720. This is also known as prime decomposition of 720.
Besides factors for 720, frequently searched terms on our website include:
We did not place any calculator here as there are already a plethora of them on the web. But you can find the factors, prime factors and the factorizations of many numbers including 720 by using the search form in the sidebar.
To sum up:
The factors, the prime factors and the prime factorization of 720 mean different things, and in strict terms cannot be used interchangeably despite being closely related.
The factors of seven hundred and twenty are: 720, 360, 240, 180, 144, 120, 90, 80, 72, 60, 48, 45, 40, 36, 30, 24, 20, 18, 16, 15, 12, 10, 9, 8, 6, 5, 4, 3, 2, 1. The prime factors of seven hundred and twenty are 2, 3, 5. And the prime factorization of seven hundred and twenty is 2 x 2 x 2 x 2 x 3 x 3 x 5. Remember that 1 is not a prime factor of 720.
No matter if you had been searching for prime factorization for 720 or prime numbers of 720, you have come to the right page. Also, if you typed what is the prime factorization of 720 in the search engine then you are right here, of course.
Taking all of the above into account, tasks including write 720 as a product of prime factors or list the factors of 720 will no longer pose a challenge to you.
If you have any questions about the factors of seven hundred and twenty then fill in the form below and we will respond as soon as possible. If our content concerning all factors of 720 has been of help to you then share it by means of pressing the social buttons. And don’t forget to bookmark us. |
# 4.4: Length of a Vector
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## Learning Objectives
• Find the length of a vector and the distance between two points in $$\mathbb{R}^n$$.
• Find the corresponding unit vector to a vector in $$\mathbb{R}^n$$.
In this section, we explore what is meant by the length of a vector in $$\mathbb{R}^n$$. We develop this concept by first looking at the distance between two points in $$\mathbb{R}^n$$.
First, we will consider the concept of distance for $$\mathbb{R}$$, that is, for points in $$\mathbb{R}^1$$. Here, the distance between two points $$P$$ and $$Q$$ is given by the absolute value of their difference. We denote the distance between $$P$$ and $$Q$$ by $$d(P,Q)$$ which is defined as $d(P,Q) = \sqrt{ \left( P-Q\right) ^{2}} \label{distance1}$
Consider now the case for $$n=2$$, demonstrated by the following picture.
There are two points $$P =\left( p_{1},p_{2}\right)$$ and $$Q = \left(q_{1},q_{2}\right)$$ in the plane. The distance between these points is shown in the picture as a solid line. Notice that this line is the hypotenuse of a right triangle which is half of the rectangle shown in dotted lines. We want to find the length of this hypotenuse which will give the distance between the two points. Note the lengths of the sides of this triangle are $$\left| p_{1}-q_{1}\right|$$ and $$\left| p_{2}-q_{2}\right|$$, the absolute value of the difference in these values. Therefore, the Pythagorean Theorem implies the length of the hypotenuse (and thus the distance between $$P$$ and $$Q$$) equals $\left( \left| p_{1}-q_{1}\right| ^{2}+\left| p_{2}-q_{2}\right| ^{2}\right) ^{1/2}=\left( \left( p_{1}-q_{1}\right) ^{2}+\left( p_{2}-q_{2}\right) ^{2}\right) ^{1/2} \label{distance2}$
Now suppose $$n=3$$ and let $$P = \left( p_{1},p_{2},p_{3}\right)$$ and $$Q = \left( q_{1},q_{2},q_{3}\right)$$ be two points in $$\mathbb{R}^{3}.$$ Consider the following picture in which the solid line joins the two points and a dotted line joins the points $$\left( q_{1},q_{2},q_{3}\right)$$ and $$\left( p_{1},p_{2},q_{3}\right) .$$
Here, we need to use Pythagorean Theorem twice in order to find the length of the solid line. First, by the Pythagorean Theorem, the length of the dotted line joining $$\left( q_{1},q_{2},q_{3}\right)$$ and $$\left( p_{1},p_{2},q_{3}\right)$$ equals $\left( \left( p_{1}-q_{1}\right) ^{2}+\left( p_{2}-q_{2}\right) ^{2}\right) ^{1/2}\nonumber$ while the length of the line joining $$\left( p_{1},p_{2},q_{3}\right)$$ to $$\left( p_{1},p_{2},p_{3}\right)$$ is just $$\left| p_{3}-q_{3}\right| .$$ Therefore, by the Pythagorean Theorem again, the length of the line joining the points $$P = \left( p_{1},p_{2},p_{3}\right)$$ and $$Q = \left( q_{1},q_{2},q_{3}\right)$$ equals $\left( \left( \left( \left( p_{1}-q_{1}\right) ^{2}+\left( p_{2}-q_{2}\right) ^{2}\right) ^{1/2}\right) ^{2}+\left( p_{3}-q_{3}\right) ^{2}\right) ^{1/2}\nonumber$ $=\left( \left( p_{1}-q_{1}\right) ^{2}+\left( p_{2}-q_{2}\right) ^{2}+\left( p_{3}-q_{3}\right) ^{2}\right) ^{1/2} \label{distance3}$
This discussion motivates the following definition for the distance between points in $$\mathbb{R}^n$$.
## Definition $$\PageIndex{1}$$: Distance Between Points
Let $$P=\left( p_{1},\cdots ,p_{n}\right)$$ and $$Q=\left( q_{1},\cdots ,q_{n}\right)$$ be two points in $$\mathbb{R}^{n}$$. Then the distance between these points is defined as $\text{ distance between }P\text{ and } Q\text{ } = d( P, Q ) = \left( \sum_{k=1}^{n}\left\vert p_{k}-q_{k}\right\vert ^{2}\right) ^{1/2}\nonumber$ This is called the distance formula. We may also write $$\left\vert P - Q \right\vert$$ as the distance between $$P$$ and $$Q$$.
From the above discussion, you can see that Definition $$\PageIndex{1}$$ holds for the special cases $$n=1,2,3$$, as in Equations $$\eqref{distance1}$$, $$\eqref{distance2}$$, $$\eqref{distance3}$$. In the following example, we use Definition $$\PageIndex{1}$$ to find the distance between two points in $$\mathbb{R}^4$$.
## Example $$\PageIndex{1}$$: Distance Between Points
Find the distance between the points $$P$$ and $$Q$$ in $$\mathbb{R}^{4}$$, where $$P$$ and $$Q$$ are given by $P=\left( 1,2,-4,6\right)\nonumber$ and $Q=\left( 2,3,-1,0\right)\nonumber$
Solution
We will use the formula given in Definition $$\PageIndex{1}$$ to find the distance between $$P$$ and $$Q$$. Use the distance formula and write $d(P,Q)= \left( \left( 1-2\right) ^{2}+\left( 2-3\right) ^{2}+\left( -4-\left( -1\right) \right) ^{2}+\left( 6-0\right)^{2}\right) ^{\frac{1}{2}} = 47\nonumber$
Therefore, $$d( P,Q) = \sqrt{47}.$$
There are certain properties of the distance between points which are important in our study. These are outlined in the following theorem.
## Theorem $$\PageIndex{1}$$: Properties of Distance
Let $$P$$ and $$Q$$ be points in $$\mathbb{R}^n$$, and let the distance between them, $$d( P, Q)$$, be given as in Definition $$\PageIndex{1}$$. Then, the following properties hold.
• $$d( P, Q) = d( Q, P)$$
• $$d( P, Q) \geq 0$$, and equals 0 exactly when $$P = Q.$$
There are many applications of the concept of distance. For instance, given two points, we can ask what collection of points are all the same distance between the given points. This is explored in the following example.
## Example $$\PageIndex{2}$$: The Plane Between Two Points
Describe the points in $$\mathbb{R}^3$$ which are at the same distance between $$\left( 1,2,3\right)$$ and $$\left( 0,1,2\right) .$$
Solution
Let $$P = \left( p_1 , p_2, p_3\right)$$ be such a point. Therefore, $$P$$ is the same distance from $$\left( 1,2,3\right)$$ and $$\left( 0,1,2\right) .$$ Then byDefinition $$\PageIndex{1}$$, $\sqrt{\left( p_1 -1\right) ^{2}+\left( p_2 -2\right) ^{2}+\left( p_3-3\right) ^{2}}= \sqrt{\left( p_1 - 0 \right)^{2}+\left( p_2-1\right) ^{2}+\left( p_3-2\right) ^{2}}\nonumber$ Squaring both sides we obtain $\left( p_1 -1\right) ^{2}+\left( p_2 -2\right) ^{2}+\left( p_3 -3\right) ^{2}=p_1^{2}+\left( p_2-1\right) ^{2}+\left( p_3 -2\right) ^{2}\nonumber$ and so $\ p_1^{2}-2p_1+14+p_2^{2}-4p_2+p_3^{2}-6p_3=p_1^{2}+p_2^{2}-2p_2+5+p_3^{2}-4p_3\nonumber$ Simplifying, this becomes $-2p_1+14-4p_2-6p_3=-2p_2+5-4p_3\nonumber$ which can be written as $2p_1+2p_2+2p_3=-9 \label{distanceplane}$ Therefore, the points $$P = \left( p_1,p_2,p_3\right)$$ which are the same distance from each of the given points form a plane whose equation is given by $$\eqref{distanceplane}$$.
We can now use our understanding of the distance between two points to define what is meant by the length of a vector. Consider the following definition.
## Definition $$\PageIndex{2}$$: Length of a Vector
Let $$\vec{u} = \left[ u_{1} \cdots u_{n} \right]^T$$ be a vector in $$\mathbb{R}^n$$. Then, the length of $$\vec{u}$$, written $$\| \vec{u} \|$$ is given by $\| \vec{u} \| = \sqrt{ u_{1}^2 + \cdots + u_{n}^2}\nonumber$
This definition corresponds to Definition $$\PageIndex{1}$$, if you consider the vector $$\vec{u}$$ to have its tail at the point $$0 = \left( 0, \cdots ,0 \right)$$ and its tip at the point $$U = \left(u_1, \cdots, u_n \right)$$. Then the length of $$\vec{u}$$ is equal to the distance between $$0$$ and $$U$$, $$d(0,U)$$. In general, $$d(P,Q)=||\vec{PQ}||$$.
ConsiderExample $$\PageIndex{1}$$. ByDefinition $$\PageIndex{2}$$, we could also find the distance between $$P$$ and $$Q$$ as the length of the vector connecting them. Hence, if we were to draw a vector $$\overrightarrow{PQ}$$ with its tail at $$P$$ and its point at $$Q$$, this vector would have length equal to $$\sqrt{47}$$.
We conclude this section with a new definition for the special case of vectors of length $$1$$.
## Definition $$\PageIndex{3}$$: Unit Vector
Let $$\vec{u}$$ be a vector in $$\mathbb{R}^{n}$$. Then, we call $$\vec{u}$$ a unit vector if it has length 1, that is if $\| \vec{u} \| = 1\nonumber$
Let $$\vec{v}$$ be a vector in $$\mathbb{R}^{n}$$. Then, the vector $$\vec{u}$$ which has the same direction as $$\vec{v}$$ but length equal to $$1$$ is the corresponding unit vector of $$\vec{v}$$. This vector is given by $\vec{u} = \frac{1}{\| \vec{v} \|} \vec{v}\nonumber$
We often use the term normalize to refer to this process. When we normalize a vector, we find the corresponding unit vector of length $$1$$. Consider the following example.
## Example $$\PageIndex{3}$$: Finding a Unit Vector
Let $$\vec{v}$$ be given by $\vec{v} = \left[ \begin{array}{rrr} 1 & -3 & 4 \end{array} \right]^T\nonumber$ Find the unit vector $$\vec{u}$$ which has the same direction as $$\vec{v}$$.
Solution
We will use Definition $$\PageIndex{3}$$ to solve this. Therefore, we need to find the length of $$\vec{v}$$ which, by Definition $$\PageIndex{2}$$ is given by $\| \vec{v} \| = \sqrt{ v_{1}^2 + v_{2}^2+ v_{3}^2}\nonumber$ Using the corresponding values we find that \begin{aligned} \| \vec{v} \| &= \sqrt{ 1^2 + \left(-3 \right)^2 + 4^2} \\ &= \sqrt{ 1 + 9 + 16} \\ &= \sqrt{26} \end{aligned} In order to find $$\vec{u}$$, we divide $$\vec{v}$$ by $$\sqrt{26}$$. The result is \begin{aligned} \vec{u} &= \frac{1}{\| \vec{v} \|} \vec{v} \\ &= \frac{1}{\sqrt{26}} \left[ \begin{array}{rrr} 1 & -3 & 4 \end{array} \right]^T \\ &= \left[ \begin{array}{rrr} \frac{1}{\sqrt{26}} & -\frac{3}{\sqrt{26}} & \frac{4}{\sqrt{26}} \end{array} \right]^T\end{aligned}
You can verify using the Definition $$\PageIndex{1}$$ that $$\| \vec{u} \| = 1$$.
Below is a video on finding the unit vector and drawing the vector and its unit vector.
Below is a video on finding the unit vector in the direction of a given vector.
This page titled 4.4: Length of a Vector is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) . |
# 1.8 Order of Operation
### ORDER OF OPERATION
Some math problems are a mixture of addition, subtraction, division, and multiplication. An operation to be performed might be true for one item but not another, so parentheses are used () for clarification.
There is a specific order to follow when making calculations.
The order in which operations are performed is:
1. Parentheses: ( )
2. Exponents: 23
3. Multiplication and division
5. Left to right
Note: It is always useful to add parentheses to clarify the order.
Example 1 - Solve 10 + 10 ÷ 10
Step 1. Division is performed before addition.
10 ÷ 10 = 1
10 + 1 = 11
10 + 10 ÷ 10 = 11
The same example can be rewritten:
10 + 10 ÷ 10
10 + (10 ÷ 10). Using the parentheses helps clarify the order of operation.
Example 2 - Solve 63 ÷ (10 - 8)2 ÷ 2 + 2
Step 1. Parentheses.
63 ÷ (10 - 8)2 ÷ 2 + 2 = 63 ÷ 22 ÷ 2 + 2
Step 2. Exponents.
216 ÷ 4 ÷ 2 + 2
Step 3. Division in order from left to right.
216 ÷ 4 = 54
54 ÷ 2 = 27 |
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### Schools Mathematics Grand Challenge
Week eight's Puzzles
Problem 22:
The twenty second problem was:
John has been given a field in the shape of a triangle. Two sides of the triangle are exactly 10 metres long. What is the largest possible area, in square metres, of John's triangular field? Give your answer as the number of square metres. There is no need to include units.
The solution was:
The area of a triangle is half the base by the perpendicular height. We get to choose which side to use as the base, so let's use one of the sides of John's field that is length 10 metres.
Since the base is now fixed, we get the biggest area by making the height be as big as possible. The height cannot be bigger than 10, otherwise the second side of length 10 would not be able to reach the corner, so we get the biggest area by putting the two sides at right angles. The area will then be 10*10/2 = 50.
Problem 23:
The twenty third problem was:
Have a look at the following picture. The circle is area 438 units and has centre C. The line from A to E goes through the centre and the line from D to B goes through the centre. The angle between CA and AB is 60 degrees.
If the area of the zig-zaggy triangle CDF is 30 units, what is the area of the bricked piece of the picture? Again, there is no need to give units.
The solution was:
We want to find the area of the bricked bit of the picture. It is the area of the segment of the circle cut out by the line BC, the line CE and the arc EB minus the area of the triange BCF. Let's figure out the area of both these parts.
We've been told that the angle CAB is 60, but CA and CB are both radii of the circle, so ACB is isosceles and the angle CBA must be 60 degrees too. Since the angle in a triangle add up to 180, ACB must be 60. Finally, since ACE is a straight line, the angle BCE must be 120. The angles are shown on the picture below.
Now 120 degrees is one third of the circle. We were told that the full area of the circle was 438, so the area of the segment BCE must be 438/3 = 146.
Now we need to figure out the area of the triangle BCF. Imagine that CB is the base and the heigth of point F is the height. CB is a radius of the circle, so it the same as CD. The point F is also the same height above DC ad CB, so BCF and CDF have the same base length and the same heigth. This means they must have the same area, which we was told was 30 units.
So, the area of the bricked bit of the picture is
(area of the segment BCE) - (area of the triangle BCF) = 438/3 - 30 = 116.
Problem 24:
The twenty fourth problem was:
In the triangle in the diagram below, the length of the side AB is the same as the length of the side AC. The point N divides the line from B to C into two equal parts. Also, the line BM divides the angle ABC into two equal parts, and the line from B to M is two times as long as the line from A to N.
What is the size, in degrees, of the angle BAC (the largest angle) in the triangle?
The solution was:
We're going to draw an extra line to make the problem a bit easier. Let's draw a line through N parallel to BM, like this.
The triangle NCP will be a scaled down version of BCM, because all the angles are the same and the BC is twice as long as NC. This means that BM is twice the length of NP, but we were told that AN was half the length of BM, so AN and NP must be the same length. So ANP is an isosceles triangle.
Now, let's mark in some angles, shown on the picture below. Let the two half-angles at the corned B be a. Then, since ABC is isosceles, the angle at C must be 2a also. Since MBC and PNC are similar, the angle at PNC must also be a. Call the angle at BAN b, then NAC is also b and because ANP is an isosceles triangle NPA must also be b. The last angle we need to mark is the one at NPC, call this c.
Now APC is a straight line we get c = 180 - b. Now, look at triangle NPC, we know the angles add up to 180, so a + c + 2a = 180, or 3a + 180 - b = 180, so 3a = b. Finally, we look at the triangle ABC we get 2a + 2b + 2a = 180, or 4a+2b = 180, or 4a + 6a = 180, so a = 18, b = 54 and the angle at the top top of the triangle is 2b = 108.
Phew! |
# 5.8 Modeling using variation (Page 2/14)
Page 2 / 14
Do the graphs of all direct variation equations look like [link] ?
No. Direct variation equations are power functions—they may be linear, quadratic, cubic, quartic, radical, etc. But all of the graphs pass through $\text{\hspace{0.17em}}\left(0,0\right).$
The quantity $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies directly with the square of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}y=24\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x=3,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is 4.
$\frac{128}{3}$
## Solving inverse variation problems
Water temperature in an ocean varies inversely to the water’s depth. The formula $\text{\hspace{0.17em}}T=\frac{14,000}{d}\text{\hspace{0.17em}}$ gives us the temperature in degrees Fahrenheit at a depth in feet below Earth’s surface. Consider the Atlantic Ocean, which covers 22% of Earth’s surface. At a certain location, at the depth of 500 feet, the temperature may be 28°F.
If we create [link] , we observe that, as the depth increases, the water temperature decreases.
$d,\text{\hspace{0.17em}}$ depth $T=\frac{\text{14,000}}{d}$ Interpretation
500 ft $\frac{14,000}{500}=28$ At a depth of 500 ft, the water temperature is 28° F.
1000 ft $\frac{14,000}{1000}=14$ At a depth of 1,000 ft, the water temperature is 14° F.
2000 ft $\frac{14,000}{2000}=7$ At a depth of 2,000 ft, the water temperature is 7° F.
We notice in the relationship between these variables that, as one quantity increases, the other decreases. The two quantities are said to be inversely proportional and each term varies inversely with the other. Inversely proportional relationships are also called inverse variations .
For our example, [link] depicts the inverse variation . We say the water temperature varies inversely with the depth of the water because, as the depth increases, the temperature decreases. The formula $\text{\hspace{0.17em}}y=\frac{k}{x}\text{\hspace{0.17em}}$ for inverse variation in this case uses $\text{\hspace{0.17em}}k=14,000.\text{\hspace{0.17em}}$
## Inverse variation
If $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ are related by an equation of the form
$y=\frac{k}{{x}^{n}}$
where $\text{\hspace{0.17em}}k\text{\hspace{0.17em}}$ is a nonzero constant, then we say that $\text{\hspace{0.17em}}y$ varies inversely with the $\text{\hspace{0.17em}}n\text{th}\text{\hspace{0.17em}}$ power of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ In inversely proportional relationships, or inverse variations , there is a constant multiple $\text{\hspace{0.17em}}k={x}^{n}y.\text{\hspace{0.17em}}$
## Writing a formula for an inversely proportional relationship
A tourist plans to drive 100 miles. Find a formula for the time the trip will take as a function of the speed the tourist drives.
Recall that multiplying speed by time gives distance. If we let $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ represent the drive time in hours, and $\text{\hspace{0.17em}}v\text{\hspace{0.17em}}$ represent the velocity (speed or rate) at which the tourist drives, then $\text{\hspace{0.17em}}vt=\text{distance}\text{.}\text{\hspace{0.17em}}$ Because the distance is fixed at 100 miles, $\text{\hspace{0.17em}}vt=100\text{\hspace{0.17em}}$ so $t=100/v.\text{\hspace{0.17em}}$ Because time is a function of velocity, we can write $\text{\hspace{0.17em}}t\left(v\right).$
$\begin{array}{ccc}\hfill t\left(v\right)& =& \frac{100}{v}\hfill \\ & =& 100{v}^{-1}\hfill \end{array}$
We can see that the constant of variation is 100 and, although we can write the relationship using the negative exponent, it is more common to see it written as a fraction. We say that time varies inversely with velocity.
Given a description of an indirect variation problem, solve for an unknown.
1. Identify the input, $\text{\hspace{0.17em}}x,\text{\hspace{0.17em}}$ and the output, $\text{\hspace{0.17em}}y.$
2. Determine the constant of variation. You may need to multiply $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by the specified power of $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to determine the constant of variation.
3. Use the constant of variation to write an equation for the relationship.
4. Substitute known values into the equation to find the unknown.
## Solving an inverse variation problem
A quantity $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ varies inversely with the cube of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}y=25\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x=2,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ when $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is 6.
The general formula for inverse variation with a cube is $\text{\hspace{0.17em}}y=\frac{k}{{x}^{3}}.\text{\hspace{0.17em}}$ The constant can be found by multiplying $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ by the cube of $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$
$\begin{array}{ccc}\hfill k& =& {x}^{3}y\hfill \\ & =& {2}^{3}\cdot 25\hfill \\ & =& 200\hfill \end{array}$
Now we use the constant to write an equation that represents this relationship.
$\begin{array}{ccc}\hfill y& =& \frac{k}{{x}^{3}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}k=200\hfill \\ y\hfill & =& \frac{200}{{x}^{3}}\hfill \end{array}$
Substitute $\text{\hspace{0.17em}}x=6\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}y.$
$\begin{array}{ccc}\hfill y& =& \frac{200}{{6}^{3}}\hfill \\ & =& \frac{25}{27}\hfill \end{array}$
root under 3-root under 2 by 5 y square
The sum of the first n terms of a certain series is 2^n-1, Show that , this series is Geometric and Find the formula of the n^th
cosA\1+sinA=secA-tanA
why two x + seven is equal to nineteen.
The numbers cannot be combined with the x
Othman
2x + 7 =19
humberto
2x +7=19. 2x=19 - 7 2x=12 x=6
Yvonne
because x is 6
SAIDI
what is the best practice that will address the issue on this topic? anyone who can help me. i'm working on my action research.
simplify each radical by removing as many factors as possible (a) √75
how is infinity bidder from undefined?
what is the value of x in 4x-2+3
give the complete question
Shanky
4x=3-2 4x=1 x=1+4 x=5 5x
Olaiya
hi can you give another equation I'd like to solve it
Daniel
what is the value of x in 4x-2+3
Olaiya
if 4x-2+3 = 0 then 4x = 2-3 4x = -1 x = -(1÷4) is the answer.
Jacob
4x-2+3 4x=-3+2 4×=-1 4×/4=-1/4
LUTHO
then x=-1/4
LUTHO
4x-2+3 4x=-3+2 4x=-1 4x÷4=-1÷4 x=-1÷4
LUTHO
A research student is working with a culture of bacteria that doubles in size every twenty minutes. The initial population count was 1350 bacteria. Rounding to five significant digits, write an exponential equation representing this situation. To the nearest whole number, what is the population size after 3 hours?
v=lbh calculate the volume if i.l=5cm, b=2cm ,h=3cm
Need help with math
Peya
can you help me on this topic of Geometry if l help you
litshani
( cosec Q _ cot Q ) whole spuare = 1_cosQ / 1+cosQ
A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So, the length of the guy wire can be found by evaluating √(90000+160000). What is the length of the guy wire?
the indicated sum of a sequence is known as
how do I attempted a trig number as a starter
cos 18 ____ sin 72 evaluate |
CLASS-6TYPES OF ALGEBRAIC EXPRESSION
TYPES OF ALGEBRAIC EXPRESSION -
Algebraic expressions can be categorized into several types based on their structure and characteristics. Here are some common types of algebraic expressions:-
1. Monomial:- A monomial is a single algebraic term consisting of a constant and/or a single variable raised to a non-negative integer exponent.
For example-
• 3x²
• 4y
• 5
• 3x
• 2x²
Above all are monomials.
2. Binomial:- A binomial is an algebraic expression composed of exactly two unlike terms connected by addition or subtraction.
For example-
• 2x + 3
• 5y - 7
• x²- 4
• 2x + 3
• 5y - 2
Above all are binomials.
3. Trinomial:- A trinomial is an algebraic expression composed of exactly three unlike terms connected by addition or subtraction.
For example-
• 3x² + 2x + 1
• x² - 4x + 7
• 5x²+ 6x - 9
• y²+ 3y - 6
• 3z²- 8y + 5
Above all are trinomial.
4. Polynomial:- A polynomial is a sum of one or more monomials, where each monomial is referred to as a term. Polynomials can have any number of terms, including zero terms (constant polynomials).
For example-
• 3x³+ 2x² + 8x - 10
• 8x³- 6x²- 3x + 7
• 15y³+ 8y²+ 5y - 9
• 3x⁶ + 2x⁵ + 8x³- 10x²+ 15
Above all are polynomial.
5. Constant Expression:- A constant expression is an algebraic expression consisting of a single constant value with no variables.
For example-
"5" or "-3.14" are constant expressions.
6. Rational Expression:- A rational expression is an algebraic expression in the form of a fraction, where both the numerator and denominator are polynomials.
For example -
(3x² + 2x + 1)
• ------------------------
(2x + 3)
(8x³- 6x²- 3x + 7)
• -------------------------------
(x² - 4x + 7)
Above all are rational expression.
7. Radical Expression:- A radical expression contains one or more square roots, cube roots, or other types of roots.
For example-
• √(x² + 9)
• ∛(8y³- 27)
8. Exponential Expression:- An exponential expression involves variables raised to exponents.
For example -
"2^x" or "3^(2y - 1)" are exponential expressions.
9. Algebraic Fraction:- An algebraic fraction is a fraction in which the numerator and denominator are both polynomials.
For example -
(5x²- 3x + 2)
• -------------------------
(2x + 3)
(2x³- 4x²- 5x + 10)
• ---------------------------------
(3x² + 2x + 5)
Above all are algebraic fraction.
10. Linear Expression:- A linear expression is a polynomial of degree 1, meaning it contains only terms with variables raised to the first power.
For example -
"3x + 2y - 5" is a linear expression.
11. Quadratic Expression:- A quadratic expression is a polynomial of degree 2, meaning it contains at least one term with a variable raised to the second power.
For example-
"ax²+ bx + c" is a quadratic expression.
These are some of the common types of algebraic expressions. Algebraic expressions are used to represent mathematical relationships, solve equations, and model real-world situations in various fields of mathematics and science. The classification of an algebraic expression into one of these types depends on its structure and the types of terms it contains. |
Prealgebra
# 4.7Solve Equations with Fractions
Prealgebra4.7 Solve Equations with Fractions
## Learning Objectives
By the end of this section, you will be able to:
• Determine whether a fraction is a solution of an equation
• Solve equations with fractions using the Addition, Subtraction, and Division Properties of Equality
• Solve equations using the Multiplication Property of Equality
• Translate sentences to equations and solve
## Be Prepared 4.7
Before you get started, take this readiness quiz. If you miss a problem, go back to the section listed and review the material.
1. Evaluate $x+4whenx=−3x+4whenx=−3$
If you missed this problem, review Example 3.23.
2. Solve: $2y−3=9.2y−3=9.$
If you missed this problem, review Example 3.61.
3. Solve: $y−3=−9y−3=−9$
If you missed this problem, review Example 4.28.
## Determine Whether a Fraction is a Solution of an Equation
As we saw in Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality, a solution of an equation is a value that makes a true statement when substituted for the variable in the equation. In those sections, we found whole number and integer solutions to equations. Now that we have worked with fractions, we are ready to find fraction solutions to equations.
The steps we take to determine whether a number is a solution to an equation are the same whether the solution is a whole number, an integer, or a fraction.
## How To
### Determine whether a number is a solution to an equation.
1. Step 1. Substitute the number for the variable in the equation.
2. Step 2. Simplify the expressions on both sides of the equation.
3. Step 3. Determine whether the resulting equation is true. If it is true, the number is a solution. If it is not true, the number is not a solution.
## Example 4.95
Determine whether each of the following is a solution of $x−310=12.x−310=12.$
1. $x=1x=1$
2. $x=45x=45$
3. $x=−45x=−45$
## Try It 4.189
Determine whether each number is a solution of the given equation.
$x−23=16:x−23=16:$
1. $x=1x=1$
2. $x=56x=56$
3. $x=−56x=−56$
## Try It 4.190
Determine whether each number is a solution of the given equation.
$y−14=38:y−14=38:$
1. $y=1y=1$
2. $y=−58y=−58$
3. $y=58y=58$
## Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality
In Solve Equations with the Subtraction and Addition Properties of Equality and Solve Equations Using Integers; The Division Property of Equality, we solved equations using the Addition, Subtraction, and Division Properties of Equality. We will use these same properties to solve equations with fractions.
## Addition, Subtraction, and Division Properties of Equality
For any numbers $a,b,andc,a,b,andc,$
$ifa=b,thena+c=b+c.ifa=b,thena+c=b+c.$ Addition Property of Equality $ifa=b,thena−c=b−c.ifa=b,thena−c=b−c.$ Subtraction Property of Equality $ifa=b,thenac=bc,c≠0.ifa=b,thenac=bc,c≠0.$ Division Property of Equality
Table 4.3
In other words, when you add or subtract the same quantity from both sides of an equation, or divide both sides by the same quantity, you still have equality.
## Example 4.96
Solve: $y+916=516.y+916=516.$
## Try It 4.191
Solve: $y+1112=512.y+1112=512.$
## Try It 4.192
Solve: $y+815=415.y+815=415.$
We used the Subtraction Property of Equality in Example 4.96. Now we’ll use the Addition Property of Equality.
## Example 4.97
Solve: $a−59=−89.a−59=−89.$
## Try It 4.193
Solve: $a−35=−85.a−35=−85.$
## Try It 4.194
Solve: $n−37=−97.n−37=−97.$
The next example may not seem to have a fraction, but let’s see what happens when we solve it.
## Example 4.98
Solve: $10q=44.10q=44.$
## Try It 4.195
Solve: $12u=−76.12u=−76.$
## Try It 4.196
Solve: $8m=92.8m=92.$
## Solve Equations with Fractions Using the Multiplication Property of Equality
Consider the equation $x4=3.x4=3.$ We want to know what number divided by $44$ gives $3.3.$ So to “undo” the division, we will need to multiply by $4.4.$ The Multiplication Property of Equality will allow us to do this. This property says that if we start with two equal quantities and multiply both by the same number, the results are equal.
## The Multiplication Property of Equality
For any numbers $a,b,a,b,$ and $c,c,$
$ifa=b,thenac=bc.ifa=b,thenac=bc.$
If you multiply both sides of an equation by the same quantity, you still have equality.
Let’s use the Multiplication Property of Equality to solve the equation $x7=−9.x7=−9.$
## Example 4.99
Solve: $x7=−9.x7=−9.$
## Try It 4.197
Solve: $f5=−25.f5=−25.$
## Try It 4.198
Solve: $h9=−27.h9=−27.$
## Example 4.100
Solve: $p−8=−40.p−8=−40.$
## Try It 4.199
Solve: $c−7=−35.c−7=−35.$
## Try It 4.200
Solve: $x−11=−12.x−11=−12.$
### Solve Equations with a Coefficient of $−1−1$
Look at the equation $−y=15.−y=15.$ Does it look as if $yy$ is already isolated? But there is a negative sign in front of $y,y,$ so it is not isolated.
There are three different ways to isolate the variable in this type of equation. We will show all three ways in Example 4.101.
## Example 4.101
Solve: $−y=15.−y=15.$
## Try It 4.201
Solve: $−y=48.−y=48.$
## Try It 4.202
Solve: $−c=−23.−c=−23.$
### Solve Equations with a Fraction Coefficient
When we have an equation with a fraction coefficient we can use the Multiplication Property of Equality to make the coefficient equal to $1.1.$
For example, in the equation:
$34x=2434x=24$
The coefficient of $xx$ is $34.34.$ To solve for $x,x,$ we need its coefficient to be $1.1.$ Since the product of a number and its reciprocal is $1,1,$ our strategy here will be to isolate $xx$ by multiplying by the reciprocal of $34.34.$ We will do this in Example 4.102.
## Example 4.102
Solve: $34x=24.34x=24.$
## Try It 4.203
Solve: $25n=14.25n=14.$
## Try It 4.204
Solve: $56y=15.56y=15.$
## Example 4.103
Solve: $−38w=72.−38w=72.$
## Try It 4.205
Solve: $−47a=52.−47a=52.$
## Try It 4.206
Solve: $−79w=84.−79w=84.$
## Translate Sentences to Equations and Solve
Now we have covered all four properties of equality—subtraction, addition, division, and multiplication. We’ll list them all together here for easy reference.
Subtraction Property of Equality:For any real numbers $a, b,a, b,$ and $c,c,$if $a=b,a=b,$ then $a−c=b−c.a−c=b−c.$ Addition Property of Equality:For any real numbers $a, b,a, b,$ and $c,c,$if $a=b,a=b,$ then $a+c=b+c.a+c=b+c.$ Division Property of Equality:For any numbers $a, b,a, b,$ and $c,c,$ where $c≠0c≠0$if $a=b,a=b,$ then $ac=bcac=bc$ Multiplication Property of Equality:For any real numbers $a, b,a, b,$ and $cc$if $a=b,a=b,$ then $ac=bcac=bc$
When you add, subtract, multiply or divide the same quantity from both sides of an equation, you still have equality.
In the next few examples, we’ll translate sentences into equations and then solve the equations. It might be helpful to review the translation table in Evaluate, Simplify, and Translate Expressions.
## Example 4.104
Translate and solve: $nn$ divided by $66$ is $−24.−24.$
## Try It 4.207
Translate and solve: $nn$ divided by $77$ is equal to $−21.−21.$
## Try It 4.208
Translate and solve: $nn$ divided by $88$ is equal to $−56.−56.$
## Example 4.105
Translate and solve: The quotient of $qq$ and $−5−5$ is $70.70.$
## Try It 4.209
Translate and solve: The quotient of $qq$ and $−8−8$ is $72.72.$
## Try It 4.210
Translate and solve: The quotient of $pp$ and $−9−9$ is $81.81.$
## Example 4.106
Translate and solve: Two-thirds of $ff$ is $18.18.$
## Try It 4.211
Translate and solve: Two-fifths of $ff$ is $16.16.$
## Try It 4.212
Translate and solve: Three-fourths of $ff$ is $21.21.$
## Example 4.107
Translate and solve: The quotient of $mm$ and $5656$ is $34.34.$
## Try It 4.213
Translate and solve. The quotient of $nn$ and $2323$ is $512.512.$
## Try It 4.214
Translate and solve The quotient of $cc$ and $3838$ is $49.49.$
## Example 4.108
Translate and solve: The sum of three-eighths and $xx$ is three and one-half.
## Try It 4.215
Translate and solve: The sum of five-eighths and $xx$ is one-fourth.
## Try It 4.216
Translate and solve: The difference of one-and-three-fourths and $xx$ is five-sixths.
## Section 4.7 Exercises
### Practice Makes Perfect
Determine Whether a Fraction is a Solution of an Equation
In the following exercises, determine whether each number is a solution of the given equation.
498.
$x−25=110:x−25=110:$
1. $x=1x=1$
2. $x=12x=12$
3. $x=−12x=−12$
499.
$y−13=512:y−13=512:$
1. $y=1y=1$
2. $y=34y=34$
3. $y=−34y=−34$
500.
$h+34=25:h+34=25:$
1. $h=1h=1$
2. $h=720h=720$
3. $h=−720h=−720$
501.
$k+25=56:k+25=56:$
1. $k=1k=1$
2. $k=1330k=1330$
3. $k=−1330k=−1330$
Solve Equations with Fractions using the Addition, Subtraction, and Division Properties of Equality
In the following exercises, solve.
502.
$y + 1 3 = 4 3 y + 1 3 = 4 3$
503.
$m + 3 8 = 7 8 m + 3 8 = 7 8$
504.
$f + 9 10 = 2 5 f + 9 10 = 2 5$
505.
$h + 5 6 = 1 6 h + 5 6 = 1 6$
506.
$a − 5 8 = − 7 8 a − 5 8 = − 7 8$
507.
$c − 1 4 = − 5 4 c − 1 4 = − 5 4$
508.
$x − ( − 3 20 ) = − 11 20 x − ( − 3 20 ) = − 11 20$
509.
$z − ( − 5 12 ) = − 7 12 z − ( − 5 12 ) = − 7 12$
510.
$n − 1 6 = 3 4 n − 1 6 = 3 4$
511.
$p − 3 10 = 5 8 p − 3 10 = 5 8$
512.
$s + ( − 1 2 ) = − 8 9 s + ( − 1 2 ) = − 8 9$
513.
$k + ( − 1 3 ) = − 4 5 k + ( − 1 3 ) = − 4 5$
514.
$5 j = 17 5 j = 17$
515.
$7 k = 18 7 k = 18$
516.
$−4 w = 26 −4 w = 26$
517.
$−9 v = 33 −9 v = 33$
Solve Equations with Fractions Using the Multiplication Property of Equality
In the following exercises, solve.
518.
$f 4 = −20 f 4 = −20$
519.
$b 3 = −9 b 3 = −9$
520.
$y 7 = −21 y 7 = −21$
521.
$x 8 = −32 x 8 = −32$
522.
$p −5 = −40 p −5 = −40$
523.
$q −4 = −40 q −4 = −40$
524.
$r −12 = −6 r −12 = −6$
525.
$s −15 = −3 s −15 = −3$
526.
$− x = 23 − x = 23$
527.
$− y = 42 − y = 42$
528.
$− h = − 5 12 − h = − 5 12$
529.
$− k = − 17 20 − k = − 17 20$
530.
$4 5 n = 20 4 5 n = 20$
531.
$3 10 p = 30 3 10 p = 30$
532.
$3 8 q = −48 3 8 q = −48$
533.
$5 2 m = −40 5 2 m = −40$
534.
$− 2 9 a = 16 − 2 9 a = 16$
535.
$− 3 7 b = 9 − 3 7 b = 9$
536.
$− 6 11 u = −24 − 6 11 u = −24$
537.
$− 5 12 v = −15 − 5 12 v = −15$
Mixed Practice
In the following exercises, solve.
538.
$3 x = 0 3 x = 0$
539.
$8 y = 0 8 y = 0$
540.
$4 f = 4 5 4 f = 4 5$
541.
$7 g = 7 9 7 g = 7 9$
542.
$p + 2 3 = 1 12 p + 2 3 = 1 12$
543.
$q + 5 6 = 1 12 q + 5 6 = 1 12$
544.
$7 8 m = 1 10 7 8 m = 1 10$
545.
$1 4 n = 7 10 1 4 n = 7 10$
546.
$− 2 5 = x + 3 4 − 2 5 = x + 3 4$
547.
$− 2 3 = y + 3 8 − 2 3 = y + 3 8$
548.
$11 20 = − f 11 20 = − f$
549.
$8 15 = − d 8 15 = − d$
Translate Sentences to Equations and Solve
In the following exercises, translate to an algebraic equation and solve.
550.
$nn$ divided by eight is $−16.−16.$
551.
$nn$ divided by six is $−24.−24.$
552.
$mm$ divided by $−9−9$ is $−7.−7.$
553.
$mm$ divided by $−7−7$ is $−8.−8.$
554.
The quotient of $ff$ and $−3−3$ is $−18.−18.$
555.
The quotient of $ff$ and $−4−4$ is $−20.−20.$
556.
The quotient of $gg$ and twelve is $8.8.$
557.
The quotient of $gg$ and nine is $14.14.$
558.
Three-fourths of $qq$ is $12.12.$
559.
Two-fifths of $qq$ is $20.20.$
560.
Seven-tenths of $pp$ is $−63.−63.$
561.
Four-ninths of $pp$ is $−28.−28.$
562.
$mm$ divided by $44$ equals negative $6.6.$
563.
The quotient of $hh$ and $22$ is $43.43.$
564.
Three-fourths of $zz$ is the same as $15.15.$
565.
The quotient of $aa$ and $2323$ is $34.34.$
566.
The sum of five-sixths and $xx$ is $12.12.$
567.
The sum of three-fourths and $xx$ is $18.18.$
568.
The difference of $yy$ and one-fourth is $−18.−18.$
569.
The difference of $yy$ and one-third is $−16.−16.$
### Everyday Math
570.
Shopping Teresa bought a pair of shoes on sale for $48.48.$ The sale price was $2323$ of the regular price. Find the regular price of the shoes by solving the equation $23p=4823p=48$
571.
Playhouse The table in a child’s playhouse is $3535$ of an adult-size table. The playhouse table is $1818$ inches high. Find the height of an adult-size table by solving the equation $35h=18.35h=18.$
### Writing Exercises
572.
Example 4.100 describes three methods to solve the equation $−y=15.−y=15.$ Which method do you prefer? Why?
573.
Richard thinks the solution to the equation $34x=2434x=24$ is $16.16.$ Explain why Richard is wrong.
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
Overall, after looking at the checklist, do you think you are well-prepared for the next Chapter? Why or why not?
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## Tuesday, November 17, 2020
Published November 17, 2020 by with 1 comment
# How Long Until My Investments Start Making Money?
Say you invest some fixed amount of money every year. How long does it take for the investments to grow faster than the amount you're putting into them?
#### Basic math problem
You invest $X per year in an account that yields R in gains. How long does it take for the gain in a year to be greater than$X?
Another way of asking this is 'when does R times the future value of investing X each year exceed X?'
The future value of a regular yearly investment where N = number of years, X = yearly investment, and R = growth rate of the investment is:
$FV = \frac{X*((1+R)^N - 1)}{R}$
What we're looking for is the number of years it takes for R times that to exceed X. That is, we want to solve:
$\frac{R*X*((1+R)^N - 1)}{R} > X$
Noticing that the R's cancel in numerator and denominator and dividing both sides by X you get:
$((1+R)^N - 1) > 1$
$((1+R)^N) > 2$
Simplifying:
$log((1+R)^N) > log(2)$
$N*log((1+R)) > log(2)$
$N > \frac{log(2)}{log(1+R)}$
This is kind of cool. You might not recognize it right away, but that says 'N is greater than the doubling time of the investment'. It's really cool that it works out that way. Because of some pretty good approximations that work out, that means that the investment growth takes over the new money moved in after roughly '72 divided by annual interest rate' years.
For a quick concrete example of what this means...say you invest $10,000 per year into an account yielding 6%. The time it takes for the 6% yield each year to exceed$10,000 is log(2)/log(1.06) which is ~12 years.
#### Simple plot
Here's a simple interactive plot showing the breakdown between money invested and money from gains for an annual \$10,000 investment using the interest rate that you enter below:
Interest rate (%)
#### 1 comment:
1. สล็อตออนไลน์ เราจะไปโฟกัสแค่เรื่องเกมอย่างเดียวไม่ได้ แต่เราต้องดูเรื่องของ เว็บเกม สล็อตออนไลน์ อันดับ 1 ที่เราเลือกใช้บริการด้วย ว่าเขามีเงื่อนไขในการเล่นอย่างไร หรือเงื่อนไขในการให้บริการแบบไหน ปกติแล้วทุกคาสิโนออนไลน์ มักจะมีโบนัสเพิ่มให้ เวลาเราฝากเงินเข้าไป เพื่อเล่นเกม สล็อตออนไลน์ |
'
# Search results
Found 1445 matches
Right Triangle (sides)
A right triangle (American English) or right-angled triangle (British English) is a triangle in which one angle is a right angle (that is, a 90-degree ... more
Worksheet 334
In a video game design, a map shows the location of other characters relative to the player, who is situated at the origin, and the direction they are facing. A character currently shows on the map at coordinates (-3, 5). If the player rotates counterclockwise by 20 degrees, then the objects in the map will correspondingly rotate 20 degrees clockwise. Find the new coordinates of the character.
To rotate the position of the character, we can imagine it as a point on a circle, and we will change the angle of the point by 20 degrees. To do so, we first need to find the radius of this circle and the original angle.
Drawing a right triangle inside the circle, we can find the radius using the Pythagorean Theorem:
Pythagorean theorem (right triangle)
To find the angle, we need to decide first if we are going to find the acute angle of the triangle, the reference angle, or if we are going to find the angle measured in standard position. While either approach will work, in this case we will do the latter. By applying the cosine function and using our given information we get
Cosine function
Subtraction
While there are two angles that have this cosine value, the angle of 120.964 degrees is in the second quadrant as desired, so it is the angle we were looking for.
Rotating the point clockwise by 20 degrees, the angle of the point will decrease to 100.964 degrees. We can then evaluate the coordinates of the rotated point
For x axis:
Cosine function
For y axis:
Sine function
The coordinates of the character on the rotated map will be (-1.109, 5.725)
Reference : PreCalculus: An Investigation of Functions,Edition 1.4 © 2014 David Lippman and Melonie Rasmussen
http://www.opentextbookstore.com/precalc/
Cosine function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
Tangent function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
Cotangent function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
Area of a triangle (by the tangent of an acute or obtuse angle of the triangle)
A triangle is a polygon with three edges and three vertices. In a scalene triangle, all sides are unequal and equivalently all angles are unequal. The area ... more
Area moments of inertia for a filled triangular area with respect to an axis through the centroid
The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a ... more
Worksheet 542
Sine function
The trigonometric functions (also called the circular functions) are functions of an angle. They relate the angles of a triangle to the lengths of its ... more
Moment of Inertia - Sphere (solid) - y axis
In physics and applied mathematics, the mass moment of inertia, usually denoted by I, measures the extent to which an object resists rotational ... more
...can't find what you're looking for?
Create a new formula
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# How do you solve 0.1p - 0.2p + 2 =-4.6?
Mar 7, 2017
See the entire solution process below:
#### Explanation:
First, subtract $\textcolor{red}{2}$ from each side of the equation to isolate the $p$ terms while keeping the equation balanced:
$0.1 p - 0.2 p + 2 - \textcolor{red}{2} = - 4.6 - \textcolor{red}{2}$
$0.1 p - 0.2 p + 0 = - 6.6$
$0.1 p - 0.2 p = - 6.6$
Next, combine like terms on the left side of the equation:
$\left(0.1 - 0.2\right) p = - 6.6$
$- 0.1 p = - 6.6$
Now, divide each side of the equation by $\textcolor{red}{- 0.1}$ to solve for $p$ while keeping the equation balanced:
$\frac{- 0.1 p}{\textcolor{red}{- 0.1}} = \frac{- 6.6}{\textcolor{red}{- 0.1}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{- 0.1}}} p}{\cancel{\textcolor{red}{- 0.1}}} = 66$
$p = 66$ |
# Difference between revisions of "2020 AMC 10A Problems/Problem 12"
## Problem
Triangle $AMC$ is isoceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$
## Solution
Since quadrilateral $UVCM$ has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that $\triangle AUV$ has $\frac 14$ the area of triangle $AMC$ by similarity, so $[UVCM]=\frac 34\cdot [AMC].$ Thus, $$\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]$$ $$72=\frac 34\cdot [AMC]$$ $$[AMC]=96\rightarrow \boxed{\textbf{(C)}}.$$
## Solution 2 (CHEATING)
Draw a to-scale diagram with your graph paper and straightedge. Measure the height and approximate the area.
## Solution 3 (Trapezoid)
$[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((-2,6)--(2,6)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); [/asy]$
We know that $\triangle AUV \sim \triangle AMC$, and since the ratios of its sides are $\frac{1}{2}$, the ratio of of their areas is $(\frac{1}{2})^2=\frac{1}{4}$.
If $\triangle AUV$ is $\frac{1}{4}$ the area of $\triangle AMC$, then trapezoid $MUVC$ is $\frac{3}{4}$ the area of $\triangle AMC$.
Let's call the intersection of $\overline{UC}$ and $\overline{MV}$ $P$. Let $\overline{UP}=x$. Then $\overline{PC}=12-x$. Since $\overline{UC} \perp \overline{MV}$, $\overline{UP}$ and $\overline{CP}$ are heights of triangles $\triangle MUV$ and $\triangle MCV$, respectively. Both of these triangles have base $12$.
Area of $\triangle MUV = \frac{x\cdot12}{2}=6x$
Area of $\triangle MCV = \frac{(12-x)\cdot12}{2}=72-6x$
Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$.
This is $\frac{3}{4}$ of the triangle, so the area of the triangle is $\frac{4}{3}\cdot{72}=\boxed{\textbf{(C) } 96}$ ~quacker88, diagram by programjames1
## Solution 4 (Medians)
Draw median $\overline{AB}$. $[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0.5, 4), E); label("B", (0, 0), S); [/asy]$
Since we know that all medians of a triangle intersect at the incenter, we know that $\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\overline{PC}:\overline{UP}=2:1$, and since the two segments sum to $12$, $\overline{PC}$ and $\overline{UP}$ are $8$ and $4$, respectively.
Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\triangle PUM$ is enough. $\overline{PC} = \overline{MP} = 8$.
The area of $\triangle PUM = \frac{4\cdot8}{2}=16$. Multiplying this by $6$ gives us $6\cdot16=\boxed{\textbf{(C) }96}$ ~quacker88
~IceMatrix |
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Percent Composition | CK-12 Foundation
# 12.3: Percent Composition
Created by: CK-12
## Lesson Objectives
The student will:
• calculate the percent composition by mass given the masses of elements in a compound.
• calculate the percent composition by mass given the formula or name of a compound.
## Vocabulary
• percent composition
## Introduction
Metals useful to man are typically extracted from ore. The ore is removed from the mine and partially purified by washing away dirt and other materials not chemically bound to the metal of interest. This partially purified ore is then treated chemically in smelters or other purifying processes to separate pure metal from the other elements. The value of the original ore is very dependent on how much pure metal can eventually be separated from it. High-grade ore and low-grade ore command significantly different prices. The ore can be evaluated before it is mined or smelted to determine what percent of the ore can eventually become pure metal. This process involves determining what percentage of the ore is metal compounds and what percentage of the metal compounds is pure metal.
## Percent Composition from Masses
Compounds are made up of two or more elements. The law of definite proportions tells us that the proportion, by mass, of elements in a compound is always the same. Water, for example, is always $11\%$ hydrogen and $89\%$ oxygen by mass. The percentage composition of a compound is the percentage by mass of each element in the compound.
Percentage composition can be determined experimentally. To do this, a known quantity of a compound is decomposed in the laboratory. The mass of each element is measured and then divided by the total mass of the original compound. This tells us what fraction of the compound is made up of that element. The fraction can then be multiplied by 100% to convert it into a percent.
Example:
Laboratory procedures show that $50.0 \ \mathrm{grams}$ of ammonia, $\text{NH}_3$, yields $41.0 \ \mathrm{grams}$ of nitrogen and $9.00 \ \mathrm{grams}$ of hydrogen upon decomposition. What is the percent composition of ammonia?
Solution:
$\% \ \text{nitrogen} = \left (\frac {41.0\ \text{grams}} {50.0\ \text{grams}}\right ) \cdot (100\%) = 82\%$
$\% \ \text{hydrogen} = \left (\frac {9.00\ \text{grams}} {50.0\ \text{grams}}\right ) \cdot (100\%) = 18\%$
Example:
The decomposition of $25.0 \ \mathrm{grams}$ of $\text{Ca(OH)}_2$ in the lab produces $13.5 \ \mathrm{grams}$ of calcium, $10.8 \ \mathrm{grams}$ of oxygen, and $0.68 \ \mathrm{grams}$ of hydrogen. What is the percent composition of calcium hydroxide?
Solution:
$\% \ \text{calcium} = \left (\frac {13.5\ \text{grams}} {25.0\ \text{grams}}\right ) \cdot (100\%) = 54.0\%$
$\% \ \text{oxygen} = \left(\frac {10.8\ \text{grams}} {25.0\ \text{grams}}\right ) \cdot (100\%) = 43.2\%$
$\% \ \text{hydrogen} = \left (\frac {0.68\ \text{grams}} {25.0\ \text{grams}}\right ) \cdot (100\%) = 2.8\%$
You should note that the sum of the percentages always adds to $100 \%$. Sometimes, the sum may total to $99 \%$ or $101 \%$ due to rounding, but if it totals to $96 \%$ or $103 \%$, you have made an error.
## Percent Composition from the Formula
Percent composition can also be calculated from the formula of a compound. Consider the formula for the compound iron(III) oxide, $\text{Fe}_2\text{O}_3$. The percent composition of the elements in this compound can be calculated by dividing the total atomic mass of the atoms of each element in the formula by the formula mass.
Example:
What is the percent composition of iron(III) oxide, $\text{Fe}_2\text{O}_3$?
Solution:
$\begin{array}{cccc} \text{Element} &\text{Atomic Mass} &\text{Number of Atoms} &\text{Product} \\& &\text{per Formula} & \\\hline\text{Fe} &55.8 \ \text{daltons} &2 &111.6 \ \text{daltons} \\\text{O} &16.0 \ \text{daltons} &3 &\underline{\ 48.0 \ \text{daltons} \ } \\& &\text{Formula mass =} &159.6 \ \text{daltons} \\\end{array}$
$\% \ \text{iron} = \left (\frac {111.6 \ \text{daltons}} {159.6 \ \text{daltons}}\right ) \cdot (100\%) = 69.9 \%$
$\% \ \text{oxygen} = \left (\frac {48.0 \ \text{daltons}} {159.6 \ \text{daltons}}\right ) \cdot (100\%) = 30.1 \%$
Example:
What is the percent composition of aluminum sulfate, $\text{Al}_2(\text{SO}_4)_3$?
Solution:
The formula mass of $\text{Al}_2(\text{SO}_4)_3$ is: $2 \cdot (27.0 \ \mathrm{daltons}) + 3 \cdot (32.0 \ \mathrm{daltons}) + 12 \cdot (16.0 \ \mathrm{daltons}) = 342.0 \ \mathrm{daltons}$
$\% \ \text{aluminum} = \left (\frac {54.0 \ \text{daltons}} {342 \ \text{daltons}}\right ) \cdot (100\%) = 15.8 \%$
$\% \ \text{sulfur} = \left (\frac {96.0 \ \text{daltons}} {342 \ \text{daltons}}\right ) \cdot (100\%) = 28.1 \%$
$\% \ \text{oxygen} = \left (\frac {192 \ \text{daltons}} {342 \ \text{daltons}}\right ) \cdot (100\%) = 56.1 \%$
## Lesson Summary
• The percent composition of a compound is the percent of the total mass contributed by each element in the compound.
• Percent composition can be determined either from the masses of each element in the compound or from the formula of the compound.
This website has solved example problems for a number of topics covered in this lesson, including the calculation of percent composition by mass.
The website below reviews how to calculate percent composition.
## Review Questions
Determine the percent composition of the following compounds.
1. $\text{BF}_3$
2. $\text{Ca(C}_2\text{H}_3\text{O}_2)_2$
3. $\text{FeF}_3$
4. $\text{CrCl}_3$
5. $(\text{NH}_4)_3\text{PO}_4$
## Date Created:
Feb 23, 2012
Nov 26, 2014
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1. ## Equation of line
Find the equation of the line through M (-1,2) with gradient m if M is the midpoint of the intercepts of the x axis and y axis
equation of line = y - y1 = m ( x - x1)
what next I am confused with the midpoint of axis.
Thanks
2. Hello Joel,
your line passes through the point $\displaystyle M (-1, 2)$, so you know that $\displaystyle 2 = -m + p$ (using the standard equation $\displaystyle y = mx + p$)
Now, let $\displaystyle x_1$ be the intercept of the y-axis, and $\displaystyle y_1$ be the intercept of the x-axis. Then we have$\displaystyle M \left (\frac{x_1}{2}, \frac{y_1}{2} \right )$. Thus we have $\displaystyle x_1 = -2$ and $\displaystyle y_1 = 4$. Since these are intercepts, we can get some more information :
$\displaystyle 0 = -2m + p$ (using $\displaystyle y = mx + p$)
Now recover the equation we found first :
$\displaystyle 2 = -m + p$
Solving the two equations simultaneously :
$\displaystyle \left \{ \begin{array}{l} 0 = -2m + p \\ 2 = -m + p \end{array} \right.$
$\displaystyle \left \{ \begin{array}{l} 2m = p \\ 2 = -m + p \end{array} \right.$
$\displaystyle \left \{ \begin{array}{l} 2m = p \\ 2 = -m + 2m \end{array} \right.$
$\displaystyle \left \{ \begin{array}{l} 2m = p \\ 2 = m \end{array} \right.$
$\displaystyle \left \{ \begin{array}{l} 4 = p \\ 2 = m \end{array} \right.$
$\displaystyle \left \{ \begin{array}{l} p = 4 \\ m = 2 \end{array} \right.$
And the equation of the line follows : $\displaystyle \boxed{y = 2x + 4}$
Plotting this line indeed shows that it passes through $\displaystyle M(-1, 2)$, and that this point is exactly midpoint between the x-intercept and y-intercept of the line
Don't hesitate to reply if anything seems confusing.
3. Hi bacterius,
After reading your first paragraph I accepted this as your proof of the x and y intercepts of the required line ie that the x intercept is -2 and y intercept is 4. So immediately y=2x+4. Was it necessary to go further. I did it by taking a straight edge and rotating it about M so that M was the midpoint between the x and y axis
bjh
4. Indeed, this is a shortcut, since we know the y-intercept we immediately know $\displaystyle p$, then we can use the point M (not the x-intercept, sorry typo) to solve for $\displaystyle m$, the gradient. I just did it the long way, I'm glad you noticed it was possible to go faster |
# Reflections on Cartesian Plane
Lesson
## Reflections
As we saw in the previous lesson on Flips, A reflection occurs when we flip an object or shape across a line. Like a mirror, the object is exactly the same size, just flipped in position. So what was on the left may now appear on the right. Every point on the object or shape has a corresponding point on the image, and they will both have the same distance from the reflection line. We can reflect points, lines, polygons on the Cartesian plane by flipping them across an axis line or another line in the plane.
If we reflect horizontally across the $y$y axis, then the y values of the coordinates remain the same and the $x$x values change sign.
In this diagram, the image is reflected across $y$y axis.
Note how the point $\left(-2,1\right)$(2,1) becomes $\left(2,1\right)$(2,1). The $y$y values have not changed and the x values have changed signs.
Similarly the point $\left(-6,3\right)$(6,3) becomes $\left(6,3\right)$(6,3). The $y$y values have not changed and the x values have changed signs.
If we reflect vertically across the $x$x axis, then the $x$x values of the coordinates remain the same and the $y$y values change sign.
In this diagram, the image is reflected across $x$x axis.
Note how the point $\left(4,3\right)$(4,3) becomes $\left(4,-3\right)$(4,3). The $x$x values have not changed and the $y$y values have changed signs.
Similarly the point $\left(0,5\right)$(0,5) becomes $\left(0,-5\right)$(0,5). The $x$x values have not changed and the $y$y values have changed signs.
Have a quick play with this interactive to further consolidate the ideas behind translations on the Cartesian plane.
Let's have a look at these worked examples.
##### Question 1
Consider the point $A\left(7,3\right)$A(7,3).
1. Plot point $A$A on the number plane.
2. Now plot point $A'$A, a reflection of point $A$A across the $x$x-axis.
##### Question 2
Consider the point $A\left(-7,-3\right)$A(7,3).
1. Plot point $A$A on the number plane.
2. Now plot point $A'$A, a reflection of point $A$A across the $y$y-axis.
##### Question 3
Consider the line segment $AB$AB, where the endpoints are $A$A$\left(-4,-2\right)$(4,2) and $B$B$\left(6,7\right)$(6,7).
1. Plot the line segment $AB$AB on the number plane.
2. Now plot the reflection of the line segment $AB$AB across the $x$x-axis.
##### Question 4
Consider the graph of the triangle and the line $x=-3$x=3.
1. The three points of the triangle, $A$A $\left(-1,7\right)$(1,7), $B$B $\left(3,-2\right)$(3,2) and $C$C $\left(0,-6\right)$(0,6) are reflected across the line $x=-3$x=3 to produce the points $A'$A, $B'$B and $C'$C.
What are the coordinates of the new points?
$A'$A$\left(-4,7\right)$(4,7),$B'$B$\left(0,-2\right)$(0,2), $C'$C$\left(-3,-6\right)$(3,6)
A
$A'$A$\left(-5,7\right)$(5,7), $B'$B$\left(-9,-2\right)$(9,2), $C'$C$\left(-6,-6\right)$(6,6)
B
$A'$A$\left(-5,-4\right)$(5,4), $B'$B$\left(-9,0\right)$(9,0), $C'$C$\left(-6,-3\right)$(6,3)
C
2. Plot the new triangle formed by reflecting the given triangle across the line $x=-3$x=3. |
# Common Core: High School - Geometry : Congruence
## Example Questions
### Example Question #71 : Congruence
Jenny has drawn a rectangle in the upper left hand corner of a piece of paper. If she folds the paper in half and traces the rectangle on the other half of the paper, what type of rigid motion has occurred?
Expansion
isometry
Reflection
Rotation
Reflection
Explanation:
This situation describes a real life application of geometry where the coordinate grid can be seen as the piece of paper. Since Jenny is tracing the rectangle after the paper is folded it can be seen that the fold is an axis and the new rectangle is congruent to the original one. Therefore, the rigid motion describing this situation is a reflection.
### Example Question #72 : Congruence
Jill and Jane sit next to each other in math class. Jill has a sheet of paper and Jane asks her for half of it to take notes on because she left her notebook in the locker. If Jill cuts the sheet of paper in half, has a rigid motion occurred?
No
Yes
No
Explanation:
Recall that a rigid motion is that that preserves the distances between points within the object while undergoing a motion in the plane. This is also called an isometry, rigid transformations, or congruence transformations and there are four different types. The piece of paper is the object in this situation. Since the paper is cut in half, it does not preserve the shape of the object and thus, is not a rigid motion.
### Example Question #1 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
What must be true if has had a rigid motion applied to it to result in ?
Only two angles from each triangle are congruent
Only corresponding angles are congruent
Both corresponding angles and sides are congruent
Both corresponding angles and sides are congruent
Explanation:
Recall that a rigid motion preserves the distance from points within a shape in the plane. Therefore, if has had a rigid motion applied to it to result in that means that all corresponding sides and angles of these two triangle are congruent.
Therefore, the statement "both corresponding angles and sides are congruent" is the correct solution to this particular question.
### Example Question #2 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Determine whether the statement is true or false:
Two triangles that are congruent have corresponding points that are separated five spaces. These two triangles describe a rigid motion.
False
True
True
Explanation:
Recall that a rigid motion preserves the distance from points within a shape in the plane. Since the statement clearly states that the two triangles are congruent, that means that the distances between the points within the shape are preserved. Also, since the corresponding points between the two triangles are separated by five spaces this is describing a translation which when combined with the preserved shape, describes a rigid motion. Therefore, the statement is true.
### Example Question #2 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
When a rigid motion is done onto A, C is the resulting image. Are triangles A and C congruent?
No
Yes
Yes
Explanation:
Recall that a rigid motion preserves the distance from points within a shape in the plane. It appears that the rigid motion was a rotation. The third angle can be identified by subtracting the two known angles from 180 degrees.
Triangle A:
Triangle C:
Since the side between two angles on triangle C is congruent to the side between those same two angles on triangle A, the two triangles are congruent.
### Example Question #3 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Determine whether the statement is true or false.
An equilateral triangle is rotated counter clockwise and transformed into an isosceles triangle is congruent to .
False
True
False
Explanation:
To determine whether this statement is true or false, first recall what it means to be "congruent". Congruent means equal or in more mathematical terms, both corresponding angles and sides of the two triangles are congruent.
The statement says that is an equilateral triangle which means all sides and angles are the same. After the triangle goes through a transformation, it becomes an isosceles triangle. To be isosceles means that two of the sides and angles of the triangle are the same. Since, not all of the angles are the same in it is not congruent to .
Therefore, the statement, " is congruent to ." is false.
### Example Question #1 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Which of the following geometric theorems cannot prove triangle congruency?
Side, Angle, Side
Hypotenuse, Leg of Right Triangle
Angle, Angle, Angle
Angle, Side, Angle
Side, Side, Side
Angle, Angle, Angle
Explanation:
To determine which theorem cannot prove triangle congruency, first recall what it means to be "congruent". Congruent means equal or in more mathematical terms, both corresponding angles and sides of the two triangles are congruent.
There are five theorems used in geometry to prove whether two triangles are congruent.
1. Side, Side, Side
2. Side, Angle, Side
3. Angle, Side, Angle
4. Angle, Angle, Side
5, Hypotenuse and One Leg from Right Triangle
Of the answers below, Angle, Angle, Angle (AAA) is not among one of the theorems to prove triangle congruency. The reason AAA does not prove triangle congruency because two triangles can have the same angles but have different side lengths thus, the triangles would not be congruent.
### Example Question #4 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Which theorem can be used to prove triangle congruency between triangle A and C?
HL
AAS
AAA
SSS
SAS
AAS
Explanation:
For this particular problem there are two geometric theorems that can prove that the triangles are similar.
The geometric theorems that could be used:
1. Angle, Side, Angle (ASA)
2. Angle, Angle, Side (AAS)
For triangle C the AAS is the most evident to use. Since ASA is not an option in the answer selections, AAS is the correct answer.
### Example Question #5 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Which theorem can be used to prove triangle congruency between triangle A and C?
ASA
SAS
AAA
HL
SSS
ASA
Explanation:
For this particular problem there are two geometric theorems that can prove that the triangles are similar.
The geometric theorems that could be used:
1. Angle, Side, Angle (ASA)
2. Angle, Angle, Side (AAS)
For triangle C the AAS is the most evident to use. Since AAS is not an option in the answer selections, ASA is the correct answer.
### Example Question #6 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Which theorem can be used to prove triangle congruency between triangle A and B?
SAS
HL
AAS
AAA
ASA
HL
Explanation:
For this particular problem there are two geometric theorems that could potentially be used to prove that the triangles are similar.
The geometric theorems that could be used:
1. Hypotenuse and One Leg of a Right Triangle (HL)
2. Side, Side, Side (SSS)
To use SSS or HL, the Pythagorean theorem will need to be used to calculate the missing side.
For these triangles the HL is the most evident to use and since SSS is not an option in the answer selections, HL is the correct answer. |
Free Essay
# Gs1140 Mod 4.4
Submitted By JackMehoffGood
Words 414
Pages 2
Jason Williams
Problem Solving
Module 4.4 geometry
1.) You have to solve for C. A = 3 m and B = 4 m. To do this you would need the formula A^2 + B ^2 = C^2. So you have 3^2 +4^2 = C^2. So you have 9+16 = C^2. So you now have 25 = C^2. Then you square root each side to get rid of the squared part. C = 5 2.) To find this you must some up the sides and the widths times 2. Prect = 2l +2w. L is 2.5m and width is 10m. (5.0*2)= 10.0m (20*2) = 40m So Prect = 50.0m 3.) You use the area formula of a circle. Acir = ( pi)r^2. D = 4M to find radius you take the diameter and divide it by 2 so radius is 2. Pi is 3.141 or 22/7. Acir = (3.141)2^2 = Area. (3.141)4 = area. Area= 12.564. or 12.571 if you use 22/7 4.) We need to find the volume of a sphere radius is 5 m. your formula is Vsph =(4/3)(pi)r^3. So you take (4/3)(3.141)5^3 or (4/3)(22/7)5^3. V = 420.59 or V = 420.839 5.) We need to find the length of the rectangular prism. Volume is 144 m^3. Width is 2 m. Height is 6 m. your formula would be L = V/H*W So L = 144/2*6 L= 144/12 So L = 12 6.) Find area of triangle. A = 3M B = 4M. Your formula is Atri = (1/2)bh. So Atri = (1/2)4*3. Atri = 2*3. Area =6 7.) You need to find the length of the hypotenuse. Y = 6m and the area is 36m^2. Your formula is 8.) Find the volume of a cylinder your radius is 3m height is 10m. the formula is Vcyl = (pi)r^2h. So it’s (3.141)3^2*10 or (22/7)3^2*10. First one is 28.269*10 V = 282.69 or 28.28*10 = V = 282.85 9.) A sphere would have more 4 units and the radius is 3 the formula for volume in a cube is A^3=V so you take 4^3 = 80.342. a sphere is (4/3)(pi)r^3. So you have (4/3)(22/7)3^3 = 4.190etc *3^3 V= 252.50etc
S |
### Consecutive Numbers
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### I'm Eight
Find a great variety of ways of asking questions which make 8.
### Calendar Capers
Choose any three by three square of dates on a calendar page. Circle any number on the top row, put a line through the other numbers that are in the same row and column as your circled number. Repeat this for a number of your choice from the second row. You should now have just one number left on the bottom row, circle it. Find the total for the three numbers circled. Compare this total with the number in the centre of the square. What do you find? Can you explain why this happens?
# Clocked
##### Stage: 3 Challenge Level:
Alex of The Mount School gave a clear explanation of a strategy and some thoughts on why her solutions worked. Thank you for your clear explanations Alex.
I have a few questions though ...
Is using steps of 7 hours the same as subtracting 5 hours each time and if so why?
What about mixing the number of hours between each point around the clock? Thus having steps of different lengths around the clock for example -starting at 1, step three to 4, then step four to 8 and so on. Dan of Bishops Stortford College made some suggestions and these follow Alex's work.
Firstly I decided that the hours 12 o'clock and 1 o'clock were 1 hour apart as they would be using time and not 11 hours apart as they would be numerically.
I decided to start with 1 o'clock at the top of the blank clock face. I then worked out the o'clock 3 hours after from this, which was 4 'o' clock. Then three hours after this which was 7 'o' clock I carried on until I came to 1, 4, 7, 10.
Then I found out that the sequence repeated itself (it went back to 1) so I decided to try the same strategy using 4 hours apart. My sequence was 1, 5, 9, .. then it started to repeat again.
I thought that this could be happening because both 3 and 4 divide in to 12 with no remainder, but 5 does not. So I decided to try 5:
My sequence this time was 1, 6, 11, 4, 9, 2, 7, 12, 5, 10, 3, 8 which worked.
This sequence can be changed on the clock face by starting with a new number at the top each time. Which gives us 12 different combinations.
Then I started from 1 o'clock again and insead of adding 5 hours I subtracted 5 hours. Which gave me the sequence: 1, 8, 3, 10, 5, 12, 7, 2, 9, 4, 11, 6. Any of these numbers can start at the top which gives us 12 more sequences. So I have 24 different combinations.
However I decided to go a bit farther and see what other numbers worked, 6 did not work as the sequence repeated itself however 7 did. By now I thought that a number that did not divide properly into 12 worked but then I came to 8. This did not work as it repeated so I deduced that the number must have a highest common factor of 1 and be a co-prime to 12, to be able to fit on the face correctly.
The last part of the problem asks, Can you convince us that you have them all? This is quite easily answered by probability. First though you have to get a correct answer:
Dan of Bishops Stortford College offered the following. There are some errors here but the approach is an excellent one. Perhaps some of you can help him out. How about using the discussion boards on AskNRICH?
The question says that each adjoining number has to be 3, 4 or 5 hours in difference to its neighbour. It says to arrange these around a clock face.
This can also be shown as a string of numbers: (x) y _ _ _ _ _ _ _ _ _ _ x (y) to make this work the end numbers have to differ by 3, 4 or 5 hours as well.
To find a solution that works you must use this table.
1 can be next to 4,5,6,10,9,8
2 can be next to 5,6,7,11,10,9
3 can be next to 6,7,8,12,11,10
4 can be next to 7,8,9,1,12,11
5 can be next to 8,9,10,1,2,12
6 can be next to 9,10,11,1,2,3
7 can be next to 10,11,12,2,3,4
8 can be next to 11,12,1,5,4,3
9 can be next to 12,1,2,6,5,4
10 can be next to 1,2,3,7,6,5
11 can be next to 2,3,4,8,7,6
12 can be next to 3,4,5,9,8,7
If you take the solution: (5),1,4,7,10,2,6,9,12,3,11,8,5,(1) (which is a correct solution obtained by trial and error). Then can see that the first figure (1) could have been 11 other things (2,3,4,5,6,7,8,9,10,11,12) making the first position out of 12 choices. You can then see that 4 could have been 5 other things (5,6,10,9,8) therefore making the second position out of 6 choices. This is now 12 choices with 6 possible choices after that (12x6). 7 could have been 4 other things (8,9,12,11) therefore making it out of 5 choices. This then makes it 12x6x5. If you do this for all 12 places you get 12x6x5x4x4x3x2x3x1x1x1x1=103,680. This means that there are 103,680 correct answers!!! Although this solution doesn?t answer the question: How many solutions can you find? It does say the number that there are. However I invite you to find all 103,680 solutions...... |
# Difference between revisions of "2016 AMC 10B Problems/Problem 23"
## Problem
In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?
$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$
## Solution 1
We draw a diagram to make our work easier: $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); label("A",A,SW); label("B",B,SE); label("C",C,ESE); label("D",D,NE); label("E",E,NW); label("F",F,WSW); label("W",W,ENE); label("X",X,ESE); label("Y",Y,WSW); label("Z",Z,WNW); [/asy]$
Assume that $AB$ is of length $1$. Therefore, the area of $ABCDEF$ is $\frac{3\sqrt 3}2$. To find the area of $WCXYFZ$, we draw $\overline{CF}$, and find the area of the trapezoids $WCFZ$ and $CXYF$.
pair A,B,C,D,E,F,W,X,Y;
A=(0,0);
B=(1,0);
C=(3/2,sqrt(3)/2);
D=(1,sqrt(3));
E=(0,sqrt(3));
F=(-1/2,sqrt(3)/2);
W=(4/3,2sqrt(3)/3);
X=(4/3,sqrt(3)/3);
Y=(-1/3,sqrt(3)/3);
Z=(-1/3,2sqrt(3)/3);
draw(A--B--C--D--E--F--cycle);
draw(W--Z);
draw(X--Y);
draw(F--C--B--E--D--A);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,ESE);
label("$D$",D,NE);
label("$E$",E,NW);
label("$F$",F,WSW);
label("$W$",W,ENE);
label("$X$",X,ESE);
label("$Y$",Y,WSW);
label("$Z$",Z,WNW);
(Error compiling LaTeX. Z=(-1/3,2sqrt(3)/3);
^
652a7fda2cdd8f5752efc512c822a61640cac1ec.asy: 15.1: no matching variable 'Z')
From this, we know that $CF=2$. We also know that the combined heights of the trapezoids is $\frac{\sqrt 3}3$, since $\overline{ZW}$ and $\overline{YX}$ are equally spaced, and the height of each of the trapezoids is $\frac{\sqrt 3}6$. From this, we know $\overline{ZW}$ and $\overline{YX}$ are each $\frac 13$ of the way from $\overline{CF}$ to $\overline{DE}$ and $\overline{AB}$, respectively. We know that these are both equal to $\frac 53$.
We find the area of each of the trapezoids, which both happen to be $\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}$, and the combined area is $\frac{11\sqrt 3}{18}^{*}$.
We find that $\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}$ is equal to $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$.
$^*$ At this point, you can answer $\textbf{(C)}$ and move on with your test.
## Solution 2 (cheap, but ingenious)
$[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label("A",A,SW); label("B",B,SE); label("C",C,ESE); label("D",D,NE); label("E",E,NW); label("F",F,WSW); label("W",W,ENE); label("X",X,ESE); label("Y",Y,WSW); label("Z",Z,WNW); [/asy]$
First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \cdot 6 = 54$ small triangles in the whole hexagon.
Thus, the answer is $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$. |
# Carpentry & Building Construction
## Math Activities :
### Circumference of a Circle
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Carpenters often have to find the distance around the outside of a circle. The distance around a circle is its circumference. To find the circumference of a circle, multiply the diameter of the circle times pi. The value of pi is approximately 3.1416. Pi is written in equations as π.
Since the diameter of a circle is equal to twice its radius, another way to find the circumference is to use the formula, 2πr, or two times pi times the radius.
Terms, Symbols, & Abbreviationscircumference – the distance around a circlediameter – the length of a straight line through the center of a circleradius – one-half the distance of the diameterpi, or π – the ratio of the circumference of a circle to its diameter, approximately 3.1416
Practice Exercise
<a onClick="window.open('/olcweb/cgi/pluginpop.cgi?it=gif::::/sites/dl/free/0078797845/249183/21_2.gif','popWin', 'width=NaN,height=NaN,resizable,scrollbars');" href="#"><img valign="absmiddle" height="16" width="16" border="0" src="/olcweb/styles/shared/linkicons/image.gif"> (3.0K)</a>
Find the circumference of the circle when the diameter is 12''.
Step 1 The formula is C = πd. Step 2 Insert the known values. C = 3.1416 × 12'' Step 3 C = 37.6992''
Find the circumference when the radius is 6''.
Step 1 The formula is C = 2πr. Step 2 Insert the known values. C = 2 × 3.1416 × 6'' Step 3 C= 37.6992''
Problem Exercises
For each problem, round the answer to the nearest hundredth.
1 Find the circumference of a circle whose diameter is 8'. A) 8.00' B) 16.00' C) 25.13' D) 50.17' 2 Find the circumference of a circle whose radius is 15'. A) 47.12' B) 94.25' C) 30.00' D) 15.00' 3 The circumference of a circle whose radius is 6' and diameter is 12' is A) 37.70'. B) 72.00' C) 56.55'. D) 226.51'. 4 A carpenter will build a platform around a tree whose circumference is 24''. She wants to leave 6'' of clearance between the platform and the tree. Find the diameter of the opening needed for the tree. A) 7.64'' B) 13.64'' C) 30.00'' D) 19.64'' 5 A deck will be built around a 14' diameter swimming pool. What is the circumference of the deck at the pool’s edge? A) 4.45' B) 43.98' C) 88.08' D) 528.48' |
# Polynomials: A Brief Description
In order to understand mathematics, first, you have to speak the language. The problem is, with mathematics, unlike French or German or any other spoken language, this dialect is replete with difficulty and at times quite incomprehensible. And in mathematics, even when you understand something, you still often feel as though you really need to understand it even more. Thus, it certainly helps if you can get off to a good start by at least understanding some of the languages. Here we give insight into what a polynomial is and then we will continue their operations like multiplying polynomials, adding, subtracting them.
A polynomial is an expression found in algebra. Technically a polynomial of degree n is an expression of the form a(n)x^n + a(n-1)x^(n-1) +… + a(1)x + a(0), where each of the a(n) terms corresponds to some integer, the n-terms following the x^ correspond to the exponents and are positive integers, and n and a(n) are not equal to 0 (if they were then this would not be a polynomial of degree n). In plain English, a polynomial is any expression such as 3x^4 + 2x^3 – x + 4, or 2x^2 – 3x + 1. The degree is the highest exponent that occurs in the expression. Thus, the first polynomial is of degree 4 and the second is of degree 2.
The first few polynomials, those of degree 1, 2, 3, 4 and 5 have special names. A first-degree polynomial is a linear function because its graph produces a line. The second, a quadratic; the third a cubic; the fourth a quartic; and the fifth, a quantic. After these, the polynomial is generally referred to by its degree.
The above-written polynomial uses the variable x, and this is most common; however, we could just as easily have written a polynomial in some other letter or variable, and some common other choices would be the letters y or t. Bear in mind that changing the letter in which the polynomial is written does not alter the nature or behaviour in any way.
Polynomials are just one kind of algebraic expression. They are very useful in modelling many real-world problems, and they occur in many formulas. In more advanced courses, polynomials are encountered to serve as substitutes for other functions for which no apparent similarity is evident. Thus, the amazing versatility of polynomials. Multiplying polynomials is very easy, and you should be knowing how to do the same. Otherwise, you will not be able to solve any algebraic problems.
As far as the pictures, or graphs, of polynomial functions, they look somewhat like roller coasters, often with many hills and valleys. These curves are “smooth” in the sense that they have no sharp turns or corners and can be drawn all in one piece. For this reason, these polynomial functions play an important role in the branch of mathematics called analysis and serve as an important tool in many other branches as well. So, do learn the world of polynomial and conquer the mathematics.
# The Basic of Multiplication
Multiplication denoted with juxtaposition, cross sign or asterisk. It is among the four elementary, arithmetic and mathematical operations while others are division, subtraction, and addition. Multiplying is a repeated addition while a multiplication of two different numbers will be adding copies of one of the two. One is said to be multiplicand which is the first value while the one is called multiplier. A multiplier is the one that it is written for the first time while the multiplicand is in the second position. The multiplication of the integers will include some negative numbers while the rational numbers like fractions or real numbers will be defined using a systematic generation for the basic definition. The multiplication may be visualized while counting the objects that are arranged into the rectangle for an entire whole number or to find an area of a rectangle where the sides are given a certain length. An area of the rectangle will not be based on the side that had been measured before the other, and this is how it illustrates commutative property. A product of two measurements will be new types of the measurement. The example is that when you multiply two sides for a rectangle, it will give a dimensional analysis.
An inverse of multiplying is dividing. The example is that 4 by 3 equals to 12 while 12 divided by 4 equals 3. Multiplication may also be defined using different numbers like complex numbers or abstract constructs like matrices. The order for which the matrices have to be multiplied in will not matter. There is a listing of different types of products which are being used in mathematics.
In arithmetic, a sign of multiplication is X found between two terms. Multiplication may also be showed by a dot sign. With algebra, the multiplication that involves variable will be written through juxtaposition. This notation may also be used for the quantities which are found within parentheses. These types of multiplication may lead to ambiguity if the concatenated variable takes place for matching a name of a new variable.
With matrix multiplication, you should know that there is a distinction between a dot symbol and a cross. A cross symbol means taking the cross products for two vectors and it leads to a vector. A dot is used to denote a dot product for two vectors, and this may lead to a scalar.
In the computer programming, an asterisk is a common notation. This is because at the beginning, the computers had limited number of small characters and there was no multiplication sign and asterisk was on each keyboard.
A common method used to multiply simple numbers using paper or pencil requires the use of the multiplication table and it has to be consulted or memorized for small numbers. Multiplying numbers with decimal places; can be error prone and tedious. The common logarithms had been invented to be used for these calculations. Multiplication has now become easy by the use of electronic computers and the latest calculators which reduce the need to multiply manually.
# Solution to your Algebra Problems
Are you fond with numbers and variables?
Math problem solving?
Algebra?
If you answered yes to all of these questions then I guess you landed on the wrong read about. But I’m guessing that the reason you’re still continually reading is because you answered No to even just one of the questions above.
Most of us claim that they hate math, but in reality they don’t really mean that. What they actually meant was I’m confused with math and I don’t want to spend another minute figuring out what to do next. This is very true especially when it comes to algebra. Since algebra is the connecting link of all the field of mathematics, just like the road systems, making it so broad, complicated and confusing? It touches and involves everything that is related to math.
There’s a study that says, when a person is confused, he or she stops. It’s the average of the norms. That’s why a lot of people especially students don’t finish solving math problems, not to mention long and complicated algebraic expressions. Until, they find a valid reason of spending too much time enduring the headache these math problems, they will not do it.
But you say that this is part of living, and studying of course. Yes that’s right and the only way to make us and the students to love solving algebraic expression is by making them want it. So how do we do that?
First, let’s identify the root cause of why they don’t like to do it? Well, because it’s complicated, takes too much time and doesn’t provide checking or verifying solutions. The same reason why it becomes so frustrating coming up to a wrong answer when you thought you were doing right all along. Then what’s next? Repeat the whole process, solve the equation and hope that this time you make it right.
Now that we’ve identified the cause, what’s the solution? Well first, as adults we have to remember that we’re already living a complicated life and as students we are already having a hard time figuring out a way to pass all of our subjects. Then why not just make life easier but looking for solutions the fast way. The internet is making a great job dealing with these fast and easy solutions.
And for our math problems we have what we call the math solver. Here, you can come up with answers the simplest possible way. Just by entering the equation in the solver then it almost automatically gives you the answer. In that way, all we have to worry about is the analysis of the problem with the given solution. Easy and simple, isn’t it? No need to recomputed and end up with the same wrong answer. We can even learn techniques on how to solve similar problem. I’m sure a lot more will be encouraged to finish up solving math problems when we know that it is this easy and there’s a way to verify our results.
# Why is Algebra Calculator the Best Resort?
Numbers are just not essential for everyday calculations; they are also essentials to define the world and make it measurable with the help of complex equations. Quantification of most of the things in life can be done with the help of numbers. You too may have studied numbers in your math class in school. But that was simple arithmetic. As you reach higher grades, you will need to deal with algebra. But, most of the students find algebra to be very difficult, and this is why you need to have algebra calculator.
## Why algebra is considered difficult?
Algebra is a branch of math that deals in solving various problems based on equations. You need to treat alphabets as numbers and do all that you do with numbers such as addition, subtraction, multiplication and almost all kinds of calculations. But, many people do not understand the concept behind adding alphabets and find the concept to be very weird. This makes it very difficult to understand the concepts behind algebra that were first discovered by the collective works of various mathematicians in different parts of the world. But, there are many fields where you will require applying the concepts of algebra as reach the higher levels of math.
### How to easily get the answers?
When you are studying algebra, you need to first understand the basics of algebra before you can start solving the problem. This is important because basics let you solve any problem that may come across your way. But many a time even if you are well versed with the basics you may not understand where to start, and this is where algebra calculator comes handy. This calculator is available online, and you can use it solve any problem based on algebra.
The online calculator has many advantages to its credit such as
• It is easy to use as compared to any calculator that you may have seen.
• You can use it any time of the day
• You can use it for all kinds of algebraic problems
• All you need to do is just type your problem in the designated space and hit enter to get the detailed answer right in front of your eyes.
##### Types of Algebraic Calculators
There are many types of algebra calculators available on the internet. If you are studying basic algebra, you can easily take the help of the simple calculator while those of you who are studying the higher levels of this branch of math can use the calculators that can solve the problems that are of higher levels. When you are required to not just solve the equation but also show all the steps involved in solving the problem then you must go in for solver that will enable you to see the solution too. This way you can also validate your answer by going through each step for yourself. Those of you who are not able to solve the problem on your own can find the solution online just by the click of the button.
# Strategies To Improve The Math
Most of the parents and students ask their teachers what are the best techniques and the tricks that can improve the skills of math. Math is a big fun and interactive subject, but some people are not interested in it, and it feels difficult for them to carry out all the concepts to the next level. So, here is the trick to improve the math, have a look at all these strategies.
1. Focus On The Things:
If you are not able to understand something, then you must have your complete focus on the thing before moving forward to the next chapter. Similar is the case when you are studying math. The step is simple, but it is highly important. If a student is going to start algebra, then he must know the basics, including the adding fractions and the dividing fractions, otherwise, he will not be able to perform the steps and if he thinks that he can learn from the lessons that are given later, then it is his complete misunderstanding as these are the steps to disaster not the steps to success.
1. Work With The Help Of Examples:
The entire process of practicing math is to do it with the help of examples as these examples make the process highly simple and easy. After learning a chapter, try to do the questions given at the end. This is the easy way to learn math. After completing all the problems, check their answers at the back of the book. When you will complete the two or three lessons like this, and you will get the correct answers, your confidence will rise ultimately, and you will have no issue in understanding further.
1. Don’t Map Out The Question To Answer Strategy:
Most of the students do the common mistake of making a strategy in their mind that involves the problem to answer mapping into the mind before writing anything about it. Let’s take the example of algebra. This is the common problem with the students at the beginning, and it is recommended that you must not draw any picture in your mind before writing the problem with the answer onto the page. Students are often tempted to solve all the answer at first place in mind, but this often leads to the wrong path that is difficult to be managed by the student later on. Or your students should also not take the help of equation solver or the simplify calculator.
1. Do Your Homework At A Quiet Place:
When you are going to study and start your homework, find a place that is quiet. The students who listen to music while doing their homework or they have some other background noise while studying; they can never study hard. With the passage of time, this thing will make its way, and you will also realize that you will only learn things that you have heard in the quiet place as it is much better to have the complete concentration on the subjects rather than dropping the attention to the background music.
# Algebra Helping Aids’ Brief Lines
On the off chance that your child asks you in helping these algebra home works, and you can’t do this home works, or you do not think about them where you have not done algebra in your secondary school days. This sort of circumstance is so rushed and with the assistance of some fabulous algebra, problem solver helps and your kids are all around arranged for coming up test. Right now, the web will resolve your issue; you can locate an extensive variety of polynomial math worksheets and some different instruments on the web, which helps the troublesome learning process. Be that as it may, these variable based math devices are an awesome approach to enhance your math expertise and some practice will give more focal points in up and coming math test. These worksheets contain a great many issues and mathematical statements where you can test yourself. What’s more, you can discover an answer key for every one of these issues in that site.
Aside from this, you can locate some other supplemental apparatus on the web is the famous variable based math solver. At the point when in correlation with adding a machine, the usefulness is the same, and this product project will give answers to hardest algebra problems too. All that you have to enter the algebra problem and the product will do whatever is left of the things. This product will give an online algebra problem solver when your kids need, and it will spare enormous cash on utilizing a mentor.
Furthermore, now an algebra problem ascended in your psyche that, how to discover the algebra problem solver. So as to discover these online algebra problem solvers, you have to do a little pursuit in the web by suing some catchphrases relying upon your requirements. In prior days, this variable based math is a creature for each tyke, however with the assistance of these algebra problem solvers, they can learn polynomial math much speedier. Furthermore, the kind will appreciate the importance of having an algebra problem solver.
# The Benefits of Online Algebra Solver Websites
We all learn math and algebra at school but it can still be difficult to do complex algebra problems when you need to. While we are at school, we often think that basic math is boring and useless, but math is actually required quite often as part of the daily routine. If you need to work something important out then you need to make sure that the problems are done accurately or else you might find yourself making a big mistake, such as incorrect factoring or simplification. There have even been a number of high profile cases of big companies making small mistakes in their math calculations which have led to major errors, including mistakes by NASA. Thankfully there is an easy way to check whether your solutions are correct: a step-by-step algebra solver that you can find online.
It’s been a long time since most adults have studied algebra, and many people forget how to do complex math because they never really had a proper understanding of it in the first place. Free math solver tools can help refresh your memory about how to do basic math and algebra, so that you will feel more confident about doing your own sums in future. Even if you never really understood how to do algebra, tools like these can help you to understand the importance of each step, so you learn why you are doing what you are doing.
A free online algebra calculator can also be useful to today’s generation of students who are in the process of learning algebra. Many children struggle in math class because they are unable to get enough one-to-one attention from their teachers. If they do not understand basic equations then it is unlikely that they will be able to cope when the class begins to move on to more complicated questions. Often, their parents are unable to offer them the additional help that they need because they also lack a good enough understanding of the math which is required. Using a step-by-step math calculator when they are studying at home can help them to see how each step of the problem is solved, so that they know what is going on at each stage. Allowing children to gain an understanding of basic algebra in this way can save parents from having to hire a costly algebra tutor to assist their child. Of course, these calculators should not be used by students as a way to cheat on their homework, because this will stop them from developing the skills that they need.
Online algebra solver tools are very easy to use, so they are accessible to both school children and adults; however some online math calculators are aimed specifically at younger students or higher level learners. If you are struggling with basic math then they are a quick and convenient solution which can help you to work out exactly what you need to. |
## Exercises3.7Exercises
###### 1.
Use the “Decoder Ring” in Figure 3.6.2 to perform the following arithmetic. Indicate whether the result is “right” or “wrong”.
1. Unsigned integers: $\hex{1 + 3}$
2. Unsigned integers: $\hex{3 + 4}$
3. Unsigned integers: $\hex{5 + 6}$
4. Signed integers: $\hex{(+1) + (+3)}$
5. Signed integers: $\hex{(-3) - (+3)}$
6. Signed integers: $\hex{(+3) - (+4)}$
Hint
For unsigned arithmetic, use the two inner rings and pay attention to passing over the top (C). For signed arithmetic, use the outer and inner rings and pay attention to passing over the bottom (V).
1. Start at the tic mark for $1\text{,}$ move $3$ tic marks CW, giving $4 = \binary{100}_{2}\text{.}$ We did not pass the tic mark at the top, so C = $\binary{0}\text{,}$ and the result is correct.
2. Start at the tic mark for $3\text{,}$ move $4$ tic marks CW, giving $7 = \binary{111}_{2}\text{.}$ We did not pass the tic mark at the top, so C = $\binary{0}\text{,}$ and the result is correct.
3. Start at the tic mark for $5\text{,}$ move $6$ tic marks CW, giving $3 = \binary{011}_{2}\text{.}$ We passed the tic mark at the top, so C = $\binary{1}\text{,}$ and the result is wrong.
4. Start at the tic mark for $+1\text{,}$ move $3$ tic marks CW, giving $-4 = \binary{100}_{2}\text{.}$ We passed the tic mark at the bottom, so V = $\binary{1}\text{,}$ and the result is wrong.
5. Start at the tic mark for $-3\text{,}$ move $3$ tic marks CCW, giving $+2 = \binary{010}_{2}\text{.}$ We passed the tic mark at the bottom, so V = $\binary{1}\text{,}$ and the result is wrong.
6. Start at the tic mark for $+3\text{,}$ move $4$ tic marks CCW, giving $-1 = \binary{111}_{2}\text{.}$ We did not pass the tic mark at the bottom, so V = $\binary{0}\text{,}$ and the result is correct.
###### 2.
Add the following pairs of 8-bit numbers (shown in hexadecimal) and indicate whether your result is “right” or “wrong.” First treat them as unsigned values, then as signed values (stored in two's complement format).
1. $\displaystyle \hex{55 + aa}$
2. $\displaystyle \hex{55 + f0}$
3. $\displaystyle \hex{80 + 7b}$
4. $\displaystyle \hex{63 + 7b}$
5. $\displaystyle \hex{0f + ff}$
6. $\displaystyle \hex{80 + 80}$
Hint
You will have two “right/wrong” answers for each sum. The computer performs only one addition, setting both C and V according to the results of the addition. It is up to the program to test the appropriate flag depending on whether the numbers are being considered as unsigned or signed in the program.
sum signed unsigned a. $\hex{ff}$ right right b. $\hex{45}$ right wrong c. $\hex{fb}$ right right
sum signed unsigned d. $\hex{de}$ wrong right e. $\hex{0e}$ right wrong f. $\hex{00}$ wrong wrong
###### 3.
Add the following pairs of 16-bit numbers (shown in hexadecimal) and indicate whether your result is “right” or “wrong.” First treat them as unsigned values, then as signed values (stored in two's complement format).
1. $\displaystyle \hex{1234 + edcc}$
2. $\displaystyle \hex{1234 + fedc}$
3. $\displaystyle \hex{8000 + 8000}$
4. $\displaystyle \hex{0400 + ffff}$
5. $\displaystyle \hex{07d0 + 782f}$
6. $\displaystyle \hex{8000 + ffff}$
Hint
You will have two “right/wrong” answers for each sum. The computer performs only one addition, setting both C and V according to the results of the addition. It is up to the program to test the appropriate flag depending on whether the numbers are being considered as unsigned or signed in the program.
sum signed unsigned a. $\hex{0000}$ right wrong b. $\hex{1110}$ right wrong c. $\hex{0000}$ wrong wrong
sum signed unsigned d. $\hex{03ff}$ right wrong e. $\hex{7fff}$ right right f. $\hex{7fff}$ wrong wrong |
## Precalculus (6th Edition) Blitzer
Published by Pearson
# Chapter 10 - Section 10.5 - The Binomial Theorum - Concept and Vocabulary Check - Page 1092: 4
#### Answer
The required solution is {{\left( x+2 \right)}^{5}}=\left( \begin{align} & 5 \\ & 0 \\ \end{align} \right){{x}^{5}}+\underline{\left( \begin{align} & 5 \\ & 1 \\ \end{align} \right)}{{x}^{4}}\cdot 2+\underline{\left( \begin{align} & 5 \\ & 2 \\ \end{align} \right)}{{x}^{3}}\cdot {{2}^{2}}+\underline{\left( \begin{align} & 5 \\ & 3 \\ \end{align} \right)}{{x}^{2}}\cdot {{2}^{3}}+\underline{\left( \begin{align} & 5 \\ & 4 \\ \end{align} \right)}x\cdot {{2}^{4}}+\underline{\left( \begin{align} & 5 \\ & 5 \\ \end{align} \right)}\cdot {{2}^{5}}
#### Work Step by Step
While writing the expansion of ${{\left( a+b \right)}^{n}}$, we note the following points: The first term of the expansion of ${{\left( a+b \right)}^{n}}$ is ${{a}^{n}}$. The exponents decrease by 1 in each successive term. And the exponents on b in the expansion of ${{\left( a+b \right)}^{n}}$ increase by 1 in each successive term. In the first term, the exponent on b is 0, as ${{b}^{0}}=1$. The last term is ${{b}^{n}}$. And the sum of the exponents on the variables in any term in the expansion of ${{\left( a+b \right)}^{n}}$ is equal to n. And the total number of terms in the polynomial expansion is one more than the power of the binomial, i.e., n. There are (n+1) terms in the expanded form of ${{\left( a+b \right)}^{n}}$. Hence, for any positive integers n,
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
Section15.1Exponentiation in Groups
Recall that $$\cdot:\Z \times \Z \to \Z$$ (multiplication) is a binary operation on the set $$\Z$$ of integers. We defined exponentiation as repeated multiplication. For $$a\in\Z$$ and $$n\in\N$$ we introduced the notation
\begin{equation*} a^n := \underbrace{a \cdot a \cdot \ldots \cdot a}_{n \text{ copies of }a } \end{equation*}
and also defined $$a^0:=1\text{.}$$ We generalize this definition of powers of integers to powers of group elements.
Following the definition of powers of integers in Definition 1.4.1, we introduce exponentiation notation for group elements as repeated application of the group operation. To be able to distinguish exponentiation with respect to different binary operations in our notation of powers of group elements we always give the binary operation next to the exponent.
Definition15.1.1.
Let $$(G,\star)$$ be a group and $$b\in G\text{.}$$
1. We set $$\gexp{b}{0}{\star}=e$$ where $$e\in G$$ is the identity of the group $$(G,\star)\text{.}$$
2. For $$n\in\N$$ we set $$\gexp{b}{n}{\star}=\underbrace{b\star b\star\dots\star b}_{n \text{ copies of }b}\text{.}$$
We read $$\gexp{b}{n}{\star}$$ as “$$b$$ to the $$n$$ by $$\star$$”. We call $$b$$ the base and $$n$$ the exponent.
It follows from the definition that in a group $$(G,\star)$$ we have $$\gexp{b}{1}{\star}=b$$ for all $$b\in G\text{.}$$ For the identity $$e$$ of any group we have $$\gexp{e}{x}{\star}=e$$ for all $$x\in\W\text{.}$$
The properties of powers of integers from Theorem 1.4.6 and Theorem 1.4.7 also hold for powers of group elements:
Both statements are proven by counting the number of copies of the base $$b$$ when rewriting the powers according to Definition 15.1.1.
1. \begin{align*} \gexp{b}{m}{\star}\star \gexp{b}{n}{\star} \amp = \underbrace{b\star\dots\star b}_{m \text{ copies of } b } \star \underbrace{b\star\dots\star b}_{n \text{ copies of } b}\\ \amp = \underbrace{b\star\dots\star b}_{m + n \text{ copies of }b} = \gexp{b}{(m+n)}{\star} \end{align*}
2. \begin{align*} \gexp{(\gexp{b}{m}{\star})}{n}{\star} \amp = \underbrace{\gexp{b}{m}{\star}\star\dots\star \gexp{b}{m}{\star}}_{n \text{ copies of }\gexp{b}{m}{\star}} \\ \amp = \underbrace{\underbrace{b\star\dots\star b}_{m \text{ copies of }b } \star\dots\star \underbrace{b\star\dots\star b}_{m \text{copies of }b}}_{n\text{ copies of } \underbrace{(b\star\dots\star b)}_{m \text{ copies of }b} } \\ \amp = \underbrace{b\star\dots\star b}_{m \cdot n\text{ copies of }b} = \gexp{b}{(m\cdot n)}{\star} \end{align*}
Note that the notation of inverses with respect to binary operations in Definition 13.4.5 is chosen such that the properties proven above also work for negative exponents. In a group $$(G,\star)$$ with identity $$e$$ for $$a\in G$$ we have
\begin{equation*} \gexp{a}{1}{\star} \star \gexp{a}{-1}{\star} = \gexp{a}{(1 + (-1))}{\star}=\gexp{a}{0}{\star}=e\text{.} \end{equation*}
Although our definition of exponentiation works in every group we restrict our examples to the groups $$(\Z_p^\otimes,\otimes)$$ where $$p$$ is a prime number where the operation $$\otimes:\Z_p^\otimes\times\Z_p^\otimes\to\Z_p^\otimes$$ is given by $$a\otimes b=(a\cdot b)\bmod p\text{.}$$
Specializing Definition 15.1.1 to the group $$(\Z_p^\otimes,\otimes)$$ we have for all $$b\in\Z_p^\otimes$$ that
\begin{equation*} \gexp{b}{0}{\otimes}=1 \end{equation*}
and for all $$b\in\Z_p^\otimes$$ and all $$n\in\N$$ that
\begin{equation*} \gexp{b}{n}{\otimes} =\underbrace{b\otimes b\otimes\dots\otimes b}_{n\text{ copies of }b }= (\underbrace{b\cdot b\cdot\dots\cdot b}_{n\text{ copies of }b })\fmod p= (b^n)\fmod p. \end{equation*}
The second equality above holds because of Theorem 3.4.14.
In $$(\Z_{11}^\otimes,\otimes)$$ where $$a\otimes b=(a\cdot b)\fmod 11$$ we have
1. $$\gexp{1}{0}{\otimes}=1$$ by Definition 15.1.1 Item 1
2. $$\gexp{2}{0}{\otimes}=1$$ by Definition 15.1.1 Item 1
3. $$\gexp{2}{1}{\otimes}=2$$ by Definition 15.1.1 Item 2
4. $$\gexp{2}{2}{\otimes}=2\otimes 2= (2\cdot 2) \bmod 11 = 4 \bmod 11 = 4$$ by Definition 15.1.1 Item 2
5. $$\gexp{2}{3}{\otimes}=2\otimes 2\otimes 2=(2\cdot 2\cdot 2) \bmod 11 = 8 \bmod 11 = 8$$ by Definition 15.1.1 Item 2
6. $$\gexp{2}{4}{\otimes}=2\otimes 2\otimes 2\otimes 2=(2\cdot 2\cdot 2\cdot 2) \bmod 11=16\bmod 11 = 5$$ by Definition 15.1.1 Item 2
When numbers become bigger the computations become easier when we compute $$\fmod$$ after each multiplication.
In the group $$(\Z_{11}^\otimes,\otimes)$$ where $$a\otimes b=(a\cdot b)\fmod 11$$ we compute
\begin{equation*} \gexp{6}{8}{\otimes}=6\otimes 6\otimes6\otimes6\otimes6\otimes6\otimes6\otimes6\text{.} \end{equation*}
We use two approaches.
1. We directly follow the definition, that is, we repeatedly apply $$a \otimes b=(a\cdot b)\fmod 11\text{.}$$ We first compute
\begin{equation*} \gexp{6}{2}{\otimes}=6\otimes 6=(6\cdot 6)\fmod 11=36 \fmod 11= 3 \end{equation*}
We now compute the other powers up to $$\gexp{6}{8}{\otimes}$$ making use of the previous result. In every step we apply Theorem 15.1.2 Item 1.
\begin{align*} \gexp{6}{3}{\otimes}\amp =\gexp{6}{2}{\otimes}\otimes 6=3\otimes 6=(3\cdot 6)\fmod 11=18\fmod 11=7\\ \gexp{6}{4}{\otimes}\amp =\gexp{6}{3}{\otimes}\otimes 6=7\otimes 6=(7\cdot 6)\fmod 11=42\fmod 11=9\\ \gexp{6}{5}{\otimes}\amp =\gexp{6}{4}{\otimes}\otimes 6=9\otimes 6=(9\cdot 6)\fmod 11=54\fmod 11=10\\ \gexp{6}{6}{\otimes}\amp =\gexp{6}{5}{\otimes}\otimes 6=10\otimes 6=(10\cdot 6)\fmod 11=60\fmod 11=5\\ \gexp{6}{7}{\otimes}\amp =\gexp{6}{6}{\otimes}\otimes 6=5\otimes 6=(5\cdot 6)\fmod 11=30\fmod 11=8\\ \gexp{6}{8}{\otimes}\amp =\gexp{6}{7}{\otimes}\otimes 6=8\otimes 6=(8\cdot 6)\fmod 11=48\fmod 11=4 \end{align*}
We have computed $$\gexp{6}{8}{\otimes}=4\text{.}$$
2. We compute $$6^8$$ in the integers and the compute the result $$\fmod 11\text{.}$$
\begin{equation*} 6^8\fmod 11=1679616\fmod 11=4 \end{equation*}
Note that we can easily conduct the computations in Item 1 by hand, but we would not want to compute $$6^8$$ without the help of a calculator. When bases and exponents are larger, the second approach is not feasible anymore as the numbers become to large for most calculators.
Computing powers as in Example 15.1.4 Item 1, where we essentially follow Definition 15.1.1, is called naive exponentiation. This is the strategy that we already had used in Algorithm 2.6.1 for computing powers of integers. Replacing the multiplication of integers by the group operation we obtain a naive exponentiation algorithm for group elements.
In the video in Figure 15.1.6 we recall the definition of exponentiation in groups and go through the steps of the naive exponentiation algorithm in detail.
Now work through computing the power of a group element in Checkpoint 15.1.7.
Naive Exponentiation
With the naive exponentiation algorithm find $$3^{16}\bmod 23\text{.}$$
Input: Base $$b :=$$ an exponent $$n:=$$ and a modulus $$m:=$$ .
let $$c:=3$$ and let $$i:=1\text{.}$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
let $$c:=(c\cdot 3)\bmod 23 =$$ and let $$i:=i+1 =$$
Because the statement $$i=16$$ is true, the loop ends here.
Output: $$3^{16} \bmod 23 = c =$$ .
$$3$$
$$16$$
$$23$$
$$9$$
$$2$$
$$4$$
$$3$$
$$12$$
$$4$$
$$13$$
$$5$$
$$16$$
$$6$$
$$2$$
$$7$$
$$6$$
$$8$$
$$18$$
$$9$$
$$8$$
$$10$$
$$1$$
$$11$$
$$3$$
$$12$$
$$9$$
$$13$$
$$4$$
$$14$$
$$12$$
$$15$$
$$13$$
$$16$$
$$13$$ |
# What Is 41/64 as a Decimal + Solution With Free Steps
The fraction 41/64 as a decimal is equal to 0.64.
A division operator is used for converting fractional quantities to decimal values. In a/b form, fractions are shown with a and b standing for the numerator and denominator, respectively.
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 41/64. The Long Division can be seen with the following procedure:
Figure 1
## 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 = 41
Divisor = 64
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 = 41 $\div$ 64
This is when we go through the Long Division solution to our problem.
## 41/64 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 41 and 64, we can see how 41 is Smaller than 64, and to solve this division, we require that 41 be Bigger than 64.
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 41, which after getting multiplied by 10 becomes 410.
We take this 410 and divide it by 64; this can be done as follows:
 410 $\div$ 64 $\approx$ 6
Where:
64 x 6 = 384
This will lead to the generation of a Remainder equal to 410 – 384 = 26. Now this means we have to repeat the process by Converting the 26 into 260 and solving for that:
260 $\div$ 64 $\approx$ 4Â
Where:
64 x 4 = 256
This, therefore, produces another Remainder which is equal to 260 – 256 = 4.
Finally, we have a Quotient generated after combining the three pieces of it as 0.64=z, with a Remainder equal to 40.
Images/mathematical drawings are created with GeoGebra. |
Categories
# Solutions of inequalities of two variables
A linear inequality in two variables x and y is of the form: ax + by ≤ c: ax + by < c: ax + by > c ax + by ≥ c where a, b and c are constants. A solution to an inequality is any pair of number x and y that satisfies the inequality.
## Example 2
Determine the solution set of 5x + 2y ≤ 17
### Solution
One solution to 5x + 2y < 17 is x =2 and y = 3 because 5(2) + 2(3) = 16, which is indeed less than 17. But the pair x = 2 and y = 3 is not the only solution. As a matter of fact, there are infinitely many solutions. If the pairs of numbers x and y is a solution, then think of this pair as a point in the plane, so the set of all solutions can be thought of as a REGION in the x –y plane.
Hence, to illustrate how to determine this region, first express y in terms of x in the inequality.
3x + 2y ≤ 17
2y ≤ -5x + 17
Y ≤ -5/2 x + 17/2
When x = 0, y = 8.5; when y = 0, x = 3 (show in a graph)
The shaded Region is the solution set. |
# 10th Maths Paper Solutions Set 1 : CBSE All India Previous Year 2012
General instructions:
1. All questions are compulsory.
2. The question paper consists of 34 questions divided into four sections A, B, C and
D.
3. Section A contains 10 questions of 1 mark each, which are multiple choices type
questions, Section B contains 8 questions of 2 marks each, Section C contains 10
questions of 3 marks each, Section D contains 6 questions of 4 marks each.
4. There is no overall choice in the paper. However, internal choice is provided in one
question of 2 marks, 3 questions of 3 marks each and two questions of 4 marks each.
5. Use of calculators is not permitted.
Q1 :
The roots of the quadratic equation 2x2x − 6 = 0 are
A.
B.
C.
D.
The given quadratic equation is 2x2x − 6 = 0.
Roots of the equation can be found by factorizing it as follows:
The roots of given quadratic equation are 2 and.
Hence, the correct answer is B.
Q2 :
If the nth term of an A.P. is (2n + 1), then the sum of its first three terms is
A. 6n + 3
B. 15
C. 12
D. 21
Given: The nth term of A.P. i.e., an = 2n + 1
To find: Sum of first three terms
On putting n = 1, 2 and 3, we obtain:
a1 = 2 × 1 + 1 = 2 + 1 = 3
a2 = 2 × 2 + 1 = 4 + 1 = 5
a3 = 2 × 3 + 1 = 6 + 1 = 7
Sum of first three terms a1+ a2+ a3= 3 + 5 + 7 = 15
Hence, the correct answer is B.
Q3 :
From a point Q, 13 cm away from the centre of a circle, the length of tangent PQ to the circle is 12 cm. The radius of the circle (in cm) is
A. 25
B.
C. 5
D. 1
The given information can be represented diagrammatically as follows:
Let O be the centre of the circle.
Given: PQ = 12 cm and OQ = 13 cm.
To find: Radius of the circle
PQ is a tangent drawn from the external point Q to the circle.
OPQ = 90° (Radius is perpendicular to the tangent at the point of contact)
On applying Pythagoras theorem in ΔOPQ, we obtain:
OQ2 = OP2 + PQ2
OP2= OQ2− PQ2
OP2= (13 cm)2− (12 cm)2
OP2= 169cm2− 144 cm2
OP2 = 25 cm2
OP = 5 cm
Thus, the radius of circle is 5 cm.
Hence, the correct answer is C.
Q4 :
In Figure 1, AP, AQ and BC are tangents to the circle. If AB = 5 cm, AC = 6 cm and BC
= 4 cm, then the length of AP (in cm) is
A. 7.5
B. 15
C. 10
D. 9
Q5 :
The circumference of a circle is 22 cm. The area of its quadrant (in cm2) is
A.
B.
C.
D.
Q6 :
A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base. The ratio of the volume of the smaller cone to the whole cone is
A. 1 : 2
B. 1 : 4
C. 1 : 6
D. 1 : 8
Q7 :
A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is
A. 45°
B. 30°
C. 60°
D. 90°
Q8 :
The Distance of the point (−3, 4) from the x-axis is
A. 3
B. −3
C. 4
D. 5
Q9 :
In Figure 2, P (5, −3) and Q (3, y) are the points of trisection of the line segment joining A (7, −2) and B (1, −5). Then y equals
A. 2
B. 4
C. −4
D.
Q10 :
Cards bearing numbers 2, 3, 4, ..., 11 are kept in a bag. A card is drawn at random from the bag. The probability of getting a card with a prime number is
A.
B.
C.
D.
Q11 :
Find the value of p for which the roots of the equation px (x − 2) + 6 = 0, are equal.
Q12 :
How many two-digit numbers are divisible by 3?
Q13 :
In Figure 3, a right triangle ABC, circumscribes a circle of radius r. If AB and BC are of lengths of 8 cm and 6 cm respectively, find the value of r.
Q14 :
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Q15 :
In Figure 4, ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is drawn at each vertex of the square and a circle of diameter 2 cm is also drawn. Find the area of the shaded region. (Use π = 3.14)
Q16 :
A solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm.
Find the number of cones so formed.
Q17 :
Find the value of k, if the point P (2, 4) is equidistant from the points A(5, k) and B (k, 7).
Q18 :
A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting
(i) a red king.
(ii) a queen or a jack.
Q19 :
Solve the following quadratic equation for x:
x2 − 4axb2 + 4a2 = 0
Q20 :
Find the sum of all multiples of 7 lying between 500 and 900.
Q21 :
Draw a triangle ABC with BC = 7 cm, B = 45° and C = 60°. Then construct another triangle, whose sides are times the corresponding sides of ΔABC.
Q22 :
In Figure 5, a circle is inscribed in a triangle PQR with PQ = 10 cm, QR = 8 cm and PR =
12 cm. Find the lengths of QM, RN and PL.
Q23 :
In Figure 6, O is the centre of the circle with AC = 24 cm, AB = 7 cm and BOD = 90°.
Find the area of the shaded region. [Use π = 3.14]
Q24 :
A hemispherical bowl of internal radius 9 cm is full of water. Its contents are emptied in a cylindrical vessel of internal radius 6 cm. Find the height of water in the cylindrical vessel.
Q25 :
The angles of depression of the top and bottom of a tower as seen from the top of a m high cliff are 45° and 60° respectively. Find the height of the tower.
Q26 :
Find the coordinates of a point P, which lies on the line segment joining the points A (−2, −2), and B (2, −4), such that .
Q27 :
If the points A (x, y), B (3, 6) and C (−3, 4) are collinear, show that x − 3y + 15 = 0.
Q28 :
All kings, queens and aces are removed from a pack of 52 cards. The remaining cards are well shuffled and then a card is drawn from it. Find the probability that the drawn card is
(i) a black face card.
(ii) a red card.
Q29 :
The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by .Find the fraction.
OR
In a flight of 2800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 100 km/h and time increased by 30 minutes. Find the original duration of the flight.
Q30 :
Find the common difference of an A. P. whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
Q31 :
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Q32 :
A hemispherical tank, full of water, is emptied by a pipe at the rate of litres per sec.
How much time will it take to empty half the tank if the diameter of the base of the tank is 3 m?
OR
A drinking glass is in the shape of the frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Q33 :
A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter
30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the
length of the canvas used in making the tent, if the breadth of the canvas is 1.5 m.
Q34 :
The angles of elevation and depression of the top and bottom of a light-house from the top of a 60 m high building are 30° and 60° respectively. Find
(i) the difference between the heights of the light-house and the building.
(ii) the distance between the light-house and the building. |
Common Core: 2nd Grade Math : Partition Circles and Rectangles into Two, Three, or Four Equal Shares: CCSS.Math.Content.2.G.A.3
Example Questions
← Previous 1 3
Example Question #1 : Partition Circles And Rectangles Into Two, Three, Or Four Equal Shares: Ccss.Math.Content.2.G.A.3
This rectangle is split into __________.
thirds
halves
fourths
thirds
Explanation:
The rectangle is split into three pieces, which means it is split into thirds.
Example Question #1 : Partition Circles And Rectangles Into Two, Three, Or Four Equal Shares: Ccss.Math.Content.2.G.A.3
This rectangle is split into __________.
fourths
halves
thirds
halves
Explanation:
The rectangle is split into two pieces, which means it is split into halves.
Example Question #3 : Partition Circles And Rectangles Into Two, Three, Or Four Equal Shares: Ccss.Math.Content.2.G.A.3
This rectangle is split into __________.
thirds
halves
fourths
fourths
Explanation:
The rectangle is split into four pieces, which means it is split into fourths.
Example Question #4 : Partition Circles And Rectangles Into Two, Three, Or Four Equal Shares: Ccss.Math.Content.2.G.A.3
This rectangle is split into __________.
fourths
thirds
halves
thirds
Explanation:
The rectangle is split into three pieces, which means it is split into thirds.
Example Question #1 : Partition Circles And Rectangles Into Two, Three, Or Four Equal Shares: Ccss.Math.Content.2.G.A.3
This circle is split into __________.
thirds
halves
fourths
thirds
Explanation:
The circle is split into three pieces, which means it is split into thirds.
Example Question #41 : Geometry
This circle is split into __________.
thirds
halves
fourths
halves
Explanation:
The circle is split into two pieces, which means it is split into halves.
Example Question #42 : Geometry
This circle is split into __________.
halves
thirds
fourths
halves
Explanation:
The circle is split into two pieces, which means it is split into halves.
Example Question #4411 : Numbers And Operations
This circle is split into __________.
fourths
halves
thirds
fourths
Explanation:
The circle is split into four pieces, which means it is split into fourths.
Example Question #9 : Partition Circles And Rectangles Into Two, Three, Or Four Equal Shares: Ccss.Math.Content.2.G.A.3
This square is split into __________.
thirds
fourths
halves
halves
Explanation:
The square is split into two pieces, which means it is split into halves.
Example Question #1 : Partition Circles And Rectangles Into Two, Three, Or Four Equal Shares: Ccss.Math.Content.2.G.A.3
This square is split into __________.
halves
fourths
thirds |
Question
1. # Perimeter Of Rhombus With Diagonals
Calculating the perimeter of a rhombus is a question that many students encounter when learning basic geometry. Rhombuses have unique properties and can look quite different from other shapes. Fortunately, finding the perimeter of a rhombus with diagonals is actually quite simple. You just need to know the length of both diagonals and then add them together. In this blog post, we will explore how to find the perimeter of a rhombus with diagonals and some helpful tips for solving similar problems in geometry.
## What is a Rhombus?
A rhombus is a quadrilateral with all sides equal in length. A rhombus is also a parallelogram with all sides equal in length. The word “rhombus” comes from the Greek word ῥόμβος (rhombos), meaning “to turn round, to roll.”
## How to calculate the perimeter of a rhombus
To calculate the perimeter of a rhombus, you will need to know the length of two of its sides. You can then use the formula:
Perimeter = 2 * (Side1 + Side2)
For example, if the length of Side1 is 5 and the length of Side2 is 3, then the perimeter would be:
Perimeter = 2 * (5 + 3)
Perimeter = 2 * 8
Perimeter = 16
## What are the benefits of calculating the perimeter of a rhombus?
When it comes to finding the perimeter of a rhombus, there are a few different methods that can be used. However, many people find that calculating the perimeter with diagonals is the most effective method. This is because it provides a more accurate measurement than other methods.
There are a few different ways to calculate the perimeter of a rhombus with diagonals. The first way is to use the formula: d1 + d2 = p. This formula is pretty straightforward and easy to remember. To use this formula, you simply need to add up the length of both diagonals.
Another way to calculate the perimeter of a rhombus with diagonals is to use the formula: 2(d1 + d2) = p. This formula is slightly more complicated than the first one but it provides a more accurate measurement. To use this formula, you need to multiply the length of both diagonals by two and then add them together.
No matter which formula you use, calculating the perimeter of a rhombus with diagonals is definitely the most accurate method. So, if you need to find an accurate measurement, this is definitely the way to go!
## How to use the perimeter of a rhombus in real life
The perimeter of a rhombus is the length of the sides of the rhombus. The sides of a rhombus are equal in length, so the perimeter is simply twice the length of one side.
You can use the perimeter of a rhombus in real life by using it to find the length of something else. For example, if you need to find the length of a piece of string or rope, you can measure it against the perimeter of the rhombus.
## Conclusion
Calculating the perimeter of a rhombus with diagonals is an important skill to have when working with shapes. By using the Pythagorean theorem and recognizing the formula for finding the length of a diagonal, you can easily calculate the perimeter quickly and accurately. With this knowledge in hand, you will be able to tackle any geometry problem involving rhombuses with precision and confidence.
2. Do you know what a rhombus is? It’s a parallelogram with four sides of equal length. A rhombus has two pairs of congruent acute angles and two pairs of congruent obtuse angles. It also has four diagonals.
So, now that you know what a rhombus is, let’s talk about how to calculate the perimeter of a rhombus with diagonals. To calculate the perimeter of a rhombus with diagonals, you’ll need to know the length of the diagonals.
First, let’s take a look at the formula:
Perimeter = 2 * (diagonal 1 + diagonal 2)
In this formula, diagonal 1 and diagonal 2 refer to the two diagonals of the rhombus. To calculate the perimeter of the rhombus, simply add the lengths of the two diagonals together, and multiply the total by two. That’s it!
For example, let’s say that the rhombus has diagonals of 6 inches and 8 inches. Using the formula, the perimeter would be 2 * (6 + 8) = 28 inches.
It’s really that simple. Now that you know how to calculate the perimeter of a rhombus with diagonals, you can use it to measure any rhombus you come across. |
Top
# Multiplication of Rational Numbers
Rational numbers are defined as one of the basis of mathematics. Otherwise called as the fractions. It also used to denote the ratio of the two quantities. It consists of both the numerator and the denominator numbers. But the denominator number will not equal to the value of 0. For example, $\frac{x}{y}$ is called as the rational numbers.
Related Calculators Multiply Rational Expressions Calculator Multiply Mixed Number Multiplying Binary Numbers Calculator Rational and Irrational Numbers Calculator
## How to Multiply Rational Numbers
The explanation for the rational numbers are shown below the following,
• By using the rational numbers we can able to do all the arithmetic operations.
• All the rules are same as the normal method, but the only difference is in this we are using the rational numbers.
Steps for multiplication of rational numbers:
The steps are given below,
• In the 1st step, we have to write the given rational numbers.
• In the 2nd step, we have to multiply the number terms present in the rational numbers.
• In the 3rd step, we have to multiply the denominator terms present in the rational numbers.
• In the 4th step,Then finally we have to simplify the following terms.
## Multiplying Rational Numbers Examples
### Solved Examples
Question 1: Multiply the given rational numbers, $\frac{2}{4}$ × $\frac{6}{9}$
Solution:
Step 1: Write the given numbers,
$\frac{2}{4}$ × $\frac{6}{9}$
Step 2: In the next step, we have to first multiply the numerator terms.
2 × 6 = 12
Step 3: In the next step, we have to multiply the denominator terms.
4 × 9 = 36
Step 4: In this step, we have to simplify the obtained result, we get,
$\frac{12}{36}$
= $\frac{1}{3}$
This is the required solution.
Question 2: Multiply the given rational numbers, $\frac{5}{4}$ × $\frac{3}{7}$
Solution:
Step 1: Write the given numbers,
$\frac{5}{4}$ × $\frac{3}{7}$
Step 2: In the next step, we have to first multiply the numerator terms.
5 × 3 =15
Step 3: In the next step, we have to multiply the denominator terms.
4 × 7 =28
Step 4: In this step, we have to simplify the obtained result, we get,
$\frac{15}{28}$
This is the required solution.
Question 3: Multiply the given rational numbers, $\frac{2}{4}$ × $\frac{6}{9}$
Solution:
Write the given numbers,
$\frac{2}{4}$ × $\frac{6}{9}$
Step 2: In the next step, we have to first multiply the numerator terms.
2 × 6 = 12
Step 3: In the next step, we have to multiply the denominator terms.
4 × 9 = 36
Step 4: In this step, we have to simplify the obtained result, we get,
$\frac{12}{36}$
= $\frac{1}{3}$
This is the required solution.
Question 4: Multiply the given rational numbers, $\frac{5}{4}$ × $\frac{3}{7}$
Solution:
Step 1: Write the given numbers,
$\frac{5}{4}$ × $\frac{3}{7}$
Step 2: In the next step, we have to first multiply the numerator terms.
5 × 3 =15
Step 3: In the next step, we have to multiply the denominator terms.
4 × 7 =28
Step 4: In this step, we have to simplify the obtained result, we get,
$\frac{15}{28}$
This is the required solution.
## Multiplying Rational Numbers Practice
Question 1: Multiply the given rational numbers, $\frac{4}{3}$ × $\frac{2}{3}$
Question 2: Multiply the given rational numbers, $\frac{3}{2}$ × $\frac{5}{4}$ |
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### High School Mathematics8.5 Inverse Relation or Inverse Functions
Inverse Relation: If R is a relation from set A into another set B, then by interchanging the first and second coordinates or ordered pairs of R we get a new relation. This relation is called the inverse relation of R and is denoted by R-|. Set builder form, R-| = {(y,x)/(x,y) Î R}. Note: Domain and range of R-| are respectively range and domin of R. Inverse Functions: To find the inverse of a function written as an equation, interchange the two variables x and y and solve for y. If the inverse is also a function, it is denoted by f-1 Example 1: Find the inverse of the function {(3,4),(5,6),(7,8),(9,10)}. State the domain and range of this inverse. State if the inverse is also a function. Original function: {(3,4),(5,6),(7,8),(9,10)} Interchange the first and second coordinates in each pair. Inverse of the function: {(4,3),(6,5),(8,7),(10,9)} domain: {4,6,8,10} range: {3,5,7,9} Since each element in the domain of the inverse maps onto one and only one element in the range, the inverse of the original function is also a function. Example 2: Find the inverse of the function y = 2x + 4. Original function: y = 2x + 4 Domain of f: {x: x is real} Range of f: {y: y is real} To find the inverse of the function interchange x and y x = 2y + 4 x/2 - 4 = 2y x/2 - 2 = y Therefore the inverse function f-1 is: y = x/2 - 2 Domain of f-1: {x: x is real} Range of f-1: {y: y is real} Directions: Choose the correct answer. Also write at least ten examples of your own.
Q 1: Find the inverse of the function: y = 2x - 5, x = -1, -2, -3y = x/2 + 5/2; x = -7, -9, -11x = y/2 + 5/2; x = 7, 9, 11y = x + 5; x = 3, 5, 7 Q 2: Find the inverse of relation or function: y = 4x - 1x = 4y + 1y = x/4 + 1/4x = 4/y - 1/y Question 3: This question is available to subscribers only! Question 4: This question is available to subscribers only! |
Thursday, October 6, 2016
Day 31: HexaSquares
With the success that we had in Math 7 yesterday with using the hexagons to talk about addition, subtraction and multiplication, I wanted to keep the process rolling. Today, we added division.
As before, I started by having them talk about the operations in the language we've been using, talking about the concepts as physical quantities. This is a much easier prospect when you have physical objects in front of you.
"15 divided by 3 is asking us how many groups of 3 can we find in 15."
Again, all of the students got their stacks of 48 hexagons.
"What is '48 divided by 8' asking of us?"
"How many groups of 8 are in 48?"
"Show me!"
And they did!
We did several different combinations, having them explain them each time.
I think I'm getting better at the kinds of questions that I'm asking and they are getting better at understanding what I'm asking. Since we have stacking, interlocking blocks, tomorrow, I'm going to ask them what 4*6*2 would look like.
It also occurred to me that I could use the hexagons to talk about irrational numbers and square roots in my Pre-Algebra class.
So I did!
Each student got a set of 50 hexagons. We discussed again what questions are being asked by squares and square roots.
"5 squared is asking us to find the area of the square that has side lengths of 5."
"The square root of 25 is asking us to find the side length of a square with an area of 25."
They played around with different combinations for a while, making squares from 16, 36 and 49 hexagons.
Near the end of the period, we began looking at how was could make a square from 17 blocks instead of 16. Could we make a square from 24?
We talked about how we couldn't do it with whole hexagons, but instead would need pieces of hexagons.
Them: "For a square with an area of 17, side lengths of 4 are too small, but side lengths of 5 are too big. It has to be somewhere in between."
Me: "So let's estimate. Do you think it's closer to 4 or closer to 5? Give me as an estimate."
Them: "4.1 or 4.2."
I built a GeoGebra applet that showed us exactly what it would be and, since we've been working with estimates for 30 days now, it was a pretty easy sell.
We also began looking at where to plot those square roots on a number line. That will continue tomorrow.
After class, a student who has been particularly distracted and distracting for the past month came up and told me how much he liked using the blocks. He was on point today and I made sure to tell him so. He said the blocks help him to understand what's going on, especially with the multiplication and division.
Nothing works for everyone, but everyone has something that works for them. |
# Rational Number
Rational numbers are all numbers that can be expressed as division of two integers. In other words rational numbers are ratio of two integers. We denote the set of rational numbers as (for “quotient”). For example, 1, 42, 1.5, -3/7 are rational numbers; π is not rational number since it cannot be expressed as division of two integers. We can define this set as follows:
Be aware of the condition a, b ∈ ℤ. It’s tempting to write
and claim that π is a rational number. It’s not because the fraction is not a valid fraction according to our definition. The numerator is not an integer thus it’s not a rational number even though it’s a fraction.
We can define a rational number another way: it is a number which decimal representation is either finite or repeating.
• 5 is a rational number since it’s decimal representation is finite.
• − 47.42 is a rational number since it’s decimal representation is finite.
• 3.666666… is a rational number. It’s decimal representation is not finite, it’s infinite, but it’s repeating and the number can be written as a fraction. In this case it’s 11/3 = 3.6666…
• π is not rational number since the decimal representation is infinite and is not repeating. Another example is √2 or Euler number.
You can see two notations of repeating numbers. The first one is the “…” symbol and the second one is a line above the repeating sequence. See examples:
## Properties of Rational Numbers
• Since rational numbers contains the integers, the set of all rational numbers is infinite set.
• Rational numbers are closed under addition. It means that if you add any two rational numbers you will always get another rational number. Try it! E. g. 5.5 + 4.5 = 10 or − 6.2 + 3.8 = − 2.4. All of them are rational numbers.
• The same goes for multiplication. We can multiply any two rational numbers and we always get another rational number. E. g. 2.2 · 7.7 = 16.94.
• The same goes for subtraction. E. g. 7.1 − 6.9 = 0.2, all of them rational numbers.
• Rational number are almost closed under division. We cannot divide by zero. Thus if we take a = 5 and b = 0, both rational numbers, we cannot compute a/b since 5/0 is not defined. As a corollary rational numbers are not closed under division. But they are closed except for division by zero. So just please… Don’t divide by zero. You know what will happen, right? |
DISCOVER
# How to convert from two's complement to decimal
Updated February 21, 2017
Two's complement is a system for representing negative binary numbers. It can also be used to implement subtraction -- to subtract 'A" from "B" convert "A" to a negative number and add. This saves having to build hardware for adding and subtracting. As long as the system for converting from a binary number to the two's complement -- and back again -- two's complement can simplify negative number representation and subtraction as well. Converting from two's complement to decimal takes two steps: convert from two's complement to binary and then convert from binary to decimal.
Represent decimals as binary numbers by continuously dividing 2 into the number and collection the remainders. For example, to convert 13 to binary, divide 13 by 2 to get 6 and first remainder of 1. Divide 2 into 6 to get 3 and second remainder 0. Divide 2 into 3 to get 1 and third remainder of 1. Divide 2 into 1 to get 0 and reminder of 1. The remainders, in reverse order of production, are 1101 and decimal 13 = binary 1101. It is easier to recognise a binary number than it is to produce it. Starting from the right, add d X 2^p where d is the binary digit and p is the position, So 1101 = (1 X 1) + (0 x 2) + (1 X 4) + (1 X 8) = 13.
Change from binary to two's complement by reversing the bits and adding 1. So binary 7 might be 00000111 and negative 7 would be 11111001 because 00000111 reversed is 11111000 and 11111000 + 1 = 11111001. The leftmost digit is the sign bit. Positive numbers have a zero sign bit and negative numbers have a 1 sign bit. One of the nice things about two's complement is that converting from two's complement to binary is exactly the same as process as converting from binary to two's compliment. For example, to convert two's complement -7 to binary 7, reverse the digits and add 1. 11111001 reversed is 00000110 and 00000110 + 1 = 00000111.
Convert from two's complement to decimal in two steps: two's complement to binary and then from binary to decimal. For example, to convert -21 in two's complement -- 11101011 -- to decimal, first convert it into binary and then convert the binary into decimal.. Reverse 11101011 to get 00010100 and add 1 to get 00010101 which is 21 in binary. Then decode the binary using the positional notation to get (0 X 128) + (0 X 64) + (0 X 32) + (1 X 16) + (0 X 8) + (1 X 4) + (0 X 2) + (1 X 1) = 21.
#### Tip
There are a couple of quick validity checks: The two's complimentary number should have a one in the leftmost digit, and If the number is even the rightmost digit is zero.
#### Warning
Is is easy to forget to add the one after flipping the bits. |
Check the below Online Education NCERT MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Permutations and Combinations Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.
Online Education for Permutations and Combinations Class 11 MCQs Questions with Answers
MCQ Questions On Permutation And Combination Class 11 Question 1.
There are 12 points in a plane out of which 5 are collinear. The number of triangles formed by the points as vertices is
(a) 185
(b) 210
(c) 220
(d) 175
Hint:
Total number of triangles that can be formed with 12 points (if none of them are collinear)
= 12C3
(this is because we can select any three points and form the triangle if they are not collinear)
With collinear points, we cannot make any triangle (as they are in straight line).
Here 5 points are collinear. Therefore we need to subtract 5C3 triangles from the above count.
Hence, required number of triangles = 12C35C3 = 220 – 10 = 210
Permutation And Combination Class 11 MCQ Question 2.
The number of combination of n distinct objects taken r at a time be x is given by
(a) n/2Cr
(b) n/2Cr/2
(c) nCr/2
(d) nCr
Hint:
The number of combination of n distinct objects taken r at a time be x is given by
nCr = n!/{(n – r)! × r!}
Let the number of combination of n distinct objects taken r at a time be x.
Now consider one of these n ways. There are e objects in this selection which can be arranged in r! ways.
So, each of the x combinations gives rise to r! permutations. So, x combinations will give rise to x×(r!).
Consequently, the number of permutations of n things, taken r at a time is x×(r!) and it is equal to nPr
So, x×(r!) = nPr
⇒ x×(r!) = n!/(n – r)!
⇒ x = n!/{(n – r)! × r!}
nCr = n!/{(n – r)! × r!}
MCQ On Permutation And Combination Class 11 Question 3.
Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is
(a) 1296
(b) 671
(c) 625
(d) 585
Hint:
No. of ways in which any number appearing in one dice = 6
No. of ways in which 2 appear in one dice = 1
No. of ways in which 2 does not appear in one dice = 5
There are 4 dice.
Getting 2 in at least one dice = Getting any number in all the 4 dice – Getting not 2 in any of the 4 dice.
= (6×6×6×6) – (5×5×5×5)
= 1296 – 625
= 671
Permutation And Combination MCQ Question 4.
If repetition of the digits is allowed, then the number of even natural numbers having three digits is
(a) 250
(b) 350
(c) 450
(d) 550
Hint:
In a 3 digit number, 1st place can be filled in 5 different ways with (0, 2, 4, 6, 8)
10th place can be filled in 10 different ways.
100th place can be filled in 9 different ways.
So, the total number of ways = 5 × 10 × 9 = 450
MCQ On Permutation And Combination Question 5.
The number of ways in which 8 distinct toys can be distributed among 5 children is
(a) 58
(b) 85
(c) 8P5
(d) 5P5
Hint:
Total number of toys = 8
Total number of children = 5
Now, each toy can be distributed in 5 ways.
So, total number of ways = 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5
= 58
Permutation And Combination MCQ Class 11 Question 6.
The value of P(n, n – 1) is
(a) n
(b) 2n
(c) n!
(d) 2n!
Hint:
Given,
Given, P(n, n – 1)
= n!/{(n – (n – 1)}
= n!/(n – n + 1)}
= n!
So, P(n, n – 1) = n!
Class 11 Maths Chapter 7 MCQ With Answers Question 7.
In how many ways can 4 different balls be distributed among 5 different boxes when any box can have any number of balls?
(a) 54 – 1
(b) 54
(c) 45 – 1
(d) 45
Hint:
Here, both balls and boxes are different.
Now, 1st ball can be placed into any of the 5 boxes.
2nd ball can be placed into any of the 5 boxes.
3rd ball can be placed into any of the 5 boxes.
4th ball can be placed into any of the 5 boxes.
So, the required number of ways = 5 × 5 × 5 × 5 = 54
Permutations And Combinations MCQ Question 8.
The number of ways of painting the faces of a cube with six different colors is
(a) 1
(b) 6
(c) 6!
(d) None of these
Hint:
Since the number of faces is same as the number of colors,
therefore the number of ways of painting them is 1
MCQ Of Chapter 7 Maths Class 11 Question 9.
Out of 5 apples, 10 mangoes and 13 oranges, any 15 fruits are to be distributed among 2 persons. Then the total number of ways of distribution is
(a) 1800
(b) 1080
(c) 1008
(d) 8001
Hint:
Given there are 5 apples, 10 mangoes and 13 oranges.
Let x1 is for apple, x2 is for mango and x3 is for orange.
Now, first we have to select total 15 fruits out of them.
x1 + x2 + x3 = 15 (where 0 ⇐ x1 ⇐ 5, 0 ⇐ x2 ⇐ 10, 0 ⇐ x3 ⇐ 13)
= (x0 + x1 + x2 +………+ x5)×(x0 + x1 + x2 +………+ x110)×(x0 + x1 + x2 +………+ x13)
= {(1- x6)/(1 – x)}×{(1- x11)/(1 – x)}×{(1- x14)/(1 – x)}
= {(1- x6)×(1- x11)×{(1- x14)}/(1 – x)³
= {(1- x6)×(1- x11)×{(1- x14)} × ∑3+r+1Cr × xr
= {(1- x11 – x6 + x17)×{(1- x14)} × ∑3+r+1Cr × xr
= {(1- x11 – x6 + x17 – x14 + x25 + x20 – x31)} × ∑2+rCr × xr
= 1 × ∑2+rCr × xr – x11 × ∑2+rCr × xr – x6 × ∑2+rCr × xr + x17 × ∑2+rCr × xr – x14 × ∑2+rCr × xr + x25 × ∑2+rCr × xr + x20 × ∑2+rCr × xr – x31 × ∑2+rCr × xr
= ∑2+rCr × xr – ∑2+rCr × xr+11 – ∑2+rCr × xr+6 + ∑2+rCr × xr+17 – ∑2+rCr × xr+14 + ∑2+rCr × xr+25 + ∑2+rCr × xr+20 – ∑2+rCr × xr+25
Now we have to find co-efficeient of x15
= 2+15C152+4C42+9C92+1C1 (rest all terms have greater than x15, so its coefficients are 0)
= 17C156C411C93C1
= 17C26C211C23C1
= {(17×16)/2} – {(6×5)/2} – {(11×10)/2} – 3
= (17×8) – (3×5) – (11×5) – 3
= 136 – 15 – 55 – 3
= 136 – 73
= 63
Again we have to distribute 15 fruits between 2 persons.
So x1 + x2 = 15
= 2-1+15C15
= 16C15
= 16C1
= 16
Now total number of ways of distribution = 16 × 63 = 1008
Permutation And Combination MCQs With Answers Question 10.
6 men and 4 women are to be seated in a row so that no two women sit together. The number of ways they can be seated is
(a) 604800
(b) 17280
(c) 120960
(d) 518400
Hint:
6 men can be sit as
× M × M × M × M × M × M ×
Now, there are 7 spaces and 4 women can be sit as 7P4 = 7P3 = 7!/3! = (7 × 6 × 5 × 4 × 3!)/3!
= 7 × 6 × 5 × 4 = 840
Now, total number of arrangement = 6! × 840
= 720 × 840
= 604800
MCQ Of Permutation And Combination Class 11 Question 11.
The number of ways can the letters of the word ASSASSINATION be arranged so that all the S are together is
(a) 152100
(b) 1512
(c) 15120
(d) 151200
Hint:
Given word is : ASSASSINATION
Total number of words = 13
Number of A : 3
Number of S : 4
Number of I : 2
Number of N : 2
Number of T : 1
Number of O : 1
Now all S are taken together. So it forms a single letter.
Now total number of words = 10
Now number of ways so that all S are together = 10!/(3!×2!×2!)
= (10×9×8×7×6×5×4×3!)/(3! × 2×2)
= (10×9×8×7×6×5×4)/(2×2)
= 10×9×8×7×6×5
= 151200
So total number of ways = 151200
MCQs On Permutation And Combination Question 12.
If repetition of the digits is allowed, then the number of even natural numbers having three digits is
(a) 250
(b) 350
(c) 450
(d) 550
Hint:
In a 3 digit number, 1st place can be filled in 5 different ways with (0, 2, 4, 6, 8)
10th place can be filled in 10 different ways.
100th place can be filled in 9 different ways.
So, the total number of ways = 5 × 10 × 9 = 450
MCQs On Permutations And Combinations Class 11 Question 13.
Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon on n sides. If Tn+1 – Tn = 21, then n equals
(a) 5
(b) 7
(c) 6
(d) 4
Hint:
The number of triangles that can be formed using the vertices of a regular polygon = nC3
Given, Tn+1 – Tn = 21
n+1C3nC3 = 21
nC2 + nC3nC3 = 21 {since n+1Cr = nCr-1 + nCr}
nC2 = 21
⇒ n(n – 1)/2 = 21
⇒ n(n – 1) = 21×2
⇒ n² – n = 42
⇒ n² – n – 42 = 0
⇒ (n – 7)×(n + 6) = 0
⇒ n = 7, -6
Since n can not be negative,
So, n = 7
Permutations And Combinations MCQ Class 11 Question 14.
How many ways are here to arrange the letters in the word GARDEN with the vowels in alphabetical order?
(a) 120
(b) 240
(c) 360
(d) 480
Hint:
Given word is GARDEN.
Total number of ways in which all letters can be arranged in alphabetical order = 6!
There are 2 vowels in the word GARDEN A and E.
So, the total number of ways in which these two vowels can be arranged = 2!
Hence, required number of ways = 6!/2! = 720/2 = 360
Permutations And Combinations Class 11 MCQ Questions Question 15.
How many factors are 25 × 36 × 52 are perfect squares
(a) 24
(b) 12
(c) 16
(d) 22
Hint:
Any factors of 25 × 36 × 52 which is a perfect square will be of the form 2a × 3b × 5c
where a can be 0 or 2 or 4, So there are 3 ways
b can be 0 or 2 or 4 or 6, So there are 4 ways
a can be 0 or 2, So there are 2 ways
So, the required number of factors = 3 × 4 × 2 = 24
Permutation And Combination Class 11 Extra Questions With Answers Question 16.
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
(a) 40
(b) 196
(c) 280
(d) 346
Hint:
There are two cases
1. When 4 is selected from the first 5 and rest 6 from remaining 8
Total arrangement = 5C4 × 8C6
= 5C1 × 8C2
= 5 × (8×7)/(2×1)
= 5 × 4 × 7
= 140
2. When all 5 is selected from the first 5 and rest 5 from remaining 8
Total arrangement = 5C5 × 8C5
= 1 × 8C3
= (8×7×6)/(3×2×1)
= 8×7
= 56
Now, total number of choices available = 140 + 56 = 196
Permutation And Combination MCQs With Answers Pdf Question 17.
Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is
(a) 1296
(b) 671
(c) 625
(d) 585
Hint:
No. of ways in which any number appearing in one dice = 6
No. of ways in which 2 appear in one dice = 1
No. of ways in which 2 does not appear in one dice = 5
There are 4 dice.
Getting 2 in at least one dice = Getting any number in all the 4 dice – Getting not 2 in any of the 4 dice.
= (6×6×6×6) – (5×5×5×5)
= 1296 – 625
= 671
Question 18.
In how many ways in which 8 students can be sated in a line is
(a) 40230
(b) 40320
(c) 5040
(d) 50400
Hint:
The number of ways in which 8 students can be sated in a line = 8P8
= 8!
= 40320
Question 19.
The number of squares that can be formed on a chess board is
(a) 64
(b) 160
(c) 224
(d) 204
Hint:
A chess board contains 9 lines horizontal and 9 lines perpendicular to them.
To obtain a square, we select 2 lines from each set lying at equal distance and this equal
distance may be 1, 2, 3, …… 8 units, which will be the length of the corresponding square.
Now, two lines from either set lying at 1 unit distance can be selected in 8C1 = 8 ways.
Hence, the number of squares with 1 unit side = 8²
Similarly, the number of squares with 2, 3, ….. 8 unit side will be 7², 6², …… 1²
Hence, total number of square = 8² + 7² + ……+ 1² = 204
Question 20.
How many 3-letter words with or without meaning, can be formed out of the letters of the word, LOGARITHMS, if repetition of letters is not allowed
(a) 720
(b) 420
(c) none of these
(d) 5040 |
Question
Find the expansion of ${{(3{{x}^{2}}-2ax+3{{a}^{2}})}^{3}}$ using binomial theorem.
Hint: So we have to find the expansion of ${{(3{{x}^{2}}-2ax+3{{a}^{2}})}^{3}}$ using binomial theorem. Take a value $a$ and $b$ from ${{(3{{x}^{2}}-2ax+3{{a}^{2}})}^{3}}$. Use binomial theorem. You will get the answer.
According to the binomial theorem, the ${{(r+1)}^{th}}$ term in the expansion of ${{(a+b)}^{n}}$ is,
${{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}$
The above term is a general term or ${{(r+1)}^{th}}$ term. The total number of terms in the binomial expansion ${{(a+b)}^{n}}$is$(n+1)$, i.e. one more than the exponent $n$.
Binomial theorem states that for any positive integer $n$, the $n$ power of the sum of two numbers a and b may be expressed as the sum of $(n+1)$ terms of the form.
The final expression follows from the previous one by the symmetry of $a$ and $b$ in the first expression, and by comparison, it follows that the sequence of binomial coefficients in the formula is symmetrical.
A simple variant of the binomial formula is obtained by substituting $1$ for $b$ so that it involves only a single variable.
In the Binomial expression, we have
${{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}$
So the coefficients ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},............,{}^{n}{{C}_{n}}$ are known as binomial or combinatorial coefficients.
You can see them ${}^{n}{{C}_{r}}$ being used here which is the binomial coefficient. The sum of the binomial coefficients will be ${{2}^{n}}$ because, we know that,
$\sum\nolimits_{r=0}^{n}{\left( {}^{n}{{C}_{r}} \right)}={{2}^{n}}$
Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even binomial coefficients and each is equal to ${{2}^{n-1}}$.
The middle term depends upon the value of $n$,
It $n$ is even: then the total number of terms in the expansion of${{(a+b)}^{n}}$ is $n+1$ (odd).
It $n$ is odd: then the total number of terms in the expansion of ${{(a+b)}^{n}}$ is $n+1$ (even).
It $n$is a positive integer,
${{(a-b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}-{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}-{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}$
For binomial expansion first, let's do a small pairing inside the bracket.
So now let$a=3{{x}^{2}}$and$b=a(2x-3a)$.
Now let's expand this as is normally done for two-digit expansion.
${{\left[ 3{{x}^{2}}-a(2x-3a) \right]}^{3}}={{(3{{x}^{2}})}^{3}}-3{{(3{{x}^{2}})}^{2}}\times a(2x-3a)+3(3{{x}^{2}})\times {{a}^{2}}{{(2x-3a)}^{2}}-{{a}^{3}}{{(2x-3a)}^{3}}$
So simplifying in a simple manner we get,
\begin{align} & {{\left[ 3{{x}^{2}}-a(2x-3a) \right]}^{3}}=27{{x}^{6}}-3(9{{x}^{4}})(2ax-3{{a}^{2}})+9{{a}^{2}}{{x}^{2}}(4{{x}^{2}}+9{{a}^{2}}-12ax)-{{a}^{3}}(8{{x}^{3}}-27{{a}^{3}}-3.4{{x}^{2}}.3a+3.9{{a}^{2}}.2x) \\ & =27{{x}^{6}}-27{{x}^{4}}(2ax-3{{a}^{2}})+36{{a}^{2}}{{x}^{4}}+81{{a}^{4}}{{x}^{2}}-108{{a}^{3}}{{x}^{3}}-8{{a}^{3}}{{x}^{3}}+27{{a}^{6}}+36{{a}^{4}}{{x}^{2}}-54{{a}^{5}}x \\ & =27{{x}^{6}}-54a{{x}^{5}}+81{{a}^{2}}{{x}^{4}}+36{{a}^{2}}{{x}^{4}}+117{{a}^{4}}{{x}^{2}}-116{{a}^{3}}{{x}^{3}}+27{{a}^{6}}-54{{a}^{5}}x \\ & =27{{x}^{6}}-54a{{x}^{5}}+117{{a}^{2}}{{x}^{4}}-116{{a}^{3}}{{x}^{3}}+117{{a}^{4}}{{x}^{2}}-54{{a}^{5}}x+27{{a}^{6}} \\ \end{align}
The Expansion ${{(3{{x}^{2}}-2ax+3{{a}^{2}})}^{3}}$is$27{{x}^{6}}-54{{x}^{5}}a+117{{a}^{2}}{{x}^{4}}-116{{a}^{3}}{{x}^{ e3}}+117{{a}^{4}}{{x}^{2}}-54{{a}^{5}}x+27{{a}^{6}}$.
Note: Read the question in a careful manner. Don’t jumble within the concepts. You should know what to select as$a$ and $b$. We had assumed $a=3{{x}^{2}}$ and $b=a(2x-3a)$. So you can assume it in your way. But keep in mind there should not be any confusion. |
Last time I focused on some basics about learning the number bonds (combinations) of 10 as well as adding 10 to any number. Today I want to show the benefits of making a 10 when adding numbers with sums greater than 10 (such as 8 + 5). Then I’ll show how to help students add up to apply that to addition and subtraction of larger numbers. I’ll model this using concrete and pictorial representations (which are both important before starting abstract forms).
Using a 10 Frame:
A ten frame is an excellent manipulative for students to experience ways to “Make a 10.” I am attaching a couple of videos I like to illustrate the point.
• Model this process with your students using 2 ten frames.
• Put 8 counters on one ten frame. (I love using 2-color counters.)
• Put 5 counters (in another color) on the second ten frame.
• Determine how many counters to move from one ten frame to the other to “make a 10.” In this example, I moved 2 to join the 8 to make a 10. That left 3 on the second ten frame. 10 + 3 = 13 (and 8 + 5 = 13).
The example below shows the same problem, but this time move 5 from the first ten frame to the second ten frame to “make a 10.” That left 3 on the first ten frame. 3 + 10 = 13 (and 8 + 5 = 13).
The making 10 strategy is also really helpful when adding 9. For example: 9 + 7 — We tell students to think 10 + 6, but showing them with the ten frame makes it less abstract when first learning.
Decomposing Numbers:
This is a pictorial representation of the same problems showing how to decompose one of the addends to make a ten.
Here are 2 good resources for this method:
I LOVE, LOVE, LOVE using the add up strategy for most double+ digit subtraction problems. Making a 10 is the essential first step. There are a few ways to show this. I’ll start with a concrete method first using place value disks.
The problem is 100 – 37.
1. Count up (Add up) from 37 to 100.
2. Ask, “What goes with 37 to make the next ten?” Answer = 3 (Using bond of 7 + 3 = 10)
3. Ask, “How many 10’s can be added to get from 40 to 100?” Ten at a time can be added while counting “50, 60, 70, 80, 90, 100.”
4. It’s easy to see that by adding up, the answer is 3 + 60.
Here’s the same problem using “The arrow way” from Eureka / Engage NY.
1. Count up (add up) from 37 to 100.
2. Write 37 and an arrow pointing to the next ten (which would be 40).
3. Above the arrow, write the amount added (3).
4. Add more arrows (or just one) to show the process of getting from 40 to 100. Depending on the students’ comfort, this can be done in one step or several.
5. Add the numbers above the arrows.
Open Number Line:
Here’s the same problem using an open number line, which is a great pictorial model showing the process. Using an open number line to count up / add up for a subtraction problem should help them see “the difference” between the two numbers.
1. Place 37 on the left side of the number line.
2. Put 100 on the right side of the number line. The task is to count up from 37 to 100.
3. Ask, “What goes with the 7 in 37 to make the next ten (40)?” Answer = 3 (because 7 + 3 = 10)
4. Ask, “How many 10’s can be added to 40 to equal 100?” This can be done in several “jumps” if needed. Best is to ask, “What goes with 4 tens to make 10 tens?” Answer = 6 tens (because we know 4 + 6 = 10)
5. Total the “jumps” made on the number line: 3 + 60
See how this strategy also works well with a missing addend problem: 3,249 + n = 6,500
1. 3249 + 1 = 3250
2. 3250 + 50 = 3300
3. 3300 + 3000 = 6300
4. 6300 + 200 = 6500 |
Surface Area of a PyramidHow the Formula is Established
The surface area of a pyramid is not so difficult to establish once your student has a clear understanding of what is going on. On this page, I will show you, every step of the way, what happens to arrive at the above formula.
It is important to know that there are in fact TWO surface area formulas for the pyramid.
The Lateral Surface Area (which is the surface area of the sides of the pyramid)
The Total Surface Area (the lateral surface area plus the base area).
I will start with the first, establish that, and then the second is merely a 'tag on'!
Before we move forward, be sure your child knows how to label the different parts of a pyramid.
For the purpose of making this section as easy as possible, I am going to use the Right Square Pyramid to illustrate each step.
The square pyramid is named after the base which it rests upon, which is a square. My example will be a Right Pyramid, as its vertex is directly above the center point of its base.
Lateral Surface Area of a Pyramid
We are first going to take a look at the triangles that make up the lateral surface of the Pyramid.
These are all isosceles triangles.We also know that the base of each triangle, is one side of the square base. If we open the pyramid at it's vertex, and reposition the triangles to be vertical, we can clearly see that the slant height (t) is in fact the perpendicular height of the triangle.We also know that the AREA OF A TRIANGLE is 1/2 the Base by the perpendicular height.The perpendicular height is t Area of triangle = (t x triangles base)/2 For this specific Pyramid, if we multiplied the Area of one Triangle by 4 we would have the surface area of a square pyramid - but a formula is only useful if it applies to ALL pyramids.For this to happen, instead of giving a specific number of sides, we introduce a little Basic Algebra, and say,ANY PYRAMID would have 'n' triangular faces.We can now rewrite the Lateral Surface area formula as:Lateral Area of Pyramid = (t x triangles base x n)/2In math WE ALWAYS try to reduce a formula down to its simplest form. If we look closely at this formula, we can see that we are not quite done.If we multiply the triangles base by the number of sides of the polygon which makes the base of the pyramid, we will in fact calculate the Perimeter of the polygon (which is the base). So we can rewrite the formula again in a simpler manner:LSA = (t x P)/2 Now for the TOTAL SURFACE AREA OF ANY PYRAMID This is simply the Lateral Surface Area + Base Surface AreaTSA = (t x P)/2 + B |
# How do you integrate int arctan(1/x) using integration by parts?
Oct 17, 2016
See the explanation section below.
#### Explanation:
$\int \arctan \left(\frac{1}{x}\right) \mathrm{dx}$
Let $\theta = \arctan \left(\frac{1}{x}\right)$.
This makes $\tan \theta = \frac{1}{x}$, so $\cot \theta = x$.
Furthermore, $\mathrm{dx} = - {\csc}^{2} \theta \text{ } d \theta$
The integral becomes:
$\int \theta \left(- {\csc}^{2} \theta\right) d \theta$
Let $u = \theta$ and $\mathrm{dv} = \left(- {\csc}^{2} \theta\right) d \theta$
So $\mathrm{du} = d \theta$ and $v = \cot \theta$
$u v - \int v \mathrm{du} = \theta \cot \theta - \int \cot \theta d \theta$
The integral can be found by substitution. We get
$\theta \cot \theta - \ln \left\mid \sin \right\mid \theta + C$
Using $\cot \theta = x$ and some trigonometry, we sind $\sin \theta = \frac{1}{\sqrt{{x}^{2} + 1}}$
Therefore
$\int \arctan \left(\frac{1}{x}\right) \mathrm{dx} = x \arctan \left(\frac{1}{x}\right) - \ln \left(\frac{1}{\sqrt{{x}^{2} + 1}}\right) + C$
$= x \arctan \left(\frac{1}{x}\right) + \frac{1}{2} \ln \left({x}^{2} + 1\right) + C$ |
## Pharmacy Math alligationAlligation Math
Here is a step-by-step example of a
Pharmacy Math alligation.
Question from Wayne in California:
If a physician orders 25% dextrose 1000mL and all you have is a Liter of 70% dextrose, how much 70% dextrose and how much sterile water will be used ?
This question is worded to attempt to trick you into thinking it's a Proportion equation. However, you would want to use
Pharmacy Math alligation to quickly solve it.
Here's how:
Based on the factors in the question, you have three numbers to work with.
1) You have 70% Dextrose to work with.
2) You have Sterile water to work with (0%).
3) Your order is for 1000mL of 25% Dextrose.
The best way to get started is to draw out a
"tic-tac-toe".
1) Take the two formulas you have and place them in the top left and bottom left positions and your desired formula in the center of the Tic Tac Toe.
70% Â Â Â 25% 0%
2) Now, do two diagonal calculations to get the difference between what you have and what you want and write the answers in the right hand top and bottom. You may have one negative number, but always write it as positive.
70% 25 Â 25% 0% 45 Â
3) What you have done is determine how many parts of each formula to use to arrive at the desired formula..
70% 25 Parts of the 70% Â 25% 00% 45 Parts of the 00% Â
4) Looking at the "Parts", you see that we get 25:45. To make things easier, go ahead and reduce that fraction down to its lowest form..
25 Parts of 70%    →   5 Parts of 70%     45 Parts of 00%    →   9 Parts of 00%
5) Now you know that you need 14 total parts.
70% 5 Parts of the 70% Â 25% 00% 9 Parts of the 00% Â = 14 Total Parts Â
6) Take the number of Parts for the first formula (the 70%) and divide it by the total number of parts.
5 parts ÷ 14 Total Parts = .36
Then, take the amount of milliliters you need and multiply it by that.
1000ml
X .36
-------
= 360ml
(of the 70%)
7) Then, to figure out the rest of the solution,
make up the difference with the other parts (the 00%)
1000 Total ml
- 360ml (of the 70%)
---------
= 640ml
(of the 00% / Sterile Water)
360 Milliliters of the 70% Dextrose 640 Milliliters of the sterile water = 1000 Milliliters of 25% Dextrose
This is how it might look on the Exam:
If a physician orders 25% dextrose 1000ml and all you have is a Liter of 70% dextrose, how much 70% dextrose and how much sterile water will be used ?
A. 750ml of 70% / 250ml Sterile Water
B. 440ml of 70% / 560ml Sterile Water
C. 360ml of 70% / 640ml Sterile Water
D. 250ml of 70% / 750ml Sterile Water
Where would you like to go now ?
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# Typical urn conditional probability question
I'm studying the book Introduction to Mathematical Statistics and its Applications by Larsen and Marx, third edition. Here my paraphrasing of question 2.4.11, and I'm puzzled by the answer they give:
An urn contains a chip that's white and another chip equally likely to be white or black. A first chip is picked out, its color recorded, and is put back in the urn. A second chip is picked out again and its color is recorded. Given that the first chip picked out is white, what's the probability that the second chip is white?
Book answer: $5/6$
Here's what I've come up with.
$P(\textrm{chip with unknown color is white})= P(\textrm{chip with unknown color is black}) = 1/2$
$P(\textrm{picking a white chip from the urn}) = P(\textrm{white chip is picked}) + P(\textrm{unknown chip is white}) \cdot P(\textrm{unknown chip is picked})$
$= 1/2 + 1/2 \cdot 1/2 = 3/4$
$P(\textrm{first chip is white}) = 3/4$
$P(\textrm{first chip is white and second chip is white}) = 3/4 \cdot 3/4$
$P(\textrm{second chip is white | first chip is white}) = P(\textrm{first chip is white and second chip is white}) / P(\textrm{first chip is white})$
$= \frac{3/4 \cdot 3/4}{3/4} = 3/4$
What's wrong with my reasoning?
-
Here is one correct way of carrying out the calculation by using your your basic strategy. If the extra ball was white, then the probability of white and then white is $1$. If the extra ball was black, the probability of white and then white is $(1/2)(1/2)$.
So the probability that the first ball is white and the second is white is $$(1/2)(1)+(1/2)(1/2)(1/2).$$
This is $5/8$. Continue! |
# Multiplication Table of 14 – Tips to Memorize 14 Times Table & Example Questions
Multiplication tables are the tool to promote number sense and develop key skills. It’s the responsibility of every school teacher or private math tutor to make memorizing multiplication tables a rewarding and enjoyable subject to master. It helps students solve mathematical problems within seconds and get rid of time-consuming ways. The process of learning 14 times is easy. In this article, we have shared a table of 14, a method to read and write, and shortcuts to learn quickly. Moreover, stick around to find the solved example questions for practice purposes.
## Table of 14
Students preparing for UPSC, GRE, SAT, and other competitive exams must memorize a table of 15 to perform quick calculations. Below, we have provided the 14 times table to its 20 multipliers. Also, this is the way we write the multiplication table of 14. Check it out
14 x 1 = 14
14 x 2 = 28
14 x 3 = 42
14 x 4 = 56
14 x 5 = 70
14 x 6 = 84
14 x 7 = 98
14 x 8 = 112
14 x 9 = 126
14 x 10 = 140
14 x 11 = 154
14 x 12 = 168
14 x 13 = 182
14 x 14 = 196
14 x 15 = 210
14 x 16 = 224
14 x 17 = 238
14 x 18 = 252
14 x 19 = 266
14 x 20 = 280
## Method to Read Table of 14
It requires a high level of dedication, encouragement, and patience to memorize each multiplication table at your fingertips. Among plenty of reading methods, we have provided a pretty easy method to read the multiplication table of 14. You can use the following to read and memorize the 14 times table:
Fourteen ones are fourteen (14)
Fourteen twos are twenty-eight (28)
Fourteen threes are forty-two (42)
Fourteen fours are fifty-six (56)
Fourteen fives are seventy (70)
Fourteen sixes are eighty-four (84)
Fourteen sevenths are ninety-eight (98)
Fourteen eighths are one hundred and twelve (112)
Fourteen nines are one hundred and twenty-six (126)
Fourteen tens are one hundred and forty (140)
Fourteen elevens are one hundred and fifty-four (154)
Fourteen twelves are one hundred and sixty-eight (168)
Fourteen thirteens are one hundred and eighty-two (182)
Fourteen fourteens are one hundred and ninety-six (196)
Fourteen fifteenths are two hundred and ten (210)
Fourteen sixteenths are two hundred and twenty-four (224)
Fourteen seventeens are two hundred and thirty-eight (238)
Fourteen eighteenths are two hundred and fifty-two (252)
Fourteen nineteens are two hundred and sixty-six (266)
Fourteen twenties are two hundred and eighty (280)
## Tips for Memorizing Multiplication Table of 14
The table of 14 doesn’t come with rules or an easy pattern which is why it is hard for students to memorize quickly compared to other multiplication tables. However, a few tips can help you memorize the results of the 14 times table, such as using the repeated addition method or doubling the table of 4.
The addition method is a bit risky because adding large numbers sometimes gets out of hand for students. Therefore, we would advise you to use double the table of 4 tips. Here is how:
Use a 4 times table and add natural numbers (multipliers) to the tens’ digit of its product to generate the results of the 14 times table.
### Example Question of Table of 14
Question: Find the value of 14 times 7 plus 14 times 9?
Solution:
Given that,
14 times 7 plus 14 times 9
In mathematical form,
= 14 x 7 + 14 x 9
Using 14 times table
= 98 + 126
= 224
Hence, the value of 14 times 7 plus 14 times 9 is 224
Question: Using the table of 14, find how many fourteens are in 56?
#### Solution:
Let p be the number,
⇒ p x 14 = 56
⇒ p = 56/14
⇒ p = 4
Hence, there are 4 fourteens in 56.
Question: Today is a chocolate day at school, and the teacher wants to distribute 2 chocolates to each student. If a class has 14 students, find how many chocolates he would need for distribution? Find the number of chocolates using the 14 times table.
#### Solution:
Given that,
Number of chocolates for each student = 2
Number of students = 14
Using 14 times table,
⇒ 14 x 2 = 28
Hence, the teacher needs a total of 28 chocolates to distribute
Question: Write a multiplication table of 14 from 2 to 8.
14 x 2 = 28
14 x 3 = 42
14 x 4 = 56
14 x 5 = 70
14 x 6 = 84
14 x 7 = 98
14 x 8 = 112
Question: Fill in the blanks using 7 multiplication table
14 x ___ = 56
___ x 6 = 84
___ x ___ = 112
14 x ___ = 98
___ x ___ = 42
14 x ___ = 126
___ x 13 = 182
___ x ___ = 252
Question: How many dollars do I need to buy 7 flowers if each flower costs \$14?
Solution:
Given that,
Total flowers = 7
Cost of each flower = \$14
Using multiplication table of 14
⇒ 7 x 14 = 98
Hence, you would need \$98 to buy 7 flowers. |
# 5.3 Counting Principles
Counting principles and permutations are essential tools in . They help us calculate the number of possible outcomes in various scenarios, from simple everyday choices to complex statistical problems.
Combinations and binomial probability build on these concepts. They allow us to determine the likelihood of specific events occurring, such as getting a certain number of successes in a series of trials. These skills are crucial for analyzing real-world data and making informed decisions.
## Fundamental Counting Principle and Permutations
### Fundamental Counting Principle application
• States that if there are $n_1$ ways to do something and $n_2$ ways to do another thing, then there are $n_1 \times n_2$ ways to do both things
• Useful for determining the total number of possible outcomes in a given situation
• Helps to break down complex counting problems into simpler, more manageable parts
• Example: If a restaurant offers 5 appetizers, 8 main courses, and 3 desserts, there are $5 \times 8 \times 3 = 120$ possible three-course meals (assuming one item from each category)
### Permutations and combinations in probability
• : An of objects in a specific order
• Number of permutations of $n$ objects taken $r$ at a time denoted as $[P(n,r)](https://www.fiveableKeyTerm:p(n,r))$ or $_{n}P_{r}$
• Formula: $P(n,r) = \frac{[n!](https://www.fiveableKeyTerm:n!)}{(n-r)!}$
• Used when the order of selection matters (arranging books on a shelf)
• Permutations with repetition: If an object can be chosen more than once, the number of permutations is $n^r$
• Example: A 4-digit PIN code using digits 0-9 (repetition allowed) has $10^4 = 10,000$ possible permutations
• : A selection of objects without regard to the order
• Number of combinations of $n$ objects taken $r$ at a time denoted as $[C(n,r)](https://www.fiveableKeyTerm:c(n,r))$, $_{n}C_{r}$, or $\binom{n}{r}$
• Formula: $C(n,r) = \frac{n!}{r!(n-r)!}$
• Used when the order of selection does not matter (selecting a committee from a group of people)
### Permutations vs combinations
• Permutations: Order matters
1. Arrange objects in a specific order
2. Use the permutation formula: $P(n,r) = \frac{n!}{(n-r)!}$
3. Example: Arranging the letters A, B, and C in different orders (ABC, ACB, BAC, BCA, CAB, CBA)
• Combinations: Order does not matter
1. Select objects without regard to the order
2. Use the combination formula: $C(n,r) = \frac{n!}{r!(n-r)!}$
3. Example: Choosing 2 toppings from a set of 5 toppings for a pizza (choosing pepperoni and mushroom is the same as choosing mushroom and pepperoni)
## Combinations and Binomial Probability
### Binomial Probability Formula calculations
• Calculates the probability of exactly $x$ successes in $n$ independent trials, each with success probability $p$
• Formula: $P(X=x) = \binom{n}{x} p^x (1-p)^{n-x}$
• $X$: Random variable representing the number of successes
• $n$: Number of trials
• $p$: Probability of success on each trial
• $x$: Number of successes
• Assumptions for the binomial probability formula:
1. Fixed number of trials $n$
2. Independent trials
3. Two possible outcomes per trial (success or failure)
4. Constant probability of success $p$ for each trial
• Example: Probability of getting exactly 3 heads in 5 coin tosses (fair coin, $p=0.5$)
• $P(X=3) = \binom{5}{3} (0.5)^3 (1-0.5)^{5-3} = 0.3125$
## Key Terms to Review (15)
Arrangement: Arrangement refers to the specific ordering or placement of a set of items, where the sequence matters. In counting principles, arrangements are crucial because they help determine the total number of possible configurations or sequences for a given set of objects, allowing for a better understanding of permutations and combinations.
C(n,r): The notation c(n,r), also known as 'n choose r', represents the number of ways to choose r elements from a set of n distinct elements without regard to the order of selection. This concept is fundamental in combinatorics and helps to calculate combinations, which are different from permutations where order matters. Understanding c(n,r) is crucial for solving problems related to probability, statistics, and various counting principles.
Combination: A combination refers to a selection of items from a larger set, where the order of selection does not matter. It’s a fundamental concept in counting principles that helps determine how many different ways a specific number of items can be chosen from a group, allowing for an understanding of probabilities and outcomes in various scenarios.
Combinatorial optimization: Combinatorial optimization refers to the process of finding the best solution from a finite set of possible solutions in situations where the objective is to optimize a particular function. It often involves selecting, arranging, or grouping items based on specific constraints and criteria, making it essential in various fields like logistics, scheduling, and network design. This concept is closely tied to counting principles, which help determine the number of feasible solutions and guide the optimization process.
Counting distinct objects: Counting distinct objects refers to the process of determining the number of unique items within a set, where duplicates are not counted multiple times. This concept is vital for understanding how to accurately assess situations involving combinations and arrangements, especially when certain elements may be indistinguishable from one another. By identifying and quantifying distinct items, one can better analyze probabilities and outcomes in various scenarios.
Dependent events: Dependent events are situations where the outcome of one event affects the outcome of another event. This relationship means that the probability of the second event occurring is influenced by whether or not the first event has occurred. Understanding how dependent events work is crucial for accurately calculating probabilities and determining outcomes in various scenarios.
Factorial: A factorial, denoted by the symbol '!', is the product of all positive integers up to a specified number. For any positive integer n, the factorial is expressed as n! = n × (n - 1) × (n - 2) × ... × 3 × 2 × 1, with the special case that 0! = 1. Factorials are essential in permutations and combinations, helping to count arrangements and selections.
Fundamental counting principle: The fundamental counting principle is a mathematical rule that states if there are 'm' ways to do one thing and 'n' ways to do another, then there are 'm × n' ways to perform both actions. This principle helps in calculating the total number of outcomes in various scenarios by multiplying the number of choices available at each step.
Independent Events: Independent events are occurrences where the outcome of one event does not affect the outcome of another event. This means that knowing the result of one event provides no information about the result of the other. Recognizing independence is crucial in probability, as it allows for simpler calculations and a clearer understanding of relationships between different events.
N!: The term n! (read as 'n factorial') represents the product of all positive integers from 1 to n. This concept is vital in counting principles, as it helps to determine the total number of ways to arrange or select items in various contexts, laying the groundwork for combinations and permutations.
P(n,r): p(n,r) represents the number of permutations of 'r' items selected from a total of 'n' distinct items. This concept is crucial in understanding how arrangements can be made, emphasizing the significance of order when selecting items. The formula for calculating p(n,r) is given by $$p(n,r) = \frac{n!}{(n-r)!}$$, where 'n!' denotes the factorial of 'n', reflecting the product of all positive integers up to 'n'.
Permutation: A permutation is an arrangement of objects in a specific order. It emphasizes the importance of order, meaning that changing the sequence of elements leads to a different permutation. This concept is crucial in various applications, such as probability and combinatorics, where the arrangement of items can impact outcomes significantly.
Probability: Probability is a branch of mathematics that deals with the likelihood of an event occurring, expressed as a number between 0 and 1. Understanding probability helps in making informed decisions based on uncertain outcomes, and it plays a crucial role in concepts such as normal distribution and z-scores, as well as the basic principles that govern random events. It provides the foundational framework for evaluating events, understanding risks, and applying various counting principles to analyze situations effectively.
Sample space: Sample space refers to the set of all possible outcomes of a random experiment or event. Understanding the sample space is crucial as it provides the foundation for calculating probabilities, using counting principles to determine outcomes, and applying various probability rules. Each outcome in the sample space represents a distinct possibility, which helps in visualizing and organizing the results of experiments.
Subset: A subset is a collection of elements that are all contained within another set. It plays a crucial role in organizing and categorizing data, allowing for the understanding of relationships between different groups of elements within a larger set. Subsets can help in various counting principles by simplifying complex problems into smaller, more manageable parts. |
# MSBSHSE Solutions For Class 8 Maths Part 2 Chapter 11 Statistics
MSBSHSE Solutions For Class 8 Maths Part 2 Chapter 11 Statistics are given here to help students with their exam preparations. As students are familiar with numbers and then comes the data interpretation and analysis, which are studied under statistics. This chapter discusses arithmetic mean, creation of sub-divided bar graphs and percentage bar graphs. All problems pertaining to these concepts are in the Maharashtra State Board Class 8 Textbooks Part 2. Our expert faculty team have created solutions to these problems in a step-by-step manner, in order to help students understand the concepts clearly. Using these solutions one can attain a strong command over the subject. For students wishing to secure an excellent score, MSBSHSE Solutions Class 8 is a must. Students can access the solutions PDF of this chapter the Maharashtra Board Solutions for Class 8 Maths Chapter 11 Statistics, from the link provided below.
## Download the PDF of Maharashtra Board Solutions For Class 8 Maths Part 2 Chapter 11 Statistics
### Access answers to Maharashtra Board Solutions For Class 8 Maths Part 2 Chapter 11 Statistics.
Practice Set 11.1 Page No: 69
1. The following table shows the number of saplings planted by 30 students. Fill in the boxes and find the average number of saplings planted by each student.
Solution:
No. of saplings (Scores) xi No. of students (Frequency) fi fi × xi 1 4 4 2 6 12 3 12 36 4 8 32 N = 30 Ʃ fixi = 84
2. The following table shows the electricity (in units) used by 25 families of Eklara village in a month of May. Complete the table and answer the following questions.
(1) How many families use 45 units electricity?
(2) State the score, the frequency of which is 5.
(3) Find N, and Æ© fixi
(4) Find the mean of electricity used by each family in the month of May.
Solution:
Electricity used (Units) xi No. of families (Frequency) fi fi × xi 30 7 210 45 2 90 60 8 480 75 5 375 90 3 270 N = 25 Ʃ fixi = 1425
(1) 2 families used 45 units od electricity.
(2) The score for which the frequency is 5 is 75.
(3) N = 25 and Æ© fixi = 1425
(4) The mean of electricity used by each family in the month of May is given by:
Mean () = Æ© fixi / N
= 1425 / 25
= 57
Thus, the mean of electricity used by each family in the month of May is 57 units.
3. The number of members in the 40 families in Bhilar are as follows:
1, 6, 5, 4, 3, 2, 7, 2, 3, 4, 5, 6, 4, 6, 2, 3, 2, 1, 4, 5, 6, 7, 3, 4, 5, 2, 4, 3, 2, 3, 5, 5, 4, 6, 2, 3, 5, 6, 4, 2. Prepare a frequency table and find the mean of members of 40 families.
Solution:
Number of members Number of families (fi) fixi 1 2 2 2 8 16 3 7 21 4 8 32 5 7 35 6 6 36 7 2 14 N = 40 Æ© fixi = 156
Now,
Mean = Æ© fixi/ N = 156/ 40 = 3.9
Thus, the mean of members of 40 families is 3.9.
4. The number of Science and Mathematics projects submitted by Model high school, Nandpur in last 20 years at the state level science exhibition is : 2, 3, 4, 1, 2, 3, 1, 5, 4, 2, 3, 1, 3, 5, 4, 3, 2, 2, 3, 2. Prepare a frequency table and find the mean of the data.
Solution:
Number of projects (xi) Frequency (fi) fixi 1 3 3 2 6 12 3 6 18 4 3 12 5 2 10 N = 20 Æ© fixi = 55
Now,
Mean = Æ© fixi / N = 55/ 20 = 2.75
Thus, the mean of the data = 2.75
Practice Set 11.2 Page No: 71
1.
(1) State the type of the graph.
(2) How much is the savings of Vaishali in the month of April?
(3) How much is the total of savings of Saroj in the months March and April?
(4) How much more is the total savings of Savita than the total savings of Megha?
(5) Whose savings in the month of April is the least?
Solution:
(1) The given is a bar graph.
(2) The savings of Vaishali in the month of April is Rs 600.
(3) The total savings of Saroj in the months March and April is Rs 800.
(4) From the table it’s seen that, the total savings of Savita = Rs 1000
And, the total savings of Megha = Rs 500
So, the difference in their savings = 1000 − 500 = 500
Hence, the total savings of Savita is Rs 500 more than the total savings of Megha.
(5) The savings of Megha in the month of April is the least i.e. Rs 200.
2. The number of boys and girls, in std 5 to std 8 in a Z.P. school is given in the table. Draw a subdivided bar graph to show the data.
(Scale: On Y axis, 1cm = 10 students)
Solution:
Given data,
Standard 5th 6th 7th 8th Boys 34 26 21 25 Girls 17 14 14 20 Total 51 40 35 45
Now,
The subdivided bar graph of the given data is as follows:
3. In the following table number of trees planted in the year 2016 and 2017 in four towns is given. Show the data with the help of subdivided bar graph.
Solution:
Given data,
Year\Town Karjat Wadgoan Shivapur Khandala 2016 150 250 200 100 2017 200 300 250 150 Total 350 550 450 250
Now,
The subdivided bar graph of the given data is as follows:
4. In the following table, data of the transport means used by students in 8th standard for commutation between home and school is given. Draw a subdivided bar diagram to show the data.
(Scale: On Y axis: 1 cm = 500 students)
Solution:
Given data,
Town → Paithan Yeola Shahapur Mean of communication ↓ Cycle 3250 1500 1250 Bus and Auto 750 500 500 On foot 1000 1000 500 Total 5000 3000 2250
Now,
The subdivided bar diagram of the given data is as follows
Practice Set 11.3 Page No: 73
1. Show the following information by a percentage bar graph.
Solution:
Given data,
Division of standard 8 A B C D Number of students securing grade A 45 33 10 15 Total number of students 60 55 40 75 Percentage of students securing grade A (45/60) x 100 = 75% (33/55) x 100 = 60% (10/40) x 100 = 25% (15/75) x 100 = 20%
Now, the percentage bar graph is:
2.
(1) State the type of the bar graph.
(2) How much percent is the Tur production to total production in Ajita’s farm?
(3) Compare the production of Gram in the farms of Yash and Ravi and state whose percentage of production is more and by how much?
(4) Whose percentage production of Tur is the least?
(5) State production percentages of Tur and gram in Sudha’s farm.
Solution:
From the given graph, it can be inferred that:
(1) The given graph is a percentage bar graph.
(2) Percent of tur production to the total production in Ajita’s farm is 60%.
(3) Production of Gram in the farm of Yash = 50%
And, the production of Gram in the farm of Ravi = 30%
So, the difference in the production = 50% – 30% =20%
Thus, Yash’s production of Gram is more and by 20%.
(4) The least production percentage of Tur is of Sudha.
(5) The production percentages of Tur and Gram in Sudha’s farm are 40% and 60% respectively.
3. The following data is collected in a survey of some students of 10th standard from some schools. Draw the percentage bar graph of the data.
Solution:
Given data,
School 1st 2nd 3rd 4th Inclination towards science stream 90 60 25 16 Inclination towards commerce stream 60 20 25 24 Total 150 80 50 40 Percentage of students having inclination towards science stream (90/150) x 100 = 60% (60/80) x 100 = 75% (25/50) x 100 = 50% (16/40) x 100 = 40% Percentage of students having inclination towards commerce stream (60/150) x 100 = 40% (20/80) x 100 = 25% (25/50) x 100 = 50% (24/40) x 100 = 60%
The percentage bar graph of the given data is as follows:
Chapter 11 Statistics is one of the easy scoring chapters in Maharashtra board textbook for Class 8. We, at BYJU’S, provide useful resources for students to help scoring full marks in their examinations.
Our subject experts have solved the difficult problems with simpler steps in easy language, so that it’s understood by students of all levels. Regular revision of important concepts and formulas over time is the best way to strengthen concepts.
## Frequently Asked Questions on Maharashtra State Board Solutions for Class 8 Maths Chapter 11 Statistics
### What are the concepts explained in the Maharashtra State Board Class 8Â Maths Chapter 11 Statistics Solutions?
These solutions help students to familiarize themselves with constructions of triangles. This chapter also deals with geometric constructions of triangles, if base, an angle adjacent to base and sum of lengths of two remaining sides are given.
### How helpful are these Maharashtra State Board Class 8 Maths Chapter 11 Solutions?
Yes, these solutions are very helpful and Students are encouraged to practise these questions first. Then they can refer back to the solutions to analyse their performance. This will also help them to rectify the mistakes, so that they can avoid making any during the board exams. These solutions also lay the foundation for board exam questions. |
# Percentage Error
It is hard for one to take a measurement and get an exact result. Percentage error is a little degree in percent at which a known length is measured above/below its actual value.
The error that might be made during the measurement could be either positive or negative, in any ways just know that error still remains error provided that it has a value order than the actual result.
For instance if the actual length of a room is 3.0m. It is said that three students on their industrial Training measured it differently and they got:
2.85m, 3.01m and 2.91m respectively. What are their errors?
Error = (Wrong value – Actual value) or (Actual value – Wrong value)
The first student’s error
= 3.0 – 2.85 = 0.15m
Second student’s error
= 3.01 – 3.0 = 0.01m
Third student’s error
= 3.0 – 2.91 = 0.09m
Note that the ones we just did above are the errors. For one to calculate the percentage error, this formula will be used:
Percentage error (PE)
= Absolute error (AE) divided by Actual value (AV) and you multiply by 100.
(PE) = (AE/AV) x 100
Now, the percentage error of the example above are:
For the first student:
AE = 0.15m.
AV = 3.0m
PE = ?
PE = AE/AV x 100
= 0.15/3 x 100
= 15/3 = 5%
For the second student:
AE = 0.01m.
AV = 3.0m
PE = ?
PE = AE/AV x 100
= 0.01/3 x 100
= 1/3% = 0.33%
For the third student:
AE = 0.09m.
AV = 3.0m
PE = ?
PE = AE/AV x 100
= 0.09/3 x 100
= 9/3 = 3%
## QUESTIONS AND SOLUTIONS:
### Q1: A candidate was to subtract 15 from a certain number, but mistakenly added 25 and his answer was 145. Find the percentage error.
Solution:
Let the certain number be y.
He was meant to subtract 15 from it and mistakenly added 25.
y + 25 = 145
y = 120 = the certain number
Actual value = 120 – 15 = 105
Absolute Error = 145 – 105 = 40
PE = 40/105 x 100
= 38.1%
### Q2: The length of a wire is 6.35, a student measured it as 6.65. What is the percentage error to 1 decimal place?
AE = 6.65 – 6.35 = 0.3
AV = 6.35
PE = ?
PE = AE/AV x 100
= 0.3/6.35 x 100
= 4.7% (1 dp)
### Q3: If the age of a man 64 years is written as 71 years, calculate the percentage error to 3 significant figures.
Solution:
AE = 71 – 64 = 7 years
AV = 64
PE = ?
PE = AE/AV x 100
= 7/64 x 100
= 10.9% (3 SF)
### Q4: Find the percentage error in a piece of wood that was measured to be 1.26m whose actual length was 1.24m.
Solution:
AE = 1.26 – 1.24 = 0.02m
AV = 1.24
PE = ?
PE = AE/AV x 100
= 0.02/1.24 x 100
= 1.6%
### Q5: A man underestimated his expenses by 6.5% but actually spent #400.00. What was his estimate?
Solution:
Let his estimate be D
AV = #400
AE = (400 – D)
PE = 6.5%
PE = [(400 – D)/400] x 100 = 6.5
Solve for D.
PE = (400 – D) = 26
D = 400 – 26 = #374
### Q6: An error of 4% was made in finding the length of rope that was actually 25m. By how many metres was the measurement wrong?
Solution:
Let the wrong value be T
AV = 25m
AE = (T – 25)m
PE = 4%
PE = [(T – 25)/25] x 100 = 4
Solve for T.
100(T – 25) = 4 x 25
100(T – 25) = 100
T – 25 = 1
T = 1 + 25 = 26m
Therefore the measurement is ±1m wrong.
Students by now will be wondering why my absolute error (AE) is = (T – 25)m and not (25 – T)m. Note that in which ever way you place it, it is still the same provided that the question comes in that nature.
Now, let’s use the reverse (25 – T)m as our AE.
PE = [(25 – T)/25] x 100 = 4
100(25 – T) = 25 x 4
100(25 – T) = 100
Solve for T
25 – T = 1
-T = 1 – 25
-T = -24 = 24m
Actual length = 25m
So the measurement is ±1m wrong.
### Q7: What is the percentage error in an area of a lawn that actually measures 750m² but found to be 690m²?
Solution:
AV = 750m²
AE = 750 – 690 = 60m²
PE = ?
PE = 60/750 x 100
= 8%
### Q8: The percentage error in the measurement of the length of a rope was 6%. If the measurement was 35m, find the actual length of the rope and by how many metres was the measurement wrongto 1 decimal place.
Solution:
Let the actual length be x.
PE = 6%
AE = (x – 35)m
[(x – 35)/x] X 100 = 6
Solve for x.
100(x – 35) = 6x
100x – 3500 = 6x
94x = 3500
x = 37.2m
So, the measurement is 37.2 – 35 = ±2.2% wrong
Solution:
AE = 15cm
AV = 165cm
PE = ?
PE = AE/AV x 100
= (15/165) x 100
= 9.1%
### Q10: The length and breath of a rectangle was mistakenly measured as 40m and 35m instead of 42.5m and 34.2m respectively. Find the percentage error in:(a) the area(b) the perimeter
Solution:
(a) the area
Actual area = 40 x 35 = 1400m²
Wrong area = 42.5 x 34.2 = 1453.5m²
AV = 1400
AE = 1453.5 – 1400 = 53.5m
PE = (53.5/1400) x 100 = 3.8%
(b) the perimeter
Perimeter of the actual rectangular = 40 + 35 = 75m
Wrong Perimeter = 42.5 + 34.2 = 76.7m
AE = 76.7 – 75 = 1.7m
Therefore:
PE = (1.7/75) x 100
= 2.3%
### Q11: The length of a file 25cm was measured as 27.5cm. Calculate the percentage error.
Solution:
Actual value = 25cm
Wrong value = 27.5cm
Absolute Error =2.5cm
Percentage error = ?
PE = (2.5/25) x 100
= 250/25 = 10% |
# Ncert Class 6 Math Decimals Exercise 8.4
Decimals Class 6 Ex. 8.4
New Ncert Class 6, Chapter 8 Math Free Solution.
Exercise 8.4
Question 1 :- Subtract:
(a) ₹18.25 from ₹20.75
(b) 202.54 m from 250 m
(c) ₹5.36 from ₹8.40
(d) 2.051 km from 5.206 km
(e) 0.314 kg from 2.107 kg
Solution 1:-
(a) ₹18.25 from 20.75
= 20.75 – 18.25
= 2.50
(b) 202.54 m from 250 m
= 250 m – 202.54 m
= 250.00 m – 202.54 m
= 47.46 m
(c) ₹5.36 from ₹8.40
= ₹8.40 – ₹5.36
= ₹3.04
(d) 2.051 km from 5.206 km
= 5.206 km – 2.051 km
= 3.155 km
(e) 0.314 kg from 2.107 kg
= 2.107 kg – 0.314 kg
= 1.793 kg
Question 2 :- Find the value of:
(a) 9.756 – 6.28
(b) 21.05 – 15.27
(c) 18.5 – 6.79
(d) 11.6 – 9.847
Solution 2:-
(a) 9.756 – 6.28
= 9.756 – 6.280
= 3.476
(b) 21.05 – 15.27
= 5.78
(c) 18.5 – 6.79
= 18.50 – 6.79
= 11.71
(d) 11.6 – 9.847
= 11.600 – 9.847
= 1.753
Question 3 :- Raju bought a book for ₹35.65 rupees. He gave ₹50 to the shopkeeper. How much money did he get-back from the shopkeeper?
Solution 3:-
Cost of book = ₹35.65
Money paid by him to the shopkeeper = ₹50
So, money got back by him
= ₹50 – ₹35.65
= ₹50.00 – ₹35.65
= ₹14.35
Question 4 :- Rani had ₹18.50. She bought one ice-cream for ₹11.75. How much money does she have now?
Solution 4:-
She bought ice-cream for ₹11.75
So, Money left with Rani = ₹ 18.50 – ₹11.75 = ₹6.75
Question 5 :- Tina had 20m 5cm long cloth. She cut 4m 50cm length of cloth from this for making curtain. How much cloth is left with her?
Solution 5:-
The total length of cloth had by Tina = 20 m 5 cm = 20.05 m
The length of cloth cut by her = 4 m 50 cm = 4.50 m
So, the length of cloth left with her = 20.05 m – 4.50 m = 15.55 m
Question 6 :- Namita travels 20km 50m every day. Out of this Namita travels 10km 200m by bus and the rest by auto. How much-distance does she travel by auto?
Solution 6:-
Distance travelled by Namita daily = 20km 50m = 20.050 km
Distance travelled by her by bus = 10km 200m = 10.200 km
So, Distance travelled by her by auto = (20.050 – 10.200) km = 9.850 km
Question 7 :- Aakash bought vegetables weighing 10kg. Out of this, 3kg 500g is onions, 2kg 75g is tomatoes and the rest is potatoes. What is the weight of the potatoes?
Solution 7:-
The total weight of vegetables bought by Aakash = 10 kg
The weight of onions bought by him = 3 kg 500 g = 3.500 kg and
The weight of tomatoes bought by him = 2 kg 75 g = 2.075 kg
So, the weight of potatoes = Weight of vegetable – (weight of onions + weight of tomatoes)
= 10.000 – (3.500 + 2.075)
= 10.000 – 5.575
= 4.425 kg |
Learning Outcomes
• Write a quadratic equation in standard form and identify the values of a, b, and c in a standard form quadratic equation.
• Use the Quadratic Formula to find all real solutions of a quadratic equation and recognize when there are no real solutions
• Solve application problems involving quadratic equations
You can solve any quadratic equation by completing the square—rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation $ax^{2}+bx+c=0$ and then solve for $x$, you find that $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. This equation is known as the Quadratic Formula.
We can derive the quadratic formula by completing the square. First, assume that the leading coefficient is positive; if it is negative, we can multiply the equation by $-1$ and obtain a positive a. Given $a{x}^{2}+bx+c=0$, $a\ne 0$, we will complete the square as follows:
1. First, move the constant term to the right side of the equal sign:
$a{x}^{2}+bx=-c$
2. As we want the leading coefficient to equal 1, divide through by a:
${x}^{2}+\frac{b}{a}x=-\frac{c}{a}$
3. Then, find $\frac{1}{2}$ of the middle term, and add ${\left(\frac{1}{2}\frac{b}{a}\right)}^{2}=\frac{{b}^{2}}{4{a}^{2}}$ to both sides of the equal sign:
${x}^{2}+\frac{b}{a}x+\frac{{b}^{2}}{4{a}^{2}}=\frac{{b}^{2}}{4{a}^{2}}-\frac{c}{a}$
4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction:
${\left(x+\frac{b}{2a}\right)}^{2}=\frac{{b}^{2}-4ac}{4{a}^{2}}$
5. Now, use the square root property, which gives
$\begin{array}{l}x+\frac{b}{2a}=\pm \sqrt{\frac{{b}^{2}-4ac}{4{a}^{2}}}\hfill \\ x+\frac{b}{2a}=\frac{\pm \sqrt{{b}^{2}-4ac}}{2a}\hfill \end{array}$
6. Finally, add $-\frac{b}{2a}$ to both sides of the equation and combine the terms on the right side. Thus,
$x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$
This formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form $ax^{2}+bx+c=0$.
The form $ax^{2}+bx+c=0$ is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it is vital that you be sure the equation is in this form. If you do not, you might use the wrong values for a, b, or c, and then the formula will give incorrect solutions.
The Quadratic Formula will work with any quadratic equation, but only if the equation is in standard form, $ax^{2}+bx+c=0$. To use it, follow these steps.
1. Put the equation in standard form first.
2. Identify the coefficients, a, b, and c. Be sure to include negative signs if the bx or c terms are subtracted.
3. Carefully substitute the values noted in step $2$ into the equation. To avoid needless errors, use parentheses around each number input into the formula.
4. Simplify as much as possible.
5. Use the $\pm$ in front of the radical to separate the solution into two values: one in which the square root is added and one in which it is subtracted.
6. Simplify both values to get the possible solutions.
That is a lot of steps. Let us try using the Quadratic Formula to solve a relatively simple equation first; then you will go back and solve it again using another factoring method.
Example
Use the Quadratic Formula to solve the equation $x^{2}+4x=5$.
You can check these solutions by substituting $1$ and $−5$ into the original equation.
$\begin{array}{r}x=1\\x^{2}+4x=5\\\left(1\right)^{2}+4\left(1\right)=5\\1+4=5\\5=5\end{array}$ $\begin{array}{r}x=-5\\x^{2}+4x=5\,\,\,\,\,\\\left(-5\right)^{2}+4\left(-5\right)=5\,\,\,\,\,\\25-20=5\,\,\,\,\,\\5=5\,\,\,\,\,\end{array}$
You get two true statements, so you know that both solutions work: $x=1$ or $-5$. You have solved the equation successfully using the Quadratic Formula!
Watch this video to see an example of how to use the quadratic formula to solve a quadratic equation that has two real, rational solutions.
Sometimes, it may be easier to solve an equation using conventional factoring methods like finding number pairs that sum to one number (in this example, $4$) and that produce a specific product (in this example $−5$) when multiplied. The power of the Quadratic Formula is that it can be used to solve any quadratic equation, even those where finding number combinations will not work.
In the next video example, we show that the quadratic formula is useful when a quadratic equation has two irrational solutions that could not have been obtained by factoring.
Most of the quadratic equations you have looked at have two solutions, like the one above. The following example is a little different.
Example
Use the Quadratic Formula to solve the equation $x^{2}-2x=6x-16$.
Again, check using the original equation.
$\begin{array}{r}x^{2}-2x=6x-16\,\,\,\,\,\\\left(4\right)^{2}-2\left(4\right)=6\left(4\right)-16\\16-8=24-16\,\,\,\,\,\,\\8=8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}$
In the next example, we will show that some quadratic equations do not have real solutions. As we simplify with the quadratic formula, we may end up with a negative number under a square root, which, as we know, is not defined for real numbers.
Example
Use the Quadratic Formula to solve the equation $x^2+x=-x-3$
Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.
A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. Note: The equation is not completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.
Example
A ball is thrown off a building from $200$ feet above the ground. Its starting velocity (also called initial velocity) is $−10$ feet per second. The negative value means it is heading toward the ground.
The equation $h=-16t^{2}-10t+200$ can be used to model the height of the ball after t seconds. About how long does it take for the ball to hit the ground?
In the next video, we show another example of how the quadratic equation can be used to find the time it takes for an object in free fall to hit the ground.
The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.
Example
Bob made a quilt that is $4$ ft $\times$ $5$ ft. He has $10$ sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)
Our last video gives another example of using the quadratic formula for a geometry problem involving the border around a quilt.
Summary
Quadratic equations can appear in different applications. The Quadratic Formula is a useful way to solve these equations or any other quadratic equation! The Quadratic Formula, $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, is found by completing the square of the quadratic equation . When you simplify using the quadratic formula and your result is a negative number under a square root, there are no real number solutions to the equation. |
# Combining Symmetries
In the last quiz, we looked at symmetries of shapes, which are rigid transformations that send the shape to itself. At the end of the quiz, we saw that we could combine symmetries by doing one after the other, and thus form a kind of “multiplication” on the symmetries.
In this quiz, we’ll explore this idea further, and in the next quiz, this will lead to the definition of a group.
To start, consider this equilateral triangle:
Let $$R$$ be clockwise rotation by $$120^{\circ},$$ and let $$R_1$$ be a reflection about the line segment $$\overline{AF}.$$ What symmetry do you get when you apply $$R_1$$ followed by $$R ?$$
Hint: Investigate where the three vertices of the triangle are sent by $$R_1$$ followed by $$R.$$
## Combining Symmetries
### Introduction
# Combining Symmetries
Now let’s combine two reflections.
Let $$R_1$$ be reflection about the line $$\overline{AF},$$ and let $$R_2$$ be reflection about the line $$\overline{CE}.$$ What symmetry do you get by applying first $$R_1$$ and then $$R_2$$?
Note: After the lines of reflection are set they do not change, even if the points they were originally based on change position.
## Combining Symmetries
### Introduction
# Combining Symmetries
In fact, we can combine any two symmetries of a shape by doing one after the other, and obtain a third symmetry of the shape. We express this with the following notation: If $$A$$ and $$B$$ are symmetries of a shape, $$A * B$$, or sometimes just $$AB,$$ denotes the symmetry you get by applying $$B$$ first, and then $$A.$$ We will call this new symmetry the product of $$A$$ and $$B.$$
For example, in the previous problem, if we let $$\phi_1$$ denote the reflection about $$\overline{AF},$$ let $$\phi_2$$ denote the reflection about $$\overline{CE},$$ and let $$R$$ denote clockwise rotation by $$120^{\circ},$$ then we can write the equation $\phi_2 * \phi_1 = R.$ To really understand the symmetries of an object, we should know not only how many symmetries there are, but what the rules are for how they combine with each other. For the equilateral triangle, we could make a 6 x 6 “multiplication table” that tells you, for any two of the six symmetries of the triangle, what their product is. In the next problems, we will do exactly that for a slightly simpler object.
## Combining Symmetries
### Introduction
# Combining Symmetries
In the next few problems, we will work out the symmetry multiplication table for the letter I. For starters, how many symmetries does this letter have? (Remember to include the identity symmetry.)
## Combining Symmetries
### Introduction
# Combining Symmetries
We’ve seen that the letter I has four symmetries: A horizontal reflection (flipping across the blue line) $$H;$$ a vertical reflection (flipping across the red line) $$V;$$ rotation by $$180^{\circ}$$ $$R;$$ and the identity transformation $$I$$.
Let’s start working out the 4 x 4 multiplication table for this object!
What are $$H * H,$$ $$V * V,$$ and $$R * R?$$
## Combining Symmetries
### Introduction
# Combining Symmetries
So far, our multiplication table for the symmetries of the letter I looks like this:
This table, once it’s filled out, will be used to read off the product of symmetries. For instance, $$H * V$$ will appear in the row labeled $$H$$ and the column labeled $$V,$$ although that entry hasn’t been filled out yet. We filled out the diagonals with $$I$$ because we saw in the last problem that $$H * H,$$ $$V * V$$ and $$R * R$$ are all $$I$$.
Let’s fill out the first row and column of the table. The first row has products like $$I * H$$ and $$I * V,$$ which result from doing the a transformation like $$H$$ or $$V$$ and then doing the identity. (Remember a product like $$I * H$$ means to do $$H$$ first.) The first column are products resulting from doing the identity first, and then some other transformation.
Which of the following correctly fills out the first row and column?
## Combining Symmetries
### Introduction
# Combining Symmetries
Now for some more interesting combinations! What are $$H * V$$ and $$V * H?$$ (Hint: When we worked out the triangle, it was useful to see where the vertices went under the product. Here, it’s useful to see where the four “outer corners” of the letter I go.)
## Combining Symmetries
### Introduction
# Combining Symmetries
We’re almost done! So far we have the following table:
Fill out the remaining four spaces by working through the remaining products, $$H * R,$$ $$V * R,$$ etc.
## Combining Symmetries
### Introduction
# Combining Symmetries
We’ve made our first multiplication table--or what we will soon call a group table:
This table completely describes all products of the four symmetries of the letter I.
## Combining Symmetries
### Introduction
# Combining Symmetries
Let’s stop for a moment and do something completely different. (At least, it will seem completely different at first…)
Suppose we only have two numbers, 0 and 1. Suppose we consider addition “mod 2” on these numbers, which means that it works exactly as normal, except that 1 + 1 = 0. We could make a little table for this operation:
Now, let’s do the same thing to ordered pairs of 0’s and 1’s, like $$(0,1)$$ and $$(1,0).$$ How do you add two such pairs? Just add the $$x$$-coordinates and $$y$$-coordinates separately. (Remember that 1 + 1 is still 0.) So $$(0,1) + (1,0) = (1,1),$$ and $$(1,1) + (1,1) = (0,0).$$
Here’s what the “addition table” for this operation on pairs looks like:
Do you see what the connection is between this addition table and the multiplication table we made for the letter I? Think about it, stare at both tables, and then turn to the next page for the answer…
## Combining Symmetries
### Introduction
# Combining Symmetries
To see the connection, suppose we rename the four ordered pairs. Let’s denote $$(0,0)$$ by $$i,$$ $$(1,0)$$ by $$a,$$ $$(0,1)$$ by $$b$$ and $$(1,1)$$ by $$c$$. Then the addition table becomes:
This is the same table as the symmetries for the letter I, just with different variable names! In other words, the algebraic structure of the symmetries of the letter I is the same as the algebraic structure of ordered pairs of 0’s and 1’s under mod 2 addition. There is some underlying mathematical object that represents this structure… and we will call that object a group.
This particular table will be represented by something called the Klein four-group. We will encounter it again, along with many other interesting examples.
## Combining Symmetries
### Introduction
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