Liz's math test included a survey question asking how many hours students spent studying for the test. The scatter plot below shows the relationship between how many hours students spent studying and their score on the test. A line was fit to the data to model the relationship.

Answer: All real numbers

Step-by-step explanation:

The domain is the set of inputs.

The volume of wood in the tree's trunk as it grows is [tex]16\pi r^3[/tex]

The capacity of a cylinder, which determines how much material it can carry, is determined by the cylinder's volume. There is a formula for the volume of a cylinder that is used in geometry to determine how much of any quantity, whether liquid or solid, may be immersed in it uniformly. A cylinder is a three-dimensional structure having two parallel, identical bases that are congruent.

Thus, the volume (V) of a right circular cylinder, using the above formula, is, [tex]V = \pi r^2h[/tex] , where

'r' is the radius of the base (circle) of the cylinder

'h' is the height of the cylinder

Given:

Height(h) = 8 x diameter = 16 x radius (r)      (diameter = 2 x radius )

∴ h = 16r.

V= [tex]\pi r^2h[/tex]

Substituting h = 16r in the volume formula we get

V = [tex]\pi r^2(16r)[/tex] =  [tex]16\pi r^3[/tex]

Thus the volume of wood in the tree's trunk as it grows is [tex]16\pi r^3[/tex].

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The equation of the parent square root function to represent the equation of the graphed function will be, y=√x-2

According to the statement

we have to show the square root function as a equation in the graphical representation.

So,

we know that the definition of a

Graph a diagram showing the relation between variable quantities, typically of two variables and it also show the relation between more than two variables.

Now, we know that the definition of Equation is a mathematical statement consisting of an equal symbol between two algebraic expressions with the same value is known as an equation.

And the equation obtained from the graph is a y=√x-2 by a some calculations in the graph.

So, The equation of the parent square root function to represent the equation of the graphed function will be, y=√x-2

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Answer:

14 3/8 ft^2

Step-by-step explanation:

Area = length * width

=   2 7/8 * 5

=  23/8 * 5

= 115/8

= 14 3/8 ft^2.

Answer:

Step-by-step explanation:

sinA(1+sin^2A) = cos^2A

sinA(2 -cos^2A) = cos^2A

Squaring both sides,

sin^2A(4-4cos^2A +cos^4A) = cos^4A

(1-cos^2A)(4-4cos^2A +cos^4A) = cos^4A

4-4cos^2A +cos^4A-4cos^2A+4cos^4A-cos^6A = cos^4A

4 -cos^6A +4cos^4A -8cos^2A = 0

cos^6A - 4 cos^4A + 8cos^2A = 4

hence proveproven

Answer:

Step-by-step explanation:

sinA+sin

3

A=cos

2

A

⇒sinA[1+sin

2

A]=cos

2

A

⇒sinA[1+1−cos

2

A]=cos

2

A

squaring on both sides.

⇒sin

2

A[4+cos

4

A−4cos

2

A]=cos

4

A

⇒(1−cos

2

A)[4+cos

4

A−9cos

2

A]=cos

4

A

⇒4+cos

4

A−4cos

2

A−64cot

2

A−cos

6

A+9cos

4

A=cos

4

A

⇒

cos

6

A−4cos

4

A+8cos

2

A=4

​

Hence Prove

4) [tex]A_{QRST} = A_{Q'R'S'T'} = 9[/tex].

5) The areas of the rectangles QRST and Q'R'S'T' are the same since the translation of the geometric loci conserved the Euclidean distance.

6) x = 1, w = 3, y = - 1/4, z = 1/3

Rigid transformations are transformations applied on geometric loci such that Euclidean distance is conserved. In this question we have applications of translations, a kind of rigid transformation.

Exercise 4

In this part we must determine the areas of rectangles QRST and Q'R'S'T':

Rectangle QRST

A = RS · QT

A = 3 · 3

A = 9

Rectangle Q'R'S'T'

Q'(x, y) = (2, - 3) + (- 3, - 3)

Q'(x, y) = (- 1, - 6)

R'(x, y) = (2, 4) + (- 3, - 3)

R'(x, y) = (- 1, 1)

S'(x, y) = (5, 4) + (- 3, - 3)

S'(x, y) = (2, 1)

T'(x, y) = (5, - 3) + (- 3, - 3)

T'(x, y) = (2, - 6)

A = R'S' · Q'T'

A = 3 · 3

A = 9

The rectangles QRST and Q'R'S'T' have both an area of 9 square units.

Exercise 5

The areas of the rectangles QRST and Q'R'S'T' are the same since the translation of the geometric loci conserved the Euclidean distance.

Exercise 6

In this case, we must solve the following equations:

PQ = A'(x, y) - A(x, y)

(4, 1) = (2 · x + 1, 4) - (- 1, w)

(4, 1) = (2 · x + 2, 4 - w)

(4, 1) = (2 · x, - w) + (2, 4)

(2, - 3) = (2 · x, - w)

(x, w) = (1, 3)

PQ = B'(x, y) - B(x, y)

(4, 1) = (3, 3 · z) - (8 · y - 1, 1)

(4, 1) = (2 - 8 · y, 3 · z)

(4, 1) = (- 8 · y, 3 · z) + (2, 0)

(2, 1) = (- 8 · y, 3 · z)

(y, z) = (- 1/4, 1/3)

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Answer:

x-2

Step-by-step explanation:

The formula for area of rectangle is length x width
And we know that width is x-16
And we know that the area is [tex]x^{2} -18x +32[/tex]

So we know that when x-2 is multiplied with  x-16, it gives off [tex]x^{2} -18x +32[/tex]

Answer:

64

Step-by-step explanation:

evaluate by substituting x = 16 and x = 0 into f(x)

f(16) = 4(16) - 8 = 64 - 8 = 56

f(0) = 4(0) - 8 = 0 - 8 = - 8

then

f(16) - f(0) = 56 - (- 8) = 56 + 8 = 64

Answer:

64

Step-by-step explanation:

f(x) = 4x - 8

First find f(16)

f(16) = 4(16) -8 =64-8=56

Then find f(0) = 4)0) -8 = -8

f(16) - f(0) = 56 - (-8) = 56+8 = 64

Answer:

[tex]\frac{1}{6}[/tex]

Step-by-step explanation:

You need to add the amount Shawna & Ryan ate together, then subtract it from 1 to find the amount of pizza that was left for Dominic to eat.

[tex]\frac{1}{3}[/tex] + [tex]\frac{1}{2}[/tex] = ?

Make all the denominators the same to add fractions.

[tex]\frac{1}{3}[/tex]  is [tex]\frac{2}{6}[/tex]

[tex]\frac{1}{2}[/tex]  is [tex]\frac{3}{6}[/tex]

[tex]\frac{2}{6}[/tex] +  [tex]\frac{3}{6}[/tex] = [tex]\frac{5}{6}[/tex]

1 -  [tex]\frac{5}{6}[/tex] = [tex]\frac{1}{6}[/tex]

Answer:

y < x

y < 4

Step-by-step explanation:

Dotted line means < or >, and since the shaded region is below the lines y = 4 and y = x, the answer is A.

Answer:

2/3

Step-by-step explanation:

1/3 cup powdered sugar

2/6 cup brown sugar

Reduce 2/6 to 1/3.

1/3 + 1/3 = 2/3

Step-by-step explanation:

When evaluating piecewise function, make sure the x value satisfies the function we use.

since 5>3, we use the second equations

That function is a constant function, this means at any x>3, exclusive, we will have a y value of -1.

So

[tex]f(5) = - 1[/tex]

Answer:

43.5 unit^2.

Step-by-step explanation:

Area of the triangle

= 1/2 * base * height

= 1/2 * AB * AC.

AB = √ [(6-1)^2 + (6-8)^2] = √29

AC = √ [(8 - (-7))^2 + (1 - (-5)^2] = √261

So the area = 1/2 * √29  * √261

                    = 43.5

Answer:

a) Height of the ledge is 1.6 m.

b) roots of the equation are: [tex]\bf d = 2[/tex] ,  [tex]\bf d = -1[/tex]

c) Maximum height reached = 1.8 m.

Step-by-step explanation:

a) The height of the ledge is the same as the height at which Jake is when he hasn't moved any horizontal distance from the ledge yet, that is, when d = 0:

[tex]h = -0.8(0)^2 + 0.8(0) + 1.6[/tex]

⇒ [tex]\bf h = 1.6 \space\ m[/tex]

∴ Height of the ledge is 1.6 m.

b) The x-intercepts occur where y = 0, that is when h = 0:

[tex]-0.8d^2 + 0.8d + 1.6 = 0[/tex]

Divide both sides of the equation by -0.8:

[tex]d^2 - d -2 = 0[/tex]

Factorizing:

⇒ [tex]d^2 -2d + d - 2 = 0[/tex]

⇒ [tex]d(d-2) +1 (d-2) = 0[/tex]

⇒ [tex](d-2)(d+1) = 0[/tex]

⇒ [tex]d - 2 = 0[/tex]    or    [tex]d + 1 = 0[/tex]

∴ roots are :  [tex]\bf d = 2[/tex]   ,   [tex]\bf d = -1[/tex]

c) To find the maximum value of a quadratic equation in the form

y = ax² + bx + c , use the formula:

max = c - (b² / 4a).

Using the formula for [tex]h = -0.8x^2 + 0.8x + 1.6[/tex] :

max h =  [tex]1.6 - ( \frac{0.8^2}{4(-0.8)} )[/tex]

       = 1.8

∴ Maximum height reached = 1.8 m.

Answer:

  a) 1.6 m

  b) -1, 2 meters

  c) 1.8 m

Step-by-step explanation:

Apparently, technology tools are allowed when solving this problem. They readily show you the x- and y-intercepts and the vertex of the graph.

The height of the ledge is the value of h when d=0. It is the constant in the given equation, and the y-intercept of the graph.

The height of the ledge is 1.6 meters.

The x-intercepts are the values of d that make h equal to zero. The graph shows them to be ...

 d = -1

  d = 2 . . . . meters

The maximum height is the "h" coordinate of the vertex (d, h).

The maximum height is 1.8 meters.

__

Additional comment

"x" is the generic independent variable. The horizontal axis of a graph is often called the "x-axis" and places where the graph crosses that axis are called "x-intercepts" even when the independent variable is named something else. Here, the independent variable is "d", not "x".

Similarly, "y" is the generic dependent variable, and the vertical axis of a graph is often called the "y-axis" even when the dependent variable is something else. This is why we refer to h when d=0 as the "y-intercept". It is the point where the graph crosses the vertical axis.

Answer:

The last answer is correct.

Step-by-step explanation:

The w stands for the number of words that she can have.  She has to have at least 150 words so the w has to be equal to 150 or greater.  The w also has to be less then or equal to 500

The ratio of the lateral surface area of cone A to the lateral surface area of cylinder B is equal to [tex]r = \frac{1}{2}\cdot \frac{\sqrt{r^{2}+h^{2}}}{h}[/tex]. (Correct choice: False)

In accordance with space geometry, the lateral areas of the cone and cylinder are described by the following equations:

Cone

[tex]A_{l} = \pi \cdot r \cdot \sqrt{r^{2}+h^{2}}[/tex]     (1)

Cylinder

[tex]A_{l} = 2\pi\cdot r\cdot h[/tex]     (2)

If we divide (2) by (1), then we have the following ratio:

[tex]r = \frac{1}{2}\cdot \frac{\sqrt{r^{2}+h^{2}}}{h}[/tex]

The ratio of the lateral surface area of cone A to the lateral surface area of cylinder B is equal to [tex]r = \frac{1}{2}\cdot \frac{\sqrt{r^{2}+h^{2}}}{h}[/tex]. (Correct choice: False)

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The correct option for the inequality is

Week 4 = Week 5. Option B. This is further explained below.

Generally,  the relationship between two expressions that do not have the same value is denoted by a sign such as, which means "not equal to," >, which means "greater than," or, which means "less than."

In conclusion, When converted to decimal form, Week 5's score comes out to -1.8, which is the same as Week 4's score.

Week 4 = Week 5

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Answer:

The Answer Is B

Step-by-step explanation:

Week 4 = Week 5

Answer:

f[g(4)] = 4

Step-by-step explanation:

Given table:

[tex]\begin{array}{| c | c | c | c | c | c |}\cline{1-6} x & -6 & -4 & 1 & 3 & 4\\\cline{1-6} f(x) & 4 & -1 & -6 & 1 & 3 \\\cline{1-6} g(x) & 1 & 4 & 3 & -4 & -6 \\\cline{1-6}\end{array}[/tex]

f[g(4)] is a composite function.

When calculating composite functions, always work from inside the brackets out.

Begin with g(4):  g(4) is the value of function g(x) when x = 4.

From inspection of the given table, g(4) = -6

Therefore, f[g(4)] = f(-6)

f(-6) is the value of function f(x) when x = -6.

From inspection of the given table, f(-6) = 4

Therefore, f[g(4)] = 4

As we can see

So

(a) The median mark is 44

(b) The number of students who scored 54 marks or less is 120

(c) The number of students who scored 60 marks or more is 20

From the question, we are to determine the median mark

The median mark for a cumulative frequency curve corresponds to the second quartile (Q₂)

Q₂ = 1/2(n+1)th mark

Where n is the total frequency

In the graph,

n = 160

∴ Q₂ = 1/2(160+1)th mark

Q₂ = 80.5th mark

The 80.5th mark is 44

∴ The median mark is 44

(b)

The number of students who scored 54 marks or less = 120 - 0

The number of students who scored 54 marks or less = 120

(c)

The number of students who scored 60 marks or more = 160 - 140

The number of students who scored 60 marks or more = 20

Hence, (a) The median mark is 44

(b) The number of students who scored 54 marks or less is 120

(c) The number of students who scored 60 marks or more is 20

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By definite integrals and area formula of a rectangle, we find that the area of the region in the first quadrant is 147 / 4 square units.

Integrals can be used to determine the area of regions bounded by curves set on Cartesian plane. The upper limit of the integral of the question is initially found:

(2 / 3) · x = 7

x = 21 / 2

The area can be defined as an rectangle in the first quadrant minus the area below the linear equation:

[tex]A = \left(\frac{21}{2} \right) \cdot (7) - \frac{2}{3} \int\limits^{\frac{21}{2} }_{0} {x} \, dx[/tex]

[tex]A = \frac{147}{2} - \frac{1}{3} \cdot \left[\left(\frac{21}{2} \right)^{2}-0^{2}\right][/tex]

A = 147/4

By definite integrals and area formula of a rectangle, we find that the area of the region in the first quadrant is 147 / 4 square units.

The statement is poorly formatted and incomplete. Correct form is shown below:

A region in the first quadrant is bounded by the line y = (2/3) · x, the y-axis and the horizontal line y = 7. Determine the area of the region.

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Answer: 768

Step-by-step explanation:

3, 6 ,12 ,24, 48, 96, 192, 384, 768

12 x 2 = 48

48 x 2 = 96

96 x 2 = 192

192 x 2 = 384

384 x 2 = 768

Answer:

[tex]\frac{4}{15}[/tex]

Step-by-step explanation:

Given the following question:

[tex]\frac{1}{5} \div\frac{3}{4}[/tex]

To find the answer we will use KCF (Keep, Change, Flip) apply it to the following question and then complete by multiplying and reducing. If needed of course.

[tex]\frac{1}{5} \div\frac{3}{4}[/tex]
Apply KCF:
[tex]\frac{1}{5} \div\frac{3}{4}=\frac{1}{5} \times\frac{4}{3}[/tex]
[tex]\frac{1}{5} \times\frac{4}{3}[/tex]
Solve:
[tex]1\times4=4[/tex]
[tex]5\times3=15[/tex]
[tex]=\frac{4}{15}[/tex]
Answer is in simplest terms.

Which means your answer is the first option or "4/15."

Hope this helps.

The equation of the graph in which the vertex (4,2) and a=-2 is

y=-2[tex](x-4)^{2}[/tex]+2.

Given that the point of the vertex is (4,2) and a=-2.

Equation is like a relationship between two or more variables expressed in equal to form. Equations of two variables look like ax+by=c. It may be a linear equation,quadratic equation, cubic equation or many more depending on the power of variable in the equation.

We know that point of vertex look like (h,k) in the equation and equation look like as under:

y=a[tex](x-h)^{2} +k[/tex].-----------1

So in order to find the equation we have to just put h=4 and k=2 and a=2 in equation 1.

y=-2[tex](x-4)^{2} +2[/tex]

Hence the equation of the graph in which the vertex (4,2) and a=-2 is

y=-2[tex](x-4)^{2}[/tex]+2.

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Based on the number of years till maturity and the required rate of return, the price of the zero-coupon bond is $852.22.

When a bond's par value is not given, assume it is $1,000.

the price of a zero coupon bond is:

= par value / (1 + rate) ^ number of years till maturity

solving gives:

= 1,000 / (1 + 3.25%)⁵

= 1,000 / 1.17341139582939453125

= $852.22

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Answer:

30000J or 30 KJ

Step-by-step explanation:

Formula for finding Gravitiational Potential Energy or GPE = mgh (or Wh since W = mg)

Given

W = 100N

h = 300m,

GPE = Wh = 100 x 300 = 30000J or 30 KJ

Answer:

That is a variable.

Step-by-step explanation:

- opposed to a constant which is known and does not change.

Transformation which can be used for mapping of one triangle to another are;

What does mapping mean in geometry?

We need to map one triangle to another ;

Hence, for mapping of one triangle to another,

The transformation which can be used from the given options are;

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Answer:

B. translation only

E. rotation then dilation

Step-by-step explanation:

B. translation only

E. rotation then dilation

a. Mean is 70.8, median is 70, and mode is 60

b. Mean describes the data best

c. Mean can be a misleading central measure in case of outliers

Calculating Mean, Median and Mode

Mean = Total sum of pries / Total number of sunglasses

Mean = (20 + 20 + 50 + 50 + 50 + 60 + 60 + 60 + 60 + 60 + 60 + 70 + 70 + 70 + 80 + 80 + 80 + 80 + 90 + 90 + 90 + 90 + 100 + 100 + 130) / 25

Mean = 70.8

Median = (n/2 + 1)th term

Here, n is the number of sunglasses.

Median = (25/2 + 1)th term

Median =13th term

Median = 70

Mode = Most occurring data

Mode = 70

Why is mean the best central measure to describe the data?

When your data distribution is continuous and symmetrical, such as when your data are normally distributed, the mean is typically the best measure of central tendency to utilize. Thus, mean describes the average price of the sunglasses here

Mean Being Misleading

In case there are outliers, the mean of the data gets highly affected. Hence, it can be misleading for analyzing average price of the data.

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Question 2

1) [tex]\overline{PR} \perp\overline{QS}, \overline{PQ} \cong \overline{PS}[/tex] (given)

2) [tex]\overline{PR} \cong \overline{PR}[/tex] (reflexive property)

3) [tex]\angle PRQ, \angle PRS[/tex] are right angles (perpendicular lines form right angles)

4) [tex]\triangle PRQ, \triangle PRS[/tex] are right triangles (a triangle with a right angle is a right triangle)

5) [tex]\triangle PRQ \cong \triangle PRS[/tex] (HL)

Question 3

1) [tex]\angle C[/tex] is a right angle, [tex]\overline{AC} \cong \overline{AE}, \overline{DE} \perp \overline{AB}[/tex] (given)

2) [tex]\angle DEA[/tex] is a right angle (perpendicular lines form right angles)

3) [tex]\triangle ACD, \triangle DAE[/tex] are right triangles (a triangle with a right triangle is a right angle)

4) [tex]\overline{AD} \cong \overline{AD}[/tex] (reflexive property)

5) [tex]\triangle ACD \cong \triangle AED[/tex] (HL)

6) [tex]\angle CAD \cong \angle DAE[/tex] (CPCTC)

7) [tex]\overline{AD}[/tex] bisects [tex]\angle BAC[/tex] (if a segment splits an angle into two congruent parts, it is an angle bisector)