At school, the teacher surveyed the color preferences of 21 students. Out of every
7 students surveyed, 4 said they liked blue the best. How many students said their favorite
color was blue?

The equation of the function in vertex form exists f(x) = -2·(x - 8)² + 6.

The given values exist

x,      f(x)

6,      -2

7,       4

8,       6

9,       4

10,      -2

The equation of the function in vertex form exists

f(x) = a(x - h)² + k

To estimate the values of a, h, and k,

When x = 6, f(x) = -2 then

-2 = a(6 - h)² + k

= (h²-12·h+36)·a + k.............(1)

When x = 7, f(x) = 4 then

4 = a( 7- h)² + k

= (h²-14·h+49)·a + k...........(2)

When x = 8, f(x) = 6

6 = a( 8- h)² + k ...........(3)

When x = 9, f(x) = 4.

4 = a( 9- h)² + k ..........(4)

When x = 10, f(x) = -2

-2 = a(10- h)² + k ...........(5)

Subtract equation (1) from (2)

4 - 2 = a( 7- h)² + k - (a(6 - h)² + k )

= 13·a - 2·a·h........(6)

Subtract equation (4) from (2)

a (9 - h)² + k - a( 7- h)² + k

32a -4ah = 0

Simplifying the equation, we get

4h = 32

h = 32/4 = 8

From equation (6) we have;

13·a - 2·a·8 = 6

-3a = 6

a = -2

From equation (1), we have;

-2 = -2 × ( 10- 8)² + k

-2 = -8 + k

k = 6

The value of k = 6.

The equation of the function in vertex form exists f(x) = -2·(x - 8)² + 6.

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The athlete that has the shortest training ride is athlete A.

The athlete that has the shorter range is Athlete B.

The athlete that has the greater median is Athlete B.

The athlete that has the greater IQR is  athlete A.

A box plot is used to study the distribution and level of a set of scores. The box plot consists of two lines and a box. the two lines are known as whiskers. The first whisker represents the minimum number and the second whisker represents the maximum number. The difference between the whiskers is known as range.

Range of Athlete A = 36 - 13 = 13

Range of Athlete B = 30 - 19 = 11

Training ride of  Athlete A = 16

Training ride of  Athlete B = 21

On the box, the first line to the left represents the lower (first) quartile. The next line on the box represents the median. The third line on the box represents the upper (third) quartile.

Median of  Athlete A = 20

Median of  Athlete B = 25

IQR of  Athlete A = 30 - 16 = 14

IQR of  Athlete B = 28 - 21 = 7

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The equation for the parabola with a focus at (-4,8) and a directrix at x=-6 is ( y - 8)² = 4( x +5 ).

Given that x = -6, it means the directrix is horizontal.

Equation of parabola that opens left or write is expressed as;

( y - k )² = 4p( x - h )

Given that;

First, we find the vertex.

The vertex( h, k ) is halfway between the directrix and the focus.

Hence, x coordinate of the vertex is;

x = ( x coordinate of focus + directrix ) / 2

x = ( -4 + (-6) ) / 2

x = -10/2

x = -5

y coordinate of the vertex will be the same as that of the focus.
y = 8

Hence vertex will be;

( -5, 8 )

Next, we find the distance p from the focus to the vertex and from the vertex to the directrix.

Subtract x coordinate of vertex from x coordinate of focus.

p = -4 - ( -5 )

p = -4 + 5

p = 1

Now, we substitute these values into the equation.

( y - k )² = 4p( x - h )

( y - (8) )² = 4(1)( x - (-5) )

( y - 8)² = 4( x +5 )

Therefore, the equation for the parabola with a focus at (-4,8) and a directrix at x=-6 is ( y - 8)² = 4( x +5 ).

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Answers

The ratio x:y:z is 10:5:3  

Ali got 20,000

===================================================

Explanation:

Let

Those amounts add to 36,000

x+y+z = 36,000

We're told that x = 2y, so let's replace x with 2y in the equation above to get these steps

x+y+z = 36,000

2y+y+z = 36,000

3y+z = 36,000

We're also told that 3y = 5z, so we can replace that previously mentioned 3y with 5z to end up with one variable, in which we can solve for like shown below.

3y+z = 36,000

5z+z = 36,000

6z = 36,000

z = (36,000)/6

z = 6,000

This tells us that Chris got 6,000

Use this to find y

3y = 5z

3y = 5*6,000

3y = 30,000

y = (30,000)/3

y = 10,000

Ben received 10,000

Lastly, determine x.

x = 2y

x = 2*10,000

x = 20,000 which is the amount Ali got.

The ratio x:y:z turns into 20,000:10,000:6,000 which fully simplifies to 10:5:3 after dividing all three parts by the GCF 2000. It might be tempting to try to reduce the 10:5 portion into 2:1, but we cannot divide by 5 since it's not a factor of that 3 at the end.

-------------------------

Check:

Ali + Ben + Chris = 20,000 + 10,000 + 6,000 = 36,000

which helps confirm the answer.

Answer:

17.1

Step-by-step explanation:

Plug in -1 for x, this gets you -4.

Solve 2^-4 and get .0625.

Add .0625 to 17 and get 17.0625.

The question says answer to the nearest tenth so you round the 6 up to a 1 giving you 17.1

[tex](\stackrel{x_1}{10}~,~\stackrel{y_1}{7})\hspace{10em} \stackrel{slope}{m} ~=~ -\cfrac{2}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{7}=\stackrel{m}{-\cfrac{2}{5}}(x-\stackrel{x_1}{10})[/tex]

Answer:5

Step-by-step explanation:

321 123

Answer:

1241

Step-by-step explanation:

∴

L.C.M. of 28, 36 and 45 = 2 × 2 × 3 × 3 × 5 × 7 = 1260

∴

the required number is 1260 - 19 = 1241

Hence, if we add 19 to 1241 we will get 1260 which is exactly divisible by 28, 36 and 45.

The quadrilaterals that describe DEFG include parallelogram, rhombus, and square

It should be noted that a quadrilateral simply means a close shape and a polygon that has four sides.

In this case, the quadrilaterals that describe DEFG include parallelogram, rhombus, and square. In a parallelogram, it should be noted that the sides are parallel to each other.

Therefore, the quadrilaterals that describe DEFG include parallelogram, rhombus, and square.

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The y-intercept of the resulting polynomial is equal to - 3.

Herein we assume that the other two zeros of the polynomial are i and - i 3, otherwise the polynomial will have at least a complex number as coefficient of the expression. This is because we need to find values from a Cartesian plan, whose ordered pairs are real numbers.

Initially, we have the following expression by algebra properties:

f(x) = (x + i) · (x - i) · (x + i 3) · (x - i 3)

f(x) = (x² + 1) · (x² + 9)

f(x) = x⁴ + 10 · x² + 9

Then, we proceed to use the two rigid transformations described in the statement. Please notice that rigid transformations are transformations applied on polynomials such that Euclidean distance is conserved:

Vertical compression

f'(x) = (1 / 3) · f(x)

f'(x) = (1 / 3) · (x⁴ + 10 · x² + 9)

f'(x) = (1 / 3) · x⁴ + (10 / 3) · x² + 3

Reflection across the x-axis

g(x) = - f'(x)

g(x) = - (1 / 3) · x⁴ - (10 / 3) · x² - 3

The y-intercept is found for x = 0:

g(0) =  - (1 / 3) · 0⁴ - (10 / 3) · 0² - 3

g(0) = - 3

The y-intercept of the resulting polynomial is equal to - 3.

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Answer:

[tex]\displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=-4 \ln |x|+5 \ln|x-1|+\dfrac{3}{x-1}+\text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given integral:

[tex]\displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x[/tex]

Factor the denominator:

[tex]\begin{aligned}\implies x^3-2x^2+x & = x(x^2-2x+1)\\& = x(x^2-x-x+1)\\& = x(x(x-1)-1(x-1))\\ & = x((x-1)(x-1))\\& = x(x-1)^2\end{aligned}[/tex]

[tex]\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int \dfrac{x^2-4}{x(x-1)^2}\:\:\text{d}x[/tex]

Take partial fractions of the given fraction by writing out the fraction as an identity:

[tex]\begin{aligned}\dfrac{x^2-4}{x(x-1)^2} & \equiv \dfrac{A}{x}+\dfrac{B}{(x-1)}+\dfrac{C}{(x-1)^2}\\\\ \implies \dfrac{x^2-4}{x(x-1)^2} & \equiv \dfrac{A(x-1)^2}{x(x-1)^2}+\dfrac{Bx(x-1)}{x(x-1)^2}+\dfrac{Cx}{x(x-1)^2}\\\\ \implies x^2-4 & \equiv A(x-1)^2+Bx(x-1)+Cx \end{aligned}[/tex]

Calculate the values of A and C using substitution:

[tex]\textsf{when }x=0 \implies -4=A(1)+B(0)+C(0) \implies A=-4[/tex]

[tex]\textsf{when }x=1 \implies -3=A(0)+B(0)+C(1) \implies C=-3[/tex]

Therefore:

[tex]\begin{aligned}\implies x^2-4 & \equiv -4(x-1)^2+Bx(x-1)-3x\\& \equiv -4(x^2-2x+1)+B(x^2-x)-3x\\& \equiv -4x^2+8x-4+Bx^2-Bx-3x\\& \equiv (B-4)x^2+(5-B)x-4\\\end{aligned}[/tex]

Compare constants to find B:

[tex]\implies 1=B-4 \implies B=5[/tex]

Substitute the found values of A, B and C:

[tex]\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-\dfrac{3}{(x-1)^2}\:\:\text{d}x[/tex]

[tex]\textsf{Apply exponent rule} \quad \dfrac{1}{a^n}=a^{-n}[/tex]

[tex]\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x=\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-3(x-1)^{-2}}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x$\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Integrating} $\dfrac{1}{x}$\\\\$\displaystyle \int \dfrac{1}{x}\:\text{d}x=\ln |x|+\text{C}$\end{minipage}}[/tex]

[tex]\boxed{\begin{minipage}{4 cm}\underline{Integrating $ax^n$}\\\\$\displaystyle \int ax^n\:\text{d}x=\dfrac{ax^{n+1}}{n+1}+\text{C}$\end{minipage}}[/tex]

[tex]\begin{aligned}\implies \displaystyle \int \dfrac{x^2-4}{x^3-2x^2+x}\:\:\text{d}x & =\displaystyle \int -\dfrac{4}{x}+\dfrac{5}{(x-1)}-3(x-1)^{-2}}\:\:\text{d}x\\\\& = \displaystyle -4\int \dfrac{1}{x}\:\:\text{d}x+5\int \dfrac{1}{(x-1)}\:\:\text{d}x-3 \int (x-1)^{-2}}\:\:\text{d}x\\\\& = \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int (x-1)^{-2}}\:\:\text{d}x\end{aligned}[/tex]

Use Integration by Substitution:

[tex]\textsf{Let }u=(x-1) \implies \dfrac{\text{d}u}{\text{d}x}=1 \implies \text{d}x=\text{d}u}[/tex]

Therefore:

[tex]\implies \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int (x-1)^{-2}}\:\:\text{d}x[/tex]

[tex]\implies \displaystyle -4 \ln |x|+5 \ln|x-1|-3 \int u^{-2}}\:\:\text{d}u[/tex]

[tex]\implies -4 \ln |x|+5 \ln|x-1|-\dfrac{3}{-1}u^{-2+1}+\text{C}[/tex]

[tex]\implies -4 \ln |x|+5 \ln|x-1|+3u^{-1}+\text{C}[/tex]

[tex]\implies -4 \ln |x|+5 \ln|x-1|+\dfrac{3}{u}+\text{C}[/tex]

Substitute back in u = (x - 1):

[tex]\implies -4 \ln |x|+5 \ln|x-1|+\dfrac{3}{x-1}+\text{C}[/tex]

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The expression that represents the height of the oblique prism is: 1/2x.

What is the Volume of a Prism?

Prism Volume = base area × height of the prism

Given that the oblique prism has:

Volume = 1/2x³ cubic units

edge length = x units

Therefore,

base area = x²

Thus:

1/2x³ = (x²)(height)

Divide both sides by x²

1/2x³ ÷ x² = height

1/2x³ × 1/x² = height

x³/2 × 1/x² = height

height = x³/2x²

height = x/2

Therefore, the expression that represents the height of the oblique prism is: 1/2x.

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The complete question is -

An oblique prism with a square base of edge length x units has a volume of x3 cubic units. Which expression represents the height of the prism?

The the /ta/ sound is the sound which listeners are likely to report hearing. The option B is correct option.

According to the statement

we have given that the The voice onset time (vot) for the sound /da/ is 17 ms, and the vot for the sound /ta/ is 91 msec. and we have to find the sound which listeners are likely to report hearing.

So, For this purpose, we know that the

Traffic announcement (TA) refers to the broadcasting of a specific type of traffic report on the Radio Data System. It is generally used by motorists, to assist with route planning, and for the avoidance of traffic congestion.

And according to this and the voice onset time is

The brief instant that elapses between the initial movement of the speech organs as one begins to articulate a voiced speech sound and the vibration of the vocal cord.

And according to this

the /ta/ sound is the sound which is the like by the listeners are likely to report hearing.

So, The the /ta/ sound is the sound which listeners are likely to report hearing. The option B is correct option.

Disclaimer: This question was incomplete. Please find the full content below.

Question:

The voice onset time (VOT) for the sound /da/ is 17 ms, and the VOT for the sound /ta/ is 91 msec. When a computer produces a sound with a VOT of 65 msec, listeners are likely to report hearing

a. the /da/ sound

b. the /ta/ sound

c. the /ja/ sound

d. a combination of /ta/ and /da/

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Area of the base = 60 inches²

Height of the prism = 4 inches

perimeter of the base = 34 inches

Surface area of the rectangular prism =  256 inches²

The surface area of a rectangular prism can be calculated as follows;

area of the base = B = 12 × 5 = 60 inches²

Height of the prism = 4 inches

Perimeter of the base is the sum of the whole side of the base of the rectangular prism.

Therefore,

perimeter of the base = 12 + 12 + 5 + 5 = 34 inches

Surface area of the rectangular prism = 2B + ph

where

Therefore,

Surface area of the rectangular prism = 2(60) + 34(4)

Surface area of the rectangular prism = 120 + 136

Surface area of the rectangular prism =  256 inches²

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The magnitude of the resulting vector, u - v, is approximately 5.83

and its angle of direction is approximately 59.04°.

We want to subtract vector v from vector u.

We are given;

v = <2, -3> = 2i - 3j

u = <5, 2> = 5i + 2j

u - v = 5i + 2j - (2i - 3j)

= 5i + 2j - 2i + 3j

= 3i + 5j

Resultant vector = √(3² + 5²)

Resultant vector = √34 ≈ 5.83

Angle of direction of resultant vector is;

tan θ = (5/3)

θ = tan⁻¹(5/3)

θ = 59.04°

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The simplified equivalent expression exists -1.9b + 7.8.

An equivalent expression exists an expression that contain the same value or worth as another expression, but does not look the same.

Combine the coefficient of the term b, we get

1.3b + 7.8–3.2b

= (1.3−3.2)⋅b + 7.8

simplifying the equation, we get

= (- 1.9) +b + 7.8

= −1.9b + 7.8

The simplified equivalent expression exists -1.9b + 7.8.

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Answer:

Step-by-step explanation:

6:1 i think

Answer:

Option (2)

Step-by-step explanation:

[tex]\mathbf{u} \cdot \mathbf{v}=(-5)(-15)+(3)(-25)=0[/tex]

So, the vectors are orthogonal.

The amount of cash needed to retire baldwin's bond early based on the brokerage will be $5427896.

It should be noted that from the information given, we are to calculate the amount of cash needed to retire baldwin's bond early based on the brokerage.

This will be:

= 6000 + (5756951 × 94.18/100)

= 6000 + 5421896.45

= 5427896

Therefore, the amount of cash needed to retire baldwin's bond early based on the brokerage will be $5427896.

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The number of classes Anna can take so the total cost for the month will be the same is 5.

The monthly cost would be  $37.50.

When the monthly cost is equal, both equations would be equal: 7.5x = 5.5x + 10

In order to determine the value of x, take the following steps:

Monthly cost when the cost is the same: 7.5 x 5 = $37.50

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Answer:

5 classes

37.50

Step-by-step explanation:

got it right on edge

Answer:

x = 1; y = 2

Step-by-step explanation:

y = 3x − 1

y = -4x + 6

In the first equation, y = 3x - 1.

Substitute 3x - 1 for y in the second equation.

3x - 1 = -4x + 6

7x = 7

x = 1

Substitute 1 for x in the first original equation.

y = 3x - 1

y = 3(1) - 1

y = 3 - 1

y = 2

Solution: x = 1; y = 2

The explicit function is (b) f(n) = -4(n - 1) + 3; the seventh term is -21

The given parameters are:

The explicit form is for an arithmetic function.

An arithmetic function has an explicit form represented as:

f(n) = a + (n -1) * d

Substitute the values of n and a in the above equation

f(n) = 3 + (n -1) * -4

Evaluate the product i.e. multiply -4 by (n - 1)

f(n) = 3 -4(n - 1)

Rewrite the equation as

f(n) = -4(n - 1) + 3

The seventh term is then calculated as:

f(n) = -4(n - 1) + 3

Substitute 7 for n in the above equation

f(7) = -4(7 - 1) + 3

Evaluate

f(7) = -21

Hence, the explicit function is (b) f(n) = -4(n - 1) + 3; the seventh term is -21

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The total amount Aisha would pay to paint her room is $54.

The first step is to determine the area of the walls. The walls have the shape of a rectangle. Thus, the formula for the area of a rectangle would be used to determine the area of the wall.

Area of the three walls without a door : 3 x (length x width)

3 x (2.4 x 5.2) = 37.44 m²

Area of the fourth wall with a door : area of the wall - area of the door

(2.4 x 5.2) - (2 x 0.8) = 10.88m²

Total area of the walls : area of the three walls + area of the wall with the door

10.88m² + 37.44 m² = 48.32 m²

The next step is to determine the number of tins that would be needed:

48.32 /12 = 4.03

5 tins would needed

Cost of the 5 tins: (4 x 12) + (0.5 x 12) = $54

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Answer:

That's a reflection in the x axis.

Step-by-step explanation:

Triangle C is the mirror image  of B with the x axis acting as a mirror.

By using properties of right triangles and trigonometric functions, we conclude that the length of the side BC is 19 units.

In this problem we can use trigonometric functions to determine the length of the side BC in three steps:

Step 1

BE = AB/tan 60°

BE = 19√2 /4

Step 2

BD = BE/cos 45°

BD = 19/2

Step 3

BC = BD/sin 30°

BC = 19

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y=5(14-x)(x) quadratic equation best models the volume of the box .

What is volume of cuboid?

Perimeter = 2(width + length) =28

Let x=width of base, then length of base = (28/2)-x = 14-x

Volume of box = length*width*height

=x(14-x)*5  ⇒ 5x(14-x)(x)

Therefore, y=5(14-x)(x) quadratic equation best models the volume of the box .

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The bags of candy that was sold is 6

The bags of cookies that was sold is 10.

$8.50c + $4.50o = 79 equation 1

o - c = 6

o = 6 + c  equation 2

Where:

Substitute for o in equation 1 using equation 2

$8.50c + $4.50(6 + c) = 79

$8.50c + 27 + 4.50c = 79

$8.50c + 4.50c = 79 - 27

$13c = 52

c = 52 / 13

c = 4

Substitute for c in equation 2

0 = 6 + 4 = 10

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