select the quadratic function with a graph that has the following features

x-intercept at (8,0)
y-intercept at (0,-32)
maximum value at (6,4)
axis of symmetry at x=6

A. f() = 1/2 x2 +6 - 32
B.f() = -x2 + 12x -32
C.f() = 1/2 x2 + 6x -16
D f() = -x2 + 12x -6​

Answer:

Step-by-step explanation:

15/25. Divide both sides by a common number which is 5 3/5 is the final answer.

The result of the given mathematical expression is [tex]\frac{3}{5}[/tex] or 0.6.

"A mathematical expression is a finite combination of symbols that is well-formed according to rules that depend on the context."

Given mathematical expression is

= (15 ÷ 25)

[tex]= \frac{15}{25}\\= \frac{3}{5}\\= 0.6[/tex]

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Answer: y= -3/2x

Explanation:

3x+ 2y- 6

→ 2y= -3x+ 6

→ y= -3/2x+ 3

Parallel to y= ax+ b is to have the same slope (ax) and have a different y- intercept.

For this equation, y= -3/2x + 3, it is y= -3/2x because the line has to pass through point (0,0).

Answer:

(n+9) (2p+q)

Step-by-step explanation:

2p(n + 9) + q(n + 9) =

Factor out the term (n+9)

(n+9) (2p+q)

This is completely factor

Answer:

Option C and option D

Step-by-step explanation:

we have that

[tex]y=x^{2}-4x+5[/tex]

This is the equation of a vertical parabola open upward

The vertex is a minimum

Convert the equation into vertex form

Complete the square

[tex]y-5=x^{2}-4x[/tex]

[tex]y-5+4=x^{2}-4x+4[/tex]

[tex]y-1=x^{2}-4x+4[/tex]

[tex]y-1=(x-2)^{2}[/tex]

[tex]y=(x-2)^{2}+1[/tex] ----> equation of the parabola in vertex form

The vertex is the point (2,1)

therefore

when the bird is 2 feet away from the first fence post, it reaches its minimum height of 1 foot

Answer: C and D

Step-by-step explanation:

Answer: C

Step-by-step explanation:

Binomial have only two terms.

A) 3x²-6x²-+x has three terms

B) 6x³-6x²+x-1 has four terms

C) 3x³-6x has two terms

D) 6x³ has only one term

Answer:

3/50 or 0.06

Step-by-step explanation:

6/100=3/50

6/100=0.06

Answer:

0.06

Step-by-step explanation:

Answer:

x=5

Step-by-step explanation:

Factoring  x2 - 2x – 15 we get

(x+3)(x-5) so x+3=0 and x-5=0 or x=-3 and x=5

Answer:

x=5

Step-by-step explanation:

x^2 - 2x – 15 = 0

Factor

What 2 numbers multiply to -15 and add to -2

-5*3 = -15

-5+3 = -2

(x-5) (x+3) = 0

Using the zero product property

(x-5) =0   x+3 =0

x-5+5 =0+5      x+3-3 = 0-3

x=5                  x=-3

Answer:  C

Step-by-step explanation:

20 people have a dog: 8 are yes dog/yes cat & 12 are yes dog/no cat

17 people have no dog: 10 are no dog/yes cat & 7 are no dog/no cat

[tex]\begin{array}{c|c|c|c}\underline{\qquad \qquad}&\underline{Cat}&\underline{No\ Cat}&\underline{TOTAL}\\Dog&8&12&20\\\underline{\ No\ Dog\ }&\underline{\ 10\ }&\underline{\quad 7\quad }&\underline{\quad 17\quad }\\TOTAL&18&19&37\end{array}[/tex]

Answer:

2(d - 3) is the equation. You cannot solve for d. You can only simplify it

Answer:

2

Step-by-step explanation:

The function is given as  [tex]f(x)=\frac{3}{(x-11)(x+4)}[/tex]

Vertical asymptotes occur when the denominator is set to 0.

Thus,

(x-11)(x+4) = 0

x = 11 or x = -4

Hence, there are 2 vertical asymptotes

Answer: 2

Step-by-step explanation:

A P E X

Step-by-step explanation:

(1)

Let the numerator be x and denominator be y. A/Q x + y = 4 + 2x → - x + y = 4

(2)

multiplying each term by 2, 2x-2y= -8

(3)

Also, (x+3) / (y+3) = 2 / 3 → 3x - 2y = -3

Subtracting (2) from (3) → x = 5 and by putting this in (1) we can get y=9. Hence, the fraction is 5 / 9

Answer:

[tex]\frac{5}{9}[/tex]

Step-by-step explanation:

let the fraction be [tex]\frac{x}{y}[/tex], then

x + y = 2x + 4 ( subtract x from both sides )

y = x + 4 → (1)

If 3 is added to numerator and denominator, then

[tex]\frac{x+3}{y+3}[/tex] = [tex]\frac{2}{3}[/tex] ( cross- multiply )

3(x + 3) = 2(y + 3) ← distribute both sides

3x + 9 = 2y + 6 ← substitute y = x + 4

3x + 9 = 2(x + 4) + 6

3x + 9 = 2x + 8 + 6 = 2x + 14 ( subtract 2x from both sides )

x + 9 = 14 ( subtract 9 from both sides )

x = 5

Substitute x = 5 into (1)

y = 5 + 4 = 9

Hence the original fraction is [tex]\frac{5}{9}[/tex]

Answer:

[tex](0, \infty)[/tex]

Step-by-step explanation:

By definition all the exponential functions of the form

[tex]f (x) = a (b) ^ x[/tex]

Where a is the main coefficient and b is the base they have range [tex](0, \infty)[/tex]

Whenever [tex]a> 0[/tex] and b> 0.

In this case the function is:

[tex]f (x) = 11(\frac{1}{3}) ^ x[/tex]

Note that for this function [tex]a = 11> 0[/tex] and [tex]b =\frac{1}{3}>0[/tex]

Therefor the range is: [tex](0, \infty)[/tex]

Answer:

The answer in the procedure

Step-by-step explanation:

we know that

The cross  section formed by the slice is a square with the same dimensions of the base of rectangular prism

The length side of the square is 6 cm

The area of the cross section is equal to

[tex]A=b^{2}[/tex]

[tex]b=6\ cm[/tex]

substitute

[tex]A=6^{2}[/tex]

[tex]A=36\ cm^{2}[/tex]

Answer:

square and 36

Step-by-step explanation:

I took the test

Answer:b

Step-by-step explanation: you graph each answer choice and see which one looks like the graph

Answer:  The correct option is

(B) [tex]y=\left(\dfrac{1}{2}\right)^{x-3}+1.[/tex]

Step-by-step explanation:  We are given to select the function that is shown in the graph.

From the graph, we know that

if the function is represented by y = f(x), then f(0) = 9. That is, the value of y at x = 0 is 9.

Option (A) :  Here, the given function is

[tex]y=\left(\dfrac{1}{2}\right)^{x+3}-1.[/tex]

So, at x = 0, the value of y is given by

[tex]y(x=0)=\left(\dfrac{1}{2}\right)^{0+3}-1=\dfrac{1}{8}-1=-\dfrac{7}{8}\neq 9.[/tex]

So, this option is not correct.

Option (B) :  Here, the given function is

[tex]y=\left(\dfrac{1}{2}\right)^{x-3}+1.[/tex]

So, at x = 0, the value of y is given by

[tex]y(x=0)=\left(\dfrac{1}{2}\right)^{0-3}+1=\left(\dfrac{1}{8}\right)^{-1}+1=8+1=9.[/tex]

So, this option is CORRECT.

Option (C) :  Here, the given function is

[tex]y=\left(\dfrac{1}{2}\right)^{x-1}+3.[/tex]

So, at x = 0, the value of y is given by

[tex]y(x=0)=\left(\dfrac{1}{2}\right)^{0-1}+3=2+3=5\neq 9.[/tex]

So, this option is not correct.

Option (D) :  Here, the given function is

[tex]y=\left(\dfrac{1}{2}\right)^{x+1}-3.[/tex]

So, at x = 0, the value of y is given by

[tex]y(x=0)=\left(\dfrac{1}{2}\right)^{0+1}-3=\dfrac{1}{2}-3=-\dfrac{5}{2}\neq 9.[/tex]

So, this option is not correct.

Thus, (B) is the correct option.

Answer:

19

Step-by-step explanation:

Plug in x = 3 into the function

f(x)=2x^2+1

f(x)=2(3)^2+1

f(x)=18+1

f(x)=19

The answer is:

f(3) = 19

Work/explanation:

To evaluate this function, plug in 3 for x:

[tex]\sf{f(x)=2x^2+1}[/tex]

[tex]\sf{f(3)=2(3)^2+1}[/tex]

[tex]\sf{f(3)=2\times9+1}[/tex]

Then, according to PEMDAS, we multiply:

[tex]\sf{f(3)=18+1}[/tex]

[tex]\sf{f(3)=19}[/tex]

Therefore, when x = 3, the function evaluates to 19.

Answer:

The maximum speed of Hal's airplane in still air is:

[tex]v= 250\ miles/h[/tex]

The speed of the wind

[tex]c = 50\ miles/h[/tex]

Step-by-step explanation:

Remember that the velocity v equals the distance d between time t.

[tex]v=\frac{d}{t}[/tex]  and [tex]t*v=d[/tex]

The distance that Hal travels when traveling with the wind is:

[tex](2\ hours)(v + c) = 600[/tex] miles

Where v is the speed of Hal and c is the wind speed.

The distance when traveling against the wind is:

[tex](3\ hours)(v-c) = 600[/tex] miles

Now we solve the first equation for v

[tex](2)(v + c) = 600[/tex]

[tex]2v + 2c = 600[/tex]

[tex]2v= 600-2c[/tex]

[tex]v= 300-c[/tex]

Now we substitute the value of v in the second equation and solve for c

[tex]3((300-c)-c) = 600[/tex]

[tex]3(300-2c) = 600[/tex]

[tex]900-6c = 600[/tex]

[tex]-6c = 600-900[/tex]

[tex]-6c = -300[/tex]

[tex]6c = 300[/tex]

[tex]c = 50\ miles/h[/tex]

Then:

[tex]v= 300-(50)[/tex]

[tex]v= 250\ miles/h[/tex]

The maximum speed of Hal's airplane in still air is:

[tex]v= 250\ miles/h[/tex]

The speed of the wind

[tex]c = 50\ miles/h[/tex]

Answer:all real number greater than 0

Step-by-step explanation:

Firstly if you input any number equal to 0 or less than 0 you will not find the defined range...

You cant use o or any negetive number as domain in the term of log or ln type math..

But if u put any value more than 0 you can find all real number as range

Such as, log(0.001)=-3

log(1)=0

log(120)=2.07

So the domain is all real number above o...but the range is all real number including 0 and negetive number..

The domain of y=logx is all real numbers greater than 0.

Hence, option b is the correct answer.

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We can either convert to standard form first or convert to a common multiplier, kinda like a common denominator.  The latter makes more sense if they wanted the result in scientific notation, but let's do it that way anyway.

a)

4.5 × 10⁴ + 3.8  × 10³ = 45 × 10³ + 3.8  × 10³ = 48.8× 10³ = 48,800

Answer:  48,800

b)

4.5 × 10⁴ - 3.8  × 10³ = 45 × 10³ - 3.8  × 10³ = 41.2× 10³ = 41,200

Answer:  41,200

c)

7.2 × 10⁻³ + 6.3  × 10⁻² = 7.2 × 10⁻³ + 63  × 10⁻³ = 70.2 × 10⁻³

       = 7.02 × 10⁻² = 0.0702

Answer: 0.0702

Answer:

46°

Step-by-step explanation:

Alternate angles from 46° and alternate angles are equal

Answer:

C. 6x - 7y = 21

Step-by-step explanation:

y=2/3 x - 3, y-intercept = -3

A. 2/3 x + 3y = -3

   3y = -2/3 x - 3

     y = -2x - 1; y-intercept = -1

B.-2/3 x + 3y = 6

   3y = 2/3 x + 6

     y = 2x + 2; y-intercept = 2

C. 6x-7y=21

    7y = 6x - 21

     y = 6/7 x - 3; y-intercept = -3

D. x+4y = 12

   4y = -x +12

     y = -1/4 + 3; y-intercept = 3

Answer is C. 6x - 7y = 21

Answer:

The roots are {-3, -2, 2, 3}.

Step-by-step explanation:

The trick here is to represent y² by some other letter, such as x.  If we do that, then y^4-13y^2+36=0 becomes x² - 13x + 36 = 0.

Recognize that the factors -4 and -9 of 36 sum up to 13.  Thus,

x² - 13x + 36 = 0 is equivalent to (x - 4)(x - 9) = 0, and x = 4 or x = 9.

Recall that x = y².

When x = 4:  y² = 4, and so y = ±2.

When x = 9, y² = 9, and so y = ±3.

The roots are {-3, -2, 2, 3}.

The roots of the bi-quadratic equation y⁴ - 13y² + 36 = 0 are,

- 2, 2, -3, 3.

A polynomial is an algebraic expression.

A polynomial of degree n in variable x can be written as,

a₀xⁿ + a₁xⁿ⁻¹ + a₂xⁿ⁻² +...+ aₙ.

Given, y⁴ - 13y² + 36 = 0.

This a bi-quadratic polynomial.

Let, x = y².

Therefore,

x² - 13x + 36 = 0.

x² - 4x - 9x + 36 = 0.

x(x - 4) - 9(x - 4) = 0.

(x - 4)(x - 9) = 0.

x = 4 Or x = 9.

Now,

For x = 4 ⇒ y² = 4 ⇒ y = ± 2.

For x = 9 ⇒ y² = 9 ⇒ y = ± 3.

So, The roots of the biquadratic equation are - 2, 2, -3, 3.

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Answer:

In graph

Step-by-step explanation:

First put it in Slope -Intercept form.

then graph the y- intercept first.

After that it's rise over run so think 1/1

up 1 over one.

Answer:

CCCCCCCCCCCCC

Step-by-step explanation:

Answer:

In general, quadratic equations have two x-intercepts. But sometimes it happens that a quadratic eqution has one x-intercept or no interepts. That's why we should fully analyze this equation:

Given the following equation: y=3x^2 + 7 + m

If y=0, then:

3x^2 + 7 + m = 0 ⇒ x^2 = (-m-7)/3

Then [tex]x =[/tex]± [tex]\sqrt{\frac{-m-7}{3}}[/tex]

Given that we can take the square root of a negative number, the only way this equation has two x-intercepts is if m<-7.

Summarizing:

The equation: y=3x^2 + 7 + m has two x-intercepts only if m is less than -7. If m equals -7, the equation has only one x-intercept, and finally, if m is greater than -7, the equation has NO x-intercepts.

Answer:

Answer is 11+6i

Step-by-step explanation:

You just have to add imaginary part together and the real part. The answer will be 11+6i

Answer: 11+6i

Step-by-step explanation:

A relation is a function if you associate exactly one output for every input. This means that, when you choose a value for x, there must be only one correspondent value for y. This only happens in the top-right parabola.

Answer:

a) y = sin x

b) y = csc x

c) y = cot x

Step-by-step explanation:

only d is even

Answer:

After these reflections, the coordinates of P′ will be (4 , -4)

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

* Lets solve the problem

- The triangle PQR with coordinates P(-4, -4), Q(-1, -3), and R(-3, -1)

- The triangle is reflected across the x-axis

∵ Δ PQR is reflected across the x-axis

∴ All y-coordinates of the vertices P, Q , R reversed their signs

∴ The new points will be (-4 , 4) , (-1 , 3) , (-3 , 1)

- The new vertices will reflected across the y-axis

∴ All x-coordinates of the new vertices reversed their signs

∴ The new points will be (4 , 4) , (1 , 3) , (3 , 1)

- The new vertices will reflected across the x-axis to form Δ P'Q'R'

∴ All y-coordinates of the new vertices reversed their signs

∴ P' = (4 , -4) , Q' = (1 , -3) , R' = (3 , -1)

* After these reflections, the coordinates of P′ will be (4 , -4)

Final answer:

After reflecting triangle PQR across the x-axis, then the y-axis, and the x-axis again, point P' will have coordinates (4, -4).

Explanation:

Reflecting a triangle across an axis involves flipping the triangle over that axis. Each reflection inverses the corresponding coordinate (x or y) of each vertex of the triangle, while the other coordinate remains the same. Starting with the first reflection across the x-axis, the y-coordinate of each point negates, but the x-coordinate remains unchanged. The second reflection is across the y-axis, which negates the x-coordinate and keeps the y-coordinate (already negated from the first reflection) the same. The third reflection across the x-axis negates the y-coordinate again, effectively returning it to its original value before the first reflection. So for point P(-4, -4), after these reflections, the new coordinates for P' will be (4, -4).

Answer:

C

Step-by-step explanation:

Use exponents property:

[tex]\dfrac{x^a}{x^b}=x^{a-b}[/tex]

1. Note that

[tex]\dfrac{x^7}{x^{11}}=x^{7-11}=x^{-4}=\dfrac{1}{x^4}[/tex]

and

[tex]\dfrac{y^6}{y^8}=y^{6-8}=y^{-2}=\dfrac{1}{y^2}[/tex]

2. Now

[tex]\sqrt{\dfrac{55x^7y^6}{11x^{11}y^8}}=\sqrt{\dfrac{5}{x^4y^2}}=\dfrac{\sqrt{5}}{\sqrt{x^4}\sqrt{y^2}}=\dfrac{\sqrt{5}}{x^2y}[/tex]

because [tex]x>0,\ y>0[/tex]

Answer:

= [tex]3ab^{\frac{2}{3} }[/tex]

Step-by-step explanation:

∛(27 a³ [tex]b^{7}[/tex])

= ∛27 · ∛a³ · ∛b³ · ∛b³ · ∛b

= 3 · a · b · b ∛b

= 3ab² [tex]b^{\frac{1}{3} }[/tex]

or

= [tex]3ab^{\frac{2}{3} }[/tex]