Solve for x. x/2≥−4 <br> a. x≥−2 <br> b. x≤−2 <br> c. x≥−8 <br> d. x≤−8

Which inequality is represented by this graph?

ohaa [14]

Answer:

B

Step-by-step explanation:

The line is solid, so eliminate A and D.

Also, the line is shaded below, so this eliminates C.

6 0

1 year ago

The cost to produce a product is modeled by the function f(x) = 5x2 − 70x 258, where x is the number of products produced. compl

ad-work [718]

The minimum cost to produce the product is $7.

What are functions?

A function from a set X to a set Y allocates exactly one element of Y to each element of X.

To determine the minimum cost:

We have to determine the minimum cost of producing this product.

Since $f(x) = 5x^{2} -70x+258$ .

Now, consider the equation $5x^{2} -70x+258= 0$ .

Dividing the above equation by 5, we get

$x^{2} -70x/5+258/5=0\\x^{2} -14x+258/5=0$

Now, considering the coefficient of 'x', dividing it by '2' and then adding and subtracting the square of the number which we got after dividing.

Since the coefficient of 'x' is 14, and half of 14 is '7'.

So, adding and subtracting from the above equation.

$x^{2} -14x+(7)^{2} -(7)^{2}+258/5=0\\x^{2} -14x+49 -49+258/5=0\\(x-7)^{2} -49+258/5=0\\(x-7)^{2} +258-245/5=0\\(x-7)^{2} +13/5=0\\5(x-7)^{2} +13=0$

Now, we have to determine the minimum cost to produce the product.

Since $f(x) = 5x^{2} -70x+258$ .

$f'(x) = 10x-70$

Now, let f'(x)=0

$10x-70=0\\10x=70$

Therefore, x=7

Now, consider which is greater than 0.

Therefore, x= 7 is the minimum cost.

Know more about functions here:

brainly.com/question/25638609

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5 0

1 year ago

I need help with number 16 please.

Ugo [173]

Do it ur self etc so don't ask me

8 0

2 years ago

What is the additive inverse of the number 3

Sever21 [200]

The inverse version of a number is it's opposite. The opposite of 3 is -3

8 0

2 years ago

Read 2 more answers

What is the answer to this?

Trava [24]

Answer:

Step-by-step explanation:

The formula of a volume of a cylinder:

$V=\pi r^2H$

r - radius

H - height

We have

$2r=4\ in\to r=2\ in\\\\H=8\ in$

Substitute:

$V_1=\pi(2^2)(8)=\pi(4)(8)=32\pi\ in^3$

The formula of a volume of a cone:

$V=\dfrac{1}{3}\pi r^2H$

We have

$r=2\ in\\\\H=0.7\ in$

Substitute:

$V_2=\dfrac{1}{3}\pi(2^2)(0.7)=\dfrac{1}{3}\pi(4)(0.7)=\dfrac{1}{3}\pi(2.8)=\dfrac{2.8\pi}{3}=\dfrac{28\pi}{30}=\dfrac{14\pi}{15}\ in^3$

The volume of the plastic object:

$V=V_1-V_2$

$V=32\pi-\dfrac{14\pi}{15}=\dfrac{480\pi}{15}-\dfrac{14\pi}{15}=\dfrac{466\pi}{15}\\\\\pi\approx3.141\\\\V\approx\dfrac{466}{15}\cdot3.141\approx97.6$

7 0

2 years ago

Solve for x. x/2≥−4 a. x≥−2 b. x≤−2 c. x≥−8 d. x≤−8