HELP ME IM A 6TH <br> GRADER

Betty Donahue is choosing between two fitness centers. Center A charges $375 per year for unlimited use of all facilities. Cente

Ronch [10]

Center A : 375 per yr

Center B : 35 + 12(12) + 5(52) = 35 + 144 + 260 = 439 per yr
I multiplied 12 * 12 because it was $ 12 a month and there are 12 months in a yr. I multiplied 5 * 52 because each aerobic class is $ 5 per class, and she went once per week, and there are 52 weeks in a yr.

so the cheaper one is Center A

3 0

3 years ago

Read 2 more answers

(06.04 MC)

Andru [333]

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

◉ $\large\bm{ -4}$

$\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}$

Before performing any calculation it's good to recall a few properties of integrals:

$\small\longrightarrow \sf{\int_{a}^b(nf(x) + m)dx = n \int^b _{a}f(x)dx + \int_{a}^bmdx}$

$\small\sf{\longrightarrow If \: a \angle c \angle b \Longrightarrow \int^{b} _a f(x)dx= \int^c _a f(x)dx+ \int^{b} _c f(x)dx }$

So we apply the first property in the first expression given by the question:

$\small \sf{\longrightarrow\int ^3_{-2} [2f(x) +2]dx= 2 \int ^3 _{-2} f(x) dx+ \int f^3 _{2} 2dx=18}$

And we solve the second integral:

$\small\sf{\longrightarrow2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot(3 - ( - 2)) }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} 2dx = 2 \int ^3_{-2} f(x)dx + 2 \cdot5 = 2 \int^3_{-2} f(x)dx10 = }$

Then we take the last equation and we subtract 10 from both sides:

$\sf{{\longrightarrow 2 \int ^3_{-2} f(x)dx} + 10 - 10 = 18 - 10}$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 8}$

And we divide both sides by 2:

$\small\longrightarrow \sf{\dfrac{2 { \int}^{3} _{2} }{2} = \dfrac{8}{2} }$

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx=4}$

Then we apply the second property to this integral:

$\small \sf{\longrightarrow 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx + 2 \int ^3_{-2} f(x)dx = 4}$

Then we use the other equality in the question and we get:

$\small\sf{\longrightarrow 2 \int ^3_{-2} f(x)dx = 2 \int ^3_{-2} f(x)dx = 8 + 2 \int ^3_{-2} f(x)dx = 4}$

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =4}$

We substract 8 from both sides:

$\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx -8=4}$

• $\small\longrightarrow \sf{2 \int ^3_{-2} f(x)dx =-4}$

7 0

1 year ago

Ayeeeeeeeeeee whats up

Ierofanga [76]

Answer:

not much

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Help pleaseee really dont know how to start with this

Usimov [2.4K]

Answer:

Slope is positive for all x, so always increasing

Step-by-step explanation:

Increasing/decreasing depends on the slope of the function, which is f'

f'(x) = 9x² + 18x + 25

If f'(x) > 0 for all x, then his claim is correct (increasing for all x)

If there's even 1 x-value for which f'(x) is not positive, his claim is incorrect

f'(x) is a quadratic function.

9x² + 18x + 25

9(x² + 2x) + 25

9(x² + 2(x)(1) + 1² - 1²) + 25

9(x + 1)² - 9 + 25

9(x + 1)² + 16

Since the minimum value of f' is 16, it's always positive.

Hence, the claim is correct

6 0

3 years ago

Profit P(x), is the difference between revenue, R(x), and cost, C(x), so P(x)=R(x)-C(x). Which expression represents P(x), if R(

siniylev [52]

Answer:

P(x)= x ^4-3x^3+x^2-4

Step-by-step explanation:

Given data

R(x) = 2x ^4-3x^3+2x-1

c(x)=x^4-x^2+2x+3

We know that

P(x)=R(x)-C(x)

Hence

P(x)= 2x ^4-3x^3+2x-1-(x^4-x^2+2x+3)

open bracket

P(x)= 2x ^4-3x^3+2x-1-x^4+x^2-2x-3

Collect like terms

P(x)= 2x ^4-x^4-3x^3+x^2-2x+2x-3-1

P(x)= x ^4-3x^3+x^2-4

4 0

3 years ago

HELP ME IM A 6TH GRADER