Ask question

Ask question

All categories

3 years ago

15

Linda's income, in thousands of dollars, is given by the function f(x) = 40x + 10, where x is the number of years. Her expenses,

in thousands of dollars, each year is given by the function g(x) = 25x + 5.
Which function best describes the amount she is able to put into savings, h, each year, in thousands of dollars?

h(x) = 65x + 15
h(x) = (40x + 10)(25x + 5)
h(x) = -15x - 5
h(x) = 15x + 5

2 answers:

Pani-rosa [81]3 years ago

7 0

the correct answer for plato is h(x) = 15x + 5.

SashulF [63]3 years ago

4 0

<span>The function best describes the amount she is able to put into savings, h, each year, in thousands of dollars is </span><span>h(x) = 15x + 5.</span>

You might be interested in

Let R be the region bounded by the following curves. Let S be the solid generated when R is revolved about the given axis. Ifpo

steposvetlana [31]

The two curves $y=x$ and $y=x^{1/15}$ intersect at $x=0$ and $x=1$ , with $x^{1/15}\ge x$ over $0\le x\le1$ .

Washer method

$\displaystyle\pi\int_0^1\left(\left(x^{1/15}\right)^2-x^2\right)\,\mathrm dx=\pi\int_0^1\left(x^{2/15}-x^2\right)\,\mathrm dx=\pi\left(\frac{15}{17}-\frac13\right)=\boxed{\frac{28\pi}{51}}$

Shell method

We have $y=x^{1/15}\implies x=y^{15}$ . The curves $x=y$ and $x=y^{15}$ intersect at $y=0$ and $y=1$ , with $y\ge y^{15}$ over $0\le y\le1$ .

$\displaystyle2\pi\int_0^1y\left(y-y^{15}\right)\,\mathrm dy=2\pi\int_0^1\left(y^2-y^{16}\right)\,\mathrm dy=2\pi\left(\frac13-\frac1{17}\right)=\boxed{\frac{28\pi}{51}}$

6 0

3 years ago

PLS HELP ITS TIMED!!!

Jlenok [28]

Answer:

it would be A

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

90/110 reduce the fraction

IrinaVladis [17]

90/110 reduces to 9/11. Good luck!

6 0

3 years ago

K divide by 6 + 8 = 5

VLD [36.1K]

Answer:

k= -18

Step-by-step explanation:

5 = (k/6) + 8 so 5-8 = k/6

-3=k/6 then -3 multiplied by 6 = k which is -18 = k

-18/6= -3

-3+8=5

5 0

3 years ago

Answer the question with explanation;

PSYCHO15rus [73]

Answer:

The statement in the question is wrong. The series actually diverges.

Step-by-step explanation:

We compute

$\lim_{n\to\infty}\frac{n^2}{(n+1)^2}=\lim_{n\to\infty}\left(\frac{n^2}{n^2+2n+1}\cdot\frac{1/n^2}{1/n^2}\right)=\lim_{n\to\infty}\frac1{1+2/n+1/n^2}=\frac1{1+0+0}=1\ne0$

Therefore, by the series divergence test, the series $\sum_{n=1}^\infty\frac{n^2}{(n+1)^2}$ diverges.

EDIT: To VectorFundament120, if $(x_n)_{n\in\mathbb N}$ is a sequence, both $\lim x_n$ and $\lim_{n\to\infty}x_n$ are common notation for its limit. The former is not wrong but I have switched to the latter if that helps.

7 0

3 years ago

Read 2 more answers