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3 years ago

Find the circumference of the circle. please help me baby

Mathematics

2 answers:

Debora [2.8K]3 years ago

7 0

Answer:
x = 43.98

step-by-step explanation:
c = 2πr
c = 2π7
c = 43.98

lesya [120]3 years ago

3 0

the circumference of the circle =43.98

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annual sales for a company are $149,999 and are increasing at a rate of 6% per year. write an exponential growth function to mod

elixir [45]

Answer:

S(t) = 149999(1.06)^t
$225543

Step-by-step explanation:

Initial value is $149999
Increase rate is 6% per year = 1.06 times

The model of the growth is:

S(t) = 149999(1.06)^t, where S- amount of annual sales, t- time in years, 1.06- growth rate per year

Annual sales after 7 years:

S(7) = 149999(1.06)^7 = $225543

4 0

3 years ago

What does -22 belong to

Len [333]

Answer:

it's part of the real number system

it's a rational number

it's an integer

Step-by-step explanation:

6 0

3 years ago

Solve this equation: please show all the work <img src="https://tex.z-dn.net/?f=x%5E%7B3%7D-y%2B9-%28xy%29" id="TexFormu

puteri [66]

$x^{3} - y + 9 - (xy)$
$(3)^{3} - (6) + 9 - (3 * 6)$
$27 - 6 + 9 - 18$
$12$

4 0

3 years ago

Find the missing length. The triangles are similar.

Marina CMI [18]

Answer:

RK= 13 SK= 12

Step-by-step explanation

First, we find the missing length of RK using the given information: 65-52=13

Then we divide 65 by 13 to get a scale factor: 65/13=5

(To check, we multipuly: 13*5=65)

We then apply the scale factor to MK to find SK: 60/5=12

(To check we multipuly: 5*12=60)

We now have our answers: RK:13, SK:12

4 0

3 years ago

6. Evaluate the function at each specified value of the independent variable and simplify. (If an answer is undefined, enter UND

Darya [45]

Complete Question

The complete question is shown on the first uploaded image

Answer:

$V(3) = 113.112$

$V(\frac{3}{2}) = 14.139$

$V(2r) = 33.514r^3$

$h(5) = 0$

$h(t) = -5.76$

$h(x+5) = x(x+5)$

Step-by-step explanation:

Considering the question 6

The function given is $V(r) = \frac{4}{3} \pi r^3$

For V(3) we have

$V(3) = \frac{4}{3} \pi 3^3$

$V(3) = \frac{4}{3} * 3.142 * 3^3$

$V(3) = 113.112$

For $V(\frac{3}{2})$

$V(\frac{3}{2}) = \frac{4}{3} * \pi * (\frac{3}{2})^3$

$V(\frac{3}{2}) = \frac{4}{3} * 3.142 * (\frac{3}{2})^3$

$V(\frac{3}{2}) = 14.139$

For V(2r)

$V(2r) = \frac{4}{3} \pi (2r)^3$

$V(2r) = \frac{4}{3} * 3.142*8r^3$

$V(2r) = 33.514r^3$

Considering the question 7

The function given is $h(t) = t^2 - 5t$

For h(5)

$h(5) = 5^2 - 5(5)$

$h(5) = 0$

For h(1.8)

$h(t) = 1.8^2 - 5(1.8)$

$h(t) = -5.76$

For h(x+ 5)

$h(x+5) = (x+5)^2 - 5(x+5)$

$h(x+5) = x^2 + 25+10x-5x-25$

$h(x+5) = x(x+5)$

7 0

3 years ago