Ask question

Ask question

All categories

3 years ago

9

Please help. !!!!! Only if you are good at college algebra

1 answer:

kicyunya [14]3 years ago

5 0

Yes I got an A in college algebra...

You might be interested in

Find the measure of each angle indicated. Round to the nearest tenth.

Rina8888 [55]

I don't know sorry anjsjdjd

4 0

3 years ago

Read 2 more answers

G C D of 130 and 150

wlad13 [49]

Answer:

GCF = 10

Step-by-step explanation:

8 0

2 years ago

f(x) = 500(1.05)x What was the average rate of change of the value of Sophia's investment from the second year to the fourth yea

hjlf

$\bf slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ f(x_2)}}-{{ f(x_1)}}}{{{ x_2}}-{{ x_1}}}\impliedby 
\begin{array}{llll}
average\ rate\\
of\ change
\end{array}\\\\
-------------------------------\\\\
f(x)=500(1.05)^x \qquad 
\begin{cases}
x_1=2\\
x_2=4
\end{cases}\implies \cfrac{f(4)-f(2)}{4-2}
\\\\\\
\cfrac{[500(1.05)^4]~-~[500(1.05)^2]}{4-2}\implies \cfrac{56.503125}{2}\implies 28.2515625$

6 0

3 years ago

Please help I’ll mark brainliest

denpristay [2]

Answer:

166.27 m2

Step-by-step explanation:

area of hexagon = $\frac{3*\sqrt{3} }{2}$ * $a^{2}$ where a = length of 1 side

this hexagon is made of 6 equivalent triangle

consider ABC triangle

let AC length be x

<em>AB = 2 * AC = 2x = a</em>

$bc^{2} = ab^{2} - ac^{2} \\(4\sqrt{3})^{2} =(2x)^{2} - x^{2} \\48=3*x^{2}\\ 16=x^{2} \\x=4$

Hence a = 8

<u>hence area of hexagon = </u> $\frac{3*\sqrt{3} }{2}$ <u>*</u> $a^{2}$ <u> = </u> $\frac{3*\sqrt{3} }{2}$ <u> * </u> $8^{2}$ <u> = 166.27 m2</u>

3 0

3 years ago

n many population growth problems, there is an upper limit beyond which the population cannot grow. Many scientists agree that t

givi [52]

Answer:

$\frac{dP}{dt} = rP(1 - \frac{P}{K}) = 0.017P(1 - \frac{P}{16})$

Step-by-step explanation:

The logistic function of population growth, that is, the solution of the differential equation is as follows:

$P(t) = \frac{KP_{0}e^{rt}}{K + P_{0}(e^{rt} - 1)}$

We use this equation to find the value of r.

In this problem, we have that:

$K = 16, P_{0} = 2, P(50) = 4$

So we find the value of r.

$P(t) = \frac{KP_{0}e^{rt}}{K + P_{0}(e^{rt} - 1)}$

$4 = \frac{16*2e^{50r}}{16 + 2*(e^{50r} - 1)}$

$4 = \frac{32e^{50r}}{14 + 2e^{50r}}$

$56 + 8e^{50r} = 32e^{50r}}$

$24e^{50r} = 56$

$e^{50r} = 2.33$

Applying ln to both sides of the equality

$50r = 0.8459$

$r = 0.017$

So

The differential equation is

$\frac{dP}{dt} = rP(1 - \frac{P}{K}) = 0.017P(1 - \frac{P}{16})$

3 0

3 years ago