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4 years ago

I’m struggling here plz help

Mathematics

1 answer:

miss Akunina [59]4 years ago

4 0

So to find h, you need to isolate that variable onto one side.

To do this, divide both sides by πr², and <u>your answer will be $\frac{V}{\pi r^2}=h$ , or the 2nd option.</u>

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4 years ago

4. (-5, 2) and (-5, 6)

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3 years ago

In 2012, the population of a city was 6.57 million. The exponential growth rate was 2.87% per year. a) Find the exponential g

icang [17]

Answer:

a) $P(t) = 6.57e ^{0.0287t}$

b) $P(6) = 7,805\ million$

c) $t = 14.64\ years$

d) $t = 24.15\ years$

Step-by-step explanation:

a) The function of exponential growth for a population has the following formula:

$P(t) = p_0e ^{rt}$

In this equation:

$p_0$ is the initial population

r is the growth rate

t is the time in years

In this problem we know that

$r = 2.87\% = 0.0287$

$p_0= 6.57$ million in the year 2012.

So the equation is:

$P(t) = 6.57e ^{0.0287t}$

Where $t = 0$ represents the year 2012

b) If $t = 0$ in 2012, then in 2018 $t = 6$

The population in 2018 is:

$P(t = 6) = 6.57e ^{0.0287(6)}\\\\P(6) = 7,805\ million$

c) To know when the population is equal to 10 million we must equal P(t) to 10 and solve for t.

$P(t) = 10 = 6.57e ^{0.0287t}\\\\\frac{10}{6.57} = e ^{0.0287t}\\\\ln(\frac{10}{6.57}) = 0.0287t\\\\t = \frac{ln(\frac{10}{6.57})}{0.0287}$

$t = 14.64\ years$

d) The function is doubled when $P(t) = 2p_0$

$P(t) = 2(6.57) = 13.14 = 6.57e ^{0.0287t}$

We solve for t.

$\frac{13.14}{6.57} = e ^{0.0287t}\\\\ln(\frac{13.14}{6.57}) = 0.0287t\\\\t = \frac{ln(\frac{13.14}{6.57})}{0.0287})$

$t = 24.15\ years$

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3 years ago

Will give brainliest

klemol [59]

Your question is calling the figure ABCD, but the image is showing PQRS

Using the letters that you wrote:

Because the figure was only rotated and not scaled , the sides with the same letters would still be the same

So AB would be the same as A'B' ( Length of AB = Length of A′B)

and

BC = B'C'

CD = C'D'

AD = A'D'

Or by the picture:

PQ = P'Q'

QR = Q'R'

RS = R'S'

PS = P'S'

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