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jek_recluse [69]

2 years ago

10

Find the radius r of the circle. Round your answer to the nearest tenth.

1 answer:

asambeis [7]2 years ago

3 0

Answer:

1.1 yd

Step-by-step explanation:

120 × π/180 = 2π/3

2.3 = r × 2π/3

=> r = 2.3 ÷ 2π/3

=> r ≈ 1.1

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The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increas

Mazyrski [523]

Answer:

The population in 40 years will be 1220.

Step-by-step explanation:

The population of a town grows at a rate proportional to the population present at time t.

This means that:

$P(t) = P(0)e^{rt}$

In which P(t) is the population after t years, P(0) is the initial population and r is the growth rate.

The initial population of 500 increases by 25% in 10 years.

This means that $P(0) = 500, P(10) = 1.25*500 = 625$

We apply this to the equation and find t.

$P(t) = P(0)e^{rt}$

$625 = 500e^{10r}$

$e^{10r} = \frac{625}{500}$

$e^{10r} = 1.25$

Applying ln to both sides

$\ln{e^{10r}} = \ln{1.25}$

$10r = \ln{1.25}$

$r = \frac{\ln{1.25}}{10}$

$r = 0.0223$

So

$P(t) = 500e^{0.0223t}$

What will be the population in 40 years

This is P(40).

$P(t) = 500e^{0.0223t}$

$P(40) = 500e^{0.0223*40} = 1220$

The population in 40 years will be 1220.

7 0

3 years ago

Find the length of side y. <br> y=_ft

jolli1 [7]

Answer:

x= 0.22947

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

2 years ago

Which statement is true of x2 + 3x-6?

eduard

(x - 2) is a factor of the polynomial

Give me feedback

5 0

3 years ago

What is the base of the triangle? Pls I need it ASAP

PIT_PIT [208]

Answer: 10 I think not sure

Step-by-step explanation: 1/2 base times height-> 1/2(6•10)

3 0

2 years ago

Read 2 more answers