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3 years ago

Help please asap aaaaaaaaaaaaaaaaaa

Mathematics

1 answer:

wel3 years ago

6 0

Answer:

Step-by-step explanation:

Your exponential formula is in the form y = ab^x. In this form, the coefficient 'a' is the initial value, the y-intercept, the value when x=0. The value 'b' is the growth factor, which is 1 more than the growth rate per increment of x. This problem is asking for the growth rate to be expressed as a percentage.

Given p(x) = 78500(1.02^x), we can compare to the exponential function form to see that ...

a = 78,500
b = 1.02 = 1 +0.02 = 1 +2%

The value of x is zero in the year 2000, so the population that year is ...

p(0) = a = 78,500

The increase per year is the value of 'b' with 1 subtracted:

growth rate = 2% per year

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I'll give brainly if right

Yanka [14]

Answer: 3.6

Step-by-step explanation:

18/5 simplified equals 3 3/5.

3/5 = 0.6

3 + 0.6 = 3.6

In conclusion, the answer is 3.6.

Hope this helps!

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3 years ago

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The equations y=-3/4x +2 and y (1/4) x +1 are shown of the graph at what x-values are the equations equal? Select each correct a

nika2105 [10]

Answer:

Step-by-step explanation:

we have to see where the two graphs intercept

it happens when x = 0 and when x = 1

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3 years ago

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Help calculus module 8 DBQ<br><br> please show work

igor_vitrenko [27]

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of $R(t)$ at the endpoints of each subinterval. The sum is then

$\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}$

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since $R$ is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval $[a,b]$ , it guarantees the existence of some $c\in(a,b)$ such that

$\dfrac{R(b)-R(a)}{b-a)=R'(c)$

Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such $c$ for which $R'(c)=0$ . I would chalk this up to not having enough information.

3. $R(t)$ gives the rate of water flow, and $R(t)\approx W(t)$ , so that the average rate of water flow over [0, 8] is the average value of $W(t)$ , given by the integral

$R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt$

If doing this by hand, you can integrate by parts, setting

$u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt$

$\mathrm dv=\mathrm dt\implies v=t$

$R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)$

For the remaining integral, consider the trigonometric substitution $t=\sqrt 7\tan s$ , so that $\mathrm dt=\sqrt 7\sec^2s\,\mathrm ds$ . Then

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}$

$R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)$

$\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}$

or approximately 3.0904, measured in gallons per hour (because this is the average value of $R$ ).

4. By the fundamental theorem of calculus,

$g'(x)=f(x)$

and $g(x)$ is increasing whenever $g'(x)=f(x)>0$ . This happens over the interval (-2, 3), since $f(x)=3$ on [-2, 0), and $-x+3>0$ on [0, 3).

5. First, by additivity of the definite integral,

$\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt$

Over the interval [-2, 0), we have $f(x)=3$ , and over the interval [0, 6], $f(x)=-x+3$ . So the integral above is

$\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}$

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3 years ago

London is deciding between two different movie streaming sites to subscribe to. Plan A costs $9 per month plus $1.50 per movie w

juin [17]

Answer:

Plan A is $3 cheaper can Plan B.

Step-by-step explanation:

first equation: 1.5t+9

second equation: .50t+38

Then substiute the 26 for x. So you get 48 for Plan A and 51 for Plan B. Then subtract so you can get 3, for Plan A is three dollars cheaper than Plan B.

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2 years ago

What are the first 50 numbers of pie?

Tomtit [17]

Answer:

3.14159265358979323846264338327950288419716939937510

Step-by-step explanation:

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3 years ago

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