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Molodets [167]
2 years ago
7

Will give brainliest!

Mathematics
1 answer:
Gala2k [10]2 years ago
6 0
Check the picture below, it hits the ground when y = 0.

\bf ~~~~~~\textit{initial velocity}\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o \\\\
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{0}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{162}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+0t+162\implies 0=-16t^2+162\implies 16t^2=162
\\\\\\
t=\cfrac{162}{16}\implies t=\cfrac{81}{8}\implies t=10\frac{1}{8}

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In 2013, the moose population in a park was measured to be 5,100. By 2018, the population was measured again to be 5,200. If the
natka813 [3]

Answer:

P(t) = 5100e^{0.0039t}

Step-by-step explanation:

The exponential model for the population in t years after 2013 is given by:

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

In which P(0) is the population in 2013 and r is the growth rate.

In 2013, the moose population in a park was measured to be 5,100

This means that P(0) = 5100

So

P(t) = 5100e^{rt}

By 2018, the population was measured again to be 5,200.

2018 is 2018-2013 = 5 years after 2013.

So this means that P(5) = 5200.

We use this to find r.

P(t) = 5100e^{rt}

5200 = 5100e^{5r}

e^{5r} = \frac{52}{51}

\ln{e^{5r}} = \ln{\frac{52}{51}}

5r = \ln{\frac{52}{51}}

r = \frac{\ln{\frac{52}{51}}}{5}

r = 0.0039

So the equation for the moose population is:

P(t) = 5100e^{0.0039t}

5 0
3 years ago
What is the approximate area of the circle?
Sophie [7]

Answer:

B

Step-by-step explanation:

because it is

5 0
2 years ago
Read 2 more answers
Write the equation in standard form for the circle with center (5,0) passing through (-1, 9/2)
andreev551 [17]

Answer:

(x - 5)^2 + y^2 = 225/4,

or you could write it as (x - 5)^2 + y^2 = 56.25.

Step-by-step explanation:

The factor form is

(x - h)^2 + (y - k)^2 = r^2  where the center is (h, k) and r = the radius.

So we have:

(x - 5)^2 + (y - 0)^2 = r^2

As the point  (-1, 9/2) is on the line:

(-1 - 5)^2 +  (9/2)^2 = r^2

r^2 =  36 +  81/4

r^2 = 225/4.

So  substituting for r^2:

(x - 5)^2 + (y - 0)^2 = 225/4

(x - 5)^2 + y^2 = 225/4 is the standard form.

3 0
3 years ago
Which construction could be used to construct an isosceles triangle ABC given line
Free_Kalibri [48]

Answer:

In the given two column proof two sides and included angles are equal.

Triangles are congruent if any pair of corresponding sides and their included angles are equal in both triangles .In the figure sides BD≅BD , AB≅ BC and <ABD ≅,BDD Therefore triangle ABD and triangle CBD are congruent by SAS property of congruence.

4 0
2 years ago
Find s(2t - 4) for s(t) = 3t - 7 <br> A)6t - 19 B) 6t - 18 C) 5t - 11 D) 5t - 19
asambeis [7]
Hello,

Answer A

s(x)=3x-7
s(2t-4)=3*(2t-4)-7=6t-12-7=6t-19
3 0
3 years ago
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