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brilliants [131]
3 years ago
6

PLEASE help me

Mathematics
1 answer:
Kay [80]3 years ago
7 0

Answer:

D) 2(4a+7)

Step-by-step explanation:

when one divides both the number by 2 you get 4a+7 so the answer is 2(4a+7)

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120 different outcomes, D. (PLZ GIVE BRANLIEST)

Step-by-step explanation:

7 0
3 years ago
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In 1992, the moose population in a park was measured to be 4260. By 1996, the population was measured again to be 3660. If the p
masha68 [24]

In 1992, the moose population in a park was measured to be 4260. By 1996, the population was measured again to be 3660. If the population continues to change linearly: A.) Find a formula for the moose population, P , in terms of t , the years since 1990. P ( t ) = B.) What does your model predict the moose population to be in 2008?

<u>Answer:</u>

Model predict the moose population to be in 2008 is 1860

<u>Solution:</u>

Let us consider this problem on graph, taking the x axis to be years since 1990 and y on the graph be the number of moose

We have two points on the graph:  (2, 4260) and (6, 3660)

Using the slope formula we can find the slope :

\mathrm{m}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

m=\frac{(3660-4260)}{(6-2)}

m = -150                                  

Now that we have the slope of the line, we can use the point-slope formula to find the equation for the line as follows:-

\begin{array}{l}{y-y_{1}=m\left(x-x_{1}\right)} \\\\ {y-4260=-150(x-2)} \\\\ {y-4260=-150 x+300} \\\\ {y=-150 x+4560}\end{array}

Since the formula that is being sought is supposed to be interms of P and t, we will replace y with P and x with t

P(t) = -150t + 4560

Number of years from 1990 to 2008 is  

2008 - 1990 = 18

So, population in 2008 will be

P(18) = -150 (18) + 4560

P(18) = -2700 + 4560

P(18) = 1860

Thus the model to predict the moose population in 2008 is found

4 0
3 years ago
,<br><img src="https://tex.z-dn.net/?f=33333%20%5Cdiv%202" id="TexFormula1" title="33333 \div 2" alt="33333 \div 2" align="absmi
Sliva [168]
Um 16,666.5 hope this help
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3 years ago
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