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2 years ago

Can you use doubling for 7 times 3

2 answers:

8 0

Well yea doubling is like getting the same number like 7×3 with equals 21
So you have to find the same number 21, but in a different way like this

7
7 7
+7 ×3
------ -----
21 21

V125BC [204]2 years ago

4 0

What do you mean doubling? You can add 7+7+7 which equals 21.

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2 years ago

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Linear regression line y=2.1x+130 predicts sales based on the money spent on advertising.

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x represents the dollars spent in advertising and y represents the company sales in dollars.

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2 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

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2 years ago

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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