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wel
3 years ago
11

Help me solve this problem: -3(x^-2 y^-2)^0

Mathematics
2 answers:
sergiy2304 [10]3 years ago
8 0

Answer:

=  - 3

I have brainlest

I have follow me

I have heart and 5 star

Alexxandr [17]3 years ago
8 0
The answer is -3 make them brainliest ^
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AlekseyPX
If (55 + x)/2 is to be 67, then 55+x = 2*67, so x = 2*67 - 55 = 79
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3 years ago
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Answer:

1: 3z+5

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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

8 0
3 years ago
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Answer:

the correct answer is 4

Step-by-step explanation:

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6 0
3 years ago
Paper clips costs $7.20 what is the unit price of the paper clips
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I could help you, but how many paper clips were there? divide 7.20 by whatever number it is...
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3 years ago
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