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2 years ago

Please Help!! Whoever helps and gets it correct gets Brainliest and 5 star rating!!

Mathematics

1 answer:

grandymaker [24]2 years ago

4 0

Answer:

the reasoning states that "all the numbers begin with a 7 or an 8"

however this is not accurate as they can be in different placements

which can make a big difference in the total estimate.

for example:

the number could've been an 8, or an 80

they both begin with an 8

however have totally different values and could have messed up the total estimated number.

hope this helps :D

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VashaNatasha [74]

Answer:

Step-by-step explanation:

x + 159 = 180

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2 years ago

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One cube has a volume of 192 cm3, and another has a volume of 375 cm3. How much longer is the edge of the larger cube than that

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3 years ago

Can anyone help me? I’ll give Brainly!!!

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2 years ago

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10 less than 5 times a number is less than 6 times the number decreased by eight

-BARSIC- [3]

Answer:

The Answer is: the Number is -2

Step-by-step explanation:

Let n = the number.

Five times the number minus 10 is less than 6 times the number minus eight:

5n - 10 = 6n - 8

-1n = -8 + 10

-n = 2

n = -2, the value of the number is -2.

Proof:

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-20 = -20

Hope this helps! Have an Awesome Day!! :-)

8 0

3 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