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3 years ago

Did i do this correctly?

Mathematics

2 answers:

den301095 [7]3 years ago

5 0

Yes u did do this correctly

Mnenie [13.5K]3 years ago

4 0

Yes you did do that correctly

your pretty smart :)

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A=10^x, b=10^yanda^y^*b*x=100then2xy=?

sergejj [24]

Answer:

Here is your answer

Step-by-step explanation:

xy = 1

Hope you like it : )

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

8 0

3 years ago

Simplify *show work!!* <br> (X^6)^1/2

Elza [17]

Answer:

$x^{3}$

Step-by-step explanation:

in powers:

when there is x to the power of 6 to the power of 6 as an example you should multiply 6x6=36

so in this one try to do it like the one up there

6* $\frac{1}{2}$ =3 so it is $x^{3}$

8 0

2 years ago

hey can someone pls help me out? i don’t understand this. find the surface area of the rectangle prism. thx hun <3

Alenkasestr [34]

Answer:1566

Step-by-step explanation:

3 0

3 years ago

25% of what number is 19

vaieri [72.5K]

Answer:

Step-by-step explanation:

Normally you would do 76 * 0.25

For this I did 19 / 0.25

I hope this helps!

4 0

3 years ago

Read 2 more answers