3 years ago

Please help will give brainliest

Mathematics

1 answer:

inessss [21]3 years ago

4 0

Answer B) is the correct answer

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Indeed help with this plz

gizmo_the_mogwai [7]

Answer:

x=\frac{6-y-z}{3}

Step-by-step explanation:

5 0

3 years ago

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A delivery truck is transporting boxes of two sizes: large and small. The combined weight of the large box and a small box is 70

andre [41]

Answer:

3.45

Step-by-step explanation:

you have to do tihis then that

3 0

3 years ago

4x2 is the GCF of this polynomial.

defon

Answer: 24x2y

Hope this helps(:

6 0

3 years ago

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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

A family is purchasing a house and needs to finance a 195,000 mortgage from the bank with an annual percentage rate (APR) of 5.3

maksim [4K]

Calculate the monthly payment:

195,000 x ( (0.05/12)x1+0.053)^12)/(1+0.053)^12-1)

Monthly payment = 1,082.84

Multiply monthly payment by 12 to get one year: 1082.84x 12 = 12994.08

Now multiply by 30 years:

12994.08 x 30 = 389822.40

The answer is $389,822

4 0

3 years ago