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3 years ago

Help please don’t understand.

Mathematics

1 answer:

sergij07 [2.7K]3 years ago

4 0

Okay so first, area is length times width, so 1/4x*x would be first = 1/4(x^2)=64. Then area or x = 16 or -16

perimeter would be width times two plus length times two. I recommend using math papa to calculate these numbers.

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For the following inequality, indicate whether the boundary line should be dashed or solid. 3x ≥ y + 1

Kay [80]

The boundary line should be solid because it have a 'great/less than or equal to' sign

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3 years ago

In November, the Sport Shop sold

Neko [114]

Answer:

March is the first month

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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

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zaharov [31]

Answer:

(-2,-5), (0,-4), (3,5) (1,1)

Solution:

Only (0,-4) is on the line

(-2,-5), (1,1), and (3,5) is in the orange

(5,5) Isn't right because it's not on the line nor in the orange

4 0

3 years ago

Identify the vertex of each function please help

Svetlanka [38]

Answer:

12. (4, 7)

13. (5, -1)

Step-by-step explanation:

Rewrite in vertex form and use this form to find the vertex (h, k).

8 0

3 years ago