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3 years ago

52 – 4 > – 29 AND – 12 + 6 > 2 Solve and give answer in interval notation

Mathematics

1 answer:

BlackZzzverrR [31]3 years ago

6 0

Answer:

wait huh?im confused lol

Step-by-step explanation:

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Parker baked 80 cookies for a bake sale. At the bake sale, 70% of his cookies sold. How many of Parker’s cookies were sold?

Doss [256]

80 * 70 percent = cookies sold

80 * .7 = cookies sold

56 = cookies sold

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3 years ago

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To the nearest hundredth, what is the length of line segment AB ?

sladkih [1.3K]

The answer is 5.83
Use the Pythagorean Theorem to calculate the 3rd side length.
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2 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

T-Riffic produces T-shirts for a $20 design fee plus an

TiliK225 [7]

0=0
5=60
10=100
15=140
17=156
33=284
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2 years ago

There is 20 people and each person get $1,255.How much money did every one get in total

mr Goodwill [35]

In total they had $25,100

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2 years ago

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