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3 years ago

13

The price of a gallon of unleaded gas has dropped to $2.83 today. Yesterday's price was $2.89. Find the percentage

1 answer:

Firlakuza [10]3 years ago

5 0

Answer:

$2.90 %Step-by-step explanation:

because if you round it it would be close to 90 and if you mutiply 2 x 2 =4 x 89

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Select the factors that can be divided out to simplify the multiplication problems: 8/15 * 5/16.

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3 years ago

Write an expression that evaluates to true if the int associated with number_of_prizes is divisible (with no remainder) by the i

hodyreva [135]

We need to find the expression for " number_of_prizes is divisible number_of_participants". Also there should not remain any remainder left. On in order words, we can say the reaminder we get after division is 0.

Let us assume number of Prizes are = p and

Number of participants = n.

If we divide number of Prizes by number of participants and there will be not remainder then there would be some quotient remaining and that quotent would be a whole number.

Let us assume that quotent is taken by q.

So, we can setup an expression now.

Let us rephrase the statement .

" Number of Prizes ÷ Number of participants = quotient".

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4 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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3 years ago