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3 years ago

How do you solve 2p+5r=q for p?

Mathematics

1 answer:

kumpel [21]3 years ago

5 0

$2p + 5r=q\\2p = q - 5r\\\\p = \frac{1}{2}q-\frac{5}{2}r$

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A truck can be rented from Company A for $100 a day plus $0.40 per mile. Company B charges $20dayplus $0.90 per mile to rent the

dangina [55]

Answer:

the number of miles is 160

Step-by-step explanation:

The computation is given below:

Let us assume the miles be x

For company A

y = $100 + $0.40x

And, for company B

y = $20 + $0.90x

In order to find out the number of miles, the rental cost of the company A should be less than the company B rental cost

SO,

$20 + $0.90x > $100 + $0.40x

Subtract $20 from both sides

0.90x > 80 + 0.40x

Now subtract 0.40x from both sides

0.50x >80

x>160

So the number of miles is 160

6 0

3 years ago

100% of will give me 93

Art [367]

Answer:

Hi Wdym m=by 100% of will give me 93

Step-by-step explanation:

5 0

2 years ago

Two angles are supplementary, and one of the measures 35 degrees. What is the measure of the other angle?

Alekssandra [29.7K]

Answer=145

this is because supplementary means the two angles equal 180
180-35=145

3 0

2 years ago

Brian invests ?1900 into a savings account. The bank gives 3.5% compound interest for the first 2 years and 4.9% thereafter. How

Scorpion4ik [409]

let's check how much is it after 2 years firstly.

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &1900\\ r=rate\to 3.5\%\to \frac{3.5}{100}\dotfill &0.035\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases} \\\\\\ A=1900\left(1+\frac{0.035}{1}\right)^{1\cdot 2}\implies A=1900(1.035)^2\implies A=2035.3275$

Brian invested the money for 6 years, so now let's check how much is that for the remaining 4 years.

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &2035.3275\\ r=rate\to 4.9\%\to \frac{4.9}{100}\dotfill &0.049\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{yearly, thus once} \end{array}\dotfill &1\\ t=years\dotfill &4 \end{cases}$

$\bf A=2035.3275\left(1+\frac{0.049}{1}\right)^{1\cdot 4}\implies A=2035.3275(1.049)^4 \\\\\\ A\approx 2464.54\implies \boxed{\stackrel{\textit{rounded up }}{A=2465}}$

4 0

3 years ago

What is the present value of an investment that will be worth $6000 at the end of five years? Assume an APR of 6% compounded mon

musickatia [10]

Answer:

4448.23

Step-by-step explanation:

$AV=PV(1+\frac{i}{n})^{nt}\\\\6000=PV(1+\frac{.06}{12})^{5*12}\\\\6000=PV*1.3488\\\\PV= 4448.23$

4 0

3 years ago