All categories

3 years ago

What is the value of s in the equation 3 r equals 10 plus 5 s, when r equals 10? 4 8 100 200

Mathematics

2 answers:

alisha [4.7K]3 years ago

6 0

Answer

Step-by-step explanation:

Given,

r = 10

Let's create an equation,

$3r = 10 + 5s$

plugging the value of r

$3 \times 10 = 10 + 5s$

Multiply the numbers

$30 = 10 + 5s$

Move 5s to L.H.S and change its sign

Similarly, Move 30 to R.H.S and change its sign.

$- 5s = 10 - 30$

Calculate

$- 5s = - 20$

The difference sign ( - ) should be cancelled on both sides

$5s = 20$

Divide both sides of the equation by 5

$\frac{5s}{2} = \frac{20}{5}$

Calculate

$s = 4$

The value of s is 4.

Hope this helps..

Best regards!!

frosja888 [35]3 years ago

6 0

Answer:

A. 4 (on edgenuity)

Step-by-step explanation:

You might be interested in

Approximately how many times greater is 6 x 10^-5 than 5 x 10^-9

8_murik_8 [283]

Answer: $6 \times 10^{-5}$ is $1.2 \times 10^4.$ times greater than $5 \times 10^{-9}$ .

Step-by-step explanation:

We are given two numbers.

First number : $6 \times 10^{-5}$ .

Second number : $5 \times 10^{-9}$ .

In order to find how many times $6 \times 10^{-5}$ is greater than $5 \times 10^{-9}$ , we need to divide first number by second number.

Therefore,

$6 \times 10^{-5}$ ÷ $5 \times 10^{-9}$

or $\frac{6 \times 10^{-5}}{5 \times 10^{-9}}$

First we would divide 6 by 5.

On dividing 6 by 5, we get 1.2.

Now, we would divide 10^-5 by 10^-9.

In order to divide them, we can apply quotient rule of exponents $\frac{a^m}{a^n} =a^{m-n}$ .

We get

$\frac{10^{-5}}{10^{-9}} = 10^{-5-(-9)} = 10^{-5+9} = 10^4$ .

Therefore,

$\frac{6 \times 10^{-5}}{5 \times 10^{-9}}$ = $1.2 \times 10^4.$

So, we could say finally $6 \times 10^{-5}$ is $1.2 \times 10^4.$ times greater than $5 \times 10^{-9}$ .

8 0

3 years ago

Lane invests $7,000, at 7% interest, compounded annually for 22 years. Calculate the compound interest for his investment.

amm1812

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$7000\\ r=rate\to 7\%\to \frac{7}{100}\dotfill &0.07\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &22 \end{cases} \\\\\\ A=7000\left(1+\frac{0.07}{1}\right)^{1\cdot 22}\implies A=7000(1.07)^{22}\implies A\approx 31012.81$