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3 years ago

Adele started her homework at 9:10.She finished at 9:45.How long did she spend on her homework

Mathematics

2 answers:

N76 [4]3 years ago

6 0

She spent 35 minutes doing her homework.

Juli2301 [7.4K]3 years ago

3 0

30 minutes is the answer

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For the inverse variation equation xy=k, what is the constant of variation,k, when x = -3and y=-2

const2013 [10]

Answer:

Step-by-step explanation:

xy=k

Value of k ; when ; x = - 3 and y = - 2

Put x = - 3 and y = - 2 into the equation :

(-3)(-2) = k

-3 * - 2 = k

6 = k

The constant of variation k = 6

8 0

3 years ago

What is the answer to 8x10^6 + 5x10^9=

Lesechka [4]

Answer:

13x10^15

Step-by-step explanation:

8+5 = 13

6+9 = 15

3 0

3 years ago

Compare the dimensions of the prisms. How many times greater is the surface area of the purple prism than the surface area of th

zysi [14]

Answer: 3 times greater

Step-by-step explanation:

Height: 4x3=12

Length: 3x3=9

Width: 3x3=9

The sides on the red cuboid times by 3 equals the sides on the purple one.

Hope this helps :)

7 0

3 years ago

How are these representations the same? How are these representations different?

Mice21 [21]

The last words are different

5 0

3 years ago

Read 2 more answers

Suppose Kaitlin places $6500 in an account that pays 12% interest compounded each year.

Leya [2.2K]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 1 year}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &1 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 1}\implies A=6500(1.12)\implies A=7280 \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~ \textit{Compound Interest Earned Amount \underline{for 2 years}} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$6500\\ r=rate\to 12\%\to \frac{12}{100}\dotfill &0.12\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, thus once} \end{array}\dotfill &1\\ t=years\dotfill &2 \end{cases}$

$\bf A=6500\left(1+\frac{0.12}{1}\right)^{1\cdot 2}\implies A=6500(1.12)^2\implies A=8153.6$

6 0

3 years ago