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3 years ago

Help with this Geometry problem please

Mathematics

1 answer:

Andre45 [30]3 years ago

8 0

The answer would be “C” because 4 goes into 8 twice and 12 3 time Than 5 goes into 10 twice and 15 3 times

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Mr. Al bought two boxes of candy with each box having 3 pieces inside of it. How many pieces candy did he have total?

Vladimir79 [104]

Answer:

Step-by-step explanation:

3 x 2 = 6

3 + 3 = 6

4 0

3 years ago

Read 2 more answers

The largest jaw breaker in a store is spherical with a radius of 2 in, and sells for $4.00. A

maw [93]

Answer:

$36

Step-by-step explanation:

1. Determine the surface area of the largest jaw breaker

Surface area of a sphere = 4πr²

4 x 3.14 x 2² = 50.24in²

2. Determine the surface area of the new jaw breaker

4 x 3.14 x 6² = 452.16 in²

3. Determine the cost per area

4/ 50.24

4. Determine the price for the new jaw breaker

4/ 50.24 x 452.16 = $36

5 0

2 years ago

Original price: 60 markup:15% what is retail price

sdas [7]

60x1.15=69
Retail price is $69

7 0

3 years ago

Determine wether the ratios are proportional: 6/36 and 1/6

dangina [55]

They are proportional. If you divide 6/36 by 6 on the numerator and the denominator, you get the same fraction of 1.6.

3 0

3 years ago

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