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2 years ago

6

2. A triangular prism has a base area of 27 square inches and a height of 13 inches. What is the

1 answer:

guajiro [1.7K]2 years ago

5 0

Answer:

A.)351 cubic inches

Step-by-step explanation:

The computation of the volume of the prism is shown below:

As we know that

Volume of the prism is

= Base × height

= 27 square inches × 13

= 351 cubic inches

hence, the first option is correct

And, the rest of the options are wrong

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Is 1/5 bigger or smaller than 1/6

MAXImum [283]

You can tell if you make the denominators the same easiest one is 30 for 6 and 5
1/5=6/30
1/6=5/30

now you can see which is bigger

4 0

2 years ago

what digit is in the ten thousands place 1.03098 which is the correct answer I know it's one of the zeros I believe which one is

elena-14-01-66 [18.8K]

Hello there

Actually, the correct answer is the 9

the first 0 is in the tenths spot, the 3 is in the hundredths, the second zero is in the thousandths, and the 9 is in the ten thousandths, while the 8 is in the hundred thousandths.

Therefore, the 9 is in the ten thousandths place

I hope this helped ^^

4 0

2 years ago

Please answer correctly !!!!!!! Will mark brainliest !!!!!!!!!!!!

LUCKY_DIMON [66]

Answer:

$=8\sqrt{15}b^{\frac{7}{2}}$

Step-by-step explanation:

$\sqrt{24b^3}\sqrt{40b^2}\sqrt{b^2}$

$=\sqrt{40}\sqrt{b^2}\sqrt{b^2}\sqrt{24b^3}$

$=\sqrt{40}b^2\sqrt{24b^3}$

$\sqrt{24b^3}$

$=\sqrt{24}\sqrt{b^3}$

$=\sqrt{24}b^{\frac{3}{2}}$

$=\sqrt{24}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3\cdot \:3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=\sqrt{2^3}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$\sqrt{2^3}$

$=2^{3\cdot \frac{1}{2}$

$=2^{3\cdot \frac{1}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{40}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3\cdot \:5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\sqrt{2^3}\sqrt{5}b^2$

$=2^{\frac{3}{2}}\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}b^{\frac{3}{2}}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^2$

$=\sqrt{3}\cdot \:2^{\frac{3}{2}+\frac{3}{2}}\sqrt{5}b^{\frac{3}{2}+2}$

$2^{\frac{3}{2}+\frac{3}{2}}$

$=2^3$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{3}{2}+2}$

$b^{\frac{3}{2}+2}$

$=b^{\frac{7}{2}}$

$=2^3\sqrt{3}\sqrt{5}b^{\frac{7}{2}}$

$=2^3\sqrt{3\cdot \:5}b^{\frac{7}{2}}$

$=8\sqrt{15}b^{\frac{7}{2}}$

4 0

2 years ago

PLZZ help me with this I really need your help

Georgia [21]

Basil yum just use ur brain and find out before asking for help, ask ur teacher or somthin

5 0

2 years ago

Read 2 more answers

HELP PLEASE I DONT KNOW WHAT TO DO

Softa [21]

The first blank should be -5. The second should be -10. The third one should 10, if it’s talking about the opposite of -10.

3 0

2 years ago