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2 years ago

A storage chest is shown. What is the volume in cubic feet?

Mathematics

1 answer:

algol132 years ago

4 0

Volume/V=length x width x height
So, V=4.5 x 2 x 2
V =18

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N/32 = 1/16

djverab [1.8K]

Answer:

2 is the correct answer dood

6 0

2 years ago

Which system of inequalities represents the graph?

vfiekz [6]

Yes the first one is the correct answer in this form of a graph

4 0

3 years ago

PLEASE HELP - QUIZ

Fittoniya [83]

Answer:

257.24 yd^2

Step-by-step explanation:

I made a video explaining, I promise it’s not anything unrelated to the question. Here it is: https://youtu.be/jHVtTCYzmzI

5 0

2 years ago

Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

Andre45 [30]

Answer:

Option d) 5 to the power of negative 5 over 6 is correct.

$\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

ie, $5^\frac{\bf -5}{\bf 6}$

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}$

$\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}$

Above equation can be written as 5 to the power of negative 5 over 6.

7 0

3 years ago

ritu can row downstream 20km in 2 hrs and upstream 4km in 2 hrs. find the speed of rowing in still water and the speed of the cu

Marina CMI [18]

Answer:

Speed of boat in still water is 6km/h

Speed of current is 4km/h

Step-by-step explanation:

We know speed = distance / time

Let the speed of boat in still water be x

Let the speed of current be y

Hence the speed of boat, down stream = $x+y$

Distance downstream = 20km

Time taken = 2 hours

We have

$x+y=\frac{20}{2}$

$x+y=10$ ......(1)

Speed of boat upstream = $x-y$

Distance upstream = 4

Time taken = 2

We have

$x-y=\frac{4}{2}$

$x-y=2$ ...(2)

Adding (1) & (2)

$2x=12$

$x=6$

Substituting in (1) we get

$y=4$

7 0

3 years ago