To make it easier, you calculate the volume of the first aquarium.
1st aquarium:
V = L x W x H
V = 8 x 9 x 13
V = 72 x 13
V = 936 in.
Rate: 936 in./2 min.
Now that you've got the volume and rate of the first aquarium, you can find how many inches of the aquarium is filled within a minute, which is also known as the unit rate. To do that, you have to divide both the numerator and denominator by their least common multiple, which is 2. 936 divided by 2 is 468 and 2 divided by 2 is 1.
So the unit rate is 468 in./1 min. Now that you've got the unit rate, you can find out how long it'll take to fill the second aquarium up by finding its volume first.
2nd aquarium:
V = L x W x H
V = 21 x 29 x 30
V = 609 x 30
V 18,270 inches
Calculations:
Now, you divide 18,270 by 468 to find how many minutes it will take to fill up the second aquarium. 18,270 divided by 468 is about 39 (the answer wasn't exact, so I said "about").
2nd aquarium's rate:
18,270 in./39 min.
As a result, it'll take about 39 minutes to fill up an aquarium measuring 21 inches by 29 inches by 30 inches using the same hose. I really hope I helped and that you understood my explanation! :) If I didn't, I'm sorry. I tried. :(
Answer:
I think 9h
Step-by-step explanation:
Since both variable are the same, you can divide the numbers. 27/3=9. Add back in the variable, and the answer would be 9h. That's how I would solve it at least
5+7-4=6
6-4=2
I’ve created two equations with no solution
Apples are gross so none. plus most people have braces so they couldnt anyway
The width of the card to the nearest tenth of an inch is 6.15 inches
<h3>Area of rectangle</h3>
- Width = x
- Length = 2x + 1
- Area = 88 square inches
Area of a rectangle = Length × Width
88 = (2x + 2) × x
88 = 2x² + 2x
2x² + 2x - 88 = 0
x = -b ± √b² - 4ac / 2a
= -2 ± √2² - 4(2)(-88) / 2(2)
= -2 ± √4 - (-704) / 4
= -2 ± √708 / 4
= -1/2 ± √177/2
x = 6.15 or -7.15 inches
The width of the rectangle can not be negative, so, 6.15 inches is the width.
Learn more about rectangle:
brainly.com/question/13048427
#SPJ1