4 years ago

What r three rectangles with an area of 300 square feet?

Mathematics

1 answer:

jonny [76]4 years ago

4 0

1x300
2x150
3x100
In terms of length width for the above .

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Yuki888 [10]

Simplified: $\frac{4}{9x} -\frac{3}{x^{2} }$

Restrictions: x CANNOT be 0 because then the denominator would become 0 and would not be possible

5 0

3 years ago

Pls help!<br><br> 3x-4+71+80=180

juin [17]

Answer:

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3 years ago

In bioligy class cher is viewing cells with a microscope. cell h measures 8.76x10^-3 microns in diameter and cell v measures 2.1

3241004551 [841]

Answer:

Larger cell 400 times bigger cell than the smaller cell

Step-by-step explanation:

Given:

diameter of the cell h = $8.76\times10^{-3}$

diameter of the cell v = $2.19 \times10^{-5}$

To Find:

How many tines larger is the bigger cell than the smaller cell?

Solution:

The cell h with diameter $8.76\times 10^{-3}$ is larger and the cell v with diameter $2.19 \times 10^{-5}$ is small. To find the number of times it is larger, we can divide the larger cell diameter by smaller cell diameter

=> $\frac{8.76\times 10^{-3}}{2.19 \times 10^{-5}}$

=> $\frac{8.76\times 10^ {-3} \times 10^{5}}{2.19 }$

=> $\frac{8.76\times 10^ {(-3+5)}}{2.19 }$

=> $\frac{8.76\times 10^2}{2.19 }$

=> $\frac{876}{2.19}$

=> 400

Larger cell is the bigger cell than the smaller cell by 400 times

6 0

3 years ago

Jada has a stand in the marketplace where she sells ground cumin. Her weekly expenses are $300, and she sells her cumin at a fix

Irina18 [472]

Profit = revenue (income) - expenses

600 = 120x - 300
600 + 300 = 120x
900 = 120x
900/120 = x
7.50 = x ....this is how much 1 kg costs

so to cover her weekly expenses of 300, she would have to sell :
300/7.5 = 40 kg of cumin <===

8 0

3 years ago

how much fence would you need to replace the fence of a 30 foot by 40 foot playground? How much sod would you need to re-sod the

Mrrafil [7]

Perimeter=sum of side lengths...

P=2L+2W

P=2(L+W), in this case L and W are 30 and 40 so

P=2(30+40)

P=2(70)

P=140

So 140 feet of fencing is needed to replace the fence.

The amount of sod needed is equal to the area which is LW

A=30(40)=1200 ft^2

3 0

3 years ago