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fenix001 [56]
4 years ago
10

Hunter raises beetles. in a square of compost 6 ft by 6 ft, he can have 500 beetles. how many beetles can he have if his square

of compost has a side length that is four times longer?
Mathematics
1 answer:
Rus_ich [418]4 years ago
7 0
<span>If Hunter raises beetles in a square of compost with a size of 6 ft by 6 ft and having 500 beetles, and then you are to find the number of beetles with a side length that is four times longer, then you can solve this by ratio and proportion.

A = s</span>²
A = 6²
A = 36 ft²

if s = 4*the original length, then the new area is 

A = 24²
A = 576 ft²

500 beetles/36 ft² = x number of beetles/576 ft²
x = 8,000 beetles
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Given vertices W(10,30), X(10,100), Y(110,100), Z(50,30) of a backyard, find distances WX, XY, YZ, ZW and XZ:

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2. XY=\sqrt{(110-10)^2+(100-100)^2}=\sqrt{100^2}=100\ ft;

3. YZ=\sqrt{(110-50)^2+(100-30)^2}=\sqrt{60^2+70^2}=10\sqrt{85}\ ft;

4. ZW=\sqrt{(50-10)^2+(30-30)^2}=\sqrt{40^2}=40\ ft;

5. XZ=\sqrt{(10-50)^2+(100-30)^2}=\sqrt{40^2+70^2}=10\sqrt{65}\ ft.

Then:

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A_{WXZ}=\sqrt{\frac{70+40+10\sqrt{65}}{2}\\\cdot (\frac{70+40+10\sqrt{65}}{2}-70)\cdot (\frac{70+40+10\sqrt{65}}{2}-40)\cdot (\frac{70+40+10\sqrt{65}}{2}-10\sqrt{65})}=\\ \\=1400\ ft^2.

2. the area

A_{XYZ}=\sqrt{\frac{100+10\sqrt{85}+10\sqrt{65}}{2}\cdot (\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-100)}\cdot\\ \\\cdot\sqrt{(\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-10\sqrt{85})\cdot (\frac{100+10\sqrt{85}+10\sqrt{65}}{2}-10\sqrt{65})}=3500\ ft^2.

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6 0
4 years ago
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