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anzhelika [568]
3 years ago
15

you are building a rectangular wading pool you want the area of the bottom to be 90 ft squared you want the length of the pool t

o be 3 ft longer than twice its width what will the dimensions of the pool be
Mathematics
1 answer:
Sindrei [870]3 years ago
5 0
L=2W+3, A=LW, using L from the first in the second gives you:

A=(2W+3)W

A=2W^2+3W, and we are told A=90 so

2W^2+3W=90

2W^2+3W-90=0

2W^2-12W+15W-90=0

2W(W-6)+15(W-6)=0

(2W+15)(W-6)=0, since W>0 for all real possibilities, 

W=6ft, and since L=2W+3

L=15ft

So the pool will be 6 ft wide by 15 ft long.
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Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
melamori03 [73]

Answer:

6+2\sqrt{21}\:\mathrm{cm^2}\approx 15.17\:\mathrm{cm^2}

Step-by-step explanation:

The quadrilateral ABCD consists of two triangles. By adding the area of the two triangles, we get the area of the entire quadrilateral.

Vertices A, B, and C form a right triangle with legs AB=3, BC=4, and AC=5. The two legs, 3 and 4, represent the triangle's height and base, respectively.

The area of a triangle with base b and height h is given by A=\frac{1}{2}bh. Therefore, the area of this right triangle is:

A=\frac{1}{2}\cdot 3\cdot 4=\frac{1}{2}\cdot 12=6\:\mathrm{cm^2}

The other triangle is a bit trickier. Triangle \triangle ADC is an isosceles triangles with sides 5, 5, and 4. To find its area, we can use Heron's Formula, given by:

A=\sqrt{s(s-a)(s-b)(s-c)}, where a, b, and c are three sides of the triangle and s is the semi-perimeter (s=\frac{a+b+c}{2}).

The semi-perimeter, s, is:

s=\frac{5+5+4}{2}=\frac{14}{2}=7

Therefore, the area of the isosceles triangle is:

A=\sqrt{7(7-5)(7-5)(7-4)},\\A=\sqrt{7\cdot 2\cdot 2\cdot 3},\\A=\sqrt{84}, \\A=2\sqrt{21}\:\mathrm{cm^2}

Thus, the area of the quadrilateral is:

6\:\mathrm{cm^2}+2\sqrt{21}\:\mathrm{cm^2}=\boxed{6+2\sqrt{21}\:\mathrm{cm^2}}

4 0
3 years ago
Will mark brainliest! Answer quickly please.
inn [45]
1400kg.
Force=ma
2100N= m*1.5
2100/1.5= 1400kg.
Please mark me brainliest :)
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2 years ago
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Answer:

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The circumcenter of an obtuse triangle lies outside the triangle.

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