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dimaraw [331]
3 years ago
11

how do you do this? i don't need it solved i just don't know how to do it. i know the note on the side says how to but i'm still

confused

Mathematics
1 answer:
olchik [2.2K]3 years ago
5 0

Answer:

Perimeter of triangle J= 18.0 units (3 s.f.)

Perimeter of triangle K= 19.0 units (3 s.f.)

Step-by-step explanation:

The length of each side of the triangle can be found using the distance formula below:

\boxed{d=  \sqrt{(y1- y2)^{2} + (x1 - x2)^{2}  } }

Please see the attached pictures for full solution.

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Step-by-step explanation:

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6 0
2 years ago
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A trough of water is 20 meters in length and its ends are in the shape of an isosceles triangle whose width is 7 meters and heig
Vaselesa [24]

Answer:

a) Depth changing rate of change is 0.24m/min, When the water is 6 meters deep

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Step-by-step explanation:

As we can see in the attachment part II, there are similar triangles, so we have the following relation between them \frac{3.5}{10} =\frac{a}{h}, then a=0.35h.

a) As we have that volume is V=\frac{1}{2} 2ahL=ahL, then V=(0.35h^{2})L, so we can derivate it \frac{dV}{dt}=2(0.35h)L\frac{dh}{dt} due to the chain rule, then we clean this expression for \frac{dh}{dt}=\frac{1}{0.7hL}\frac{dV}{dt} and compute with the knowns \frac{dh}{dt}=\frac{1}{0.7(6m)(20m)}2m^{3}/min=0.24m/min, is the depth changing rate of change when the water is 6 meters deep.

b) As the width of the top is 2a=0.7h, we can derivate it and obtain \frac{da}{dt}=0.7\frac{dh}{dt}  =0.7*0.24m/min=0.17m/min The width of the top of the water is changing, When the water is 6 meters deep at this rate

8 0
3 years ago
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gregori [183]
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Answer:

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Hope this helps!

3 0
3 years ago
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nata0808 [166]
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