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3 years ago

3 in 3.62 stands for 3

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

3 0

The answer is false
3 in 3.62 doesn’t go ahead

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Please help due tomorrow Fri

ch4aika [34]

5 feet is the answer

3 0

3 years ago

Read 2 more answers

a pyramid whose altitude is 5ft weighs 800lbs. at what distance from its vertex must it be cut by a plane parallel to its base s

slavikrds [6]

A structure which has a square base and four triangular sides meeting at a point is called pyramid.

At distance of 3.97 feet from its vertex , pyramid is cut by plane so that the two solids of equal weight will be formed.

<u>It is assumed that weight of pyramid is proportional to its volume.</u>

So, $w = k V$ , where V is volume of original pyramid and v is volume of small pyramid and k is constant.

Let us consider that at h distance, pyramid is cut from its vertex. So a small pyramid is also formed.

Assume that base area of original pyramid is A and base of small pyramid is a.

Volume of original pyramid is, $V= \frac{1}{3} *A* 5$

So, weight of original pyramid, $W = k *(\frac{1}{3} *A* 5) = 800$

Volume of small pyramid is, $v = \frac{1}{3}* a* h$

So, weight of small pyramid, $w = k*(\frac{1}{3}* a* h)=400$

<u>Since, base and height of small pyramid and original pyramid are in proportion.</u>

So, $\frac{a}{A} = (\frac{h}{5}) ^{2}$

$a = (\frac{h}{5} )^{2}A$

Substituting value of a in equation $k*(\frac{1}{3}* a* h)=400$

So, $k*(\frac{1}{3}* \frac{h^{2} }{25}A * h)=400$

$(k*\frac{1}{3}* A*5)*\frac{1}{5} *\frac{h^{2} }{25} * h)=400$

Since, $(k *\frac{1}{3} *A* 5) = 800$ , substitute in above equation.

So, $800*\frac{h^{3} }{125}=400\\\\h^{3}=\frac{125}{2}\\\\h=\sqrt[\frac{1}{3} ]{62.5}\\\\h=3.97 feet$

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6 0

2 years ago

Read 2 more answers

(Full question above)

zlopas [31]

Answer:

B. 300

Step-by-step explanation:

Final figure volume = big cuboid volume - small cuboid volume

Volume of a cuboid = height x width x length

Final figure = (6 x 6 x 15) - (4 x 4 x 15)

Final figure = 540 - 240

Final figure = 300mm^2

5 0

3 years ago

Pls answer quick pls pls pls

xxTIMURxx [149]

Answer:

I got 2 km per hour

Step-by-step explanation:

7 0

2 years ago

How do you do this trig problem.

loris [4]

Consider a right triangle in which one of the angles $\theta$ satisfies

$\sin\theta=\dfrac35\implies\theta=\sin^{-1}\dfrac35$

That is, the side opposite $\theta$ occurs in a ratio of 3 to 5 with the hypotenuse. The side adjacent to $\theta$ then occurs in a ratio of 4 to 5 with the hypotenuse. In other words,

$\cos\theta=\dfrac45$

because

$\sin^2\theta+\cos^2\theta=\dfrac9{25}+\dfrac{16}{25}=1$

Then in this triangle,

$\cot\theta=\cot\left(\sin^{-1}\dfrac35\right)=\dfrac{\cos\theta}{\sin\theta}=\dfrac{\frac45}{\frac35}=\dfrac43$

7 0

2 years ago