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3 years ago

What is k in terms of n?

Mathematics

2 answers:

soldier1979 [14.2K]3 years ago

4 0

Answer:

i think its newton.

Step-by-step explanation:

topjm [15]3 years ago

4 0

My answer is Newton

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

Learn more:

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6 0

2 years ago

Read 2 more answers

Which of the following is an asymptote of y = sec(x)?

babunello [35]

Answer: option d. x = 3π/2

Solution:

function y = sec(x)

1) y = 1 / cos(x)

2) When cos(x) = 0, 1 / cos(x) is not defined

3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...

4) limit of sec(x) = lim of 1 / cos(x).

When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.

So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).

Answer: 3π/2

The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).

8 0

3 years ago

Read 2 more answers

Simplify 4/(√5+ √2) - 3/(√5- √2)

MakcuM [25]

Step-by-step explanation:

$\frac{4}{ \sqrt{5} + \sqrt{2}} - \frac{3}{ \sqrt{5} - \sqrt{2} } = \\ = \frac{4( \sqrt{5} - \sqrt{2} ) - 3( \sqrt{5} + \sqrt{2} ) }{( \sqrt{5} + \sqrt{2} )\times ( \sqrt{5} - \sqrt{2} )} = \\ = \frac{4 \sqrt{5} - 4 \sqrt{2} - 3 \sqrt{5} - 3 \sqrt{2} }{5 - \sqrt{10} + \sqrt{10} - 2 } = \\ = \frac{ \sqrt{5} - 7 \sqrt{2} }{3}$