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3 years ago

I want the answer please!!!!

Mathematics

2 answers:

Ira Lisetskai [31]3 years ago

7 0

The second choice i is the answer

erastova [34]3 years ago

3 0

the second choice

the third choice

the fifth choice

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Which is a unit rate?

Angelina_Jolie [31]

The answer is

6 hours : 1 book

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3 years ago

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The Pythagorean Theorem can only be used with which type of triangle?

valkas [14]

Right triangle because a2 + b2 = c2

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3 years ago

Pyramid A has a triangular base where each side measures 4 units and a volume of 36 cubic units. Pyramid B has the same height,

omeli [17]

Answer:

The volume of pyramid B is 81 cubic units

Step-by-step explanation:

Given

Pyramid A

$s = 4$ -- base sides

$V = 36$ -- Volume

Pyramid B

$s = 6$ --- base sides

Required

Determine the volume of pyramid B [Missing from the question]

From the question, we understand that both pyramids are equilateral triangular pyramids.

The volume is calculated as:

$V = \frac{1}{3} * B * h$

Where B represents the area of the base equilateral triangle, and it is calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where s represents the side lengths

First, we calculate the height of pyramid A

For Pyramid A, the base area is:

$B = \frac{1}{2} * s^2 * sin(60)$

$B = \frac{1}{2} * 4^2 * \frac{\sqrt 3}{2}$

$B = \frac{1}{2} * 16 * \frac{\sqrt 3}{2}$

$B = 4\sqrt 3$

The height is calculated from:

$V = \frac{1}{3} * B * h$

This gives:

$36 = \frac{1}{3} * 4\sqrt 3 * h$

Make h the subject

$h = \frac{3 * 36}{4\sqrt 3}$

$h = \frac{3 * 9}{\sqrt 3}$

$h = \frac{27}{\sqrt 3}$

To calculate the volume of pyramid B, we make use of:

$V = \frac{1}{3} * B * h$

Since the heights of both pyramids are the same, we can make use of:

$h = \frac{27}{\sqrt 3}$

The base area B, is then calculated as:

$B = \frac{1}{2} * s^2 * sin(60)$

Where

$s = 6$

So:

$B = \frac{1}{2} * 6^2 * sin(60)$

$B = \frac{1}{2} * 36 * \frac{\sqrt 3}{2}$

$B = 9\sqrt 3$

So:

$V = \frac{1}{3} * B * h$

Where

$B = 9\sqrt 3$ and $h = \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9\sqrt 3 * \frac{27}{\sqrt 3}$

$V = \frac{1}{3} * 9 * 27$

$V = 81$

6 0

3 years ago

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Can someone tell me if it’s correct?

mart [117]

Yes yes it is correct

8 0

3 years ago

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Find an equation for the line tangent to the curve at the point defined by the given value of d²y/dx².

mars1129 [50]

Answer:

Step-by-step explanation:

Given:

x = 2cost,

t = (1/2)arccosx

y = 2sint

dy/dx = dy/dt . dt/dx

dy/dt = 2cost

dt/dx = -1/√(1 - x²)

dy/dx = -2cost/√(1 - x²)

Differentiate again to obtain d²y/dx²

d²y/dx² = 2sint/√(1 - x²) - 2xcost/(1 - x²)^(-3/2)

At t = π/4, we have

(√2)/√(1 - x²) - (√2)x(1 - x²)^(3/2)

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2 years ago