Using trigonometry. This picture shows a ‘not to scale’ sketch of the following question. Use Sin 30. Sin 30= opposite/hypo tenuse. That is simply dividing 6 ft over Sin 30 that is 12 feet.

3 0

3 years ago

He base of a pentagonal prism has an area of 2.4 m2. The prism has a volume of 4.8 m3. What is its height? Use sketches, formula

77julia77 [94]

Answer:

The height of the prism is 2m

Step-by-step explanation:

Given;

The volume of pentagonal prism, V = 4.8 m^3

Area of the prism, A = 2.4 m^2

To determine the height of the prism, we consider the following;

Volume of any prism = Area of the prism x height of the prism

Height of the prism = volume of the prism / Area of the prism

Height of the prism = 4.8 / 2.4

Height of the prism = 2 m

Therefore, the height of the prism is 2m

3 0

4 years ago

The standard formula for the volume of a rectangular pyramid is . If the pyramid is scaled proportionally by a factor of k, its

PolarNik [594]

Answer:

The answer in the procedure

Step-by-step explanation:

we know that

The volume of a rectangular pyramid is equal to

$V=\frac{1}{3}LWH$

where

L is the length of the rectangular base

W is the width of the rectangular base

H is the height of the pyramid

If the pyramid is scaled proportionally by a factor of k

then

the new dimensions are

L=kL

W=kW

H=kH

substitute and find the new Volume V'

$V'=\frac{1}{3}(kL)(kW)(kH)$

$V'=\frac{1}{3}(k^{3})LWH$

$V'=(k^{3})\frac{1}{3}LWH$

$V'=(k^{3})V$

The new volume is equal to the scale factor k elevated to the cube multiplied by the original volume

8 0

3 years ago

Read 2 more answers

What is true about the solution above.

Masja [62]

ANSWER

$x = \pm \sqrt{3}$

and they are actual solutions.

EXPLANATION

The given equation is:

$\frac{ {x}^{2} }{2x - 6} = \frac{9}{6x - 18}$

Cross multiply

${x}^{2} (6x - 18) = 9(2x -6 )$

This implies;

${x}^{2} (6x - 18) - 9(2x - 6) = 0$

$3{x}^{2} (2x - 6) - 9(2x - 6) = 0$

Factor

$(3 {x}^{2} - 9)(2x - 6) = 0$

$3 {x}^{2} - 9 = 0 \: or \: 2x - 6= 0$

$3 {x}^{2} = 9 \: or \: 2x = 6$

${x}^{2} = 3\: or \: x = 3$

${x} = \pm \sqrt{3} \: or \: x = 3$

The domain of the given equation is

$x \ne3$

Therefore the actual solutions are

$x = \pm \sqrt{3}$

NB: x=3 is not in the domain of the given equation. It cannot be an extraneous solution.

7 0

3 years ago