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3 years ago

What is the area of the triangle?

Mathematics

1 answer:

LenaWriter [7]3 years ago

4 0

Answer:

area of triangle is 44.45ft^2

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WZ = 32, YZ = 6, and X is the midpoint of WY. Find WX.

GrogVix [38]

We are given the length of two segments:

WZ = 32

YZ = 6

and we are told that x is the midpoint of the segment WY

We are asked to find the length of the segment WX

Notice that the total length of the segment WZ is 32. from the point Y to the point Z we have 6 units. therefore, between W and Y there is 32 - 6 = 26 units.

SInce X is the midpoint of the distance between W and Y, then it has to cut the segment WY (26 units long) in two equal parts, each of length 13 units (half of 26).

Therefore, WX must be of length 13 units.

3 0

1 year ago

A cone has a diameter of 18 units and height of 8 units. What is its volume?

m_a_m_a [10]

Answer:72

Step-by-step explanation:Given : A cone has a diameter of 18 units and height of 8 units.

We have to determine the volume of the given cone.

We know Volume of cone = 1/3 $\pi \\$ $r^{2}$ $h$

Where r is radius of cone

h is the height of cone

Given : diameter = 18units so, radius is half of diameter = 9 units

height = 8 units

Substitute, we have,

Volume of cone = 1/3 $n\\$ $(9^{2} ) 8$

Simplify, we have,

Volume of cone = $72\pi$

Thus, the volume of given cone is 72π cubic units.

3 0

3 years ago

A survey of 100 doctors revealed that 2 out of every 5 doctors recommended Brand M deodorant. Based on these results, how many d

kotegsom [21]

Answer:

Step-by-step explanation:

5 x 5 is 25 so 2 x 5, 20 times makes 40

4 0

3 years ago

How to solve log and exponential equations

IceJOKER [234]

To be honest idk I’m jus tryna get 100%

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3 years ago

Read 2 more answers

Consider the function on the interval (0, 2π). f(x) = sin x + cos x (a) Find the open intervals on which the function is increas

Annette [7]

Answer:Increasing in x∈(0,π/4)∪(5π/4,2π) decreasing in(π/4,5π/4)

Step-by-step explanation:

given f(x) = sin(x) + cos(x)

f(x) can be rewritten as $\sqrt{2} [\frac{sin(x)}{\sqrt{2} }+\frac{cos(x)}{\sqrt{2} } ]..................(a)\\\\\ \frac{1}{\sqrt{2} } = cos(45) = sin(45)\\\\$

Using these result in equation a we get

f(x) = $\sqrt{2} [ cos(45)sin(x)+sin(45)cos(x)]\\\\= \sqrt{2} [sin(45+x)]..........(b)$

Now we know that for derivative with respect to dependent variable is positive for an increasing function

Differentiating b on both sides with respect to x we get

f '(x) = $f '(x)=\sqrt{2} \frac{dsin(45+x)}{dx}\\ \\f'(x)=\sqrt{2} cos(45+x)\\\\f'(x)>0=>\sqrt{2} cos(45+x)>0$

where x∈(0,2π)

we know that cox(x) > 0 for x∈[0,π/2]∪[3π/2,2π]

Thus for cos(π/4+x)>0 we should have

1) π/4 + x < π/2 => x<π/4 => x∈[0,π/4]

2) π/4 + x > 3π/2 => x > 5π/4 => x∈[5π/4,2π]

from conditions 1 and 2 we have x∈(0,π/4)∪(5π/4,2π)

Thus the function is decreasing in x∈(π/4,5π/4)

5 0

3 years ago