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DochEvi [55]
3 years ago
6

Help with this question please!

Mathematics
1 answer:
vova2212 [387]3 years ago
7 0

<em>V = 120 in^3</em>

<u><em>Here is why:</em></u>

The volume formula for a pyramid is:

V=\frac{1}{3}Bh

The B stands for the area of the base, which in this case is a right triangle. We will find the area of the right triangle first, then plug it into the equation.

The area formula for a triangle is:

A=\frac{1}{2}bh

Plug in the numbers.

A=\frac{1}{2}*9*10

A = 45

Now we have found the area of B, we can plug it into the volume formula for the pyramid.

V=\frac{1}{3}*45*8

V = 120 in^3

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Describe does Glide Reflection from the preimage triangle and black to the image triangle in red
AveGali [126]

Answer:

d

Step-by-step explanation:

Lets look at V which is at (2,-4).

We will also look at it's image V' which is at (-2,-1).

Let's go through the choices:

a. translation left 4 units and down 1 unit reflection x axis

Translating V left 4 and down 1 would give us (2-4,-4-1)

(-2,-5)

Reflecting that across the x-axis would give us

(-2,5).

So this is not (-2,-1) so let's move on...

b. translation left 2 units and up 2 units reflection line x=1

Translating V left 2 and up 2 gives us

(2-2,-4+2)

(0,-2)

Now reflecting this over x=1 gives us

(2,-2). ( Imagine this x=1 which is verical going right smack between points (0,-2) and (2,-2).)

We still did not reach our goal image.

c. translation left 2 units up and up 3 units reflection y axis

Starting at V(2,-4) and moving left 2 & up 3 gives (2-2,-4+3)=(0,-1).

Now reflecting this over the vertical axis, the y-axis, gives us (0,-1).

So we did not reach our (-2,-1) point yet.

d. translation left 2 units and 3 units reflection line x=-1.

​Starting at (2,-4) then moving left 2 and up 3 gives us (2-2,-4+3)=(0,-1).

Now we are reflecting over x=-1 which means we want to choose a point on the side making x=-1 the middle. Thuis point on the other side of (0,-1) of x=-1 would be (-2,-1) which is the goal.

8 0
3 years ago
Please someone explain to me how did we multiply the denominator in this
AleksandrR [38]

ok so what you wanna do is use multiplication

8 0
2 years ago
Which is the correct simplified form of the expression? (HELP)
aksik [14]
The answer should be option C! I think? I hope it helps :)
8 0
3 years ago
How to solve this :') please help
blsea [12.9K]

Answer:

3/2

Step-by-step explanation:

sin(3π/4 - β) = sin(3π/4)cosβ - cos(3π/4)sinπ =

\sin(\frac{3\pi}{4}-\beta)=\sin(\frac{3\pi}{4})\cos\beta-\cos\frac{3\pi}{4}\sin\beta\\=\frac{1}{\sqrt{2}}\cos\beta-(-\frac{1}{\sqrt{2}})\sin\beta\\\\=\frac{1}{\sqrt{2}}(\cos\beta +\sin\beta)\\\\\sin^2(\frac{3\pi}{4}-\beta)=\frac{1}{2}\cos\beta +\sin\beta)^2=\frac{1}{2}(\cos^2\beta +\sin^2\beta+2\sin\beta\cos\beta\\=\frac{1}{2}(1+\sin2\beta)=\frac{1}{2}(1-\frac{1}{5}) = \frac{2}{5}\\

Use \cot^2x=\csc^2x-1=\frac{1}{\sin^2x}-1

so

\cot^2(\frac{3\pi}{4}-\beta)=\frac{1}{\sin^2(\frac{3\pi}{4}-\beta)}-1 = \frac{1}{\frac{2}{5}}-1=\frac{5}{2}-1=\frac{3}{2}

7 0
2 years ago
A students claims that 8 to the 3rd power and 8 to the negative 5th power is greater than 1. Explain whether the student is corr
Serga [27]
<span>7s over 5t to the negative 3rd power 2 to the negative 3rd power times X to the 2nd power times Z to the negative 7th power 7s to the zero power times T to the negative 5th power over 2 to the negative 1st power times M to the 2nd power If you would please simplify these </span>
8 0
3 years ago
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