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KonstantinChe [14]
3 years ago
8

Find the value of each expression. Show your work.

Mathematics
1 answer:
Verdich [7]3 years ago
4 0

Answer:

A) (1.2)^{2} = 1.44

B)2^ 3+17-(3\times 4) = 13

C)\frac{(9)^2}{(3)^3} = 3

Step-by-step explanation:

Here, the given expressions are:

A) (1.2)^{2}

Solving this, we get

(1.2)^{2}  = 1.2  \times 1.2 = 1.44

⇒(1.2)^{2}  = 1.44

B) 2^ 3+17-(3\times 4)

Now, solving this, we get

2^ 3+17-(3\times 4) = (2 \times 2 \times2) + (17 -12)\\=8 + 5 = 13

⇒2^ 3+17-(3\times 4) = 13

C) \frac{(9)^2}{(3)^3}

Simplifying this, we get

\frac{(9)^2}{(3)^3} = \frac{(9\times9)}{(3 \times 3\times3)}  = 3

⇒ \frac{(9)^2}{(3)^3} = 3

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Answer:

https://ohsrehak.weebly.com/uploads/5/4/6/9/54699399/5-1_bisectors_of_triangles_solutions.pdf   paste the click

Step-by-step explanation:

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3 years ago
Find the distance between P1(1, -91°) and P2(3, 335°) on the polar plane.
Dmitrij [34]

Answer:

2.749

Step-by-step explanation:

Use the polar distance formula:

\sqrt{r_{1}^{2}+{r_{2}^{2}-2 r_{1} r_{2}cos(theta_{1} -theta_{2})

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5 0
3 years ago
What is the surface area of a square prism with sides that measure 8 units?
V125BC [204]

Answer:

<em>The surface area of square prism = 384 units squared</em>

Step-by-step explanation:

<u><em>Explanation</em></u>:-

<em>Given the square prism side 'a' = 8 units</em>

<em>The surface area of square prism = 2 a² + 4 a h</em>

Given square prism length , width and height also equal identical sides

The surface area of square prism = 2 a² + 4 a h

                                                        = 2(8)² + 4 (8)(8)

                                                        = 2(64) + 4(64)

                                                       = 384 units squared

<u><em>Final answer</em></u>:-

<em>The surface area of square prism = 384 units squared</em>

5 0
3 years ago
Find x<br><br><br><br><br> Thanks for the help in advance!
Marina86 [1]

I'm not sure if this is the easiest way of doing this, but it surely work.

Let the base of the triangle be AB, and let CH be the height. Just for reference, we have

AH=2,\quad HB=6,\quad AC=x

Moreover, let CH=y and BC=z

Now, AHC, CHB and ABC are all right triangles. If we write the pythagorean theorem for each of them, we have the following system

\begin{cases}4+y^2=x^2\\36+y^2=z^2\\x^2+z^2=64\end{cases}

If we solve the first two equations for y squared, we have

y^2=x^2-4\\y^2=z^2-36

And we can deduce

z^2 = x^2+32

So that the third equation becomes

x^2+x^2+32=64 \iff 2x^2 = 32 \iff x^2=16 \iff x=4

(we can't accept the negative root because negative lengths make no sense)

7 0
3 years ago
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