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bixtya [17]
3 years ago
10

What is the simplified form of 3 Start Root 5 c End Root times Start Root 15 c cubed End root? (worth 30 brainily points)

Mathematics
1 answer:
aleksley [76]3 years ago
8 0

Answer:

A. 15 c squared Start Root 3 End Root

Step-by-step explanation:

3\sqrt{5c} *\sqrt{15c^3} \\

Multiply the terms inside the square roots together

3\sqrt{75c^4}

Look for items that are perfects squares

3\sqrt{25*3*c^4}

We can separate terms inside the square root into separate square roots

3\sqrt{25} *\sqrt{3}*\sqrt{c^4}

3*5 *\sqrt{3} *c^2

Combine like terms

15c^2\sqrt{3}

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qwelly [4]

Answer:

8

Step-by-step explanation:

sin theta = opposite side/ hypotenuse

sin 30 = 4/x

Multiply each side by x

x sin 30 = 4

Divide each side by sin 30

x sin 30 /sin 30 = 4 sin 30

x = 4 /sin 30

We know that sin 30 = 1/2

x = 4 / 1/2

x = 8

7 0
3 years ago
Read 2 more answers
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aleksandr82 [10.1K]
Answer: should be D
I think so
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2 years ago
Will give brainly
ad-work [718]

Answer:

The wheels have to rotate 500 times

Step-by-step explanation:

The wheel is 1 meter in diameter so we need to find the circumference.

C = 2\pir

C = 2\pi1

C= 2\pi

2meters = .002kilometers

1000meters = 1kilometer

1000/2 = 500

6 0
3 years ago
What is the value of x? (will give brainliest if you show your work) PLEASE HELP
andrew-mc [135]
Check the picture below.

\bf \cfrac{8}{12}=\cfrac{x+4}{2x+1}\implies 16x+8=12x+48\implies 4x=40&#10;\\\\\\&#10;x=\cfrac{40}{4}\implies x=10

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Use Gauss-Jordan elimination to solve the following linear system.
marysya [2.9K]
Just put the coefients in to a matrix

1x-6y-3z=4
-2x+0y-3z=-8
-2x+2y-3z=-14

\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\-2&2&-3|-14\end{array}\right]
mulstiply 2nd row by -1 and add to 3rd
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&2&0|-6\end{array}\right]
divde last row by 2
\left[\begin{array}{ccc}1&-6&-3|4\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 2rd row by 6 and add to top one
\left[\begin{array}{ccc}1&0&-3|-14\\-2&0&-3|-8\\0&1&0|-3\end{array}\right]
multiply 1st row by -1 and add to 2nd
\left[\begin{array}{ccc}1&0&-3|-14\\-3&0&0|6\\0&1&0|-3\end{array}\right]
divide 2nd row by -3
\left[\begin{array}{ccc}1&0&-3|-14\\1&0&0|-2\\0&1&0|-3\end{array}\right]
mulstiply 2nd row by -1 and add to 1st row
\left[\begin{array}{ccc}0&0&-3|-12\\1&0&0|-2\\0&1&0|-3\end{array}\right]
divide 1st row by -3
\left[\begin{array}{ccc}0&0&1|4\\1&0&0|-2\\0&1&0|-3\end{array}\right]

rerange
\left[\begin{array}{ccc}1&0&0|-2\\0&1&0|-3\\0&0&1| 4\end{array}\right]

x=-2
y=-3
z=4
(x,y,z)
(-2,-3,4)

B is answer
7 0
3 years ago
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