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kodGreya [7K]
3 years ago
8

Factor completely 5c5 + 60c4 + 180c3.

Mathematics
1 answer:
Mrac [35]3 years ago
8 0
5c^5 + 60c^4 + 180c^3

find the GCF, 5c³

5c³(5c^5 + 60c^4 + 180c^3/ 5c^3)

5c³(c² + 12c + 36)

5c³(c² + 2(c)(6) + 6²)

5c³(c + 6)² <<< the answer.

hope this helps, God bless!
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Grover decides to swim downstream to search for new bones. He can swim 30 miles down stream in 3 hours. To return home swimming
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You need to divide 30 miles by 5 hours,so your average speed is going to be 6 miles an hour
6 0
3 years ago
Find the measure of angle DFC​
Jlenok [28]
DFC equals 180 because all triangles add up to 180. And their sides are the same so 180/3=60
7 0
3 years ago
Read 2 more answers
Which of the expressions are equivalent to the one below? (14+4)/6
Rufina [12.5K]

Answer:

(A)

Step-by-step explanation:

This sign -  / Is the same as this sign - ÷

14+4 = 4+14

This is why (4+14)÷6 = (14+4)/6

Hope this Helps!!! :)

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7 0
3 years ago
Read 2 more answers
the diameter of a piston for an automobile is 3 11/16in with tolerence of 1/64in. find the upper and lower limits of the pistons
padilas [110]

Upper Tolerance

Remark

The 11/16 is the only thing that will be affected. The three won't go up or down when we add 1/64 so we should just work with the 11/16. We need only add 11/16 and 1/64 together to see what the upper range is. Later on we can add 3 into the mix.

Solution

<u>Upper Limit</u>

\dfrac{11}{16} +  \dfrac{1}{64}

Now change the 11/16 into 64. Multiply numerator and denominator or 11/16 by 4

\dfrac{11*4}{16*4} +  \dfrac{1}{64}

\dfrac{11*4}{16*4} +  \dfrac{1}{64} Which results in

\dfrac{44}{64} +  \dfrac{1}{64}

With a final result for the fractions of 45/64

So the upper tolerance = 3 45/64

<u>Lower Tolerance</u>

Just follow the same steps as you did for the upper tolerance except you subtract 1/64 like this.

\dfrac{11}{16} - \dfrac{1}{64}

Your answer should be 3 and 43/64


4 0
3 years ago
Find all real solutions and show work: <br> 2x^3-3x^2+18x-27=0
Amiraneli [1.4K]

Given:

The given equation is

2x^3-3x^2+18x-27=0

To find:

All the real solutions.

Solution:

We have,

2x^3-3x^2+18x-27=0

It can be written as

x^2(2x-3)+9(2x-3)=0

(2x-3)(x^2+9)=0

Using zero product property, we get

(2x-3)=0 and (x^2+9)=0

2x=3 and x^2=-9

x=\dfrac{3}{2} and x=\pm\sqrt{-9}

We know that x=\pm\sqrt{-9} is an imaginary number because there is a negative sign under the square root.

Therefore, x=\dfrac{3}{2} is the only real solution of the given equation.

5 0
3 years ago
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