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mrs_skeptik [129]
3 years ago
9

Can somebody help me please?

Mathematics
2 answers:
Julli [10]3 years ago
8 0

Answer: the answer is E, 8000

Step-by-step explanation:

V=L×W×H

Oliga [24]3 years ago
8 0

Answer:

i believe the answer is E.

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What is the length of the red segment in the graph below?
Anon25 [30]

Answer: b 1.20

Step-by-step explanation:

b 1.20

7 0
1 year ago
Evaluate the function at the indicated values. f(x)= 6x^2+ 5x -12
seropon [69]

Step-by-step explanation:

We need to find the value of the function at the indicated value.

f(x)= 6x^2+ 5x -12

Indicated values are : f(-7), f(0), f(6), f(7)

To find f(-7), put x = -7 in the given function

f(-7)= 6(-7)^2+ 5(-7) -12=247

To find f(0), put x = 0 in the given function

f(0)= 6(0)^2+ 5(0) -12=-12

To find f(6), put x = 6 in the given function

f(6)= 6(6)^2+ 5(6) -12=234

To find f(7), put x = 7 in the given function

f(7)= 6(7)^2+ 5(7) -12=317

Hence, this is the required solution.

8 0
3 years ago
Which table does not represent a function?
andrew-mc [135]

the correct option is answer choice D, because there are multiple 2's in the x column with different f(x) values.

4 0
3 years ago
Which expression is equivalent to (4g3h2k4)3
RSB [31]

Answer:

d) 8g^{6}h^{4}  k^{12}  - (h^{25} k^{15} )

      \frac{(4g^{3} h^{2}k^{4} )^{3}  }{8g^{3}h^{2}  } - (h^{5} k^{3} )^{5}    = 8g^{6}h^{4}  k^{12}  - (h^{25} k^{15} )

Step-by-step explanation:

<u><em>Explanation</em></u>

Given expression

           =     \frac{(4g^{3} h^{2}k^{4} )^{3}  }{8g^{3}h^{2}  } - (h^{5} k^{3} )^{5}

By using

        (ab)ⁿ = aⁿbⁿ

         \frac{a^{m} }{a^{n} } = a^{m-n}

   =     \frac{(4)^{3} g^{9} h^{6}k^{12} )  }{8g^{3}h^{2}  } - (h^{5} k^{3} )^{5}

After simplification , we get

  = 8g^{9}g^{-3} h^{6} h^{-2} k^{12}  - (h^{5} k^{3} )^{5}

 = 8g^{9-3}h^{6-2}  k^{12}  - (h^{5} k^{3} )^{5}

= 8g^{6}h^{4}  k^{12}  - (h^{25} k^{15} )

3 0
2 years ago
Find the difference <br> (6ab+4a-3) - (2ab-4)
Y_Kistochka [10]
This would be 8(ab) sooo yaaaa
3 0
3 years ago
Read 2 more answers
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