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Ivanshal [37]
3 years ago
14

I need help. With this p

Mathematics
1 answer:
iogann1982 [59]3 years ago
6 0

a_n=a_1+(n-1)d - the formula of n-th term of an arithmetic sequence.

We have:

a_5=12.4;\ a_9=22.4

a_9-a_5=(9-5)d\to4d=22.4-12.4\\\\4d=10\ \ \ \ |:4\\\\d=2.5

a_{31}=a_9+(31-9)d\to a_{31}=22.4+22\cdot2.5=77.4

Answer: C. 77.4

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Pls help!!!! Rewrite the expression. <br> 4+4+5x2x5+(3+3+3)x6x6+2+2
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Choose one of the factors of 500x3 + 108y18
sammy [17]

Answer:

Option C

Therefore, one of the factors of 500x^3 +108y^\left (18\right ) is (5x+3y^6).

Step-by-step explanation:

Given: 500x^3 +108y^\left (18\right )

the common factor from 500x^3 and 108y^\left (18\right ) is 4.

therefore, 4\cdot \left ( 125x^3+27y^\left (18\right ) \right )

4\cdot \left ( (5x)^3+(3y^6)^3 \right))

Now, use the formula for above expression: (a^3+b^3)=(a+b)(a^2-ab+b^2)

here, a=5x and b=3y^6

( (5x)^3+(3y^6)^3 \right))=(5x+3y^6)(25x^2-15xy^6+9y^12)

Therefore, we have

500x^3 +108y^\left (18\right )=4\cdot (5x+3y^6)\left (25x^2-15xy^6+9y^\left ( 12 \right )  \right )

Therefore, one of the factors of 500x^3 +108y^\left (18\right ) is (5x+3y^6).








7 0
3 years ago
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