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Tresset [83]
2 years ago
15

Which expression is equivalent to 486 – 9 + 6 + 33 × 2? A . [(486 – 9) + (6 + 33)] × 2 B . 486 – [(9 + 6 + 33) × 2] C . (486 – 9

) + 6 (33 × 2) D . (486 – 9 + 6) + (33 × 2)
Mathematics
1 answer:
hichkok12 [17]2 years ago
3 0

Answer:

D . (486 – 9 + 6) + (33 × 2)

Step-by-step explanation:

Which expression is equivalent to 486 – 9 + 6 + 33 × 2?

486 – 9 + 6 + 33 × 2

486 – 9 + 6 + 66

477 + 6 + 66

549

D . (486 – 9 + 6) + (33 × 2)

      477 + 6 + 66

               549

Hope this helps!

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Answer:

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Step-by-step explanation:

-18h = 17 = xh/6

Subtract xh/6 from both sides of the equation.

-18 - xh/6 = 17

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Use induction to prove: For every integer n &gt; 1, the number n5 - n is a multiple of 5.
nignag [31]

Answer:

we need to prove : for every integer n>1, the number n^{5}-n is a multiple of 5.

1) check divisibility for n=1, f(1)=(1)^{5}-1=0  (divisible)

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3) Induction,

f(k+1)=(k+1)^{5}-(k+1)

=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1

=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k

Now, f(k+1)-f(k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)

f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k

f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k

Take out the common factor,

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add both the sides by f(k)

f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)

We have proved that difference between f(k+1) and f(k) is divisible by 5.

so, our assumption in step 2 is correct.

Since f(k) is divisible by 5, then f(k+1) must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number n^{5}-n is a multiple of 5.

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Answer: 3(9b+8)

Step-by-step explanation: Have a brilliant day of learning!-Lily ^-^

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