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3 years ago

Given that QN = QM, find NQM. A. 40° B. 80° C. 100° D. 140°

Mathematics

2 answers:

Helga [31]3 years ago

6 0

Hello there.

Given that QN = QM, find NQM.

D. 140°

labwork [276]3 years ago

4 0

Thank you for posting your question here at brainly. Feel free to ask more questions.

The best and most correct answer among the choices provided by the question is D. 140°.
 
Hope my answer would be a great help for you.

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Use induction to prove: For every integer n > 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)

2) Assume that $f(k)$ is divisible by 5, $f(k)=(k)^{5}-k$

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,

$f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)$ (divisible by 5)

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.

3 0

3 years ago

Am I correct or not???

pentagon [3]

Yes that is correct.Well done!:D

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142 x 12 in algorithm standard and in area model

Katen [24]

Answer:

0333

Step-by-step explanation:

000

0000

1112222

______

11122222 ans

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SOMEONE PLEASE EXPLAIN THIS TO ME!!!! Ty! (Problems like 8-15)!!!!

Vilka [71]

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cricket20 [7]

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Step-by-step explanation:

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