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leva [86]
3 years ago
14

What is the area, in square centimeters, of the parallelogram

Mathematics
1 answer:
Vsevolod [243]3 years ago
6 0

Answer: 96cm² or C

Step-by-step explanation:

Area of parallelogram = b × h

b = 12

h = 8

12 × 8 = 96

The area of this parallelogram is 96cm²

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There are 5 times as many males as females on the maths course at university. What fraction of the course are female? Give your
koban [17]

Answer: 1/6

Step-by-step explanation: Let the no of females be x

The no of males be 5x

The total no of males and females in the math course be 5x+x= 6x

The fraction of females is x/6x = 1/6

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3 years ago
Explain why 3(x+4)=3(x-5) has no solution. Choose the best response below.
Evgesh-ka [11]
Its either A or C i know that 
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3 years ago
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Luke has 48 books.He puts 7 books in a box.How many boxes can he fill?explain how you interpreted the remainder
Alex17521 [72]
48 books divided by 7 books per box. So since 7 can go into 48 6 times to make 42, the remainder is 6 books are left out. 6 boxes are filled. I interpreted the remainder by subtracting 42 from 48. And that's all I did. :)

I hope you like this answer and have a good night! :D
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3 years ago
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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)

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
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