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aev [14]
3 years ago
6

In a survey of market, it is found that 143 persons use white toothpaste and 135 use red toothpaste. If 70 of them use both the

toothpaste then find the number of persons who use at least one of the toothpaste​
Mathematics
1 answer:
denpristay [2]3 years ago
6 0

Answer:

let W and R represent white toothpaste and red toothpaste represently

n(W)=143

n(R)=135

n(WnR)=70

n( WuR)=?

Now,

n(WuR)=n(W)+n(R)_n(WnR)

=143+135_70

=208

that's all don't forget to write therefore.

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Oki so help plz and show proof or evidence
muminat

Answer:

Option B because the initial fee stays the same which means that there will still be + 70 and then they added 60.

60t + 70

Step-by-step explanation:

5 0
2 years ago
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
2 years ago
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6 0
3 years ago
Read 2 more answers
2 and 2 3rds divided by 1 and 3 5ths
il63 [147K]

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Hope this helps!

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3 years ago
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Answer:D


Step-by-step explanation:


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3 years ago
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