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3 years ago

The Smith family went to the theater

Mathematics

1 answer:

Ivenika [448]3 years ago

7 0

Answer:

3 adult tickets and 4 child tickets

Step-by-step explanation:

a + k = 7

a = 7 - k

12a + 9k = 72

12(7 - k) + 9k = 72

84 - 3k = 72

3k = 12

k = 4

a = 3

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Step-by-step explanation:

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Use mathematical induction to prove that for each integer n ≥ 4, 5^n ≥ 2 2^n+1 + 100

anastassius [24]

The given Statement which we have to prove using mathematical induction is

$5^n\geq 2*2^{n+1}+100$

for , n≥4.

⇒For, n=4

LHS

$=5^4\\\\5*5*5*5\\\\=625\\\\\text{RHS}=2.2^{4+1}+100\\\\=64+100\\\\=164$

LHS >RHS

Hence this statement is true for, n=4.

⇒Suppose this statement is true for, n=k.

$5^k\geq 2*2^{k+1}+100$

-------------------------------------------(1)

Now, we will prove that , this statement is true for, n=k+1.

$5^{k+1}\geq 2*2^{k+1+1}+100\\\\5^{k+1}\geq 2^{k+3}+100$

LHS

$5^{k+1}=5^k*5\\\\5^k*5\geq 5 \times(2*2^{k+1}+100)----\text{Using 1}\\\\5^k*5\geq (3+2) \times(2*2^{k+1}+100)\\\\ 5^k*5\geq 3\times (2^{k+2}+100)+2 \times(2*2^{k+1}+100)\\\\5^k*5\geq 3\times(2^{k+2}+100)+(2^{k+3}+200)\\\\5^{k+1}\geq (2^{k+3}+100)+3\times2^{k+2}+400\\\\5^{k+1}\geq (2^{k+3}+100)+\text{Any number}\\\\5^{k+1}\geq (2^{k+3}+100)$

Hence this Statement is true for , n=k+1, whenever it is true for, n=k.

Hence Proved.

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3 years ago