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mash [69]
2 years ago
7

the admission fee at a small fair is $1.50 for children and $4.00 for adults. on a certain day, $5,050 is collected. if $1,00 ad

ults attended the fair, write and solve linear equation to find the number of children that attended
Mathematics
1 answer:
maks197457 [2]2 years ago
5 0

Answer:

↓

Step-by-step explanation:

Ok here:

(5050-[100·4])÷1.50=k(kids)

(5050-400)÷1.50=k

3100=k

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Use mathematical induction to prove that for each integer n > 4,5" > 2^2n+1 + 100.
Flura [38]

Answer:

The inequality that you have is 5^{n}>2^{2n+1}+100,\,n>4. You can use mathematical induction as follows:

Step-by-step explanation:

For n=5 we have:

5^{5}=3125

2^{(2(5)+1)}+100=2148

Hence, we have that 5^{5}>2^{(2(5)+1)}+100.

Now suppose that the inequality holds for n=k and let's proof that the same holds for n=k+1. In fact,

5^{k+1}=5^{k}\cdot 5>(2^{2k+1}+100)\cdot 5.

Where the last inequality holds by the induction hypothesis.Then,

5^{k+1}>(2^{2k+1}+100)\cdot (4+1)

5^{k+1}>2^{2k+1}\cdot 4+100\cdot 4+2^{2k+1}+100

5^{k+1}>2^{2k+3}+100\cdot 4

5^{k+1}>2^{2(k+1)+1}+100

Then, the inequality is True whenever n>4.

3 0
3 years ago
Allison’s cupcake and cookie budget for her party is $105. She would like to have at least 40 cupcakes and cookies combined, and
FinnZ [79.3K]

Answer:

a + b ≥ 40

a + 5 ≤ b

2.50a + 1.50b ≤ 105

Step-by-step explanation:

We are told she would like to have at least 40 cupcakes and cookies combined.

If a is number of cupcakes and b is number of cookies, then this inequality is;

a + b ≥ 40

Secondly, we are told she would like to have at most 5 more cupcakes than cookies.

Thus, the inequality is;

a + 5 ≤ b

Lastly, we are told that cupcakes sell for $2.50 and cookies for $1.50

Thus, the inequality is;

2.50a + 1.50b ≤ 105

Finally, the 3 inequalities are;

a + b ≥ 40

a + 5 ≤ b

2.50a + 1.50b ≤ 105

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Step-by-step explanation:

I think it's B

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