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LuckyWell [14K]
3 years ago
13

A real world situation that can be model with 3b - 5=22

Mathematics
2 answers:
STatiana [176]3 years ago
5 0
First add plus 5 to -5 and 22. Then -5 cancels outdo now u have 3b=27. Divide 3 to 3b and 27. Then you will have b=9
Serhud [2]3 years ago
5 0

Jon had put $5 in the bank. He was getting Interest from the bank (the bank was paying him the longer he kept his money in the bank) it was $3. Some time went by and he now had $22.


b = 9 (HE GOT $9 A WEEK! BUT YOU TAKE AWAY 5 BECAUSE HE PUT THAT MONEY IN NOT THE BANK)

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Which brand of rice is the better buy? Explain your reasoning.
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So you have to divide the amount by how many you’re buying and get the unit rate per each ounce so the answer is b
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Doe's anyone understand this?
Kitty [74]

just subtracting by 2x on both sides

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If Babe Ruth average 46 home runs per year, how many home runs did he hit in 12 years?
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Which of the following numbers could be the sides of a right triangle. A)8,15,17. B)6,8,12 C)9,11,21 D)4,13,16
ArbitrLikvidat [17]

Answer:

A) 8, 15, 17.

Step-by-step explanation:

Right triangle obey the Pythagorean theorem. Thus, we choose the two smaller numbers (being the cathetus) and if after applying the P. Theorem we get the biggest of each option  (the hypotenuse) that means that those numbers could be the sides of a right triangle.

The Pythagorean theorem states that: a^2+b^2=c^2

Thus:

\sqrt{a^2+b^2}=c

Option A:

\sqrt{8^2+15^2}=17 → 17 = 17 OK!

Option B:

\sqrt{6^2+8^2}  = 10 → 10 ≠ 12 NO

Option C:

\sqrt{9^2+11^2} = [tex]\sqrt{202}[/tex] → \sqrt{202} ≠ 21 NO

Option D:

\sqrt{4^2+13^2} = \sqrt{185} → \sqrt{185} ≠ 16 NO

3 0
3 years ago
Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.
NISA [10]

Answer:

\frac{\sqrt{\pi}}{4}

Step-by-step explanation:

You are going to integrate the following function:

g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}  (1)

furthermore, you know that:

\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}

lets call to this integral, the integral Io.

for a general form of I you have In:

I_n=\int_0^{\infty}x^ne^{-ax^2}dx

furthermore you use the fact that:

I_n=-\frac{\partial I_{n-2}}{\partial a}

by using this last expression in an iterative way you obtain the following:

\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}} (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}

5 0
3 years ago
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