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lakkis [162]
3 years ago
9

A and b are positive integers. Find

gn="absmiddle" class="latex-formula"> if a^3b^2=72
Mathematics
1 answer:
lord [1]3 years ago
7 0

A could be 2 while B could be 3, so -2a+3b turns into -4+9, which equals 5.

From what I know you can't really solve a a single equation with two-variables so it's just a matter of trial and error.

Just try plugging in a small number like 2 for a just to try it and you get 8b^2=72.

Divide everything by 8 to isolate b and you get that b^2=9.

Square root everything and you'll find that b=3. This is just one possible combination, I'm sure there are many more but this is obviously the one that was intended to be found.

Now that we know that a=2 and b=3 just plug them into the equation.

-2(2)+3(3)=?

-4+9=?

5

Sorry about having to use this ^ symbol, the equation maker is not working.

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I have a question that has already been answered. I just need someone to please explain the steps.
Jobisdone [24]

Answer:

Step-by-step explanation:

Using brackets will really help.

y = (x^2 + 4x + ... ) - 5       You are trying to complete the square. The square in this case is a trinomial that is squared.

To do that, you take 1/2 the linear term (4x)/2, drop the x (4/2), and square the result (4/2)^2. The number is 2^2 which is 4.

So far what you have is

y = (x^2 + 4x + 4) - 5  

Now you just can't add 4 without adjusting it somehow. If you do, the whole question will change it's value. Because you added 4 inside the brackets, you must subtract 4 outside the brackets.

What that means is 4  inside - 4 outside. So it looks like this

y = (x^2 + 4x + 4) - 5 - 4

Now you continue on

y = (x + 2)^2 - 9  The 4 combines with the 5 to make nine.

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2 years ago
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Step-by-step explanation:

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3 years ago
In a certain Algebra 2 class of 29 students, 7 of them play basketball and 14 of them
Mariulka [41]

Answer:

<em>Two possible answers below</em>

Step-by-step explanation:

<u>Probability and Sets</u>

We are given two sets: Students that play basketball and students that play baseball.

It's given there are 29 students in certain Algebra 2 class, 10 of which don't play any of the mentioned sports.

This leaves only 29-10=19 players of either baseball, basketball, or both sports. If one student is randomly selected, then the propability that they play basketball or baseball is:

\displaystyle P=\frac{19}{29}

P = 0.66

Note: if we are to calculate the probability to choose one student who plays only one of the sports, then we proceed as follows:

We also know 7 students play basketball and 14 play baseball. Since 14+7 =21, the difference of 21-19=2 students corresponds to those who play both sports.

Thus, there 19-2=17 students who play only one of the sports. The probability is:

\displaystyle P=\frac{17}{29}

P = 0.59

3 0
3 years ago
- 30 = 12 – 6r<br><br> Pls help
oee [108]

Answer: r = 7

Step-by-step explanation:

Subtract 12 from both sides to isolate the r variable. You have -42 = -6r. Divide both sides by -6 to get r by itself and you get r = 7. Verify by substituting 7 as the r value and solving the equation.

8 0
2 years ago
Read 2 more answers
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