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NISA [10]
3 years ago
9

HELP PLEASE WHICH ONE IS CORRECT?

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
7 0

I got chu UwU, the answer is B cause it is smaller

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R=5(CA−0.3) solve for c
Maurinko [17]
R = 5(CA - 0.3)

<span>Multiply everything in the parenthesis by 5.</span>

R = 5CA - 1.5

Add 1.5 to both sides

1.5 + R = 5CA

Divide 5A on both sides.

C = 1.5 + R / 5A

Hope this helps!

4 0
3 years ago
Read 2 more answers
Need help with 7+2=16-?
N76 [4]

Answer:

7

Step-by-step explanation:

7+2= 9

16-9=7

7 0
3 years ago
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Solve for x<br> 1-1=1-7<br><br> O no solution<br> O x = 5 or x = 10<br> O x = 5<br> O x = 10
Triss [41]

Answer:

7 9. If you have any fractions, get rid of those first by multiplying ALL ... to model and solve 3n 6 15. Then solve the equation. 10. 13 x = 32 + 5 x 3.

Step-by-step explanation:

3 0
3 years ago
The graph below shows the number of visitors at a theme park. How many more visitors in total went to the park in June, July and
Olin [163]
From the information given by the graph

Year 1 visitors
June = 9450
July = 9500
August = 9600

Year 2 visitors
June = 9450
July = 9600
August = 9800

Total visitor Year 1 = 9450 + 9500 + 9600 = 28550
Total visitor Year 2 = 9450 + 9600 + 9800 = 28850

Visitor in Year 2 is 28850-28550 = 300 more than visitor in Year 1
8 0
4 years ago
A factory has three machines making energy drinks. When all three machines are running, 520 drinks are made each hour. When only
Vinil7 [7]

Let A, B and C be the amount machines A, B and C, respectively, produce each hour alone.

With all three machines together produce 520 drinks each hour, we know that the addition of the amount that each produces each hour is equal to 520, that is.

A+B+C=520_{}

Now, similarly, machines A and B produce together 320 drinks each hour, so the sum of the amounts A and B produce each hour is equal to 320, so:

A+B=320

If machine C produce 30 more drinks than B each hour, than the amount of B plus 30 will be the amount of C:

C=B+30

So, we got the system of equations:

\begin{gathered} A+B+C=520 \\ A+B=320 \\ C=B+30 \end{gathered}

To solve, notice that A + B is known, because of the second equation, so we can substitute A + B by 320 in the first equation:

\begin{gathered} (A+B)+C=520 \\ 320+C=520 \\ C=520-320=200 \end{gathered}

Now, we can substitute C into the third equation and solve for B:

\begin{gathered} C=B+30 \\ 200=B+30 \\ B+30=200 \\ B=200-30=170 \end{gathered}

And, finally, we can substitute B into the second equation and solve for A:

\begin{gathered} A+B=320 \\ A+170=320 \\ A=320-170=150 \end{gathered}

Answers:

First box:

A+B+C=520

Second box:

A+B=320

Third box:

C=B+30

Fourth box:

150

3 0
1 year ago
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