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Drupady [299]
3 years ago
8

Solve Q=3a+5ac for a

Mathematics
1 answer:
lilavasa [31]3 years ago
8 0
Here is how you solve... 

Q= 3a+5ac 

take out the common variable.
Q= a(3+5c)

get A by its self so divide each side by (3+5c).
Q/a(3+5c)= a(3+5c)/(3+5c)

Q/a(3+5c)=A

So your answer is A= Q/a(3+5c)

hope that helped.
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-3 is one answer and -5 is another answer
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3 years ago
How many solutions does the system have?
alisha [4.7K]

Answer:

y = 3/2 , x = 3/2

Step-by-step explanation:

Solve the following system:

{x + y = 3 | (equation 1)

10 x = 15 | (equation 2)

Divide equation 2 by 5:

{y + x = 3 | (equation 1)

0 y+2 x = 3 | (equation 2)

Divide equation 2 by 2:

{y + x = 3 | (equation 1)

0 y+x = 3/2 | (equation 2)

Subtract equation 2 from equation 1:

{y+0 x = 3/2 | (equation 1)

0 y+x = 3/2 | (equation 2)

Collect results:

Answer: {y = 3/2 , x = 3/2

6 0
3 years ago
Is this correct?<br><br><br><br><br> pls help math geniuses!!
lukranit [14]

Answer:

option B is correct

Brainliest Plzzz

6 0
2 years ago
Read 2 more answers
Convert 150 base 10 to a binary <br>number.
STALIN [3.7K]

Answer:

10010110

Step-by-step explanation:

A binary number is basically a number in base 2. So, each place represents a power of two. The ones place represents 1 or 2^0, the tens represents 2 or 2^1, the hundreds represents 4 or 2^2 and so on. To find the highest place needed to make in the number in binary, we find the greatest power of two that is below 150. In this case, it is 128. We can start by writing 128 in binary:

10000000

Now, we subtract 128 from 150 and find the largest power of two that is smaller than the difference. The largest power of 2 smaller than 22 is 16, so we put a one in the place for 2^4:

10010000

We can subtract 16 from 22 to get what is leftover to continue the process. We get 6, and the largest power of 2 smaller is 4. We can place a one in the hundreds place to represent it:

10010100

What is left is 6-4 or 2, which is just 2^1, so we can place it in the tens place:

10010110

8 0
2 years ago
I don't know how to work this problem out
madam [21]
1/5(x-y) = 1
x+y = 9

x-y = 5 (x5)
x+y = 9

First you want to create both equations so at least one of the variables (x or y) are the same in both equations, so that is why I multiplied the top one by 5, you multiply the whole equation (both sides)

2x = 14
x = 7

Then you either Plus or minus one from the other, I plused the top one onto the bottom one, then solved for x

7 + y = 9
y = 2

Then put x back into one of the first two equations to get Y


4 0
3 years ago
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