Question

if A+A=30, B+B=20 and C+C=8. Then A+B×C=??

Answer 1

Answer:

The value of A+B×C = 55 .

Step-by-step explanation:

As given

if A+A=30, B+B=20 and C+C=8.

Take

A+A=30

2A = 30

$A= \frac{30}{2}$

A = 15

Take

B+B=20

2B = 20

$B= \frac{20}{2}$

B = 10

Take

C+C=8

2C = 8

$C= \frac{8}{2}$

C = 4

Thus

A+B×C= 15 + 10 × 4

          = 15 + 40

         = 55

Therefore the value of A+B×C = 55 .

Answer 2

A+BxC= 55 if A+A=30 , B+B=20 , C+C=8

Given:

• A+A=30
• B+B=20
• C+C=8

To Find:

• A+BxC=

Solution:

Step 1:

Solve for A

A + A = 30

=> 2A = 30

=> A = 15

Step 2:

Solve for B

B + B = 20

=> 2B = 20

=> B = 10

Step 3:

Solve for C

C + C = 8

=> 2C = 8

=> C = 4

Step 4:

Substitute values of A , B and C in A + B x C  and apply BODMAS

15 + 10 x  4

First Multiply the numbers

= 15 + 40

Then Add the numbers

= 55

Hence A+BxC= 55

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