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kramer
3 years ago
11

A=11+4b-4c. Solve for C.

Mathematics
1 answer:
devlian [24]3 years ago
6 0
Hey Joshua!

The correct answer is option A

a= 11 + 4b - 4c
-4c = a + 11 + 4b
Divide both sides by -4
-4c/-4 = (a + 11 + 4b)/-4
c = a -4b - 11/-4


I hope I helped!
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Three consecutive odd integers have a sum of 27. Find the integers.
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\sf \bf {\boxed {\mathbb {GIVEN:}}}

Sum of three consecutive odd integers = 27

\sf \bf {\boxed {\mathbb {TO\:FIND:}}}

The values of the three integers.

\sf \bf {\boxed {\mathbb {SOLUTION:}}}

\sf\purple{The\:three\:consecutive \:odd\:integers\:are\:7,\:9\:and\:11.}

\sf \bf {\boxed {\mathbb {STEP-BY-STEP\:\:EXPLANATION:}}}

Let us assume the three consecutive odd integers to be x, (x+2) and (x+4).

As per the condition, we have

Sum \:  \:  of \:  \:  the  \:  \: three \:  \:  consecutive \:  \:  odd \:  \:  integers  = 27

➺ \: x + (x + 2) + (x + 4) = 27

➺ \: x + x + 2 + x + 4 = 27

Now, collect the like terms.

➺ \: (x + x + x) + (2 + 4) = 27

➺ \: 3x + 6 = 27

➺ \: 3x = 27 - 6

➺ \: 3x = 21

➺ \: x =  \frac{21}{3} \\

➺ \: x = 7

Therefore, the three consecutive odd integers whose sum is 27 are \boxed{  7  }, \boxed{ 9   } and \boxed{ 11   } respectively.

\sf \bf {\boxed {\mathbb {TO\:VERIFY :}}}

⇢ 7 + 9 + 11 = 27

⇢ 27 = 27

⇢ L. H. S. = R. H. S.

\sf\blue{Hence\:verified.}

\red{\large\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese \: \: Mystique35ヅ}}}}}

4 0
3 years ago
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