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katovenus [111]
3 years ago
11

That’s what i’ve done so far

Mathematics
1 answer:
Sati [7]3 years ago
8 0

Answer:

• for x

{ \tt{(4x + 1) + 2(x + 5) = 20}} \\ { \tt{4x + 1 + 2x + 10 = 20}} \\  { \tt{6x + 11 = 20}} \\ { \tt{6x = 20 - 11}} \\{ \tt{6x = 9}} \\{ \boxed { \tt{x =  \frac{3}{2} = 1.5 }}}

• for A:

{ \tt{2x + 4 = A}} \\  { \tt{A = 2( \frac{3}{2}) + 4 }} \\ { \tt{A = 3 + 4}} \\ { \boxed{ \tt{A = 7}}}

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What is the solution set for the given inequality if the replacement set for r is {5, 6, 7, 8, 9, 10}? 2r ≤ 3r – 8 A. {8, 9, 10}
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2r\leq3r-8\\
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3 years ago
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VLD [36.1K]

Step-by-step explanation:

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combine like terms

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8 0
3 years ago
Read 2 more answers
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