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ira [324]
4 years ago
13

2x < - x + 20 + 7 I don't get how to answer this question

Mathematics
1 answer:
dimulka [17.4K]4 years ago
6 0

ANSWER

X < 9

Solve

2x < - x + 20 + 7

2x < - x + 27 change the signs

2x + x < 27 add the like terms

3x < 27 divide by 3

x < 9

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3 years ago
BRAINLIEST ✨
inna [77]

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, or</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, or</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, or</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.Check: (2 + 2/3), (2 + 3*2/3), (2 + 6*2/3)</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.Check: (2 + 2/3), (2 + 3*2/3), (2 + 6*2/3)= 8/3, 4, 6</em>

<em>Solution: T2 = a+d, T4 =a+3d. T7 = a+6d.In the GP: (a+d)(a+6d) = (a+3d)^2, ora^2+7ad+6d^2 = a^2+6ad+9d^2, orad = 3d^2, ora = 3d.a = 2, d = 2/3.The common ratio of the GP is 3/2. Answer.Check: (2 + 2/3), (2 + 3*2/3), (2 + 6*2/3)= 8/3, 4, 6Common ratio: 4/(8/3) = 3/2 same as 6/4 = 3/2. Correct.</em>

Step-by-step explanation:

Am I right? just commet if I'm wrong

then <em>TANKYOU>^_^<!</em>

7 0
3 years ago
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