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sergejj [24]
3 years ago
5

How to solve this equation 15(6-g)

Mathematics
1 answer:
NikAS [45]3 years ago
7 0
<span>15<span>(<span>6−g</span>)</span></span><span>=<span><span>(15)</span><span>(<span>6+<span>−g</span></span>)</span></span></span><span>=<span><span><span>(15)</span><span>(6)</span></span>+<span><span>(15)</span><span>(<span>−g</span>)</span></span></span></span><span>=<span>90−<span>15g</span></span></span><span>=<span><span>−<span>15g</span></span>+<span>90</span></span></span>
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Sabendo que a soma de três termos consecutivos de uma progressão aritmética é 18 e que o seu produto é 16, determina esses númer
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Step-by-step explanation:

6 0
3 years ago
HELP PLEASE! BRAINLIEST
V125BC [204]
Since AED forms a straight line, all involved angles sum to 180°. Therefore AEB + BED = 180. Also, since EC bisects BED, BEC = CED, and BED = 2× CED. Now to substitute the first equation:
AEB + BED = 180
AEB + 2×CED = 180
11x-12 + 2(4x+1) = 180
11x-12+8x+2 = 180
19x-10 = 180
19x-10+10 = 180+10
19x = 190
x = 10
So what is m<AEC?? It is the sum of AEB + BEC, and since BEC = CED we can say that:
AEC = AEB + CED
AEC = 11x-12 + 4x+1 = 15x-11 = 15(10)-11 = 150-11
m<AEC = 139°
3 0
4 years ago
3
atroni [7]
I’m not sure what the answer is , hopefully some ine can help you 66
6 0
3 years ago
Simplify and Show your work
BlackZzzverrR [31]

1.\frac{ 5( {4x}^{3}  {y}^{2} ) ^{3}}{( {4x}^{5} {y}^{3}) ^{4}   } \\  =  \frac{5 \times  {4}^{3}  \times {x}^{3 \times 3}  \times  {y}^{2 \times 3}  }{ {4}^{4}  \times  {x}^{5 \times 4} \times  {y}^{3 \times 4}   }  \\  =  \frac{5 \times 64 \times  {x}^{9} \times  {y}^{6}  }{256 \times  {x}^{20} \times  {y}^{12}  }  \\  =  \frac{320 {x}^{9} {y}^{6}  }{256 {x}^{20}  {y}^{12} }  \\  =  \frac{5 {x}^{9 - 20}  {y}^{6 - 12} }{4}  \\  =  \frac{5}{4 {x}^{11}  {y}^{6} }

2. {3}^{5x - 4}  =  {9}^{2x + 12}  \\  =  > {3}^{5x - 4}  =  {3}^{2(2x + 12)}  \\ =  >  {3}^{5x - 4}  =  {3}^{4x + 24}  \\ =  > 5x - 4 = 4x + 24 \\  =  > 5x - 4x = 24  + 4 \\  =  > x = 28

Hope you could understand.

If you have any query, feel free to ask.

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3 years ago
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