How Do You Solve (8x+7)^3?
2 answers:
<span><span>(<span><span>8x</span>+7</span>)</span>*<span>(<span><span>8x</span>+7</span>)</span></span>*<span>(<span><span>8x</span>+7</span><span>) </span></span><span>(<span><span>8x</span>+7</span>)</span><span>(<span><span><span>64<span>x^2</span></span>+<span>112x</span></span>+49</span><span>) </span></span><span><span><span><span><span><span><span>(<span>8x</span>)</span><span>(<span>64<span>x^2</span></span>)</span></span>+<span><span>(<span>8x</span>)</span><span>(<span>112x</span>)</span></span></span>+<span><span>(<span>8x</span>)</span><span>(49)</span></span></span>+<span><span>(7)</span><span>(<span>64<span>x^2</span></span>)</span></span></span>+<span><span>(7)</span><span>(<span>112x</span>)</span></span></span>+<span><span>(7)</span><span>(49) </span></span></span><span><span><span><span><span>512<span>x^3</span></span>+<span>896<span>x^2</span></span></span>+<span>392x</span></span>+<span>448<span>x^2</span></span></span>+<span>784x</span></span>+<span>343 </span>Answer: <span><span><span>512<span>x^3</span></span>+<span>1344<span>x^2</span></span></span>+<span>1176x</span></span>+<span>343</span>
I would use pascal's triangle goes like this 0 1 1 1 1 2 1 2 1 3 1 3 3 1 the numbers on the left are the power of the binomial baseically for the third power (a+b)^3=1a^3b^0+3a^2+b^1+3a^1+b^2+1a^0+b^3 so you have (8x+7)^3=1(8x)^3(7)^0+3(8x)^2(7)^1+3(8x)^1(7)^2+1(8x)^0(7)^3= 512x^3+1344x^2+1176x+343
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Never is the right answer .
Really true. just wanting points is not a reason to mess with someone else.
<span>This is an isosceles triangle. If each side is 1, we can find the hypotenuse using the Pythagorean theorem. a^2 + b^2 = c^2 1^2 + 1^2 = c^2 1 + 1 = c^2 2 = c^2 root 2 = c Thus, the answer is root 2. Hope this helps :)</span>
Answer: 21, 22, 23
Step-by-step explanation:
you divide 66 with 3, that equals 22, then you have three 22. Then you have to take 1 from one of the 22s and give it to the other one, making it 21, 22 and 23