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kvv77 [185]
2 years ago
14

7 and a number equals 21​

Mathematics
1 answer:
torisob [31]2 years ago
5 0

Answer:

7x3=21

Step-by-step explanation:

7+7+7=21, 7x3=21

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Just 24 plz, I am struggling
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check the pictured below.


now, notice that APB is just a flat-line and therefore those two angles there add up to 180°.

also notice that since ∡ACB is 90°, then those two angles add up to that much.


\bf \stackrel{\measuredangle ACP}{(3x+2y)}+\stackrel{\measuredangle BCP}{(3x+4y)}=90\implies 6x+6y=90\implies \stackrel{\textit{common factor}}{6(x+y)}=90
\\\\\\
x+y=\cfrac{90}{6}\implies x+y=15\implies \boxed{x=15-y}
\\\\[-0.35em]
~\dotfill


\bf \stackrel{\measuredangle APC}{(7x+3)}+\stackrel{\measuredangle BPC}{(16y)}=180\implies 7x+16y=177\implies 7\left( \boxed{15-y} \right)+16y=177
\\\\\\
105-7y+16y=177\implies 9y=72\implies y=\cfrac{72}{9}\implies \blacktriangleright y=8 \blacktriangleleft
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
\measuredangle BPC=16(8)\implies \measuredangle BPC=128

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3 years ago
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\dfrac{-14x^3}{x^3-5x^4}\qquad\text{where}\ x^3-5x^4\neq0\\\\x^3-5x^4\neq0\qquad\text{distributive}\\\\x^3(1-5x)\neq0\iff x^3\neq0\ \wedge\ 1-5x\neq0\\\\x\neq0\ \wedge\ x\neq\dfrac{1}{5}\\\\\dfrac{-14x^3}{x^3(1-5x)}\qquad\text{cancel}\ x^3\\\\=\dfrac{-14}{1-5x}\qquad\text{where}\ x\neq\dfrac{1}{5}

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3 years ago
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