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Ket [755]
3 years ago
15

(x - 2)(-5x2 + x) = (x)(-5x2)+(*)(x) + (-2)(-5x2)+(-2)(x) is an example of:

Mathematics
1 answer:
IgorLugansk [536]3 years ago
3 0

Answer:

x=0 or x=1

Step-by-step explanation:

(x−2)(−5x2+x)=x(−5x2)+x−2(−5x2)−2x

−5x3+11x2−2x=−5x3+10x2−x

Step 1: Add 5x^3 to both sides.

−5x3+11x2−2x+5x3=−5x3+10x2−x+5x3

11x2−2x=10x2−x

Step 2: Subtract 10x^2-x from both sides.

11x2−2x−(10x2−x)=10x2−x−(10x2−x)

x2−x=0

Step 3: Factor left side of equation.

x(x−1)=0

Step 4: Set factors equal to 0.

x=0 or x−1=0

x=0 or x=1

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The ratio of defective ones to new ones tested is 8:50 or 4:25 or 16% .
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2 years ago
Solve using the method of elimination and determine if the system has 1 solution, no solutions, or infinite solutions
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\begin{array}{rrrrr} 10x&-&18y&=&2\\ -5x&+&9y&=&-1 \end{array}~\hfill \implies ~\hfill \stackrel{\textit{second equation }\times 2}{ \begin{array}{rrrrr} 10x&-&18y&=&2\\ 2(-5x&+&9y&)=&2(-1) \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{rrrrr} 10x&-&18y&=&2\\ -10x&+&18y&=&-2\\\cline{1-5} 0&+&0&=&0 \end{array}\qquad \impliedby \textit{another way of saying \underline{infinite solutions}}

if we were to solve both equations for "y", we'd get

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8 0
2 years ago
I did all the work, please just check my answers.
hammer [34]
Yes uu did it correctly
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2 years ago
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