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ELEN [110]
3 years ago
14

Six multiplication problems involved in solving 325 x 89

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer:

(300*80) + (300*9) + (20*80) + (20*9) + (5*80) + (5*9) = 28925

Step-by-step explanation:

Given;

325 x 89

Expand 325 = 300 + 20 + 5

Expend 89.0 = 80 + 9

now multiple to two expanded numbers;

(300 + 20 + 5) x (80 + 9)

distribute the number accordingly as follows;

(300*80) + (300*9) + (20*80) + (20*9) + (5*80) + (5*9), now we have Six multiplication problems.

= 24,000 + 2700 + 1600 + 180 + 400 + 45

= 28925

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Solve this quadratic equation using the quadratic formula. 3x2 + 5x + 1 = 0
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Zarrin [17]

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Step-by-step explanation:

You can do this using the quadratic equation:

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Then:

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Now we plug in our coefficients and solve:

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x =\dfrac{-12+ \sqrt{12^{2}-4(4)(-135)}}{2(4)}              x =\dfrac{-12- \sqrt{12^{2}-4(4)(-135)}}{2(4)}

x =\dfrac{-12+ \sqrt{144+2160}}{8}                        x =\dfrac{-12- \sqrt{144+2160}}{8}

x =\dfrac{-12+ \sqrt{2304}}{8}                                 x =\dfrac{-12- \sqrt{2304}}{8}

x =\dfrac{-12+ 48}{8}                                        x =\dfrac{-12- 48}{8}

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So if you are looking for a positive solution, just take the positive one as x.

8 0
3 years ago
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