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Sonja [21]
3 years ago
7

What is 13\15-1\5 and what is the simplified answer

Mathematics
1 answer:
valentinak56 [21]3 years ago
4 0
13/15 - <span>1/5 </span>
<span>common denominator is 15 </span>
<span>Multiply numerator and denominator of 1/5 by 3 = 3/15 </span>
<span>13/15 - 3/15 = 10/15 </span><span>= 2/3

2/3 is your answer.</span>
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Answer:

\pmb{3x^4-3x^3-3x^2+2x+2 }

<h3>Step to step explanation:</h3>

we have to subtract 5×⁴-4׳+3ײ-2×from 8×⁴-7׳+2

8x^4-7x^3+2-(5x^4-4x^3+3x^2-2x)\\\\8x^4-7x^3+2-5x^4+4x^3-3x^2+2x\\\\3x^4-3x^3-3x^2+2x+2

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If α and β are the zeroes of the polynomial 6y 2 − 7y + 2, find a quadratic polynomial whose zeroes are 1 α and 1 β .
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Answer:

2y^2-7y+6=0

Step-by-step explanation:

We are given that \alpha and \beta are the zeroes of the polynomial 6y^2-7y+2

y^2-\frac{7}{6}y+\frac{1}{3}

We have to find a quadratic polynomial whose zeroes are 1/\alpha and 1/\beta.

General quadratic equation

x^2-(sum\;of\;zeroes)x+ product\;of\;zeroes

We get

\alpha+\beta=\frac{7}{6}

\alpha \beta=\frac{1}{3}

\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}

\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7/6}{1/3}

\frac{1}{\alpha}+\frac{1}{\beta}=\frac{7}{6}\times 3=7/2

\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{\alpha \beta}

\frac{1}{\alpha}\times \frac{1}{\beta}=\frac{1}{1/3}=3

Substitute the values

y^2-(7/2)y+3=0

2y^2-7y+6=0

Hence, the quadratic polynomial whose zeroes are 1/\alpha and 1/\beta is given by

2y^2-7y+6=0

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