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Pavlova-9 [17]
3 years ago
9

According to the Rational Root Theorem, which is a factor of the polynomial f(x) = 60x^4 + 86x^3 - 46x^2 - 43x + 8?

Mathematics
1 answer:
Natalija [7]3 years ago
3 0
Factors of 8 over factorsof 60, +/- one of them ay be a root
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Solve for (y)(x)<br> x-y=1<br> 1/2x+y=5
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Using addition

x-y+1/2x+y=1+5
x+1/2x+0y=6
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\frac{3}{2}x=6
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Using the value of x
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With x as 4 and y as 3, their product is 12
And can be written as (4,3)
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Select the procedure that can be used to show the converse of the Pythagorean theorem using side lengths chosen from 3 feet, 4 f
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Using Pythagoras theorem,  we know that 3² + 4² = 5², draw the 3-foot side and the 4-foot side with a right angle between them. The 5-foot side will fit to form a right triangle

<h3>How to use Pythagoras theorem to prove a right triangle?</h3>

The Pythagoras theorem states that the sum of two squares equals the squared of the longest side.

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\huge \boxed{\mathfrak{Question} \downarrow}

  • Expand & simplify ⇨ ( \sqrt{10}  -  \sqrt{2} ) ^{2}. Give your answer in the form b - c \:  \sqrt{5} where b & c are integers.

\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}

( \sqrt { 10 } - \sqrt { 2 } ) ^ { 2 }

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{10}-\sqrt{2}\right)^{2}.

\left(\sqrt{10}\right)^{2}-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}

The square of \sqrt{10} is 10.

10-2\sqrt{10}\sqrt{2}+\left(\sqrt{2}\right)^{2}

Factor 10=2\times 5. Rewrite the square root of the product \sqrt{2\times 5} as the product of square roots \sqrt{2}\sqrt{5}.

10-2\sqrt{2}\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}

Multiply \sqrt{2} and \sqrt{2} to get 2.

10-2\times 2\sqrt{5}+\left(\sqrt{2}\right)^{2}

Multiply -2 and 2 to get -4.

10-4\sqrt{5}+\left(\sqrt{2}\right)^{2}

The square of \sqrt{2} is 2.

10-4\sqrt{5}+2

Add 10 and 2 to get 12.

\boxed{ \boxed{\bf\:12-4\sqrt{5} }}

  • Here, b & c are integers where \boxed{ \sf \: b = 12 \: and \: c = 4}
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