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Aleksandr [31]
3 years ago
8

GIVEN: AB, BC, and AC are congruent

Mathematics
1 answer:
andrew-mc [135]3 years ago
4 0
Answer:
A,B ,C are not collinear so false
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CAN SOMEONE SHOW ME HOW TO DO THIS PLEASE????????????????
morpeh [17]
For (h+g)(x) you just add the two functions:

(h+g)(x) = 4x + 2x^2

For (h•g)(x) you multiply them:

(h•g)(x) = 4x • 2x^2 = 8x^3

For (h-g)(x) you subtract them:

(h-g)(x) = 4x - 2x^2


For (h-g)(-2) you sub -2 into the equation we just created:

(h-g)(-2) = 4(-2) - 2(-2)^2
(h-g)(-2) = -8 - 2(4)
(h-g)(-2) = -8 - 8
(h-g)(-2) = -16
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3 years ago
Evaluate: 14/2 - 3 + 6/3
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Step-by-step explanation:

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FinnZ [79.3K]

\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}
7 0
3 years ago
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3d one

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The diagonals of a square are___.
soldier1979 [14.2K]

Answer:

The diagonals of a square intersect cross in a 90 degree angle. This means that the diagonals of a square are perpendicular. The diagonals of a square are the same length congruent.

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