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2 years ago

Some help please please

Mathematics

2 answers:

Ksenya-84 [330]2 years ago

7 0

Answer:

oh snap i have this question too sorry cant help :(

Step-by-step explanation:

but i can provide with advice bc is the side down and bd is across us this to your advantage

Tpy6a [65]2 years ago

5 0

Answer:

$8\sqrt{2}$

Step-by-step explanation:

To find the length of an isosceles triangle, you know that the angles of the triangle are 45-45-90, so let's use that triangle theorem.

Therefore, the length of the hypotenuse (BD) is $x\sqrt{2}$ , and x is equal to 8 (length of BC).

$x\sqrt{2}$

x=8

Therefore, $8\sqrt{2}$ would be the length of side BD.

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Read 2 more answers

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Answer:

Answer image is attached.

Step-by-step explanation:

Given rational expressions:

$1.\ \dfrac{x^2+x+4}{x-2}\\2.\ \dfrac{x^2-x+4}{x-2}\\3.\ \dfrac{x^2-4x+10}{x-2}\\4.\ \dfrac{x^2-5x+16}{x-2}$

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$(x-2)+\dfrac{6}{x-2}\\(x+3)+\dfrac{10}{x-2}\\(x+1)+\dfrac{6}{x-2}\\(x-3)+\dfrac{10}{x-2}$

We have to match the rewritten terms with the given expressions.

Let us consider the rewritten terms and let us solve them one by one by taking LCM.

$(x-2)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x-2)^{2}+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+4+6 }{x-2}\\\Rightarrow \dfrac{x^2-4x+10}{x-2}$

So, correct option is 3.

$(x+3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x+3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{(x^2+3x-2x-6)+10}{x-2}\\\Rightarrow \dfrac{x^2+x+4}{x-2}$

So, correct option is 1.

$(x+1)+\dfrac{6}{x-2}\\\Rightarrow \dfrac{(x+1)(x-2)+6}{x-2}\\\Rightarrow \dfrac{x^{2} +x-2x-2+6}{x-2}\\\Rightarrow \dfrac{x^{2} -x+4}{x-2}$

So, correct option is 2.

$(x-3)+\dfrac{10}{x-2}\\\Rightarrow \dfrac{(x-3)(x-2)+10}{x-2}\\\Rightarrow \dfrac{x^2-3x-2x+6+10}{x-2}\\\Rightarrow \dfrac{x^2-5x+16}{x-2}$

So, correct option is 4.

The answer is also attached in the answer area.

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Answer:

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Step-by-step explanation:

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