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ivann1987 [24]
2 years ago
10

2. Draw what might be the missing figure in

Mathematics
1 answer:
adoni [48]2 years ago
8 0

Answer:

<em><u>PLEASE </u></em><em><u>ATTACH</u></em><em><u> </u></em><em><u>THE </u></em><em><u>FIGURE </u></em><em><u>LIKE </u></em><em><u>THIS.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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36:38<br> Write the ratio as a fraction in simplest form.
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it will 8:9 hope it helps

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Is the center of dilation colinear with the hypotenuses of the right triangles shown?
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Step-by-step explanation:

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What is the measure of angle J in the triangle? Drawing is not to scale
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How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the
miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

<h3>8d.</h3>

\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}

___

<h3>9b.</h3>

\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}

___

<h3>10b.</h3>

\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}

Neither solution makes any denominator be zero, so both are good solutions.

8 0
3 years ago
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