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professor190 [17]
3 years ago
8

Use the Pythagorean Theorem to solve for x. *

Mathematics
2 answers:
hichkok12 [17]3 years ago
6 0

Answer:

x=8

Step-by-step explanation:

c²- b² = a²

c² = 10² = 100

b² = 6² = 36

100 - 36 = 64

The square root of 64 is 8, therefore x = 8.

Y_Kistochka [10]3 years ago
6 0

The other person provided a great answer but i do it a little different....                            

                                        \left[\begin{array}{ccc}A^2=C^2-B^2\\A^2=10^2-6^2\\A^2=100-36\\A^2=64\\\sqrt{64}=8 \end{array}\right]

All i did was use the formula A^2=C^2-B^2. I Plugged in the numbers to the formula and got A^2=10^2-6^2. Next, I found what 10^2 equals by doing 10*10 which equals 100. Then, i found out what 6^2 equals by doing 6*6 which equals 36. Next, what i did was subtracted 100 from 36 and got 64. Finally i found the square root of the number 64 which equal 8 since 8x8=64.  

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tresset_1 [31]

Answer:

D

Step-by-step explanation:

y_2 - y_1 / x_2 - x _1

4 - 2 / 3 - (-1) = 2/4 = 1/2

y +4 = 1/2 (x + 3) distributive property

y + 4 = 1/2x + 1.5

y= 1/2x - 5/2 subtraction (you subtract 4 from both sides) (-5/2 = -2.5)

therefore your equation is y = 1/2x - 5/2

hope this helps :)

7 0
3 years ago
I have the answer, 2-(3/x+2)
Marat540 [252]

Start with the answer format we want, and work your way toward forming a single fraction like so

a + \frac{b}{x+2}\\\\a*1+\frac{b}{x+2}\\\\a*\frac{x+2}{x+2}+\frac{b}{x+2}\\\\\frac{a(x+2)}{x+2}+\frac{b}{x+2}\\\\\frac{a(x+2)+b}{x+2}\\\\\frac{ax+2a+b}{x+2}\\\\\frac{ax+(2a+b)}{x+2}\\\\

Compare that last expression to (2x+1)/(x+2). Notice how the ax and 2x match up, so a = 2 must be the case.

Then we have 2a+b as the remaining portion in the numerator. Plugging in a = 2 leads to 2a+b = 2*2+b = 4+b. Set this equal to the +1 found in (2x+1)/(x+2) to have the terms match.

So, 4+b = 1 leads to b = -3

Therefore, a = 2 and b = -3

------------------------------------------------

An alternative route:

\frac{2x+1}{x+2}\\\\\frac{2x+1+0}{x+2}\\\\\frac{2x+1+4-4}{x+2}\\\\\frac{(2x+4)+1-4}{x+2}\\\\\frac{2(x+2)-3}{x+2}\\\\\frac{2(x+2)}{x+2}+\frac{-3}{x+2}\\\\2-\frac{3}{x+2}\\\\

I added and subtracted 4 in the third step so that I could form 2x+4, which then factors to 2(x+2). That way I could cancel out a pair of (x+2) terms toward the very end.

------------------------------------------------

Other alternative methods involve synthetic division or polynomial long division. They are slightly separate but related concepts.

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3 years ago
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