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schepotkina [342]
3 years ago
8

Solve the quadratic equation x2+8x−30=0 by completing the square.

Mathematics
1 answer:
diamong [38]3 years ago
5 0

Answer:

               \bold{x=-4\pm\sqrt{46}}

Step-by-step explanation:

(a+b)^2=a^2+2ab+b^2\\\\\\x^2+8x-30=0\\\\\underbrace{x^2+2\cdot x\cdot 4+4^2}-4^2-30=0\\\\{}\qquad(x+4)^2\,-\,16-30=0\\\\{}\qquad\ \ (x+4)^2=46\\\\ {}\qquad\ \ x+4=\pm\sqrt{46}\\\\ {}\qquad\ \ x=-4\pm\sqrt{46}

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PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!
yan [13]

Answer:

b. f(n) = 2n^2 + 1

Step-by-step explanation:

Thanks for explaining the nth thing to me. It all makes sense now :'D

I'll be honest I think you're at least a grade level above me in math.

It's b. because

f(n) = 2(1)^2 + 1

        2(1) + 1

        2 + 1 = 3 (first term)

f(n) = 2(2)^2 + 1

        2(4) + 1

        8 + 1 = 9 (second term)  ...and so on. :)

8 0
3 years ago
The location of point J is (8,-6). The location of point L is (-2,9). Determine the location of point K which is 1/5 of the way
AURORKA [14]

Answer:

(6 , -3)

Step-by-step explanation:

Given in the question,

point J(8,-6)

x1 = 8

y1 = -6

point L(-2,9)

x2 = -2

y2 = 9

Location of point K which is 1/5 of the way from J to L

which means ratio of point K from J to L is 1 : 4

a : b

1 : 4

xk = x1+\frac{a}{a+b}(x2-x1)

yk = y1+\frac{a}{a+b}(y2-y1)

Plug values in the equation

xk = 8 + (1)/(1+4) (-2-8)

xk = 6

yk = -6 (1)/(1+4)(9+6)

yk = -3

8 0
4 years ago
How to find the angle?
kumpel [21]
If everything on inside of a line is 180->
(Line) --------/---------- everything on top equals 180 and there should be a number on either side. For instance 180-50 for the top would be 130, the wide angle's angle. I answer everything I like and come across so sorry I am four weeks late.
4 0
3 years ago
Please help ASAP thank you
Kobotan [32]

Answer:

Scalene, Acute

Step-by-step explanation:

scalene because the sides are of different length, and acute because it's less than 90°

3 0
3 years ago
How do you solve :<br><br><img src="https://tex.z-dn.net/?f=x%20%7B%7D%5E%7B2%7D%20%20%3D%20%20%5Cfrac%7B2%7D%7B3%7D%20" id="Tex
Neporo4naja [7]

x^2=\dfrac{2}{3}\to x=\pm\sqrt{\dfrac{2}{3}}\\\\x=-\dfrac{\sqrt{2}}{\sqrt3}\ \vee\ x=\dfrac{\sqrt2}{\sqrt3}\\\\x=-\dfrac{\sqrt2\cdot\sqrt3}{\sqrt3\cdot\sqrt3}\ \vee\ x=\dfrac{\sqrt2\cdot\sqrt3}{\sqrt3\cdot\sqrt3}\\\\\boxed{x=-\dfrac{\sqrt6}{3}\ \vee\ x=\dfrac{\sqrt6}{3}}

5 0
3 years ago
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