Find the distance between (20, -2) and (19, 5)
any would be appreciated thx! ❤❤❤​

answer: 5 root 2

step-by-step explanation:

using pyth. theorem

x^2= 1^2 +7^2

x^2 = 50

x = 5 root 2

answer:

square root of 50 approx.   7.07

step-by-step explanation:

do the distance formula, which is d= sqrt of (xsub2-xsub1)squared + (ysub2-xsub1)squared.

it would be d= sqrt of   (19-20)^2 +())^2 = sqrt of 50

answer:

8x^2+8x

(5x^2 -7x-6)    

the restrictions are that any of the fractions cannot be zero

x≠2

x≠-1

x≠0

step-by-step explanation:

16/ (x-2)

4/(x+1) +6/x

the common denominator for all the fractions is (x-2)*(x+1)*x

so multiply by (x-2)(x+1)x over (x-2)(x+1)x

16/ (x-2)             (x-2)(x+1)x

*

4/(x+1) +6/x       (x-2)(x+1)x

16/ (x-2) *       (x-2)(x+1)x

4/(x+1)   (x-2)(x+1)x + 6/x * (x-2)(x+1)x

canceling terms

16   (x+1)x

4   (x-2)x + 6 * (x-2)(x+1)

distributing in the denominator

16   (x+1)x

(4x^2 -8x)     + 6 * (x^2-2x+x-2)

16   (x+1)x

(4x^2 -8x)     + (6x^2-6x-12)

16x^2+16x

(10x^2 -14x-12)    

divide the top and bottom by 2

8x^2+8x

(5x^2 -7x-6)    

the restrictions are that any of the fractions cannot be zero

x-2 ≠0

x≠2

x+1≠0

x≠-1

x≠0

answer:

n = 7     n=-7

step-by-step explanation:

n^2 - 49 = 0

this is the difference of squares

(a^ - b^2) = (a-b) (a+b)

n^2 -49   = (n-7) (n+7) =0

using the zero product property

n-7 = 0   n+7 =0

n-7+7 =0+7       n+7-7 = 0-7

n = 7     n=-7