$\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\qquad 
B(\stackrel{x_2}{4}~,~\stackrel{y_2}{1})\qquad \qquad 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AB=\sqrt{[4-(-1)]^2+[1-5]^2}\implies AB=\sqrt{(4+1)^2+(1-5)^2}
\\\\\\
AB=\sqrt{5^2+(-4)^2}\implies AB=\sqrt{41}$

4 0

3 years ago

PLEASE HELP For what value of the variable will the value of 2x+1 be 20 greater than 8x+5?

wlad13 [49]

It will be a negative.

for an example: -20 for x would mqke 2x + 1 into -39 and 8x + 5 into -155. that's 116 apart, so maybe -2(8) + 5 = -11

-2(2) + 1 = -3

but that's only 8 apart, so it'd be higher.

-4(2) + 1 = -7

-4(8) + 5 = -27

BOOM

Tadaaaa,

You probably thought positive, but sometimes it's negative, you know what I mean?

6 0

3 years ago

Help solve this matrix question

Anna007 [38]

Answer:

The correct answer is A -6/17

7 0

3 years ago

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