Y=x^2-2x-19 y+4x=5 solution set system of equation is (-6,29) and ______.

1 answer:

6 0

Hello :
<span>Y=x²-2x-19 ...(1)
y+4x=5.....(2)
by (2) : y = 5-4x ....(*)
subsct in (1) : 5-4x</span>=x²-2x-19
x²+2x-24 =0
<span>Δ = 2²-4(1)(-24) = 100=10²
x1 = (-2-10)/2 =-6
x2= (-2+10)/2=4
if : x =-6 by(*) : y= 5-4(-6)=29
if x= 4 y = 5-4(4)=-11
</span><span>solution set system of equation is (-6,29) and (4,-11)______.</span>

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Answer:

The answer is

<h2> $\sqrt{41} \: \: or \: \: \: 6.403 \: \: \: units$ </h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{( {x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-6, -5) and (2, 0)

The distance is

$d = \sqrt{ ({ - 6 - 2})^{2} + ({ - 5 - 0})^{2} } \\ = \sqrt{ ({ - 4})^{2} + ({ - 5})^{2} } \\ = \sqrt{16 + 25} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$