Step-by-step explanation:
Hey there!
Given;
<u>Solve</u><u> </u><u>it by</u><u> </u><u>using</u><u> </u><u>mid-term</u><u> </u><u>factorization</u><u>. </u>
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<u>Take</u><u> </u><u>common</u><u> </u><u>and</u><u> </u><u>simplify</u><u> </u><u>them to get </u><u>answer</u><u>. </u>
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<u>Therefore the</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>option </u><u>D</u><u>.</u>
<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>
<span>The discriminant of a quadratic equation is the b^2-4ac portion that the square root is taken of. If the discriminant is negative, then the function has 2 imaginary roots, if the discriminant is equal to 0, then the function has only 1 real root, and finally, if the discriminant is greater than 0, the function has 2 real roots. So let's look at the equations and see which have a positive discriminant.
f(x) = x^2 + 6x + 8
6^2 - 4*1*8
36 - 32 = 4
Positive, so f(x) has 2 real roots.
g(x) = x^2 + 4x + 8
4^2 - 4*1*8
16 - 32 = -16
Negative, so g(x) does not have any real roots
h(x) = x^2 – 12x + 32
-12^2 - 4*1*32
144 - 128 = 16
Positive, so h(x) has 2 real roots.
k(x) = x^2 + 4x – 1
4^2 - 4*1*(-1)
16 - (-4) = 20
Positive, so k(x) has 2 real roots.
p(x) = 5x^2 + 5x + 4
5^2 - 4*5*4
25 - 80 = -55
Negative, so p(x) does not have any real roots
t(x) = x^2 – 2x – 15
-2^2 - 4*1*(-15)
4 - (-60) = 64
Positive, so t(x) has 2 real roots.</span>
Hello,
1) We must find the equation of the line RS:
R=(-1;6)S=(5;5)
y-6=(x+1)(5-6)/(5+1)y=-1/6*(x+1)+6
y=-x/6+35/6 Slope=-1/6
The perpendicular has the slope 6.
2) Find M the middle of [RS]M=((5-1)/2;(6+5)/2)=(2:11/2)
3) Equation of the perpendicular bisector:
y-11/2=(x-2)*6y=6x-12+11/2
So y=6x-13/2
And sorry for my poor english
3)
Answer:
169
Step-by-step explanation:
