X - 2y = 3
<span>4x^2 - 5xy + 6y = 3
lets solve for x the first and substitute in the second:
x = 3 + 2y
4(</span>3 + 2y)^2 - 5(3 + 2y)y + 6y = 3
4(9 + 12y + 4y^2) - 15y - 10y^2 = 3
36 + 48y +16y^2<span> - 15y - </span><span>10y^2 = 3
6y^2 + 33y + 33 = 0
we can solve using the general quadratic formula:
y = (-33 +- </span>√(33^2 - 4*6*33)<span>)/12
</span>y = (-33 +- √(297)<span>)/12
</span>so there are 2 solutions for y:
y1 = (-33 + √(297)<span>)/12
</span>y2 = (-33 - √(297)<span>)/12
</span>pick one and then substitute the y value in the first equation to find x
Answer:
6. D
7. B
Step-by-step explanation:
Nancy takes a test that has two sections: essay and true/false. She scores 56 points on the essay. Each true/false question is 4 points. So x number of true/false is worth 4 points each or 4x.
Together her score will be 56 + 4x.
She wants to score more than 80 or > 80.
So her test can be represented as the inequality 56 + 4x > 80.
This is solved using inverse operations.
56 + 4x > 80
4x > 24
x > 6
This should be graphed as an open circle at 6 on the umber line pointing and shaded to the the right for numbers greater than 6.
Answer: 6
Step-by-step explanation:
You would divide the number by the other number to get the awnser
Answer:
x <= 0 or 1 = < x <= 5.
Step-by-step explanation:
First we find the critical points:
x(x - 1)(x - 5) = 0
gives x = 0, x = 1 and x = 5.
Construct a Table of values:
<u> x < 0 </u> <u>x = 0 </u> 0<u>< x < 1</u> <u>1 =< x <= 5</u> <u>x = 5</u>
x <0 0 >0 <0 0
x - 1 <0 -1 >0 <0 0
x - 5 < 0 0 > 0 <0 0
x(x-1)(x-5) < 0 0 >0 <0 0
So the answers are x =< 0 or 1 =< x <= 5.
Answer:(2,3)
Step-by-step explanation:
y2-y1/x2-x1
