The zeros of a quadratic function are found where the graph intersects the x-axis. If the graph interects the x-axis in 2 places, we have 2 real solutions; if the graph intersects--or just touches--the x-axis in one place we have one real solution multiplicity 2; if the graph doesn't go through the x-axis at all we have 2 imaginary solutions. Ours goes through the x-axis in 2 places so we have 2 real solutions. Choice A.
Answer:
AB= 0.625 units (3 s.f.)
∠BAC= 52.9° (1 d.p.)
∠ABC= 32.1° (1 d.p.)
Step-by-step explanation:
Please see the attached pictures for full solution.
- Find AB using cosine rule
- find ∠BAC using sine rule
- find ∠ABC using angle sum of triangle property
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
The answer is 3.6 i know because i am a nerd
Distribute
10+4x+4+5
Combine like terms
19+4x
Final answer: 4x+19
