Answer:
- 8) 4 + 2q²/p² - 4r/p + r²/p²
- 9) (3/4, -9/4)
- 10) (3/8, 41/16)
Step-by-step explanation:
8. ============
Given
- α and β are roots of px² + qx + r = 0
The sum of the roots is α + β = -q/p, the product of then roots αβ = r/p
- (2 + α²)(2 + β²) =
- 4 + 2(α² + β²) + (αβ)² =
- 4 + 2((α + β)² -2αβ) + (αβ)² =
- 4 + 2((-q/p)² - 2r/p) + (r/p)² =
- 4 + 2q²/p² - 4r/p + r²/p²
------------------------------
9. ============
<u>Given function</u>
The minimum point is reached at vertex
<u>The vertex is:</u>
- x = -b/2a
- x = -(-3)/2*2 = 3/4
<u>The corresponding y-coordinate is:</u>
- y = 2(3/4)² - 3(3/4) - 1 = 9/8 - 9/4 - 1 = 1/8(9 - 18 - 9) = - 18/8 = - 9/4
<u>So the point is: </u>
---------------
10. ============
<u>Given function</u>
The maximum is reached at vertex
<u>The vertex is:</u>
- x = -b/2a
- x = -(-3)/2(-4) = -3/8
<u>The corresponding y-coordinate is:</u>
- y = 2 - 3(-3/8) -4(-3/8)² = 2 + 9/8 - 9/16 = 1/16(32 + 18 - 9) = 41/16
<u>So the maximum point is:</u>
Answer:
what grade is this
Step-by-step explanation:
I'm very confused is this algebra?
Answer: A. 31.5
Step-by-step explanation:
Using law of cosine because it is side angle side
a=16
b=21
angle = 116
c^2=a^2+b^2-2ab cosC, let C be the angle given
c^2=(21)^2+(16)^2-2(16)(21) cos(116)
c^2=441+256-672 cos(116)
i used a ti-89 and answered the right side and got
(you can find a ti-89 simulator online, but for problems like this make sure it is in degree mode)
991.585
c^2=991.585
take square root of both sides
=31.4894, which if you round it up it would be 31.5, so the answer would be A
therefore:
a=16
b=21
c=31.5
Answer:
32/4 7 3/4
Step-by-step explanation:
