The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at
(-1.5, -3.25).
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For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
Answer:
<u>Given</u>
- tanθ = 3.454
- θ is in the III quadrant
We know in the III quadrant both sine and cosine are negative.
<u>Use the following identities to get values of sinθ and cos θ</u>
- sinθ = - tanθ/√(1 +tan²θ)
- cosθ = - 1/√(1 +tan²θ)
<u>Substitute the value of tanθ and find sine and cosine:</u>
- sinθ = - 3.454/√(1 + 3.454²) = - 0.961
- cosθ = - 1/√(1 + 3.454²) = - 0.278
Answer:
y = (x - 4)² - 25
Step-by-step explanation:
The equation of a parabola in vertex form is
y = a(x - h)² + k
where (h, k) are the coordinates of the vertex and a is a multiplier
To obtain this form use the method of completing the square.
Given
y = (x + 1)(x - 9) ← expand factors using FOIL, thus
y = x² - 8x - 9
To complete the square
add/subtract ( half the coefficient of the x- term )² to x² - 8x
y = x² + 2(- 4)x + 16 - 16 - 9
= (x - 4)² - 25 ← in vertex form
<span>0.00146 is the answer
Hope this helps</span>
<span>Given
∠FHJ = 60°
</span><span>∠FGH = 28°
</span>∠FHG = 180 - ∠FHJ
∠FHG = 180 - 60
∠FHG = 120
∠HFG = 180 - (28 + 120)
∠HFG = 180 - 148
∠HFG = 32
