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Blababa [14]
1 year ago
9

Does anyone know the minim value of the function f(x) =2x^2 -4x -6?. I’m online due to my quarantine and am looking for extra he

lp. Here is a picture. Nvm :/

Mathematics
1 answer:
navik [9.2K]1 year ago
4 0

Answer:

f(x) = - 8

Explanation:

The given function is

f(x) =2x^2 -4x -6

The first step is to find the derivative of the function. Recall, if

y = ax^b

y' = abx^(b - 1)

Thus,

f'(x) = 4x - 4

We would equate f'(x) to zero and solve for x. We have

4x - 4 = 0

4x = 4

x = 4/4

x = 1

We would substitute x = 1 into the original function and solve for f(x) or y. It becomes

f(1) =2(1)^2 -4(1) - 6 = 2 - 4 - 6

f(1) = - 8

Thus, the minimum value is f(x) = - 8

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SashulF [63]

Answer:

15/4

Step-by-step explanation:

The formula for slope is:

(y₂-y₁)/(x₂-x₁)

Now we plug in the coordinates given to us into the equation:

(5-(-10))/(-1-(-5))

15/4

3 0
3 years ago
Please, quickly. Sorry for the poor picture quality.
Nimfa-mama [501]

Answer:

<2 is 50º

Step-by-step explanation:

you first want to find what x is, so you would put the two equations in an (equation)=(equation) and get x and the number across from each other, that would be x=15º. you then plug in the 15 to the equation, and solve that, resulting in 130º. The angle is going to be 180 in total, so subtract the 130º from 180º and you would get 50º

3 0
3 years ago
What is the distance between 4.4 and -2.2 on a number line?
SSSSS [86.1K]

Answer:

2.2

Step-by-step explanation:

4.4 + -2.2 = 2.2

....................................................................

4 0
3 years ago
78 times 56 plz help me
rewona [7]
Your answer is 4,368
7 0
3 years ago
Read 2 more answers
Find all points of intersection of the given curves. (Assume 0 ≤ θ ≤ 2π and r ≥ 0. Order your answers from smallest to largest θ
DerKrebs [107]

Answer:

∅1=15°,∅2=75°,∅3=105°,∅4=165°,∅5=195°,∅6=255°,∅7=285°,

∅8=345°

Step-by-step explanation:

Data

r = 8 sin(2θ), r = 4 and r=4

iqualiting; 8.sin(2∅)=4; sin(2∅)=1/2, 2∅=asin(1/2), 2∅=30°, ∅=15°

according the graph 2, the cut points are:

I quadrant:

0+15° = 15°

90°-15°=75°

II quadrant:

90°+15°=105°

180°-15°=165°

III quadrant:

180°+15°=195°

270°-15°=255°

IV quadrant:

270°+15°=285°

360°-15°=345°

No intersection whit the pole (0)

7 0
3 years ago
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