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sdas [7]
3 years ago
13

What is the range of the cubic function f(x)=x^3-15​

Mathematics
1 answer:
Nina [5.8K]3 years ago
3 0
B

The end behavior of a positive cubic function states that as x approaches infinity, y approaches infinity.

Similarly, as x approaches negative infinity, y approaches negative infinity.

This means that when written in interval notation, the range is (-inf, +inf).
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A line passes through (4, 5) and (8, 9). Which equation best represents the line?
S_A_V [24]
Here, Coordinates: (4, 5) & (8, 9)
Calculation of Slope, m = y2-y1 / x2-x1
m = 9-5 / 8-4
m = 4 / 4
m = 1

Now, Take any coordinate: 
y - y1 = m(x - x1)
y - 5 = 1(x - 4)
y - 5 = x - 4
y = x + 1

In short, Your Answer would be: y = x + 1

Hope this helps!
4 0
3 years ago
Read 2 more answers
Let f(x)=x^2+2 and g(x)=1−3x. Find each function value: (f+g)(−1)
kykrilka [37]

Answer:

it's 1+2+1+3 = 7 ......

5 0
3 years ago
An object is launched from a platform. Its height (in meters), xx seconds after the launch, is modeled by: h(x)=-5(x-4)^2+180 Ho
kozerog [31]

Answer:

x=10

Step-by-step explanation:

h(x)=-5(x-4)^2+180

Set y = 0 to find when it hits the ground

0 = -5(x-4)^2+180

Subtract 180 from each side

-180 =-5(x-4)^2+180-180

-180 =-5(x-4)^2

Divide by -5

-180/-5 =-5/-5(x-4)^2

36 = (x-4)^2

Take the square root of each side

±sqrt(36) = sqrt( (x-4)^2)

±6 =  (x-4)

Add 4  to each side

4±6 =  (x-4)+4

4±6=x

x = 10 or -2

But time cannot be negative so x=10

4 0
3 years ago
Read 2 more answers
"We might think that a ball that is dropped from a height of 15 feet and rebounds to a height 7/8 of its previous height at each
tatyana61 [14]

Answer:

Total Time = 4.51 s

Step-by-step explanation:

Solution:

- It firstly asks you to prove that that statement is true. To prove it, we will need a little bit of kinematics:

                             y = v_o*t + 0.5*a*t^2

Where,   v_o : Initial velocity = 0 ... dropped

              a: Acceleration due to gravity = 32 ft / s^2

              y = h ( Initial height )

                             h = 0 + 0.5*32*t^2

                             t^2 = 2*h / 32

                             t = 0.25*√h   ...... Proven

- We know that ball rebounds back to 7/8 of its previous height h. So we will calculate times for each bounce:

1st : 0.25*\sqrt{15}\\\\2nd: 0.25*\sqrt{15} + 0.25*\sqrt{15*\frac{7}{8} } + 0.25*\sqrt{15*\frac{7}{8} } = 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} }\\\\3rd: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 2*0.25*\sqrt{15*(\frac{7}{8} })^2\\\\= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2\\\\4th: 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 2*0.25*\sqrt{15*(\frac{7}{8} })^3 \\\\

= 0.25*\sqrt{15} + 0.5*\sqrt{15*\frac{7}{8} } + 0.5*\sqrt{15*(\frac{7}{8} })^2 + 0.5*\sqrt{15*(\frac{7}{8} })^3

- How long it has been bouncing at nth bounce, we will look at the pattern between 1st, 2nd and 3rd and 4th bounce times calculated above. We see it follows a geometric series with formula:

  Total Time ( nth bounce ) = Sum to nth ( \frac{1}{2}*\sqrt{15*(\frac{7}{8})^(^i^-^1^) }  - \frac{1}{4}*\sqrt{15})

- The formula for sum to infinity for geometric progression is:

                                   S∞ = a / 1 - r

Where, a = 15 , r = ( 7 / 8 )

                                   S∞ = 15 / 1 - (7/8) = 15 / (1/8)

                                   S∞ = 120

- Then we have:

                                  Total Time = 0.5*√S∞ - 0.25*√15

                                  Total Time = 0.5*√120 - 0.25*√15

                                  Total Time = 4.51 s

5 0
3 years ago
Which postulate or theorem can be used? look at that picture if you answer I’ll give you best brainliest
o-na [289]

Answer:

Step-by-step explanation:popote

5 0
3 years ago
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