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3 years ago

SOMEONE PLEASE ANSWER ASAP FOR BRAINLIEST!!!!!

Mathematics

1 answer:

arsen [322]3 years ago

6 0

Answer:

D. the ball has been traveling for 5 seconds when it reaches the bottom of the ramp

Step-by-step explanation:

f(x) = x²- 10x + 25

We can factor this.

What 2 numbers multiply to 25 and add to -10

-5 *-5 = 25

-5+-5 = -10

f(x) =(x-5) (x-5)

Using the zero product property, we know that it will cross the axis at x=5

At 5 seconds f(x) =0. This means the height is zero at 5 seconds

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PLZZZZZZZZZZZZ HELP THANKS :D

svetoff [14.1K]

Sdfff bhbhhb. Dddffff

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3 years ago

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A tunnel is in the shape of a parabola. The maximum height is 16 m and it is 16 m wide at the base, as shown below.

NeX [460]

Vertex at the origin and opening down → y=ax^2
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4

y=ax^2→y=-(1/4)x^2

7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4

Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m

Please, see the attached file.

Answer: Third option 15.75 m

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3 years ago

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Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u

soldier1979 [14.2K]

$x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }$

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

$p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}$

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

$\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}$

$\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}$

$\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}$

Therefore:

$\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]$

Recall that at t=25, $p(25)=\dfrac { 5800 } {13} \approx 446.15$

Therefore:

$\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009$