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3 years ago

HELP MEEEEEE BIG POINTS AVAILIABLE

Mathematics

2 answers:

Zolol [24]3 years ago

4 0

There’s no picture tho??????

ozzi3 years ago

3 0

Answer:

help u with what no attachment

Step-by-step explanation:

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Tom throws a ball into the air. The ball travels on a parabolic path represented by the equation , where represents the height o

tigry1 [53]

Answer:

2.5 second

Step-by-step explanation:

The equation is missing in the question.

The equation is, $h=-8t^2+40t$ , where 'h' is the height and 't' is time measured in second.

Now we know to reach its maximum height, h in t seconds, the derivative of h with respect to time t is given by :

$\frac{dh}{dt} =0$

Now the differentiating the equation with respect to time t, we get

$\frac{dh}{dt}=\frac{d}{dt}(-8t^2+40t)$

$\frac{dh}{dt}=-16t+40$

For maximum height, $\frac{dh}{dt} =0$

So, $-16t+40=0$

$\Rightarrow 16t=40$

$\Rightarrow t=\frac{40}{16}$

$\Rightarrrow t = 2.5$

Therefore, the ball takes 2.5 seconds time to reach the maximum height.

7 0

3 years ago

Which ordered pair is a solution to the equation? y=x3/4

Leni [432]

Answer:

(8,6)

Step-by-step explanation:

y=x3/4

y=8(3/4)

y=24/4

y=6

when you input 8 for x, you get 6 for y.

6 0

3 years ago

Read 2 more answers

Can someone help me with this!!! Will give you all my points and brainliest.

Angelina_Jolie [31]

The function is shifted to the right 3 units

a shift to the right is subtracted inside the function

g(x) = f(x-3)

7 0

3 years ago

What figure has properties of all sides congruent and all right angles?

andre [41]

A. Rectangle

Here bestie

7 0

2 years ago

A boat is pulled toward a dock by means of a rope wound on a drum that is located 6 ft above the bow of the boat. If the rope is

Monica [59]

Answer:

Step-by-step explanation:

Given that:

The height of the dock (h) = 6

Let represent d to be the distance between the boat and the dock

Let the length of the rope between the boat and the drum be denoted by (l)

Then, the rate of change for the length of the rope be:

dl/dt = -5 ft/s

Using Pythagoras rule to determine the relationship between these values, we have:

$l^2 = h^2 +d^2$

$l^2 = 6^2 + d^2$

$l^2 = 36 + d^2$

We relate to: $2l * \dfrac{dl}{dt} = 2d* \dfrac{dd}{dt}$

From the question;

l = 34,

So to find $\dfrac{dd}{dt}$ , we get;

$d = \sqrt{l^2 - 36}$

$d = \sqrt{34^2 - 36}$

$d = \sqrt{1156- 36}$

$d = \sqrt{1120}$

d = 33.46

So, we have:

$\dfrac{dd}{dt}= \dfrac{l}{d} \times \dfrac{dl}{dt}$

$\dfrac{dd}{dt}= \dfrac{34}{33.46} \times - 5$

$\dfrac{dd}{dt}= 1.016 \times - 5$

$\dfrac{dd}{dt}=-5.08 \ ft/sec$

4 0

2 years ago