All categories

3 years ago

Plz tell me how to do question 18

Mathematics

1 answer:

bagirrra123 [75]3 years ago

8 0

Answer:

750kmph

Step-by-step explanation:

Average is sum of all parts divided by the number of parts.

Let Stot be average speed for total trip (800kmph).

Let S1 be speed for first quarter of trip (950kmph).

Let S2, S3, and S4 be the speeds of each of the remaining quarters respectively.

Let Sans be the average speed for the remaining three quarters of the trip (a.k.a. The Answer!).

Stot = (S1 + S2 + S3 + S4) / 4

800kmph = (950kmph + S2 + S3 + S4) / 4

3200kmph = 950kmph + S2 + S3 + S4

3200kmph - 950kmph = 2250kmph = S2 + S3 + S4

Divide both sides by 3.

2250kmph / 3 = (S2 + S3 + S4) / 3 = Sans

Sans = 750kmph

You might be interested in

Chris labels every lattice point in the coordinate plane with the square of the distance from the point to the origin (a lattice

stiv31 [10]

Answer:

4 times

Step-by-step explanation:

A lattice point may be defined as the point of intersection of two grid lines or more than two grid lines that is placed in a regularly spaced points arrays. This is called a lattice point.

In the context, Chris tries to label every lattice point in a coordinate plane with its square of distance from the point to its origin. The lattice points means that the numbers are both the integers. So for number 25, Chris has to label 4 times

i.e. (55),(-5,5),(5,-5),(-5,-5)

4 0

3 years ago

A ball is thrown straight out at 80 feet per second from an upstairs window that's 15 feet off the ground. Find the ball's horiz

Alex73 [517]

Answer:

Step-by-step explanation:

In order to find the horizontal distance the ball travels, we need to know first how long it took to hit the ground. We will find that time in the y-dimension, and then use that time in the x-dimension, which is the dimension in question when we talk about horizontal distance. Here's what we know in the y-dimension:

a = -32 ft/s/s

v₀ = 0 (since the ball is being thrown straight out the window, the angle is 0 degrees, which translates to no upwards velocity at all)

Δx = -15 feet (negative because the ball lands 15 feet below the point from which it drops)

t = ?? sec.

The equation we will use is the one for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in:

$-15=(0)t+\frac{1}{2}(-32)t^2$ which simplifies down to

$-15=-16t^2$ so

$t=\sqrt{\frac{-15}{-16} }$ so

t = .968 sec (That is not the correct number of sig fig's but if I use the correct number, the answer doesn't come out to be one of the choices given. So I deviate from the rules a bit here out of necessity.)

Now we use that time in the x-dimension. Here's what we know in that dimension specifically:

a = 0 (acceleration in this dimension is always 0)

v₀ = 80 ft/sec

t = .968 sec

Δx = ?? feet

We use the equation for displacement again, and filling in what we know in this dimension:

Δx = $(80)(.968) +(0)(.968)^2$ and of course the portion of that after the plus sign goes to 0, leaving us with simply: