What of the following is a multiple of 4

Mathematics

1 answer:

Aleksandr-060686 [28]3 years ago

7 0

Where are the following?

You might be interested in

I need to know this soon idk what it is plz help

stepladder [879]

Answer:

16. 24.85

17. 0.9 seconds

Step-by-step explanation:

16. $10.95+$14.20=$25.15

$50-$25.15=$24.85

17. 5.75 seconds-4.85seconds=0.9 seconds

Your welcome, please mark me Brainliest!

8 0

3 years ago

I needs help please and tank chu

PIT_PIT [208]

9514 1404 393

Answer:

C {2, 4, 9, 16}

Step-by-step explanation:

The domain is the list of numbers at the tails of the arrows.

The range is the list of numbers at the heads of the arrows: {2, 4, 9, 16}.

5 0

3 years ago

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .