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Lady_Fox [76]
3 years ago
15

The bullet leaves a rifle with a speed of 699m/s how much time elapses before it strikes a target 1,000m away

Mathematics
2 answers:
Dafna1 [17]3 years ago
5 0

Answer:

I think u need to subtract 1000-699

Mashutka [201]3 years ago
3 0
It will take the bullet 1.301 seconds to strike the target
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Solve Equation<br> 1. 6 - 4x = 6x - 8x + 2
Helga [31]

Answer:

X=2

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

6−4x=6x−8x+2

6+−4x=6x+−8x+2

−4x+6=(6x+−8x)+(2)(Combine Like Terms)

−4x+6=−2x+2

−4x+6=−2x+2

Step 2: Add 2x to both sides.

−4x+6+2x=−2x+2+2x

−2x+6=2

Step 3: Subtract 6 from both sides.

−2x+6−6=2−6

−2x=−4

Step 4: Divide both sides by -2.

−2x/−2=−4/−2

Step 5 :Answer

X=2

LEAVE A LIKE IF THIS BELPED PLEASE.

4 0
4 years ago
- Hose A fills a water truck at the constant rate of 60 gallons every 15 minutes. Hose B fills a water truck at a constant rate
joja [24]

Answer:

"The rate of Hose A is 1 gallon per minute more than the rate of Hose B"

Step-by-step explanation:

Hose A:

60 gallons every 15 minutes means

60/15 = 4 gallons per minute

Hose B:

y = 3x

y is gallons

x is minutes

we put x = 1 and get:

y = 3*1 = 3

That means 3 gallons per minute

Thus,

the correct choice is C), or third option. 4 is 1 greater than 3 so Hose A is 1 gallons per minute more than the rate of Hose B.

8 0
3 years ago
Read 2 more answers
Charlie ate 2/3 of his chocolate bar. Kelsey ate
Contact [7]
I think he’s because 3/6 is half and 2/3 is half
3 0
3 years ago
Read 2 more answers
The chart given here lists the amount of rainfall received in centimeters in each month. By how much is the combined rainfall re
wlad13 [49]
It is B:23 centimeters 54+56=110. 63+70=133 133-110= 23c
4 0
3 years ago
True or False (&amp; Justify): If a ball is thrown horizontally from the top edge of a 150m building at a speed of 25 m/sec, the
kap26 [50]

Answer:

38.6 m/s

Step-by-step explanation:

The motion of the ball is a projectile motion, which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

Therefore we have to analyze the horizontal and vertical motion separately.

Along the horizontal direction, the velocity is constant during the motion, since there are no forces acting in this direction. So the horizontal velocity 3 seconds after the launch will be the same as the velocity at the launch:

v_x = v_0 = 25 m/s

The vertical velocity instead changes according to the suvat equation:

v_y = u_y - gt

where

u_y=0 is the initial vertical velocity

g=9.8 m/s^2 is the acceleration due to gravity

t is the time

Therefore, after t = 3 s,

v_y=0-(9.8)(3)=-29.4 m/s

So the velocity after 3 seconds is < 25, -29.4 > m/s. The magnitude of the velocity is

v=\sqrt{v_x^2+v_y^2}=\sqrt{25^2+(-29.4)^2}=38.6 m/s

3 0
4 years ago
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