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Bingel [31]
3 years ago
11

Y=2x+1 y=-x+4 It's a system of equation

Mathematics
1 answer:
MrRissso [65]3 years ago
8 0
Since y=y. set the equations equal to each then solve
2x + 1 = -x + 4
+x +x
3x + 1 = 4
-1 -1
3x = 3
/3 /3
x = 1
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If the equation y = 3x + 2 is changed to y = -3x + 2, how will the graph of the line change?
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3 years ago
Sam attempts a 42-yard field goal in a football game. For his attempt to be a success, the football needs to pass through the up
Zigmanuir [339]

Answer:

The correct option is;

The kick is good! The ball clears the crossbar by nearly 15 feet.

Step-by-step explanation:

Here we have;

Velocity of ball = 70 ft/s

Angle of motion = 50°

Therefore;

Vertical component of velocity, v_y = v×sinθ = v×sin×50 = 53.62 ft/s

v_y = u_y - g·t

At maximum height v_y = 0, therefore;

u_y = g·t and

t =  u_y/g

Where:

u_y = Initial vertical velocity = 53.62 ft/s

g = Acceleration due to gravity = 32.1740 ft/s²

∴ t = 53.62/32.1740 = 1.66666 s

Total time of flight = 2 × 1.66666 = 3.33332 s

Max height is given by the following relation;

v_y² = u_y² - 2·g·s

v_y = 0 ft/s at maximum height, therefore;

u_y² = 2·g·s

53.62² = 2×32.1740×s

s = 53.62² ÷(2×32.1740) = 44.69 ft

Horizontal velocity = 70×cos(50) = 44.995 ft/s

Hence the time it takes the ball to reach 42 yards is given by the following relation;

42 yards = 126 ft

Time, t = \frac{Distance}{Velocity} = \frac{126}{44.995} = 2.8 \, s

Height of ball at 2.8 s is given by;

Direction of ball at 2.8 s = downwards

Hence time  after peak = 2.8 - 1.666 = 1.1336429 s

Position at 2.8 s is given by

s = u_y ·t + 1/2·g·t²

u_y = 0 ft/s as ball is on second half of flight

∴ s = 1/2·g·t² = 1/2×32.1740×1.1336429²= 20.674 ft below maximum height

∴ Height above ground =  Maximum height - 20.674 ft = 44.69 - 20.674

Height above ground = 24.01 ft

Hence the ball clears the crossbar by 24.01 - 10 = 14.01 ft

The best option is therefore;

The kick is good! The ball clears the crossbar by nearly 15 feet.

8 0
4 years ago
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