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Brut [27]
3 years ago
7

What value of x makes the equation true? 16 + (18+ 10) = (x +18) +10 x=​

Mathematics
1 answer:
ikadub [295]3 years ago
8 0

Answer:

x=16

Step-by-step explanation:

16+18+10=44

18+10+x=28+x

44-28=x

x=16

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(x + 1)^2 = 3x – 1 solving Quadratics ?
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Step-by-step explanation:

(x + 1)^{2}  = 3x - 1 \\  \\  \therefore \:  {x }^{2}  + 2x + 1 = 3x - 1 \\  \\ \therefore \:  {x }^{2}  + 2x + 1  -  3x  + 1 = 0\\  \\ \therefore \:  {x }^{2}  + 2x  -  3x + 1 - 1 = 0 \\  \\ \therefore \:  {x }^{2}  - x  = 0 \\  \\ \therefore \:  x(x - 1) = 0 \\  \\ \therefore \: x = 0 \:  \: or \:  \: x - 1 = 0 \\  \\ \therefore \: x = 0 \:  \: or \:  \: x = 1 \\  \\  \huge{ \red{ \boxed{\therefore \: x =  \{0, \:  \: 1 \}}}}

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1 unit to the right

Step-by-step explanation:

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On a coordinate plane, 5 squares are shown. Square L M N P has points (negative 3, 1), (negative 1, 1), (negative 1, negative 1)
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A ball is launched from 8 yards off the ground and travels in parabolic motion, landing in a net 80 yards away and also 8 yards
Masteriza [31]

Answer:

The answer to this question is the wall is 45.95 yards tall

Step-by-step explanation:

To solve this, we list out the given variables and the unknowns thus

Height of ball at launch = 8 yards

Distance of net from the ball = 80 yards

Distance of the wall down the path = 75%

Maximum height of the ball= 80 yards

equation of Motion of the ball = parabolic motion =

v² = u² - 2gS

S = 80 - 8 = 72 yards

at maximum height v = 0 thus u² = 2×9.81×72 =1412.64

u = 37.59 m/s

also v = u - gt and again at max height v = 0

Therefore 37.59 = 9.81×t or t = 3.83 s

If the motion of the ball is free of obstruction then time of flight before the ball just reaches the 8 yards off the ground = 3.83×2 = 7.66 seconds

Taking the initial velocity as zero at maximum height and from the equation

S = ut + 0.5×gt² we get, where S is the heigt of the ball from touching the actual field ground which is 80 yards we have

80 = 0.5×9.81×t²

so that t² = 2×80÷9.81 = 16.31 or t = 4.04s

Therefore the total time of flight = 4.04 + 3.83 = 7.87 seconds

if the ball is considered as having a constant horizontal velocity, therefore

at 75% of the way the time it took will be 0.75×7.87 = 5.9 seconds

However time it took  the  ball to reach maximum height and then starts descent = 3.83s, and the time at which the ball is directly over the wall = 2.07 seconds on the second half just after reaching mximum height

Thus at 2.07 seconds the distance trvelled from the maximum height is

S = ut +0.5gt² as before where u = 0

hence S = 0.5×9.81×2.07² = 21.05 yards or (80 -21.05) yards off the ground =  58.95 yards

As stated in the question, the ball cleared the wall by 13 yards therefore the height of the wall is 58.95 - 13 = 45.95 yards

7 0
3 years ago
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