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frez [133]
4 years ago
5

Solve the following equation: 3(2x - 8) + 4x = -2(12 - 7x) - 4x

Mathematics
1 answer:
dezoksy [38]4 years ago
6 0

Answer:

Step-by-step explanation:

3(2x - 8) + 4x = -2(12 - 7x) - 4x

6x - 24 + 4x = - 24 + 14x - 4x

10x + 24 = 24 + 10 x

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Simplify the expression.<br> 40 -2.32<br> A. 4<br> B. 342<br> C.<br> 22<br> D. 28
Troyanec [42]

Answer:

37.68

Step-by-step explanation:

40 -2.32

= 37.68

7 0
3 years ago
Laura's gross annual salary is $44184. What is the maximum amount of rent she can afford to pay?
fgiga [73]
D is the answer I believe it is.
6 0
3 years ago
In the drawing above two rugby players start 41 meters apart at rest. Player 1 accelerates at 2.2 m/s^2 and player 2 accelerates
Anettt [7]

Answer:

The time that elapses before the players collide is 4.59 secs

Step-by-step explanation:

The distance between the two players is 41 meters. Hence, they would have total distance of 41 meters by the time they collide.

We can write that

S₁ + S₂ = 41 m

Where S₁ is the distance covered by Player 1 before collision

and S₂ is the distance covered by Player 2 before collision

From one of the equations of kinematics for linear motion,

S = ut + \frac{1}{2} at^{2}

Where S is distance

u is the initial velocity

a is acceleration

and t is time

Since the players collide at the same time, then time spent by player 1 before collision equals time spent by player 2 before collision

That is, t₁ = t₂ = t

Where t₁ is the time spent by player 1

and t₂ is the time spent by player 2

For player 1

S = S₁

u = 0 m/s ( The player starts at rest)

a = 2.2 m/s²

Then,

S₁ = 0(t) + \frac{1}{2} (2.2)t^{2}

S₁ = 1.1t^{2}

t^{2} = \frac{S_{1} }{1.1}

For player 2

S = S₂

u = 0 m/s ( The player starts at rest)

a = 1.7 m/s²

Then,

S₂ = 0(t) + \frac{1}{2} (1.7)t^{2}

S₂ = 0.85t^{2}

t^{2} = \frac{S_{2} }{0.85}

Since the time spent by both players is equal, We can  write that

t^{2} = t^{2}

Then,

\frac{S_{1} }{1.1} =  \frac{S_{2} }{0.85}

0.85S_{1} = 1.1S_{2}

From, S₁ + S₂ = 41 m

S₂ = 41  - S₁

Then,

0.85S_{1} = 1.1 ( 41 -S_{1} )

0.85S_{1} = 45.1 - 1.1S_{1}

0.85S_{1} + 1.1S_{1} = 45.1 \\1.95S_{1} = 45.1\\S_{1} = \frac{45.1}{1.95} \\S_{1}  = 23.13 m

This is the distance covered by the first player. We can then put this value into t^{2} = \frac{S_{1} }{1.1} to determine how much time elapses before the players collide.

t^{2} = \frac{S_{1} }{1.1}

t^{2} = \frac{23.13 }{1.1}

t^{2} = 21.03\\t = \sqrt{21.03} \\t = 4.59 secs

Hence, the time that elapses before the players collide is 4.59 secs

3 0
4 years ago
Given s left parenthesis t right parenthesis equals 5 t squared plus 5 ts(t)=5t2+5t​, find
finlep [7]

Answer:

The velocity function is v(t)=10t+5.

The acceleration function is a(t)=10.

When t = 44​, the velocity is v(44)=445 \:\frac{ft}{s}.

When t = 44​, the acceleration is a(44)=10\: \frac{ft}{s^2}.

Step-by-step explanation:

We know that the position function is given by

s(t)=5t^2+5t

Velocity is defined as the rate of change of position or the rate of displacement. If you take the derivative of the position function you get the instantaneous velocity function.

v(t)=\frac{ds}{dt}

Acceleration is defined as the rate of change of velocity. If you take the derivative of the instantaneous velocity function you get the instantaneous acceleration function.

a(t)=\frac{dv}{dt}

The instantaneous velocity function is given by

v(t)=\frac{d}{dt} s(t)=\frac{d}{dt}(5t^2+5t)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(5t^2\right)+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}\\\\v(t)=10t+\frac{d}{dt}\left(5t\right)\\\\\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dt}\left(t\right)=1\\\\v(t)=10t+5

The instantaneous acceleration function is given by

a(t)=\frac{dv}{dt} =\frac{d}{dt}(10t+5)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\a(t)=\frac{d}{dt}\left(10t\right)+\frac{d}{dt}\left(5\right)\\\\a(t)=10

To find the velocity and acceleration when t = 44, we substitute this value into the velocity and acceleration functions

v(44)=10(44)+5\\v(44)=445 \:\frac{ft}{s}

a(44)=10\: \frac{ft}{s^2}

6 0
4 years ago
Factor the polynomial below X cubed +12 X squared +36X
lyudmila [28]
I’m sorry but I have no idea
3 0
3 years ago
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