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Mathematics

2 answers:

frez [133]3 years ago

5 0

Let x be the unknown number of all 3 equations
1.) 3x - 2 = x + 10 x=6
2.) x + 4 = 19 - 2x x=5
3.) 8x - 20 = x + 15 x=5

trasher [3.6K]3 years ago

3 0

1) 3x-2=x+10

Now, lets solve this step-by-step.

3x - 2 = x + 10
-x -x

3x - x = 2x 2x - 2 = 10

2x - 2 = 10
2x - 2+2 = 10+2 The inverse operation helps isolate the variable.

2x = 12

2x/2 = 12/2

x = 12/2
x = 6

Answer: x = 6

2) x + 4 = 19 - 2x 

We can solve this much faster

x + 4 = 19 - 2x
+2x

x+2x = 3x

3x + 4 = 19
-4 -4

19-4 = 15

3x= 15
3x/3 = 15/3
x= 5

Answer: x = 5

3) 8x - 20 = x + 15

Much faster now!!

8x - 20 = x + 15
-x
8x-x = 7x

7x - 20 = 15
+20 +20
15+20 = 35

7x = 35
7x/7 = 35/7
x = 35/7
x = 5

Answer: x= 5

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A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4

Vladimir79 [104]

Answer:

The time interval when $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

$V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)$

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

$V_Q(t) \geq 60$ between $0 \leq t \leq 4$

The schematic free body graphical representation of the above illustration was attached in the file below and the point when $V_Q(t) \geq 60$ is at 4 is obtained in the parabolic curve.

So, $V_Q(t) \geq 60$ is at $1.866 \leq t \leq 3.519$

Taking the integral of the time interval in order to determine the distance; we have:

distance = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$