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3 years ago

I need these ASAP………..

Mathematics

1 answer:

RoseWind [281]3 years ago

4 0

Step-by-step explanation:

1) 3x - 4 = 86 + 24

→ 3x - 4 = 110

→ 3x = 110 + 4 = 114

→ x = 114/3 = 38

therefore, value of x is 38.

2) -10x = 30 + 90

→ -10x = 120

→ x = 120/10 = 12

→ x = 12

therefore, value of x is 12.

3) -8x + 23 + 51 = 130

→ -8x + 74 = 130

→ -8x = 130 - 74

→ -8x = 56

→ x = 56/8

→ x = 7

therefore, value of x is 7.

4) -9 + 2x = 59 + 90

→ -9 + 2x = 149

→ 2x = 149 + 9

→ 2x = 158

→ x = 158/2 = 79

→ x = 79

therefore, value of x is 79.

hope this answer helps you dear...take care !

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

= $\int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt$

= By using the Scientific calculator notation;

distance = 106.109 m

4 0

3 years ago

Help!!! Choose the product . - 6p ^ 3 * (3p ^ 2 + 5p - 1)

Lunna [17]

Answer:

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Step-by-step explanation:

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<em>hope this helps.....</em>

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3 years ago

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DaniilM [7]

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4 years ago

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ivolga24 [154]

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4 years ago