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3 years ago

The pair of points (6,y) and (10,-1) lie on a line with slope 1/4 ?

Mathematics

1 answer:

cupoosta [38]3 years ago

5 0

slope = (y2-y1)/(x2-x1)

1/4 = (y--1)/(6-10)

1/4 = (y+1)/(-4)

multiply by -4 on each side

-1 = y+1

subtract 1 from each side

-2 = y

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3 years ago

Solve this equation-6x^2=-144q

joja [24]

Answer:

4.9

Step-by-step explanation:

-6x2 = -144q

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3 years ago

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Answer:

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2 years ago

Read 2 more answers

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

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3 years ago