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3 years ago

Help please

Mathematics

1 answer:

Contact [7]3 years ago

8 0

Answer:

A.( The option a is correct

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40 POINTSSS AND BRAINLIEST 40 POINTS

My name is Ann [436]

Answer:

the x intercept is x=-8

the y intercept is y=12

Step-by-step explanation:

3 0

3 years ago

What is the solution

algol [13]

Hey there!

Always get the variable by itself first.

4+∛2y=6

Subtract 4 on both sides.

∛2y= 2

Cube both sides.

2y= 8

Divide both sides by 2.

y=4

I hope this helps!

~kaikers

5 0

4 years ago

Solve for x in the image below

mestny [16]

$\dfrac{7}{8} - \dfrac{1}{x} = \dfrac{3}{4} \\\\ \dfrac{7x - 8}{8x} = \dfrac{6x}{8x} \\\\ 7x - 8 = 6x \\\\ x = 8$

Answer : x = 8

Hope this helps. - M

8 0

3 years ago

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