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3 years ago

Please help on this one ? :)

Mathematics

1 answer:

rusak2 [61]3 years ago

7 0

Answer:

C. All real numbers

Step-by-step explanation:

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Write the expression: Add 2 and 4, then subtract 9

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The expression is
(2+4)-9

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2 years ago

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The length of a rectangle is 3yd longer than its width. If the perimeter of the rectangle is 50yd, find it’s length and width

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2 years ago

Find the geometric mean of 5 and 320. A. 35 B. 40 C. 45 D. 50

Softa [21]

Geometric mean of a and b is: $\sqrt{ab}$
$\sqrt{5*320}= \sqrt{1600}$ = 40
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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 = $\int\limits^{3.519}_{1.866} {V_Q(t)} \, dt$

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

= By using the Scientific calculator notation;

distance = 106.109 m

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3 years ago

What is the volume of the rectangular prism?

lara31 [8.8K]

1/2^3=1/8
we have 30 of them so
30.1/8=30/8 => 3 and 6/8 or 3 and 3/4 if simplified

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1 year ago