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3 years ago

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A museum has an aquarium in the shape of a right rectangular prism that is 22.9 meters long, 7.5 meters wide, and 4.6 meters hig

h. What is the volume, round to the nearest cubic meter, of the aquarium

1 answer:

Vedmedyk [2.9K]3 years ago

8 0

Answer:

790

Step-by-step explanation:

just multiply all of the values

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PLEASE HELP ME ANSWER THESE TWO QUESTIONS!!!!!!!

Fittoniya [83]

1. x=50 2. x=4 hope this helps:)

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3 years ago

What are the first 6 terms for this arithmetic equation?

Ipatiy [6.2K]

Answer:

3, - 2, - 7, - 12, - 17, - 22

Step-by-step explanation:

To find the first 6 terms substitute n = 1, 2, 3, 4, 5 into the formula, that is

t(1 + 1) = t(1) - 5 ⇒ t(2) = 3 - 5 = - 2

t(2 + 1) = t(2) - 5 ⇒ t(3) = - 2 - 5 = - 7

t(3 + 1) = t(3) - 5 ⇒ t(4) = - 7 - 5 = - 12

t(4 + 1) = t(4) - 5 ⇒ t(5) = - 12 - 5 = - 17

t(5 + 1) = t(5) - 5 ⇒ t(6) = - 17 - 5 = - 22

3 0

3 years ago

Kyle wants to put fencing around his rectangular shaped yard. The length is 12.5 feet and the width is 8.4 feet. How much fencin

Kay [80]

Answer:

20.9

Step-by-step explanation:

The length and width would be 20.9 if you add then divide the formula give its different sometimes

3 0

3 years ago

The water from a fire hose follows a path described by y equals 2.0 plus 0.9 x minus 0.10 x squared (units are in meters). If

Alecsey [184]

Answer:

The resultant velocity is 12.21 m/s.

Step-by-step explanation:

We are given that the water from a fire hose follows a path described by y equals 2.0 plus 0.9 x minus 0.10 x squared (units are in meters).

Also, v Subscript x is constant at 10.0 m/s.

The water from a fire hose follows a path described by the following equation below;

$y=2.0 + 0.9x-0.10x^{2}$

The velocity of the $x$ component is constant at = $v_x=10.0 \text{ m/s}$

and the point at which resultant velocity has to be calculated is (9.0,2.0).

Let the velocity of x and y component be represented as;

$v_x=\frac{dx}{dt} \text{ and } v_y=\frac{dy}{dt}$

Now, differentiating the above equation with respect to t, we get;

$y=2.0 + 0.9x-0.10x^{2}$

$\frac{dy}{dt} =0 + 0.9\frac{dx}{dt} -(0.10\times 2)\frac{dx}{dt}$

$\frac{dy}{dt} = 0.9\frac{dx}{dt} -0.2\frac{dx}{dt}$

$v_y = 0.9v_x -0.2v_x$

$v_y = 0.7v_x$

Now, putting $v_x=10.0 \text{ m/s}$ in the above equation;

$v_y = 0.7 \times 10.0$ = 7 m/s

Now, the resultant velocity is given by = $v=\sqrt{v_x^{2}+v_y^{2} }$

$v=\sqrt{10^{2}+7^{2} }$

= $\sqrt{149}$ = 12.21 m/s

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3 years ago

Given two different circles O and Q, which of the following cases presents two circles with no (zero) common tangents?

borishaifa [10]

Concentric circles because if a line is tangent to one circle, it would only intersect teh other

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3 years ago

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