All categories

3 years ago

Replace variables with values and

Mathematics

1 answer:

Verizon [17]3 years ago

7 0

Answer:

480

Step-by-step explanation:

Substitute in the numbers for your letters, the length being 4, width being 10 and height being 12. To make things easier, multiply 10 by 12 which gets you 120. Then multiply 120 by 4 and you get 480.

You might be interested in

Patty spent 3 1/2 times as much as Sandy. if sandy spent $20.50, how much did patty spend

katrin2010 [14]

12.6$Answer:

Step-by-step explanation:

1$ - 20$ = 19$ / 3 = 6.3$

6.3$ x 2 = 12.6$

6 0

3 years ago

26) Jenny drove 48 mph for 65 minutes. Which is true? a. Jenny drove exactly 48 miles b. Jenny drove more than 48 miles c. Jenny

shutvik [7]

Answer:

B- Jenny must have driven more than 48 miles.

Step-by-step explanation:

If Jenny rode for 60 at 48 mph, it would be exactly 48 miles. Though there is only 60 minutes in an hour, and Jenny rode 5 minutes longer than that, her drive being 65 minutes. To find how much she rode, 5 extra minutes into an hour would divide equally into 60, resulting in 12. Divide 48 by 12, and Jenny would have driven 4 miles in the 5 minutes drive after 60 minutes.

(Jenny basically driven 52 miles in 65 minutes).

6 0

2 years ago

Fill in the blanks to complete the equation: 1 3/5= blank+3/5

Galina-37 [17]

The answer is..

1 3/5=1+3/5

5 0

3 years ago

3. A rocket is launched from a height of 3 meters with an initial velocity of 15 meters per second.

Vikki [24]

Let the rocket is launched at an angle of \theta with respect to the positive direction of the x-axis with an initial velocity u=15m/s.

Let the initial position of the rocked is at the origin of the cartesian coordinate system where the illustrative path of the rochet has been shown in the figure.

As per assumed sine convention, the physical quantities like displacement, velocity, acceleration, have been taken positively in the positive direction of the x and y-axis.

Let the point P(x,y) be the position of the rocket at any time instant t as shown.

Gravitational force is acting downward, so it will not change the horizontal component of the initial velocity, i.e. $U_x=U cos\theta$ is constant.

So, after time, t, the horizontal component of the position of the rocket is

$x= U \cos\theta t \;\cdots(i)$

The vertical component of the velocity will vary as per the equation of laws of motion,

$s=ut+\frac12at^2\;\cdots(ii)$ , where,s, u and a are the displacement, initial velocity, and acceleration of the object in the same direction.

(a) At instant position P(x,y):

The vertical component of the initial velocity is, $U_y=U sin\theta$ .

Vertical displacement =y

So, s=y

Acceleration due to gravity is $g=9.81 m/s^2$ in the downward direction.

So, $a=-g=-9.81 m/s^2$ (as per sigh convention)

Now, from equation (ii),

$y=U sin\theta t +\frac 12 (-9.81)t^2$

$\Rightarrow y=U \sin\theta \times \frac {x}{U \cos\theta} +\frac 12 (-g)\times \left(\frac {x}{U \cos\theta} \right)^2$

$\Rightarrow y=U^2 \tan\theta-\frac 1 2g U^2 \sec^2 \theta\;\cdots(iii)$

This is the required, quadratic equation, where $U=15 m/s$ and $g=9.81 m/s^2$ .

(b) At the highest point the vertical velocity,v, of the rocket becomes zero.

From the law of motion, $v=u+at$

$\Rightarroe 0=U\sin\theta-gt$

$\Rightarroe t=\frac{U\sin\theta}{g}\cdots(iv)$

The rocket will reach the maximum height at $t= 1.53 \sin\theta s$

So, from equations (ii) and (iv), the maximum height, $y_m$ is

$y_m=U\sin\theta\times \frac{U\sin\theta}{g}-\frac 12 g \left(\frac{U\sin\theta}{g}\right)^2$

$\Rightarrow y_m=23 \sin\theta -11.5 \sin^2\theta$

In the case of vertical launch, $\theta=90^{\circ}$ , and

$\Rightarrow y_m=11.5 m$ and $t=1.53 s$ .

Height from the ground= 11.5+3=14.5 m.

(c) Height of rocket after t=4 s:

$y=15 \sin\theta \times 4- \frac 12 (9.81)\times 4^2$

$\Rightarrow y=15 \sin\theta-78.48$

$\Rightarrow -63.48 m >y> 78.48$

This is the mathematical position of the graph shown which is below ground but there is the ground at $y=-3m$ , so the rocket will be at the ground at t=4 s.