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Inessa [10]
3 years ago
6

Adam built a toy box for his children’s wooden blocks.

Mathematics
1 answer:
navik [9.2K]3 years ago
4 0

Answer:

A. 162

B. 128

Step-by-step explanation:

We have wooden cubes of 2 in of side.

A. If the box is 18 x 12 x 6 in., in the base (18 x 12), we can fit 54 cubes:

\dfrac{18}{2}\cdot \dfrac{12}{2}=9\cdot 6=54

As the height is 6 in, we can fit 6/2=3 layers of 54 cubes in the box, se we can fit 162 cubes:

B=3\cdot54=162

B. If we have a box with measures 16 x 9 x 9 in., and the base is 16x9 in, we can fit (16/2)*(9/2)=8*4=32 cubes per layer.

Note that in the 9 in side we can only put 4, as we need 10 in to fit 5 cubes (we can not cut them).

In the same way, as we have 9 in of height, we can fit 4 layers of 32 cubes, so this is a total of 128 cubes:

B=4\cdot32=128

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Your line in slope-intercept form is
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3 years ago
A plane takes off from an airport and flies at a speed of 400km/h on a course of 120° for 2 hours. the plane then changes its co
butalik [34]

Answer:

Distance from the airport = 894.43 km

Step-by-step explanation:

Displacement and Velocity

The velocity of an object assumed as constant in time can be computed as

\displaystyle \vec{v}=\frac{\vec{x}}{t}

Where \vec x is the displacement. Both the velocity and displacement are vectors. The displacement can be computed from the above relation as

\displaystyle \vec{x}=\vec{v}.t

The plane goes at 400 Km/h on a course of 120° for 2 hours. We can compute the components of the velocity as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_1}=\ km/h

The displacement of the plane in 2 hours is

\displaystyle \vec{x_1}=\vec{v_1}.t_1=.(2)

\displaystyle \vec{x_1}=km

Now the plane keeps the same speed but now its course is 210° for 1 hour. The components of the velocity are

\displaystyle \vec{v_2}=

\displaystyle \vec{v_2}=km/h

The displacement in 1 hour is

\displaystyle \vec{x_2}=\vec{v_2}.t_2=.(1)

\displaystyle \vec{x_2}=km

The total displacement is the vector sum of both

\displaystyle \vec{x_t}=\vec{x_1}+\vec{x_2}=+

\displaystyle \vec{x_t}=km

\displaystyle \vec{x_t}=

The distance from the airport is the module of the displacement:

\displaystyle |\vec{x_t}|=\sqrt{(-746.41)^2+492.82^2}

\displaystyle |\vec{x_t}|=894.43\ km

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Step-by-step explanation:

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