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3 years ago

Aarón made a picture frame with the dimensions show in the figure what is the area of the picture

Mathematics

1 answer:

tatuchka [14]3 years ago

4 0

Answer:

Area of the picture is $72\textrm{ }cm^{2}$

Step-by-step explanation:

Given:

Outer rectangle dimensions are 10 cm by 12 cm.

Inner rectangle dimensions are 6 cm by 8 cm.

Area of the picture given as:

$Area=\textrm{Outer rectangle area}-\textrm{Inner rectangle area}$

Outer rectangle area = $10\times 12=120\textrm{ }cm^{2}$

Inner rectangle area = $6\times 8=48\textrm{ }cm^{2}$

Therefore, the area of the picture is given:

$Area=120-48=72\textrm{ }cm^{2}$

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6x + 15xy<br> Looking For Factor

padilas [110]

Answer:

x(6 + 15y)

Step-by-step explanation:

In this instance, you can divide by a factor of x.

6x/x = 6

15xy/x = 15y

Therefore, we are left with 6 + 15y.

6 0

3 years ago

Describe in detail how you would create a number line with the following points: 1, 3.25, the opposite of 2, and – (–4fraction o

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3 years ago

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Vesna [10]

Answer:

14 rides

Step-by-step explanation:

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3 years ago

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

$x=0,y=2\implies z=-2\implies(0,2,-2)$

$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(6-6v,2v-2uv,-8+6v+4uv)\cdot(20v,8v,4v)\,\mathrm du\,\mathrm dv$

$=\displaystyle8\int_0^1\int_0^1(11v-10v^2)\,\mathrm du\,\mathrm dv=\boxed{\frac{52}3}$

8 0

3 years ago