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3 years ago

Hank estimated the width of the door to his class room in feet what is a reasonable estimate

1 answer:

4 0

My reasonable estimate might be about 2 1/2 feet. I know that because taking three rulers up to your bedroom or a normal room and get about 2 or 2 1/2 feet

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8 millions + 6 thousands + 9 hundreds + 7 ones = 806. 907 A True. b.false

suter [353]

Answer:

true

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Is (0,0) a solution of the graphed inequality?

sweet [91]

No because it is not within the shaded area

5 0

3 years ago

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What is the scale factor in the dilation if the coordinates of A prime are (–7, 6) and the coordinates of C prime are (–4, 3)?

olchik [2.2K]

Option A: $\frac{1}{3}$ is the dilation factor

Explanation:

The Coordinates of the square ABCD are A(-21,18), B(-12,18), C(-12,9), D(-21,9)

Also, the coordinates of the square A'B'C'D' are A'(–7, 6), B'(-4,6), C'(-4,3), D'(-8,3)

Now, we shall determine the scale factor in the dilation if the coordinates A' and C'.

Since, the dilation is the enlargement of the image in the same shape but different size.

To determine the scale factor, let us divide A'C' by AC

Thus, we have,

$\begin{aligned}A^{\prime} C^{\prime} &=\sqrt{(-4+7)^{2}+(3-6)^{2}} \\&=\sqrt{3^{2}+(-3)^{2}} \\&=\sqrt{9+9} \\&=\sqrt{18}\\&=3\sqrt{2} \end{aligned}$

Also,

$\begin{aligned}A C &=\sqrt{(-12+21)^{2}+(9-18)^{2}} \\&=\sqrt{9^{2}+(-9)^{2}} \\&=\sqrt{81+81} \\&=\sqrt{162}\\&=9\sqrt{2} \end{aligned}$

Dividing A'C' by AC, we have,

$\frac{A'C'}{AC} =\frac{3\sqrt{2} }{9\sqrt{2}} =\frac{1}{3}$

Thus, $\frac{1}{3}$ is the dilation factor

3 0

3 years ago

Read 2 more answers

Which expression has a value of -6 A. -4 • 2 - (-14) B. -16 divided by (-2) - (-2) C. -2 - (-8) divided by (-1) D. -12 divided b

tatiyna

Answer:

its B. -16 divided by (-2) - (-2)

8 0

3 years ago

A special liquid is held in a tank described as (x 2 + y 2 ) ≤ z ≤ 1 in a Cartesian coordinate system. Assume that the density o

vivado [14]

Since $\rho=\dfrac mV$ (density = mass/volume), we can get the mass/weight of the liquid by integrating the density $\rho(x,y,z)$ over the interior of the tank. This is done with the integral

$\displaystyle\int_{-1}^1\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int_{x^2+y^2}^1(2-z^2)\,\mathrm dz\,\mathrm dy\,\mathrm dx$

which is more readily computed in cylindrical coordinates as

$\displaystyle\int_0^{2\pi}\int_0^1\int_{r^2}^1(2-z^2)r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=\boxed{\frac{3\pi}4}$

7 0

3 years ago