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Naily [24]
3 years ago
13

15. Bradley has a goal to work 28 hours each week at the pizza shop. So far he has

Mathematics
1 answer:
Lina20 [59]3 years ago
7 0
16

28-12=16

He needs 16 more hours of work to complete
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Heba ate 1/12 of a box of cereal. Now the box is 3/4 full
vaieri [72.5K]

Answer:

5/6

Step-by-step explanation:

1/12 + 3/4 = 9/12

9/12 + 1/12 = 10/12

simplify 10/12

= 5/6

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2 years ago
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Order these units from smallest to largest:
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The units are ordered smallest to largest from top to bottom:
Picometer (pm) = 1 x 10⁻¹² meters
Nanometer (nm) = 1 x 10⁻⁹ meters
Micrometer (um) = 1 x 10⁻⁶ meters
Millimeter (mm) = 1 x 10⁻³ meters
Centimeter (cm) = 1 x 10⁻² meters
Decimeter (dm) = 1 x 10⁻¹ meters
Meters (m) = 1 meter
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7 0
3 years ago
An initial investment of $60.00 increases in value by 15% each year. Which of the following statements are true? Select all that
aivan3 [116]
The situations can be represented by the exponential function f(x)=60x1.15^x
After 7.86 years the value of the investment will be three times the initial value (If you round to the nearest dollar)
After 8 years the value of the investment will be $184.00 (If you round to the nearest dollar)
3 0
3 years ago
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Sketch the domain D bounded by y = x^2, y = (1/2)x^2, and y=6x. Use a change of variables with the map x = uv, y = u^2 (for u ?
cluponka [151]

Under the given transformation, the Jacobian and its determinant are

\begin{cases}x=uv\\y=u^2\end{cases}\implies J=\begin{bmatrix}v&u\\2u&0\end{bmatrix}\implies|\det J|=2u^2

so that

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\iint_{D'}\frac{2u^2}{u^2}\,\mathrm du\,\mathrm dv=2\iint_{D'}\mathrm du\,\mathrm dv

where D' is the region D transformed into the u-v plane. The remaining integral is the twice the area of D'.

Now, the integral over D is

\displaystyle\iint_D\frac{\mathrm dx\,\mathrm dy}y=\left\{\int_0^6\int_{x^2/2}^{x^2}+\int_6^{12}\int_{x^2/2}^{6x}\right\}\frac{\mathrm dx\,\mathrm dy}y

but through the given transformation, the boundary of D' is the set of equations,

\begin{array}{l}y=x^2\implies u^2=u^2v^2\implies v^2=1\implies v=\pm1\\y=\frac{x^2}2\implies u^2=\frac{u^2v^2}2\implies v^2=2\implies v=\pm\sqrt2\\y=6x\implies u^2=6uv\implies u=6v\end{array}

We require that u>0, and the last equation tells us that we would also need v>0. This means 1\le v\le\sqrt2 and 0, so that the integral over D' is

\displaystyle2\iint_{D'}\mathrm du\,\mathrm dv=2\int_1^{\sqrt2}\int_0^{6v}\mathrm du\,\mathrm dv=\boxed6

4 0
2 years ago
Nemos aquarium is filled with 2400 cm of water the base of the aquarium is 20 cm long and 12 cm wide what is the height of the w
SSSSS [86.1K]

The answer to this question is 10 cm

In this question, you asked the height and given the volume of water(2400cm3), aquarium length(20cm), and aquarium width(12cm).

Then you need to divide the volume with the length and width. The equation would be:

Aquarium height= volume / length / width

Aquarium height= 2400 cm3 / 20cm / 12 cm = 10cm.

7 0
2 years ago
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