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Scrat [10]
2 years ago
12

Ashley bought bag of oranges at the grocery store she gave half of the oranges to David then she gave Erin 6 oranges and Kept th

e 2 remaining oranges for herself how many did Ashley buy??
Mathematics
2 answers:
KengaRu [80]2 years ago
8 0
1/2 to david is 8, 6 to erin, 2 for herself 8+8=16

andreev551 [17]2 years ago
5 0
The answer would be 16. She gave Erin 6 and kept two. That would be 8 oranges between the two of them and David got half, making the total 16 oranges.

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Step-by-step explanation:25790=/,svxy

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Answer:

This is quite simple when you consider it. go to -4 on the x axis and find where it goes on the y which in this case is 1. so the output for y is 1

Step-by-step explanation:

4 0
3 years ago
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Rzqust [24]

Answer:

y = -2^x

Step-by-step explanation:

If the equation of a function is in the form of y = h(x)

When the graph of this function is reflected across x-axis,

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y = -h(x)

Further reflected across y-axis, then the transformed function will be,

y = -h(-x)

By this rule,

Given equation when reflected across x-axis,

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5 0
2 years ago
Mandy and her mom are making homemade play-dough. Besides the flour, the recipe calls for 1/4 cup of salt for every 3/4 cup of w
prisoha [69]

Answer:

about a half teaspoon

Step-by-step explanation:

4 0
2 years ago
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
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