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lbvjy [14]
3 years ago
12

If Roberto does a job in 11 hours and with the help of Sarah they can do it together in 2 hours, how long would it take Sarah to

do it alone?
Mathematics
1 answer:
Levart [38]3 years ago
3 0
I think it is seven but im probaly wrong
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The dress store is having a sale where all merchandise is 1/4 off. A woman buys $48 of merchandise at a sale price.
Fantom [35]

Answer:$36 depending on what question is i just assuming how much she has to pay

Step-by-step explanation:

48 divded by 4 is 12. $48-$12 is $36. The $12 is the 1/4 discount.

7 0
3 years ago
Why does it take 3 copies of 1/6 to show the same amount as 1 copy of 1/2?
melisa1 [442]
Because 1/2 of 6 =1/3 and 1/2×1/3=1/6
7 0
3 years ago
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Please help me I need a good grade I really need this or ,y mom will spank me please help me ¡!!!
AURORKA [14]

Answer:

A D and E

Step-by-step explanation:

3 0
3 years ago
What is the length of segment AB?<br> Consider the diagram.<br> 7,9,18,25
Mrrafil [7]

Answer:

9

Step-by-step explanation:

It is congruent to the other side.

5 0
3 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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