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Talja [164]
2 years ago
15

Answer? I’m struggling

Mathematics
1 answer:
Dennis_Churaev [7]2 years ago
3 0

a. 3\frac{11}{12}

b.\frac{4}{3}

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Convert 66°F to degrees Celsius.
HACTEHA [7]

Answer: I got 18.889 degrees celsius

Step-by-step explanation:

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3 years ago
Make ”a” the subject of the formula 9a-8=2(a-b)
beks73 [17]

Answer:

Step-by-step explanation:

9a - 8 = 2a - 2b

7a - 8 = -2b

7a = 8 - 2b

a = (8 - 2b)/7

3 0
3 years ago
Work out the percentage change to 2 decimal places when a price of £198 is increased to £209.99.
bija089 [108]

Answer:

The percentage change = 6.05%

Step-by-step explanation:

New value = £209.99

Old value = £198

Percentage change = [New Value - Old Value] / [Old Value] × 100

                                 = [209.99 - 198] / [198] × 100

                                 = [11.99] / [ 198] × 100

                                 = 0.06 × 100

                                 = 6.05%

Therefore, the percentage change = 6.05%

5 0
2 years ago
For f(x) = 0.01(2)x, find the average rate of change from x = 12 to x = 15.
loris [4]
The correct answer is C. i think
6 0
3 years ago
Read 2 more answers
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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