Can i please get this answered

Mathematics

1 answer:

forsale [732]2 years ago

5 0

Answer:

0 is the answer . The explanation is in the attachment . 

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Sin(x2 + y2) da r , where r is the region in the first quadrant between the circles with center the origin and radii 2 and 5

sweet [91]

Converting to polar coordinates gives

$\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_{\theta=0}^{\theta=\pi/2}\int_{r=2}^{r=5}\sin(r^2)r\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\pi\int_{r=2}^{r=5}2r\sin(r^2)\,\mathrm dr$
$=\displaystyle\pi\int_{r^2=4}^{r^2=25}\sin(r^2)\,\mathrm d(r^2)$
$=-\cos(r^2)\bigg|_{r^2=4}^{r^2=25}$
$=\pi(\cos4-\cos25)$

3 0

3 years ago

Match the following equation to the correct situation.

uranmaximum [27]

Answer:

The equation given in the question matches the situation C.

Step-by-step explanation:

A. The amount that Cindy paid for books was 15% of p i.e. it is only $\frac{15p}{100} = 0.15p$

B. The amount is given to a charity after saving 15% of p. That means if p is the amount of charity, then there were a savings of 15% of p.

Therefore, the total charity payment after savings = p - 15% of p = $p - \frac{15p}{100} = p - 0.15p = 0.85p$

C. The amount that Sandy paid for a shirt with 15% tax on p.

Therefore, if the price of a shirt is p, then total payment of Sandy after tax is = p + 15% of p = $(p + \frac{15p}{100}) = p + 0.15p = 1.15p$

Hence, the equation given in the question matches the situation C. (Answer)

8 0

3 years ago

A ball, with an initial position of x = 25.89 meters, undergoes a displacement of 32.2 meters. What is it’s final position?

Snowcat [4.5K]

Answer:

The answer is 58.09

Step-by-step explanation:

To get the final position, you add the displacement to the initial position.

Final position = 25.89 + 32.2

Final position = 58.09metres

4 0

3 years ago

The vertical asymptote of the function f(x) = 3 log(x + 3)

ipn [44]

A function has a vertical asymptote where it is not defined, and the logarithm is defined only if its argument is strictly positive: it has a vertical asymptote if its argument is zero, and is not defined if its argument is negative.

So, you have a vertical asymptote where

$x+3=0 \iff x=-3$