$x^2-12x=-36\ \ \ |add\ 36\ to\ both\ sides\\\\x^2-12x+36=0\\\\x^2-2x\cdot6+6^2=0\ \ |use\ (a-b)^2=a^2-2ab+b^2\\\\(x-6)^2=0\iff x-6=0\ \ \ |add\ 6\ to\ both\ sides\\\\\boxed{x=6}$

6 0

3 years ago

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The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati

makkiz [27]

Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

$y = \sqrt{x} \,\,\,\,, 0\leq x \leq 1 \\$

And the second one would be

$y = 2-x \,\,\,\,\,, 1 \leq x \leq 2$ .

If you rotate the first region around the "y" axis you get that

$A_1 = 2\pi \int\limits_{0}^{1} x\sqrt{x} dx = \frac{4\pi}{5} = 2.51$

And if you rotate the second region around the "y" axis you get that

$A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188$

And the sum would be 2.51+4.188 = 6.698

If you revolve just the outer curve you get

If you rotate the first region around the x axis you get that

$A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708$

And if you rotate the second region around the x axis you get that

$A_2 = \pi \int\limits_{1}^{2} (2-x)^2 dx = \frac{\pi}{3} = 1.0472$

And the sum would be 1.5708+1.0472 = 2.618

7 0

3 years ago

What is the equation of the midline of the sinusoidal function?<br> [Enter the answer in the box]

Contact [7]

Answer:

$y=0$

Step-by-step explanation:

We are asked to find the equation of mid-line of the given sinusoidal function.

Since the mid-line of a sinusoidal function is the line that runs between the maximum and minimum y-values of the function. We can consider it the middle y-value.

$\text{Mid-line}=\frac{\text{Maximum value+Minimum value}}{2}$