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2 years ago

Someone help me on this plzzz!!

Mathematics

1 answer:

agasfer [191]2 years ago

7 0

Answer:

y = (3/2)x + 2

Step-by-step explanation:

The equation of a straight line is given by:

y = mx + b;

where y, x are variables, m is the slope of the line and b is the y intercept (value of y when x is 0).

From the graph, we can see that the y intercept is 2 units. Also, the line passes through point (2, 5) and (-2, -1). Therefore the slope of the line is:

$slope(m)=\frac{y_2-y_1}{x_2-x_1}=\frac{-1-5}{-2-2}=\frac{3}{2}$

Using the equation a straight line:

y = mx + b; substituting gives:

y = (3/2)x + 2

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Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
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So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
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etc.

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2 years ago

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Step-by-step explanation:

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The semi-circle has a diameter of 2 + 4 + 2 = 8. The area this figure uses the radius which is half the diameter. The radius is 4. To find the area substitute r = 4 into $\pi r^2 = \pi 4^2 = 16\pi$ . However the semi-circle has a smaller circle cut out of it with radius 2. The area of the smaller circle is $\pir^2 = \pi 2^2=4\pi$ . The semi circle in the shape is the areas subtracted which equals 12π.

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The area of the rectangle is found using A = b*h = 2*5 = 10.

The total area is 12π + 10 meters squared.

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2 years ago