HELP ME PLEASE

Mathematics

1 answer:

zhannawk [14.2K]2 years ago

7 0

Answer:(a) getting the equation:

We'll assume that the altitude is represented by the x-axis and that the boiling point is represented by the y-axis.

Therefore, we have two given points:

(8400 , 200.4) and (4200 , 206.7)

Since the relation is linear, therefore, the graph forms of straight line with the general equation:

y = mx + c where m is the slope and c is the y-intercept.

First, we will calculate the slope:

m = (y2-y1) / (x2-x1) = (206.7 - 200.4) / (4200 - 8400) = -0.0015

Therefore, the equation of the line now is:

y = -0.0015x + c

Then, we will need to calculate the y-intercept. In order to do so, we will use any give point and substitute in the previously obtained equation as follows:

y = -0.0015x + c

206.7 = -0.0015(4200) + c

c = 213

Based on the above calculations, the equation of the line is:

y = -0.0015x + 213

(b) getting the boiling point at altitude = 2100 ft:

Now, in order to calculate the boiling point at altitude 2100, we will substitute in the equation of the line as follows:

y = -0.0015(2100) + 213 = 209.85 degrees Fahrenheit

Step-by-step explanation:

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Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect

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$4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1$

defines an ellipse centered at $(0,0)$ with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the $x$ -axis. So if cross sections are taken perpendicular to the $x$ -axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at $x=x_0$ will have a base of length

$\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}$
$\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}$

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a $\sqrt2:1$ ratio with either of the remaining legs. So if the hypotenuse is $\dfrac43\sqrt{9-{x_0}^2}$ , then either leg will have length $\dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}$ .

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is $\dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}$ .

Finally, $x_0$ can be chosen from any value in $-3\le x_0\le3$ . We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

$\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)$

and so the volume would be

$\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx$
$=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}$
$=16$