-PLEASE HELP- Find the slope of the line passing through the points (2,5) and (8,-4)

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1.The plate, including the bound

rjkz [21]

Answer:

We have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . and the hottest value is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

Step-by-step explanation:

We need to take the derivative with respect of x and y, and equal to zero to find the local minimums.

The temperature equation is:

$T(x,y)=x^{2}+2y^{2}-\frac{3}{2}x$

Let's take the partials derivatives.

$T_{x}(x,y)=2x-\frac{3}{2}=0$

$T_{y}(x,y)=4y=0$

So, we can find the critical point (x,y) of T(x,y).

$2x-\frac{3}{2}=0$

$x=\frac{3}{4}$

$4y=0$

$y=0$

The critical point is (3/4,0) so the temperature at this point is: $T(\frac{3}{4},0)=(\frac{3}{4})^{2}+2(0)^{2}-(\frac{3}{2})(\frac{3}{4})$

$T(\frac{3}{4},0)=-\frac{9}{16}$

Now, we need to evaluate the boundary condition.

$x^{2}+y^{2}=1$

We can solve this equation for y and evaluate this value in the temperature.

$y=\pm \sqrt{1-x^{2}}$

$T(x,\sqrt{1-x^{2}})=x^{2}+2(1-x^{2})-\frac{3}{2}x$

$T(x,\sqrt{1-x^{2}})=-x^{2}-\frac{3}{2}x+2$

Now, let's find the critical point again, as we did above.

$T_{x}(x,\sqrt{1-x^{2}})=-2x-\frac{3}{2}=0$

$x=-\frac{3}{4}$

Evaluating T(x,y) at this point, we have:

$T(-(3/4),\sqrt{1-(-3/4)^{2}})=-(-\frac{3}{4})^{2}-\frac{3}{2}(-\frac{3}{4})+2$

$T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$

Now, we can see that at point (3/4,0) we have the coldest value of temperature $T(\frac{3}{4},0) = -9/16$ . On the other hand, at the point $-(3/4),\frac{\sqrt{7}}{4})$ we have the hottest value of temperature, it is $T(-(3/4),\frac{\sqrt{7}}{4})=\frac{5}{16}$ .

I hope it helps you!

4 0

2 years ago

5)

masya89 [10]

Answer:

9.95

Step-by-step explanation:

you add 4 miles + 5.2 miles which is 9.2 miles then u add the 3/4 mile which is .75 in a decimal form which equals 9.95

7 0

3 years ago

Is my answer right? I think it’s A, but I don’t want to get it wrong.

never [62]

The answer is 16 if you solve it, because the square root of 64 is 8 and 2 * 8 = 16

5 0

3 years ago

What is the slope of the line that passed through the points (8,6) and (-2, 10)

lbvjy [14]

Slope: -4/10 = 4/-10

Work: y2-y1/x2-x1

1) 10-6/-2-8

2)4/-10

8 0

3 years ago

A football is thrown into the air. The height, h(t), of the ball, in metres, after t seconds is modelled by h(t) 4.9(t1.25)2

kolbaska11 [484]

Answer: how high off the ground was the ball when it was thrown

Step-by-step explanation:

7 0

3 years ago