$\bf ~~~~~~\textit{parabola vertex form}
\\\\
\begin{array}{llll}
y=a(x- h)^2+ k\\\\
x=a(y- k)^2+ h
\end{array}
\qquad\qquad
vertex~~(\stackrel{}{ h},\stackrel{}{ k})\\\\
-------------------------------\\\\
\stackrel{\textit{zeros\qquad \qquad \qquad }}{
\begin{cases}
x=-4\implies &x+4=0\\
x=2\implies &x-2=0
\end{cases}}
\\\\\\
\stackrel{\textit{factored vertex form}}{y=a(x+4)(x-2)}\qquad (6,10)\textit{ we also know that }
\begin{cases}
x=6\\
y=10
\end{cases}
\\\\\\
10=a(6+4)(6-2)$

3 0

3 years ago

Read 2 more answers

Daniel bought four bags of potatoes. The weight of the first bag was 2.6 pounds. The second bag weighed 0.4 pound less than twic

ipn [44]

Answer:

Average weight of the bags of potatoes is 3.8 pounds.

Step-by-step explanation:

Given:

Weight of the first bag, $W_{1}=2.6$ pounds.

Weight of the second bag is 0.4 pounds less than twice the weight of first bag. This means,

$W_{2}=2W_{1}-0.4=2(2.6)-0.4=5.2-0.4=4.8\textrm{ pounds}$

Weight of the third bag is 0.6 pounds more than that of the first bag. This means,

$W_{3}=W_{1}+0.6=2.6+0.6=3.2\textrm{ pounds}$

Weight of the fourth bag is 0.3 pounds less than that of the second bag. This means,

$W_{4}=W_{2}-0.3=4.8-0.3=4.5\textrm{ pounds}$

Therefore, the average weight of the bags of potatoes is given as the sum of all the weights and then divide the sum by 4. Therefore,

$W_{average}=\frac{W_{1}+W_{2}+W_{3}+W_{4}}{4}=\frac{2.6+4.8+3.2+4.5}{2}=\frac{15.1}{4}=3.775\approx 3.8$

Therefore, average weight of the bags of potatoes is 3.8 pounds.

8 0

3 years ago

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

3 years ago