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neonofarm [45]
2 years ago
5

Omg I need help on this

Mathematics
1 answer:
beks73 [17]2 years ago
6 0

Answer:

y = -3x - 5

Step-by-step explanation:

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Glven the line segment RT, with endpolnt R(6,9) and<br> midpoint A(-3,7)
Rzqust [24]
It would be (-12, 5)
7 0
3 years ago
A rectangular area is to be fenced in with 300 feet of chicken wire. find the maximum area that can be enclosed.
GarryVolchara [31]
Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.

2(x+y) = 300
x+y = 150
y = 150-x

A=x(150-x) <--(substitution)

The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150

So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.

A=75(150-75)
A=75*75
A=5625

So the maximum area that can be enclosed is 5625 square feet.
6 0
3 years ago
What is the answer I’m not sure I’m right
trasher [3.6K]

Answer:

x=0

Step-by-step explanation:

If you combine like terms it is x^2 +6x -27=-27

x^2 +6x =0

x^2=0

x=0

3 0
3 years ago
Find an equation of the tangent plane to the given parametric surface at the specified point.
Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

Step-by-step explanation:

Given equation

      r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}---(1)

Normal vector  tangent to plane is:

\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}

\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}

Normal vector  tangent to plane is given by:

r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]

Expanding with first row

\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\

at u=5, v =π/3

                  =\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k} ---(2)

at u=5, v =π/3 (1) becomes,

                 r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

                r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}

From above eq coordinates of r₀ can be found as:

            r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})

From (2) coordinates of normal vector can be found as

            n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)  

Equation of tangent line can be found as:

  (\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}

5 0
3 years ago
The manager of a store wants to have a sales promotion where she gives prizes to the first several people through the door. She
Viktor [21]
The answer is 60 people. Each person would get <span>4 pins, 6 ornaments and 9 mugs.</span>
4 0
3 years ago
Read 2 more answers
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