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mojhsa [17]
2 years ago
10

A ladder reaches a window 20 feet above the ground. The foot of the ladder is 4 feet from the wall. Which of the following measu

res correspond to the length to the laddder? Round to the nearest hundredth

Mathematics
1 answer:
VMariaS [17]2 years ago
6 0
20.40 ft^{2}

When this situation is drawn out, it forms a right triangle (as shown in the attached image). The sides that are given are the two legs of the triangle. The side that is missing is the hypotenuse. The missing side can be found using the pythagorean theorem: a^{2} +  b^{2} =  c^{2}

A = 20 
B= 4
C = ?

This process is shown in the attached image. The length of the missing side (ladder) rounded to the nearest hundredth is 20.40ft^{2}

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The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 88/(2 + x2 + y2), where T is measured in °C and x,
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Answer:

D_uT(3,1)=-\frac{44}{9}*\frac{1}{\sqrt{2} } \approx-3.46

Step-by-step explanation:

To find the rate of change of temperature with respect to distance at the point (3, 1) in the x-direction and the y-direction we need to find the Directional Derivative of T(x,y). The definition of the directional derivative is given by:

D_uT(x,y)=T_x(x,y)i+T_y(x,y)j

Where i and j are the rectangular components of a unit vector. In this case, the problem don't give us additional information, so let's asume:

i=\frac{1}{\sqrt{2} }

j=\frac{1}{\sqrt{2} }

So, we need to find the partial derivative with respect to x and y:

In order to do the things easier let's make the next substitution:

u=2+x^2+y^2

and express T(x,y) as:

T(x,y)=88*u^{-1}

The partial derivative with respect to x is:

Using the chain rule:

\frac{\partial u}{\partial x}=2x

Hence:

T_x(x,y)=88*(u^{-2})*\frac{\partial u}{\partial x}

Symplying the expression and replacing the value of u:

T_x(x,y)=\frac{-176x}{(2+x^2+y^2)^2}

The partial derivative with respect to y is:

Using the chain rule:

\frac{\partial u}{\partial y}=2y

Hence:

T_y(x,y)=88*(u^{-2})*\frac{\partial u}{\partial y}

Symplying the expression and replacing the value of u:

T_y(x,y)=\frac{-176y}{(2+x^2+y^2)^2}

Therefore:

D_uT(x,y)=(\frac{1}{\sqrt{2} } )*(\frac{-176x}{(2+x^2+y^2)^2} -\frac{176y}{(2+x^2+y^2)^2})

Evaluating the point (3,1)

D_uT(3,1)=(\frac{1}{\sqrt{2} } )*(\frac{-176(3)-176(1)}{(2+3^2+1^2)^2})=(\frac{1}{\sqrt{2} })* (-\frac{704}{144})=(\frac{1}{\sqrt{2} }) ( - \frac{44}{9})\approx -3.46

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2 years ago
Arthur wrote that 15 – 14.7 = 3.
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3 is incorrect because 14.7 + 3 = 17.7
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0.96 or 19/20

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