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1 year ago

11

Bruno uses a piece of wrapping paper with dimensions 1 and 1/4 feet by 3 feet to wrap a gift. What is the total area of the pape

r used to wrap the gift?

2 answers:

GrogVix [38]1 year ago

6 0

Area is determined by using the formula A=lw Area= Length x Width

A= 1 1/4 x 3 convert to an improper fraction

A= 5/4 x 3/1 multiply

A=15/4 divide

A=3 3/4 square feet

Leto [7]1 year ago

4 0

Answer: $A=3\dfrac{3}{4}\ ft^2$

Step-by-step explanation:

Given: The dimensions of the wrapping paper :

$1\text{ and}\dfrac{1}{4}\text{ feet }\text{ by 3 feet}\\\\\text{i.e }1+\dfrac{1}{4}\text{ feet }\text{ by 3 feet}\\\\\text{i.e }\dfrac{5}{4}\text{ feet }\text{ by 3 feet}$

Now, we know that the area of a rectangle is given by :-

$A=length*width$

Now, the area of the wrapping paper is given by :-

$A=\dfrac{5}{4}\times3=\dfrac{15}{4}=3\dfrac{3}{4}\ ft^2$

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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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