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ollegr [7]
2 years ago
14

A square hole 7.50 cm along each side is cut in a sheet of copper. (a) Calculate the change in the area of this hole resulting w

hen the temperature of the sheet is increased by 49.0 K. cm2(b) Does this change represent an increase or a decrease in the area enclosed by the hole
Mathematics
1 answer:
Katena32 [7]2 years ago
3 0

Answer:

it increases the the hole

Step-by-step explanation:

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Ilu<br> y = x2 – 16x + 63?
Neporo4naja [7]

Answer:

y=-14x+63

Step-by-step explanation:

this is the most simplified equation

7 0
3 years ago
A doctor has an annual income of $152,125. The income tax the doctor has to pay is 6% , what is the amount of income tax in doll
kupik [55]
I believe the amount of income tax the doctor has to pay is $9,127.5
5 0
3 years ago
Read 2 more answers
How does the graph of f(x)=3|x+2|+4 relate to its parent function
Mariana [72]
If parent functin is f(x)=|x|
it is moved to the left 2 units
vertically streched by a factor of 3
and moved up by 4 units in that order

because
to move a function to left c units, add c to every x
to vertically strech function by factor of c, multiply whole function by c
to move funciotn up c units, add c to whole function


so it is 2 to the left, verteically streched by a factor of 3 then moved up 4 units
5 0
3 years ago
Read 2 more answers
Use calculus to find the largest possible area for a rectangular field that can be enclosed with a fence that is 500 meters long
rewona [7]
Let x and y be the sides of the rectangle. The perimeter is given to be 500m, so we are maximizing the area function A(x,y)=xy subject to the constraint 2x+2y=500.

From the constraint, we find

2x+2y=500\implies x+y=250\implies y=250-x

so we can write the area function independently of y:

A(x,y)=\hat A(x)=x(250-x)=250x-x^2

Differentiating and setting equal to zero, we find one critical point:

\dfrac{\mathrm d\hat A(x)}{\mathrm dx}=250-2x=0\implies x=125

which means y=250-125=125, so in fact the largest area is achieved with a square fence that surrounds an area of A(125,125)=125^2=15625\text{ m}^2.
7 0
2 years ago
The Length of a standard jewel case is 7cm more than its width. The area of the rectangular top of the case is 408cm. Find the l
salantis [7]

Answer:

The length of the case is 24 cm and its width is 17cm.

Step-by-step explanation:

The Length of a standard jewel case is 7cm more than its width.

Let the length be represented by L and the width be represented by W, this means that:

L = 7 + W

The area of the rectangular top of the case is 408cm². The area od a rectangle is given as:

A = L * W

Since L = 7 + W:

A = (7 + W) * W = 7W + W²

The area is 408 cm², hence:

408 = 7W + W²

Solving this as a quadratic equation:

=> W² + 7W - 408 = 0

W² + 24W - 17W - 408 = 0

W(W + 24) - 17(W + 24) = 0

(W - 17) (W + 24) = 0

=> W = 17cm or -24 cm

Since width cannot be negative, the width of the case is 17 cm.

Hence, the length, L, is:

L = 7 + 17 = 24cm.

The length of the case is 24 cm and its width is 17cm.

4 0
3 years ago
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