PLZ HELP WILL GIVE BRAINLIEST!!!!!!!!

Mathematics

2 answers:

liubo4ka [24]3 years ago

8 0

Yeah its d for sure it is easy if u think

Nimfa-mama [501]3 years ago

5 0

C is the best one I'm pretty sure it makes the most sense

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Use the function f(x) = 2x3 - 3x2 + 7 to complete the exercises.

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Answer:

what are the excersises

Step-by-step explanation:

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A classroom has the dimensions shown the dimensions are the length which is 9.1 the width which is 10.3 and the height which is

melisa1 [442]

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Among all the rectangles whose perimeters are 100 feet, find the dimensions of the one with the maximum area. explain step by st

kompoz [17]

<span>We let x be the length and y be the width of the rectangle. Then,

Perimeter = 2x + 2y
100 = 2x + 2y
50 = x + y
y = 50 - x

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We then take the derivative; set it equal to zero:

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2 years ago

An equilateral triangle has a perimeter of 15x +3. what is the length of each side of the triangle?

xxTIMURxx [149]

The length of each side is 5x + 1

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3 years ago

Find a formula for the described function. A rectangle has perimeter 16 m. Express the area A of the rectangle as a function of

Alex17521 [72]

Part A:

Let the length of one of the sides of the rectangle be L, then the length of the other side is obtained as follow.

Let the length of the other side be x, then

$2(L+x)=16 \\ \\ \Rightarrow L+x=8 \\ \\ \Rightarrow x=8-L$

Thus, if the length of one of the side is x, the length of the other side is 8 - L.

Hence, the area of the rectangle in terms of L is given by

$L(8 - L) = 8L-L^2$

Part B:

To find the domain of A

Recall that the domain of a function is the set of values which can be assumed by the independent variable. In this case, the domain is the set of values that L can take.

Notice that the length of a side of a rectangle cannot be negative or 0, thus L cannot be 8 as 8 - 8 = 0 or any number greater than 8.

Hence the domain of the area are the set of values between 0 and 8 not inclusive.

Therefore,

$dom(Area)=0\ \textless \ L\ \textless \ 8 \ or \ (0, 8)$

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3 years ago