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Alexandra [31]
3 years ago
11

Which of the following are the correct properties of a slope? Check all that apply. A. A negative slope means that the line move

s down from left to right. B. The slope of a flat line should be exactly 0. C. The slope of a flat line is smaller than the slope of a steep line with a positive slope. D. The slope of a flat line is undefined. E. A negative slope means that the line moves up from left to right.
Mathematics
1 answer:
Svetllana [295]3 years ago
7 0

Answer:A and D

Step-by-step explanation: I am smart trust me :)

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D Community property of addition

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Step-by-step explanation:

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A homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his pro
Anestetic [448]

Answer:

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

Step-by-step explanation:

Given that, a homeowner plans to enclose a 200 square foot rectangle playground.

Let the width of the playground be y and the length of the  playground be x which is the side along the boundary.

The perimeter of the playground is = 2(length +width)

                                                          =2(x+y) foot

The material costs $1 per foot.

Therefore total cost to give boundary of the play ground

=$[ 2(x+y)×1]    

=$[2(x+y)]  

But the neighbor will play one third of the side x foot.

So the neighbor will play  =\$(\frac13x)

Now homeowner's total cost for the material is

=\$[ 2(x+y)-\frac13x]

=\$[2x+2y-\frac13x]

=\$[2y+x+x-\frac13x]

=\$[2y+x+\frac{3x-x}{3}]

=\$[2y+x+\frac{2}{3}x]

=\$[2y+\frac53x]

\therefore C(x)=2y+\frac53x  

where C(x) is total cost of material in $.

Given that the area of the playground is 200.

We know that,

The area of a rectangle is =length×width

                                           =xy square foot

∴xy=200

\Rightarrow y=\frac{200}{x}

Putting the value of y in C(x)

\therefore C(x)=2(\frac{200}{x})+\frac53x

The domain of C is(0,\infty ).

\therefore C(x)=2(\frac{200}{x})+\frac53x

Differentiating with respect to x

C'(x)= - \frac{400}{x^2}+\frac53

Again differentiating with respect to x

C''(x) = \frac{800}{x^3}

To find the critical point set C'(x)=0

\therefore 0= - \frac{400}{x^2}+\frac53

\Rightarrow \frac{400}{x^2}=\frac{5}{3}

\Rightarrow x^2 =\frac{400\times 3}{5}

\Rightarrow x=\sqrt{240}

\Rightarrow x=15.49 \approx15.5

Therefore

\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}>0

Therefore at x= 15.5 , C(x) is minimum.

Putting the value of x in y=\frac{200}{x} we get

\therefore y=\frac{200}{15.5}

    =12.9

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

                                             

6 0
3 years ago
What is the y-coordinate of the plotted point?
Nana76 [90]

Answer:

(3,2)

Step-by-step explanation:

run before you jump

3 0
2 years ago
Read 2 more answers
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