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nevsk [136]
2 years ago
12

Help me, please??????

Mathematics
1 answer:
julsineya [31]2 years ago
3 0

Answer:

B

Step-by-step explanation:

The slope of this function is obviously decreasing, and the point where it intersects the y axis, is (0,-2)

Answer B contains this information

Hope this helps!

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A piece of wire 28 ft long is cut into two pieces. One piece is used to form a square, and the remaining piece is used to form a
marusya05 [52]

Answer:

Wire should be cut in two parts with the length = 15.69 ft and 12.31 ft

Step-by-step explanation:

Length of the wire = 28 ft has been cut in two pieces.

One piece is used to form a square and remaining piece to form a circle.

Let the length of the wire which forms the square is 'l' ft.

Area of the square = (side)²

Perimeter of the square = 4(side) = l

Length of one side = \frac{l}{4}

So, the area of the square = \frac{l^{2}}{16} ft²

Now length of the remaining part = perimeter of the circle = (28 - l) ft

2πr = (28 - l)

r = \frac{28-l}{2\pi }

Area of the circle formed = πr²

= \frac{\pi(28-l)^{2} }{4(\pi )^{2} }

= \frac{(28-l)^{2}}{4\pi }

Combined area of the square and circle

= \frac{l^{2}}{16} + \frac{(28-l)^{2}}{4\pi }

= \frac{l^{2}}{16} + \frac{(28-l)^{2}}{4\pi }

Now to maximize the area we will find the derivative of the area with respect to l.

\frac{dA}{dl}=\frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28-l)^{2}}{4\pi}]

= \frac{d}{dl}[\frac{l^{2}}{16}+\frac{(28)^{2}+l^{2}-56l }{4\pi }]

= \frac{2l}{16}+\frac{(2l-56)}{4\pi }

Now equate the derivative to zero.

\frac{2l}{16}+\frac{(2l-56)}{4\pi } = 0

(2l-56)=-\frac{2l}{16}\times 4\pi

(2l-56)=-\frac{\pi l}{2}

2l+\frac{\pi l}{2}=56

\frac{(4l+l\pi )}{2}=56

l(4 + π) = 112

l(4 + 3.14) = 112

l = \frac{112}{7.14}

l = 15.69 ft

Length of the other part = 28 - 15.69 = 12.31 ft

Therefore, wire should be cut in two parts with the length = 15.69 ft and 12.31 ft

8 0
3 years ago
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