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Allushta [10]
3 years ago
5

The length of a rectangle is 3meters longer than the width. If the area is 34 square meters, find the rectangle's dimensions. Ro

und to the nearest tenth of a meter.
Mathematics
1 answer:
m_a_m_a [10]3 years ago
3 0

Answer:

Step-by-step explanation:

The length of a rectangle is 3meters longer than the width. If the area is 34 square meters, find the rectangle's dimensions. Round to the nearest tenth of a meter.

Area of the rectangle = Length × Width

L = w + 3

A = 34 m³

Hence,

34 = (w + 3)w

34 = w² + 3w

w² + 3w - 34 = 0

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Write an equation of a line that passes through (-1,6) and has a slope of 4.
jeka94

Answer:

y-6=4(x+1) or y=4x+10

Step-by-step explanation:

Point: (-1,6)

m=4

Slope-point form is y-y1=m(x-x1)

y1=6

x1=-1

Now, plug it in!

y-6=4(x-(-1)) which is y-6=4(x+1)

Of course, if you want you could always simplify it down to y=4x+10 if you wanted to

Hope this helps!

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1. X+4x+6x need help ASAP 10 minutes remaining ​
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A ribbon 105 centimeters long is cut into two pieces. One of the pieces is four times longer than the other. What is the length,
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Answer:

  21 cm

Step-by-step explanation:

The two pieces have a length ratio of 1 : 4, so the shorter piece is 1/(1+4) = 1/5 of the total length.

 105 cm/5 = 21 cm

The shorter piece is 21 cm.

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4 0
3 years ago
A piece of wire of length 5050 is​ cut, and the resulting two pieces are formed to make a circle and a square. Where should the
algol [13]

Answer:

x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.

Step-by-step explanation:

Let x be the length of wire that is cut to form a circle within the 5050 wire, so 5050 - x would be the length to form a square.

A circle with perimeter of x would have a radius of x/(2π), and its area would be

A_c = \pir^2 = \pi (\frac{x}{2pi})^2 = \frac{x^2}{4\pi}

A square with perimeter of 5050 - x would have side length of (5050 - x)/4, and its area would be

A_s = (\frac{5050 - x}{4})^2 = \frac{(5050 - x)^2}{16}

The total combined area of the square and circles is

\sum A = A_c + A_s = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16}

To find the maximum and minimum of this, we just take the 1st derivative, and set it to 0

A' = \frac{2x}{4\pi} + \frac{-2(5050-x)}{16} = 0

\frac{x}{2\pi} - \frac{5050 - x}{8} = 0

Multiple both sides by 8π and we have

4x - 5050\pi + x\pi = 0

x(4 + \pi) = 5050\pi

x = \frac{5050\pi}{4 + \pi} = 2221.5

At x = 2221.5:

A = \frac{x^2}{4\pi} + \frac{(5050 - x)^2}{16} = 392720 + 500026 = 892746 [/tex]

At x = 0, A = 5050^2/16 = 1593906

At x = 5050, A = \frac{5050^2}{4\pi} = 2029424

As 892746 < 1593906 < 2029424, x should be cut at 2221.5 to minimize the total combined area, and at 5050 to maximize it.

3 0
3 years ago
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