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2 years ago

10

A rectangle has a length of 12.6 m and a width of 6.3 m. A similar rectangle has a width of 2.1 m. What is the length of the sim

ilar rectangle?

1 answer:

Oxana [17]2 years ago

5 0

Step-by-step explanation:

Please see the photo for the method

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A line segment connecting any two opposite Vertex of a quadrilateral

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Answer:

As applied to a polygon, a diagonal is a line segment joining any two non-consecutive vertices. Therefore, a quadrilateral has two diagonals, joining opposite pairs of vertices. For any convex polygon, all the diagonals are inside the polygon, but for re-entrant polygons, some diagonals are outside of the polygon.

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2 years ago

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2 years ago

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3 years ago

Read 2 more answers

fred has two types of peanut butter,three types of jelly, and one type of bread. how many different peanut butter and jelly sand

nydimaria [60]

Answer:

6

Step-by-step explanation:

fred has 2 peanut butter and 3 jelly

type 1 peanut butter can go with all three types of jelly = 3 pb&js where ther is one constant peanutbutter and 3 different jellys

type 2 peanut butter can go with all three jellies = 3 pb&js where there is one constant peanutbutter and 3 different jellies

3+3=6

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2 years ago

A cylinder shaped can needs to be constructed to hold 400 cubic centimeters of soup. The material for the sides of the can costs

LenKa [72]

Answer:

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

Step-by-step explanation:

Volume of the Cylinder=400 cm³

Volume of a Cylinder=πr²h

Therefore: πr²h=400

$h=\frac{400}{\pi r^2}$

Total Surface Area of a Cylinder=2πr²+2πrh

Cost of the materials for the Top and Bottom=0.06 cents per square centimeter

Cost of the materials for the sides=0.03 cents per square centimeter

Cost of the Cylinder=0.06(2πr²)+0.03(2πrh)

C=0.12πr²+0.06πrh

Recall: $h=\frac{400}{\pi r^2}$

Therefore:

$C(r)=0.12\pi r^2+0.06 \pi r(\frac{400}{\pi r^2})$

$C(r)=0.12\pi r^2+\frac{24}{r}$

$C(r)=\frac{0.12\pi r^3+24}{r}$

The minimum cost occurs when the derivative of the Cost =0.

$C^{'}(r)=\frac{6\pi r^3-600}{25r^2}$

$6\pi r^3-600=0$

$6\pi r^3=600$

$\pi r^3=100$

$r^3=\frac{100}{\pi}$

$r^3=31.83$

r=3.17 cm

Recall that:

$h=\frac{400}{\pi r^2}$

$h=\frac{400}{\pi *3.17^2}$

h=12.67cm

The dimensions of the can that will minimize the cost are a Radius of 3.17cm and a Height of 12.67cm.

3 0

3 years ago