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3 years ago

A square piece of cheese covers 10 square inches of a plate. What is the approximate length of one side of the piece of cheese?

Mathematics

1 answer:

Lelu [443]3 years ago

8 0

Answer:

Step-by-step explanation:

If its a square all the sides are equal so therefor if it took up ten inches of the plate the length is 10 by 10 so 100 because the equation for length is base times height.

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kristin drove to her friend's house and back. on the trip there she drove 40 km/h and on the return trip she went 50 km/h. how l

Leno4ka [110]

Answer:

50km/he=4hr

40km/he=?

40*4/50

=3hr12min

4hr+3hr 12min

=7hr12min

3 0

3 years ago

What is the surface area, in centimeters, of a cube with a side length, s. of 1/3cm?

jek_recluse [69]

6/9 = 2/3 cm^2
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3 years ago

A farmer wishes to enclose two identical adjoining rectangular pens against the side of a large barn. She wants each pen to have

Harman [31]

Answer:

L(min) = 125 ft

Dimensions of the pens

biggest side x /2 = 31.225 ft

shorter side w = 20.82 ft

Step-by-step explanation:

Let call w width of the pen ( the shorter side )

Let call x the biggest side of the two pen.

The configuration we get is that we will use 3 times w + just one side of x

Then total length of fence is

L = 3*w + x (1)

And area of one pen is A = (x/2)* w and that area have to be 650 ft²

650*2 = x*w ⇒ w = 1300/x

So to get L as function of x we plugg this value in equation (1)

L(x) = 3* 1300/x + x

Taking derivatives on both sides of the equation

L´(x) = -3900/x² + 1

L´(x) = 0 ⇒ x² = 3900 ⇒ x = 62.45 ft

and w = 1300/x w = 20.82 ft

And the least amount of fence is

L(min) = 3* 20.82 + 62,45

L(min) = 62.45 + 62.45

L(min) = 125 ft

3 0

4 years ago

Hexagonal tiles of side 20 cm are used to tile a room which measures 6.25 m by 4.85 m. Assuming we complete the edges by cutting

ch4aika [34]

Answer: 292 tiles

Step-by-step explanation:

To find the number of tiles needed , we need to know the area of the hexagonal tiles and also the area of the room.

Area of hexagon

The area of hexagon is given as :

Area = $\frac{3\sqrt{3}}{2}$ $a^{2}$

Where a is the length of side.

Therefore:

A = $\frac{3\sqrt{3}}{2}$ $20^{2}$

A = $\frac{3\sqrt{3}}{2}$ X 400

A = (400 x 3 x $\sqrt{3}$ ) / 2

A = $\frac{2078.460969}{2}$

A = 1039.23 $cm^{2}$

Also to find the area of the room , we must first convert from m to cm

6.25 m = 6.25 x 100 cm

6.25m = 625 cm

4.85m = 485cm

Therefore: Area of the room = length x breadth

Area = 625 x 485

Area = 303125 $cm^{2}$

Therefore , the number of tiles needed = area of the room / area of hexagon

Number of tiles = 303125/ 1039.23

N = 291.6823032

Therefore , the number of tiles needed ≈ 292

4 0

4 years ago

PLEASE HELP ASAP PLEEAASSEEE

STatiana [176]

I think It’s A took the test

8 0

4 years ago

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