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ale4655 [162]
2 years ago
13

A piece of fabric is 1/3 yard by 3 5/8 yards.

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

Answer:

sorry if this is wrong

Step-by-step explanation:

a. 3

b.15

c.8

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Law of cosines.Anyone good in Trigonometry?​
marin [14]

Answer:

c=11.8\ units

Step-by-step explanation:

we know that

The formula of law of cosines is equal to

c^2=a^2+b^2-2(a)(b)cos(C)

where

a, b and c are sides. C is the angle opposite side c

In this problem we have

a=3\ units\\b=10\ units

C=120^o

substitute the given values

c^2=3^2+10^2-2(3)(10)cos(120^o)

c^2=109-(60)cos(120^o)

c^2=109-(60)(-0.5))

c^2=109+30

c^2=139

c=11.8\ units

3 0
3 years ago
If someone understands this please help
katrin [286]

Answer:

A= 4

B= 4

C= -1

D= 4

E= -1

F= 4

G= -1

Step-by-step explanation:

The expected value equation is the probability of something happening multiplied by the amount of times it happens. In this case you have four equal sized sections, so you have a one in four chance to land in any of these sections. A, B, D, and F represent the four for this one in four chance. C, E, and G represent the amount of points you get when you land on those sections, in this case -1

3 0
3 years ago
A hollow toy, with the dimensions shown in the figure, is to be stuffed with rigid foam. What is the maximum amount of foam that
Fudgin [204]

Option B: 67.02 cubic inches

<h2>Given that:</h2>

The diagram consists of two parts:

  • First a cone on the top with 12 inches height and 4 inches diameter.
  • Second a semi sphere with diameter of 4 inches.

<h2>Calculations:</h2>

The amount of foams that can be stuffed with foam is total volume of both the containers.

Volume of the given cone of 12 inches height and 4 inches diameter can be given as:

\dfrac{1}{3} \times \pi \times 2^2 \times 12\\= 50.265 \: \rm inch ^3

Volume of the semi sphere of 4 inches diameter can be given as:

\dfrac{2}{3} \pi \times 2^3\\= 16.755 \: \rm inch^3

Thus total volume is 50.265 + 16.755 = 67.02 cubic inches.

Thus option B : 67.02 cubic inches is the needed volume.

Learn more here:

brainly.com/question/1315822

7 0
2 years ago
Find the lateral area and total surface area of each of the following. GEOMETRY!!! HELP PLS
AlexFokin [52]

(a) The lateral surface area of the prism is 720 sq units and the total surface area is 752 sq units.

(b) The lateral surface area of the cone is 136π sq units and the total surface area is 200π sq units.

(c) The lateral surface area of the cylinder is 242π sq units and the total surface area is 484π sq units.

<h3>Lateral surface area of the prism</h3>

L.S.A = Ph

where;

  • P is perimeter of the base
  • h is height of the prism

h² = 17² - 8²

h² = 225

h = 15

L.S.A = (3 x 16) x 15 = 720 sq units

<h3>Total surface area of the prism</h3>

T.S.A = PH + 2B

T.S.A = 720 + 2(16) = 752 sq units

<h3>Lateral surface area of the cone</h3>

L.S.A = πrt

where;

  • t is the slant height = 17

r² = 17² - 15²

r² = 64

r = 8

L.S.A = π(8)(17) = 136π sq units

<h3>Total surface area of the cone</h3>

T.S.A = πrt + πr²

T.S.A = 136π sq units + π(8)²

T.S.A = 200π sq units

<h3> Lateral surface area of the cylinder</h3>

L.S.A = 2πrh

where;

  • r is the radius of the cylinder = 11
  • h is height of the cylinder = 11

L.S.A = 2π(11 x 11) = 242π sq units

<h3>Total surface area of the cylinder</h3>

T.S.A = 2πrh + 2πr² = 2πr(r + h)

T.S.A = 2π(11)(11 + 11)

T.S.A = 484π sq units.

Thus, the lateral surface area of the prism is 720 sq units and the total surface area is 752 sq units.

  • the lateral surface area of the cone is 136π sq units and the total surface area is 200π sq units.
  • the lateral surface area of the cylinder is 242π sq units and the total surface area is 484π sq units.

Learn more about surface area here: brainly.com/question/76387

#SPJ1

3 0
2 years ago
Write an equation for the vertical translation y=-2/9 |x| -7; 1 units down
klasskru [66]

Answer:

y = -2/9 |X| - 8

Step-by-step explanation:

I'm not certain whether the -7 is part of the |X| or it's just out of the equation. But shifting an equation up or down is very intuitive.

If you want to move an equation up a certain unit, you just add that many units (positive) onto the end of an equation. The same goes for moving a unit down, where you subtract that many units you want to go down.

8 0
3 years ago
Read 2 more answers
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