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satela [25.4K]
3 years ago
12

A flower pot, in the shape of a cylinder,l has a circumference of 66 in. and a height of 30 in.

Mathematics
1 answer:
Vaselesa [24]3 years ago
6 0
The amount of soil that the pot can hold is best approximated by calculating for the volume of the cylinder through the equation,
                                V = πr²h
We first calculate for the radius, r, given the circumference, C
                      C = 2πr;    r = C/2π = 66 in/2(π) = 10.5 in
Then, substituting the known values. 
                             V = π(10.5 in)²(30 in) 
                                 V = 10,390.81 in³
Thus, the volume of the soil is approximately 10,400. The answer is the third set.
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Triangles 1 and 2 have the angle measures
lawyer [7]

Answer:

C

Step-by-step explanation:

Both triangles have three angles of the same value.

Remember that the angles in all triangles add up to 180°.

Let's use that to find out the unknown angles.

For the first triangle:

180 - 82 - 43 = 55°

55° is also in the second triangle.

Let's check with the second triangle:

180 - 82 - 55 = 43°

43° is also in the first triangle.

Therefore, both triangles are similar as the angles in both triangles are the same - 82°, 43° and 55°.

Hence, C.

7 0
3 years ago
F(x)=−2x+16<br> g(x)=4x+10<br> please give me the intersection points
vagabundo [1.1K]

Answer:

  • (1, 14)

Step-by-step explanation:

The intersection is when both functions have same coordinates

  • f(x) = g(x)

Substitute to get

  • - 2x + 16 = 4x + 10
  • 4x + 2x = 16 - 10
  • 6x = 6
  • x = 1

The y- coordinate is

  • y = - 2(1) + 16 = -2 + 16 = 14

So the intersection point is (1, 14)

7 0
2 years ago
Read 2 more answers
during the 2011 season, a quarterback passed for 302 yards per game. he played in all 16 regular season games that year. how man
MariettaO [177]
302 times 16 because he passed 302 yards for every game 4832
7 0
3 years ago
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Please show working out thanks
IgorLugansk [536]

Answer:

10m x 15m

Step-by-step explanation:

You are given some information.

1. The area of the garden: A₁ = 150m²

2. The area of the path: A₂ = 186m²

3. The width of the path: 3m

If the garden has width w and length l, the area of the garden is:

(1) A₁ = l * w

The area of the path is given by:

(2) A₂ = 3l + 3l + 3w + 3w + 4*3*3 = 6l + 6w + 36

Multiplying (2) with l gives:

(3) A₂l = 6l² + 6lw + 36l

Replacing l*w in (3) with A₁ from (1):

(4) A₂l = 6l² + 6A₁ + 36l

Combining:

(5) 6l² + (36 - A₂)l +6A₁ = 0

Simplifying:

(6) l² - 25l + 150 = 0

This equation can be factored:

(7) (l - 10)*(l - 15) = 0

Solving for l we get 2 solutions:

l₁ = 10, l₂ = 15

Using (1) to find w:

w₁ = 15, w₂ = 10

The two solutions are equivalent. The garden has dimensions 10m and 15m.

3 0
3 years ago
Could use some help with this. Ty to all who help!
Gekata [30.6K]
106 they add up to 180
6 0
3 years ago
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