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deff fn [24]
2 years ago
9

What is the surface area of an ice cream cone if the radius is 2 inches

Mathematics
1 answer:
PtichkaEL [24]2 years ago
5 0
It goes 19 1 and 4ths to the area surface
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Resuelve el siguiente problema un buzo en una laguna decendio 8m en una hora.Si cada hora bojo la misma cantidad de metros, ¿cua
S_A_V [24]

Answer:

X = 32 meters.

Step-by-step explanation:

  • Let the unknown distance be X.

<u>Given the following data;</u>

  • Distance = 8 meters per hour
  • Time = 4 hours

To find how many meters he would cover in four hours;

1 hour = 8 meters

4 hours = X meters

Cross-multiplying, we have;

X = 8 * 4

<em>X = 32 meters.</em>

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3 years ago
In preparation for a conference, Toby is setting up some stations where people can create
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Answer:

9

Step-by-step explanation:

I believe it should be 9 becasue the GCF of 9 and 18 is 9.

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3 years ago
In a photograph, a stadium measures 7 inches across by 2 inches high. If the actual stadium measures 1000 feet across, which equ
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Answer:

x/500 = 2/8.

Step-by-step explanation:

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2 years ago
Which of the following equations describes the line shown below , check all that apply
Bond [772]
The line is passing through the y-axis at 3. The slope of the line is 3.
So your equation would be y = 3x + 3.

Now solve A through F to see which equation(s) matches the line.

A. y + 6 = 3(x + 3)
     y + 6 = 3x + 9
     y = 3x + 3

B. y + 6 = 3(x + 1)
     y + 6 = 3x + 3
     y = 3x - 3

C. y - 6 = 3(x - 3)
     y - 6 = 3x - 9 
     y = 3x - 3

D. y - 6 = 3(x - 1)
     y - 6 = 3x - 3
     y = 3x + 3

E. y = 3x + 2

F. y = 3x + 3

Which letters equaled y = 3x +3?
A, D, and F.

The equations that describe the line are A, D, and F.
5 0
3 years ago
JKL has vertices J(3, -5) K(-8, -1) and L(5,1). JKL will be reflected across the line y= -1 to form J'K'L'. Which quadrant will
katen-ka-za [31]

Answer:

The correct answer is second quadrant.

Step-by-step explanation:

Vertices of J ≡ (3 , -5) which shows J is in fourth quadrant. When J is reflected across the line y = -1, the point J moves to first quadrant as J' with coordinates (3 , 3).

Vertices of K ≡ (-8 , -1) which shows K is in third quadrant. When K is reflected across the line y = -1, the point K remains where it is (in the third quadrant) as it is on the line y = -1 only.

Vertices of L ≡ (5 , 1) which shows L is in first quadrant. When L is reflected across the line y = -1, the point L moves to fourth quadrant as L' with coordinates (5 , -3).

Thus each of the first, fourth and third quadrant contains a vertex except the second quadrant.

4 0
3 years ago
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