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notsponge [240]
3 years ago
14

Find the lateral area for the cylinder with the given measurement. r = 4, h = 5

Mathematics
1 answer:
Ivanshal [37]3 years ago
4 0
AL≈125.66<span><span><span>r-Radius</span><span>h-Height</span></span></span>
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PLZ ANSWER CORRECTLY 1 2⁄6 ÷ 4 1⁄3
JulijaS [17]

Answer:

0.30715935334

Step-by-step explanation:

8/6 ÷ 13/3

1.33÷ 4.33

0.30715935334

5 0
3 years ago
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Please help please.......​
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2 years ago
Fill in each blank for the function f(x) = 3x - 6.
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Step-by-step explanation:

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3 years ago
For A (1, –1), B (–1, 3), and C (4, –1), find a possible location of a fourth point, D, so that a parallelogram is formed using
katrin [286]

The possible coordinates for the vertex D of the parallelogram could be (-4, 3)

It is given that A(1, –1), B (–1, 3), and C (4, –1) are the three vertices of the parallelogram.

Let us assume that the vertices are in the order A, C, B, and D where the coordinates of D are (a, b).

In this scenario, if we join AB and CD, they will become the diagonals of the parallelogram ABCD.

According to the properties of a parallelogram, diagonals bisect each other.

Hence, mid-point of AB = mid-point of CD

Now, according to the mid-point theorem, if mid-point of AB is given as (x,y), then,

x = (x₁ + x₂)/2 and y = (y₁ + y₂)/2

Here, for AC,

x₁ = 1, y₁ = -1

x₂ = -1, y₂ = 3

Then, x = ( 1 - 1)/2 and y = (-1 + 3)/2

(x, y) ≡ (0, 1)  ............. (1)

Since (x, y) is also the mid-point of CD, we also have,

x₁ = 4, y₁ = -1

x₂ = a, y₂ = b

Then, x = (4 + a)/2 and y = (-1 + b)/2

(x, y) ≡ ((4 + a)/2, (-1 + b)/2) ................... (2)

From (1) and (2),

(4+a)/2 = 0 and (-1+b)/2 = 1

4+a = 0 and (-1+b) = 2

a = -4 and b = 3

Hence, the fourth vertex D of the parallelogram can be possibly located at (-4, 3)

Learn more about a parallelogram here:

brainly.com/question/1563728

#SPJ1

8 0
2 years ago
What are the solution(s) to the quadratic equation 40 − x2 = 0? x = ±2 x = ±4 x = ±2 x = ±4
hichkok12 [17]

Answer: x=\±2\sqrt{10}

Step-by-step explanation:

To find the solutions of the quadratic equation given in the problem, you can apply the proccedure shown below:

- Subtract 40 from both sides of the equation.

- Multiply both sides of the equation by -1

- Apply square root to both sides of the equation.

Therefore, by applyin the steps above, you obtain:

40-x^2-40=0-40\\-x^2=-40\\(-1)(-x^2)=(-40)(-1)\\\\x^2=40\\\\\sqrt{x^2}=\±\sqrt{40}\\\\x=\±2\sqrt{10}

3 0
3 years ago
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