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nlexa [21]
3 years ago
14

What is the surface area of a cylinder with base radius 2 and height 9

Mathematics
1 answer:
ch4aika [34]3 years ago
7 0

The surface area of a cylinder with a base radius and a height of 9 is 138.23

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3 is the common difference
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I only need the answers to 8 an 9
musickatia [10]

Answer:

#8: distance is 3 units

#9: distance is 1 unit

Step-by-step explanation:

for #8 the x values are the same and there's a 3 unit difference between the y value

for #9 the y values are the same and there's a 1 unit difference between the x values

hope this helps chu :3

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without building the graph, find the coordinates of the point of intersection of the lines given by the equation y=3x-1 and 3x+y
DaniilM [7]
<h2><u>1. Determining the value of x and y:</u></h2>

Given equation(s):

  • y = 3x - 1
  • 3x + y = -7

To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.

<h3 /><h3><u>Method-1: Substitution method</u></h3>

Given value of the y-variable: 3x - 1

Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.

\implies 3x + y = -7

\implies3x + (3x - 1) = -7

\implies3x + 3x - 1 = -7

Combine like terms as needed;

\implies 3x + 3x - 1 = -7

\implies 6x - 1 = -7

Add 1 to both sides of the equation;

\implies 6x - 1 + 1 = -7 + 1

\implies 6x = -6

Divide 6 to both sides of the equation;

\implies \dfrac{6x}{6}  = \dfrac{-6}{6}

\implies x = -1

Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.

\implies y = 3(-1) - 1

\implies     \ \ = -3 - 1

\implies     = -4

Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h3 /><h3><u>Method 2: System of equations</u></h3>

Convert the equations into slope intercept form;

\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.

\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.

Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.

\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.

\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.

\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

\implies 0 = (6x) + (6)

\implies0 = 6x + 6

This problem is now an algebraic problem. Isolate "x" to determine its value.

\implies 0 - 6 = 6x + 6 - 6

\implies -6 = 6x

\implies -1 = x

Like done in method 1, substitute the value of x into the first equation to determine the value of y.

\implies y = 3(-1) - 1

\implies y = -3 - 1

\implies y = -4

Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h2><u>2. Determining the intersection point;</u></h2>

The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.

⇒ (x, y) ⇒ (-1, -4)

Therefore, the point of intersection is (-1, -4).

<h3>Graph:</h3>

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