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klasskru [66]
3 years ago
11

Find the total surface area of this cone.

Mathematics
1 answer:
xxMikexx [17]3 years ago
5 0

Answer:

SA = 360π in²

Step-by-step explanation:

The slant height l is the hypotenuse of the right triangle inside the cone.

Using Pythagoras' identity to find l

l² = 24² + 10² = 576 + 100 = 676 ( take the square root of both sides )

l = \sqrt{676} = 26

Then

SA = πrl + πr²

      = (π × 10 × 26) + (π × 10²)

      = 260π + 100π

      = 360π in²

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y = \frac{1}{2x^{2} - 4}
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0(2x^{2} - 4) = (2x - 4)(\frac{1}{2x^{2} - 4})
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0 - 0 = 1
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8 0
3 years ago
Find the coordinates of the midpoint of the segment whose endpoints are H(5, 13) and K(7, 5). (12, 18) (9, 7) (2, 8) (6, 9)
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Answer: D. (6, 9)

<u>Step-by-step explanation:</u>

Midpoint is the "average" of the x's and y's:

Given: (5, 13) and (7, 5)

Midpoint: (\dfrac{5+7}{2},\dfrac{13+5}{2})

             = (\dfrac{12}{2},\dfrac{18}{2})

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3 years ago
After five weeks of dating, Terri concludes she knows next to nothing about her boyfriend, James, even though she has shared a w
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6 0
3 years ago
Solving a Two-Step Matrix Equation<br> Solve the equation:
Cloud [144]

Answer:

\boxed {x_{1} = 3}

\boxed {x_{2} = -4}

Step-by-step explanation:

Solve the following equation:

\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]

-In order to solve a pair of equations by using substitution, you first need to solve one of the equations for one of variables and then you would substitute the result for that variable in the other equation:

-First equation:

3x_{1} + 2x_{2} + 1 = 2

-Second equation:

5x_{1} + 5x_{2} + 2 = -3

-Choose one of the two following equations, which I choose the first one, then you solve for x_{1} by isolating

3x_{1} + 2x_{2} + 1 = 2

-Subtract 1 to both sides:

3x_{1} + 2x_{2} + 1 - 1 = 2 - 1

3x_{1} + 2x_{2} = 1

-Subtract 2x_{2} to both sides:

3x_{1} + 2x_{2} - 2x_{2} = -2x_{2} + 1

3x_{1} = -2x_{2} + 1

-Divide both sides by 3:

3x_{1} = -2x_{2} + 1

x_{1} = \frac{1}{3} (-2x_{2} + 1)

-Multiply -2x_{2} + 1 by \frac{1}{3}:

x_{1} = \frac{1}{3} (-2x_{2} + 1)

x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}

-Substitute -\frac{2x_{2} + 1}{3} for x_{1} in the second equation, which is 5x_{1} + 5x_{2} + 2 = -3:

5x_{1} + 5x_{2} + 2 = -3

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

Multiply -\frac{2x_{2} + 1}{3} by 5:

5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3

-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3

-Combine like terms:

-\frac{10}{3}x_{2} + \frac{5}{3} + 5x_{2} + 2 = -3

\frac{5}{3}x_{2} + \frac{11}{3} = -3

-Subtract \frac{11}{3} to both sides:

\frac{5}{3}x_{2} + \frac{11}{3} - \frac{11}{3} = -3 - \frac{11}{3}

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\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}

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x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

-Multiply -\frac{2}{3} and -4:

x_{1} = -\frac{2}{3}(-4) + \frac{1}{3}

x_{1} = \frac{8 + 1}{3}

-Since both \frac{1}{3} and \frac{8}{3} have the same denominator, then add the numerators together. Also, after you have added both numerators together, reduce the fraction to the lowest term:

x_{1} = \frac{8 + 1}{3}

x_{1} = \frac{9}{3}

\boxed {x_{1} = 3}

5 0
3 years ago
Help plzzzzzzzzzzzzzzzzz
beks73 [17]
6^3

5•23


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