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Sever21 [200]
3 years ago
9

HELP ITS DUE TODAY!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
RoseWind [281]3 years ago
4 0

Answer/Step-by-step explanation:

x + 74° + 20° = 180° (sum of interior angles of a triangle)

x + 94° = 180°

x = 180° - 94° (subtraction property of equality)

x = 86°

y = 74° + 20° (exterior angle theorem of a triangle)

y = 94°

y + z + 32° = 180° (sum of interior angles of a ∆)

Plug in the value of y

94 + z + 32 = 180

z + 126 = 180

z = 180 - 126

z = 54°

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For a given input value nnn, the function ggg outputs a value mmm to satisfy the following equation.
Mademuasel [1]

For a given input value n, the function g outputs a value m to satisfy the following equation.

3m-5n=11

Write a formula for g(n), in terms of n.

g(n)=

Answer:

3m-5n=11

the function g outputs a value m. It means we need to solve for m in terms of n

solve the given equation for m

3m-5n=11

Inverse of -5n is +5n so we add 5n on both sides

3m-5n + 5n=11 + 5n

3m = 11 + 5n

Divide by 3 on both sides

m = \frac{11+5n}{3}

The function g outputs the value m

So g(n) = \frac{11+5n}{3}

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4 years ago
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Napkins come in a variety of packages. A package of 40 napkins sells for $4.59, and a package of 10 napkins sells for $2.30. Fin
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Jacob throws an acorn into the air. It lands in front of him. The acorn's path is
Burka [1]

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  • hits the ground at x = -0.732, and x = 2.732
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Step-by-step explanation:

The acorn will hit the ground where the value of x is such that y=0. We can find these values of x by solving the quadratic using any of several means.

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<h3>graphing</h3>

The attachment shows a graphing calculator solution to the equation

  -3x^2 + 6x + 6 = 0

The values of x are -0.732 and 2.732. The negative value is the point where the acorn would have originated from if its parabolic path were extrapolated backward in time. Only the positive horizontal distance is a reasonable solution.

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<h3>completing the square</h3>

We can also solve the equation algebraically. One of the simplest methods is "completing the square."

  -3x^2 +6x +6 = 0

  x^2 -2x = 2 . . . . . . . . divide by -3 and add 2

  x^2 -2x +1 = 2 +1 . . . . add 1 to complete the square

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  x -1 = ±√3  . . . . . . . take the square root

  x = 1 ±√3 . . . . . . . add 1; where the acorn hits the ground

The numerical values of these solutions are approximately ...

  x ≈ {-0.732, 2.732}

The solutions to the equation say the acorn hits the ground at a distance of -0.732 behind Jacob, and at a distance of 2.732 in front of Jacob. The "behind" distance represents and extrapolation of the acorn's path backward in time before Jacob threw it. Only the positive solution is reasonable.

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