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k0ka [10]
2 years ago
9

Mr. Toshio lent out 11 rulers at the beginning of class, collected 4 rulers in the middle of class, and gave out 7 at the end of

class. He had 18 at the end of the day. How many rulers did he start with?
Mathematics
2 answers:
Drupady [299]2 years ago
8 0
Mr. Toshio had 32 rulers. 32-11 = 21 rulers. 21+4 = 25 rulers. and 25-7 = 18 rulers
astra-53 [7]2 years ago
7 0

Answer:

Mr. Toshio started with 32 rulers

Step-by-step explanation:

Let x be the number of rulers he start with

Mr. Toshio lent out 11 rulers at the beginning of class

Number of rulers - 11

x-11

He  collected 4 rulers in the middle of class. So we add 4

x-11+4

He gave out 7 at the end of class. So we subtract 7

x-11+4-7

He had 18 at the end of the day.

x-11+4-7=18

x-14=18

Add 14 on both sides

x=32

Mr. Toshio started with 32 rulers

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We are given the set of points
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2 years ago
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34kurt

Answer:

f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252

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<em>f(x) and g(x) and not inverse functions</em>

Step-by-step explanation:

Given

f(x) = 3x^2 + 9

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Required

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Determine g(f(x))

Determine if both functions are inverse:

Calculating f(g(x))

f(x) = 3x^2 + 9

f(g(x)) = 3(\frac{1}{3}x^2 - 9)^2 + 9

f(g(x)) = 3(\frac{1}{3}x^2 - 9)(\frac{1}{3}x^2 - 9) + 9

Expand Brackets

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f(g(x)) = \frac{1}{3}x^4 - 18x^2 + 252

Calculating g(f(x))

g(x) = \dfrac{1}{3}x^2 - 9

g(f(x)) = \frac{1}{3}(3x^2 + 9)^2 - 9

g(f(x)) = \frac{1}{3}(3x^2 + 9)(3x^2 + 9) - 9

g(f(x)) = (x^2 + 3)(3x^2 + 9) - 9

Expand Brackets

g(f(x)) = x^2(3x^2 + 9) + 3(3x^2 + 9) - 9

g(f(x)) = 3x^4 + 9x^2 + 9x^2 + 27 - 9

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Checking for inverse functions

f(x) = 3x^2 + 9

Represent f(x) with y

y = 3x^2 + 9

Swap positions of x and y

x = 3y^2 + 9

Subtract 9 from both sides

x - 9 = 3y^2 + 9 - 9

x - 9 = 3y^2

3y^2 = x - 9

Divide through by 3

\frac{3y^2}{3} = \frac{x}{3} - \frac{9}{3}

y^2 = \frac{x}{3} - 3

Take square root of both sides

\sqrt{y^2} = \sqrt{\frac{x}{3} - 3}

y = \sqrt{\frac{x}{3} - 3}

Represent y with g(x)

g(x) = \sqrt{\frac{x}{3} - 3}

Note that the resulting value of g(x) is not the same as g(x) = \dfrac{1}{3}x^2 - 9

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Answer:

see below

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