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3 years ago

11

If i put a penny in a jar every day for a year how much money will i have

1 answer:

katovenus [111]3 years ago

7 0

Hi there!

A year is 365 days, correct?
If we save our pennies up from January 1st to December 31st, you'd have 3 dollars (since 100 pennies is 1 dollar) and 65 pennies left over. If it was a leap year, however, you'd have 3 dollars and 66 cents. :)

Hope this helps! :D

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HELP WITH ALL <br> Determine the composite functions of the following set of functions

elixir [45]

QUESTION 1

i) Given $f(x)=5x-2$ and $g(x)=x^2$ .

$(f\circ g)=f(g(x))$

$(f\circ g)=f(x^2)$

We plug in $x^2$ into $f(x)=5x-2$

$(f\circ g)=5x^2-2$

ii) Given $f(x)=5x-2$ and $g(x)=x^2$ .

$(g\circ f)=g(f(x))$

$(g\circ f)=g(5x-2)$

We plug in $5x-2$ into $g(x)=x^2$

$(g\circ f)=(5x-2)^2$

$(g\circ f)=25x^2-10x-10x+4$

$(g\circ f)=25x^2-20x+4$

QUESTION 2

i) Given $g(x)=x^2-4$ and $h(x)=\sqrt{2x+1}$ .

$(h\circ g)=h(g(x))$

$(h\circ g)=h(x^2-4)$

We plug in $x^2-4$ into $h(x)=\sqrt{2x+1}$

$(h\circ g)=\sqrt{2(x^2-4)+1}$

$(h\circ g)=\sqrt{2x^2-8+1}$

$(h\circ g)=\sqrt{2x^2-7}$

ii) Given $g(x)=x^2-4$ and $h(x)=\sqrt{2x+1}$ .

$(g\circ h)=g(h(x))$

$(g\circ h)=g(\sqrt{2x+1})$

We plug in $\sqrt{2x+1$ into $g(x)=x^2-4$

$(g\circ h)=(\sqrt{2x+1})^2-4$

$(g\circ h)=2x+1-4$

$(g\circ h)=2x-3$

QUESTION 3

i) Given $g(x)=x+5$ and $h(x)=\sqrt{x-4}$ .

$(g\circ h)=g(h(x))$

$(g\circ h)=g(\sqrt{x-4})$

We plug in $\sqrt{x-4}$ into $g(x)=x+5$

$(g\circ h)=\sqrt{(x-4)+5}$

ii) Given $g(x)=x+5$ and $h(x)=\sqrt{x-4}$ .

$(g\circ g)=g(g(x))$

$(g\circ g)=g(x+5)$

We plug in $x+5$ into $g(x)=x+5$

$(g\circ g)=(x+5)+5}$

$(g\circ g)=x+10}$

QUESTION 4

The given functions are;

$f(x)=-2x+1$ and $g(x)=-x^2$

$f\circ f=f(f(x))$

$f\circ f=f(-2x+1)$

$f\circ f=-2(-2x+1)+1)$

$f\circ f=4x-2+1)$

$f\circ f=4x-1)$

ii)

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$g\circ g=g(-x^2)$

$g\circ g=-(x^2)^2$

$g\circ g=-x^4$

QUESTION 5

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$h\circ g\circ f=h\circ (g(5x-2))$

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$h\circ g\circ f=h\circ (25x^2-20x+4)$

$h\circ g\circ f=h(25x^2-20x+4)$

$h\circ g\circ f=-3(25x^2-20x+4)$

$h\circ g\circ f=-75x^2+60x-12)$

ii.

$f\circ f\circ g=f\circ (f(g(x))$

$f\circ f\circ g=f\circ (f(x^2)$

$f\circ f\circ g=f\circ (5x^2-2)$

$f\circ f\circ g=f(5x^2-2)$

$f\circ f\circ g=5(5x^2-2)-2$

$f\circ f\circ g=25x^2-10-2$

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Answer:

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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The word “<em>more than”</em> goes with addition.

Happy studying,

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