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3 years ago

Set up the inequality of $3.000 and $15

Mathematics

1 answer:

Lilit [14]3 years ago

4 0

Answer:

$3.00 < $15.00

Step-by-step explanation:

It is unusual to see a dollar amount with three digits to the right of the decimal point. (This sometimes happens when prices are expected to be multiplied by large quantities.) On the other hand, it sometimes happens that a period gets substituted for a comma. So, some relevant inequalities might be ...

$3.00 < $15

$3,000 > $15

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If the parent function f(x) = x^2 is fatter translated 11 units to the left, then translated 5 units down, write the resulting f

nikklg [1K]

The quadratic function given by:

$f(x)=a(x-h)^2+k, \ \ \ a\neq 0$

is in vertex form. The graph of $f$ is a parabola whose axis is the vertical line $x=h$ and whose vertex is the point $(h, k)$ . So:

To translate the graph of a function to the right, left, upward or downward we have:

$For \ a \ positive \ real \ number \ c. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:\\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ g(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ g(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{right}: \\ g(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{left}: \\ g(x)=f(x+c)$

By knowing this things, we can solve our problem as follows:

FIRST.

Translating 11 units to the left:

$g(x)=f(x+11) \\ \\ \therefore g(x)=(x+11)^2$

Then translating 5 units down:

$g(x)=f(x)-c \\ \\ \therefore g(x)=(x+11)^2-5$

Since the new function is fatter, the factor we need to multiply the term $(x+11)^2$ must be less than 1, to make the graph fatter. So, according to our options, there are two factors 1/2 and 2.