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3 years ago

A few years ago, the Colorado Rockies played 160 games and won 64. for every two games they won, how many games did they lose?

Mathematics

2 answers:

Monica [59]3 years ago

7 0

The answer is A 3.

In order to find this answer, we first have to find the number of games that they lost. To do so, we subtract the wins from the total.

160 - 64 = 96.

Now that we have the number of losses, we can divide wins by losses to get a ratio.

64/96 = 2/3

So for every 2 wins, we get 3 losses.

dusya [7]3 years ago

4 0

Answer:

A. 3

Step-by-step explanation:

Given: Colorado Rockies played 160 games and won 64.

To find: For every two games they won, how many games did they lose?

Solution: In order to find for every two games they won, how many games did they lose, we first need to find the number of games they lose.

So, we have

total games $=160$

games won $=64$

games lose $=160-64=96$

Now, we need to find the ratio of number of games won to the number of games they lose

$=\frac{64}{96}$

$=\frac{2}{3}$

Therefore, the ratio is 2:3.

So, for every 2 games they won, they lose 3 games.

Hence, option A. 3 is correct.

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2 years ago

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horrorfan [7]

Answer:

A) See attached for graph.

B) (-3, 0) (0, 0) (18, 0)

C) (-3, 0) ∪ [3, 18)

Step-by-step explanation:

Piecewise functions have <u>multiple pieces</u> of curves/lines where each piece corresponds to its definition over an <u>interval</u>.

Given piecewise function:

$g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad \textsf{if }x\geq 3\end{cases}$

Therefore, the function has two definitions:

$g(x)=x^3-9x \quad \textsf{when x is less than 3}$

$g(x)=-\log_4(x-2)+2 \quad \textsf{when x is more than or equal to 3}$

When <u>graphing</u> piecewise functions:

Use an open circle where the value of x is <u>not included</u> in the interval.
Use a closed circle where the value of x is <u>included</u> in the interval.
Use an arrow to show that the function <u>continues indefinitely</u>.

<u>First piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

$\implies g(3)=(3)^3-9(3)=0 \implies (3,0)$

Place an open circle at point (3, 0).

Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.

<u>Second piece of function</u>

Substitute the endpoint of the interval into the corresponding function:

$\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)$

Place an closed circle at point (3, 2).

Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.

See attached for graph.

The x-intercepts are where the curve crosses the x-axis, so when y = 0.

Set the <u>first piece</u> of the function to zero and solve for x:

$\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}$

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.

Set the <u>second piece</u> to zero and solve for x:

$\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}$

$\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b$

$\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}$

Therefore, the x-intercept for the second piece is (18, 0).

So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).

From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.

Interval notation: (-3, 0) ∪ [3, 18)

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