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dlinn [17]
2 years ago
14

Last week, Riley worked 7.5 hours per day for 3 days and earned a total of $306.45.

Mathematics
2 answers:
maria [59]2 years ago
8 0

Answer:

13,62$

Step-by-step explanation:

The total hours Riley worked last week: 7.5*3= 22.5 (hours)

Riley's hourly rate of pay: 306.45/ 22.5= 13.62$

saul85 [17]2 years ago
8 0
7.5 x 3 = 22.5

306.45 / 22.5 = 13.62
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Find a function where f(0)=2 and f(1)=2
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Answer:

Do you want to be extremely boring?

Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?

f(x) = 2 is a valid solution.

Want something more fun? Why not a parabola? f(x)= ax^2+bx+c.

At this point you have three parameters to play with, and from the fact that f(0)=2 we can already fix one of them, in particular c=2. At this point I would recommend picking an easy value for one of the two, let's say a= 1 (or even a=-1, it will just flip everything upside down) and find out b accordingly:f(1)=2 \rightarrow 1^2+b+2=2 \rightarrow b=-1

Our function becomes

f(x) = x^2-x+2

Notice that it works even by switching sign in the first two terms: f(x) = -x^2+x+2

Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: f(x) = A cos (kx)

Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need A= 2, and at that point the first condition is guaranteed; using the second to find k we get 2= 2 cos (k1) = cos k = 1 \rightarrow k = 2\pi

f(x) = 2cos(2\pi x)

Or how about a sine wave that oscillates around 2? with a similar reasoning you get

f(x)= 2+sin(2\pi x)

Sky is the limit.

8 0
2 years ago
Make a table with the domain of {2,3,4,5,6} and draw a graph of the absolute value function y = 2|x-4| + 3.
Jet001 [13]

Answer:

<h2>In the attachment.</h2>

Step-by-step explanation:

|a|=\left\{\begin{array}{ccc}a&for\ a\geq0\\-a&for\ a

Put each value of x from the set {2, 3, 4, 5, 6}

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x = 2 → y = 2|2 - 4| + 3 = 2|-2| + 3 = 2(2) + 3 = 4 + 3 = 7 → (2, 7)

x = 3 → y = 2|3 - 4| + 3 = 2|-1| + 3 = 2(1) + 3 = 2 + 3 = 5 → (3, 5)

x = 4 → y = 2|4 - 4| + 3 = 2|0| + 3 = 2(0) + 3 = 0 + 3 = 3 → (4, 3)

x = 5 → y = 2|5 - 4| + 3 = 2|1| + 3 = 2(1) + 3 = 2 + 3 = 5 → (5, 5)

x = 6 → y = 2|6 - 4| + 3 = 2|2| + 3 = 2(2) + 3 = 4 + 3 = 7 → (6, 7)

Mark the points in the coordinates system.

The domain is only five numbers, therefore the graph of this function is only five points.

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Answer:

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Step-by-step explanation:

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