Brenda has $30. She gets $5 every time she takes out the trash. Which inequality can be used to find x, the number of times Bren
1 answer:
Brenda has to take out the trash more than 4 times to have more than $50.
Step-by-step explanation:
Given,
Amount Brenda already have = $30
Amount she gets each time for throwing trash = $5
Amount she wants to have = More than $50
Let,
x be the number of times she throws trash.
Amount for throwing trash * Number of times trash thrown + Amount already have > Amount she wants to have
Dividing both sides by 5
Brenda has to take out the trash more than 4 times to have more than $50.
Keywords: variable, inequality
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2
Suggesting that you want this in standard form, in terms of quadratic equations, you would technically follow a process similar if not almost exactly like the 2 - step equation method with the exception of separating the (x)s and the equations to find x and then plug it in and what-not.
With that being said you would subtract 5 in (x+5) from said 5 in the second equation and -10 in the first equation in order to get 2x^2+7x-15, you would continue to do the same for the x by subtracting it from both ends making the 7x a 6x because there is a 1 at the beginning of each x if there is no number that is shown already. Which finally gives you the equation (y= 2x^2+6x-15)
Suggesting that you want this in standard form, in terms of quadratic equations, you would technically follow a process similar if not almost exactly like the 2 - step equation method with the exception of separating the (x)s and the equations to find x and then plug it in and what-not.
With that being said you would subtract 5 in (x+5) from said 5 in the second equation and -10 in the first equation in order to get 2x^2+7x-15, you would continue to do the same for the x by subtracting it from both ends making the 7x a 6x because there is a 1 at the beginning of each x if there is no number that is shown already. Which finally gives you the equation (y= 2x^2+6x-15)
Answer:
Step-by-step explanation:
Looking at the graph, we can see the domain to be from (0 , 2π).
Now we have to find one period that corresponds to cos(x).
The half-period of cos(x) for this graph appears to be pi/3 and adding another pi/3 gets us 2pi/3 to be our cosine period.
b = 2pi/3
a is the same range as cos(x). Range: (0,0)
y = [a] * cos ([b]*x)
y = [1] * cos([2pi/3]x)