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Tasya [4]
3 years ago
8

tony cashed a check for $673 at quick cash .the fee to cash a check is11% of the amount of the check. how much did tony pay to c

ash his check?
Mathematics
1 answer:
murzikaleks [220]3 years ago
4 0
673 x .11 (11%) - 74.03

Tony paid $74.03 to cash his check. 
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True or false when the x intercept of a graph occurs when the x=0
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True, hope this helps you.

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Find the slope between the pair of points: (-4,3) and (6,-2)
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(-2-3)/(6--4)

-5/10

-1/2
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Use the following federal tax table for biweekly gross earnings of a single person to help answer the question below.
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A biologist has been observing a tree's height. This type of tree typically grows by 0.22 feet each month. Fifteen months into t
Lyrx [107]

Data:

x: number of months

y: tree's height

Tipical grow: 0.22

Fifteen months into the observation, the tree was 20.5 feet tall: x=15 y=20.5ft (15,20.5)

In this case the slope (m) or rate of change is the tipical grow.

m=0.22

To find the line's slope-intercep equation you use the slope (m) and the given values of x and y (15 , 20.5) in the next formula to find the y-intercept (b):

\begin{gathered} y=mx+b \\ 20.5=0.22(15)+b \\ 20.5=3.3+b \\ 20.5-3.3=b \\ 17.2=b \end{gathered}

Use the slope(m) and y-intercept (b) to write the equation:

\begin{gathered} y=mx+b \\  \\ y=0.22x+17.2 \end{gathered}A) This line's slope-intercept equation is: y=0.22x+17.2

B) To find the height of the tree after 29 months you substitute in the equation the x for 29 and evaluate to find the y:

\begin{gathered} y=0.22(29)+17.2 \\ y=6.38+17.2 \\  \\ y=23.58 \end{gathered}Then, after 29 months the tree would be 23.58 feet in height

C) In this case as you have the height and need to find the number of moths you substitute the y for 29.96feet and solve the equation for x, as follow:

\begin{gathered} 29.96=0.22x+17.2 \\ 29.96-17.2=0.22x \\ 12.76=0.22x \\ \frac{12.76}{0.22}=x \\  \\ 58=x \end{gathered}Then, after 58 months the tree would be 29.96feet tall
3 0
1 year ago
Which of the following statements is true for the logistic differential equation?
solong [7]

Answer:

All of the above

Step-by-step explanation:

dy/dt = y/3 (18 − y)

0 = y/3 (18 − y)

y = 0 or 18

d²y/dt² = y/3 (-dy/dt) + (1/3 dy/dt) (18 − y)

d²y/dt² = dy/dt (-y/3 + 6 − y/3)

d²y/dt² = dy/dt (6 − 2y/3)

d²y/dt² = y/3 (18 − y) (6 − 2y/3)

0 = y/3 (18 − y) (6 − 2y/3)

y = 0, 9, 18

y" = 0 at y = 9 and changes signs from + to -, so y' is a maximum at y = 9.

y' and y" = 0 at y = 0 and y = 18, so those are both asymptotes / limiting values.

8 0
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