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SpyIntel [72]
3 years ago
9

Question 1 (10 points)

Mathematics
2 answers:
matrenka [14]3 years ago
7 0

Answer:

Question 1  10 / 10 points

A summary of selected ledger accounts appear below for S. Ball for the current calendar year.

Answer questions 1 through 4 based on this information.

 

1. What was the total amount of withdrawals for the year?  

$6000

Question 2

2.What was the net income?

$4250

Question 3

 

3.What was the total revenue?

$17,000

4.What were the total expenses?

$12.750

5.

__2__  

Unearned Revenue

__5__  

Brandon Jones, Capital

__1__  

Cash

__3__  

Equipment

__4__  

Mortgage

1.  

Current Assets

2.  

Current Liabilities

3.  

Fixed Assets

4.  

Long-term Liability

5.  

Equity

Question 6  

At the end of the closing process, Income Summary will hold a balance.

Question options:

b) False

Step-by-step explanation:

I just took this quiz and got a 100%

Reika [66]3 years ago
4 0

First answer is Ball. capital

Second answer is 10/31 3,000

Thrid answer is 123

Forth answer is 6000

hope helps

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Pairs that satisfy function y=2x+1
evablogger [386]

Test a pair from each table by substituting their values into the given equation and solving.

A) y = 2x + 1 .......... 2 = 2(0) + 1 .......... 2 ≠ 1

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C) y = 2x + 1 .......... -1 = 2(0) + 1 .......... -1 ≠ 1

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Linda goes water-skiing one sunny afternoon. After skiing for 15 min, she signals to the driver of the boat to take her back to
trasher [3.6K]
60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2


abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.


d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
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