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RUDIKE [14]
3 years ago
5

Tonya took out a loan to help pay for her house. She borrowed $40,000 for 15 years at a yearly simple interest rate of 5%. How m

uch interest will she end up paying the bank?
Mathematics
2 answers:
lesya692 [45]3 years ago
5 0
40,000*0.05= 2,000
2,000*15=
30,000
OLEGan [10]3 years ago
3 0
I may be wrong but it could be 30,000.
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The slope of a line can be found by graph using rise over run (\frac{rise}{run}). Look at the image below to see what this mean.

Rise = vertical

run = horizontal

I took two clear points on the line as seen below and drew a right angle that formed a right triangle with the line.

Now let's count how many units the rise is (look at the second image). This is 3 units

rise = 3

Count the units of the run (third image). This is 1 unit

run = 1

Let's put these two numbers together in the \frac{rise}{run} form:

\frac{3}{1} or simplified = 3

However, this is not the final answer. This is a negative slope, meaning that when you look at the line from left to right (like when you are reading words) the line reaches from the top left corner to the bottom right corner. A positive slope (reading left to right) reaches from the bottom left corner to top right corner.

To signify that the slope is negative you simply add a negative sign to it:

-3

Let me know if you have any further questions.

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2 years ago
If (7^2)x=1 what is x
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1/49 ..................
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Find the equation of the line that is perpendicular to the line y = (-1/3)x -1 and passes through the point (1, 5)?
Anit [1.1K]

bearing in mind that perpendicular lines have negative reciprocal slopes, so


\bf \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}~\hspace{10em}\stackrel{slope}{y=\stackrel{\downarrow }{-\cfrac{1}{3}}x-1} \\\\[-0.35em] ~\dotfill


\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{1}\implies 3}}


so we're really looking for a line whose slope is 3 and runs through (1,5)


\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{5})~\hspace{10em} slope = m\implies 3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-5=3(x-1) \\\\\\ y-5=3x-3\implies y=3x+2

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