# Some problems that ave been giving me trouble

• Nov 29th 2007, 08:28 AM
Xplicit
Some problems that ave been giving me trouble
16-X
------ = 2+x (thats a fraction lol)
5

Find the equation of the line joining the points (0,5) and (2,11)

Calculate

4 and 1 fifth (multiplied by) 1 and 1 seventh

(not sure how to type that out)

Nuno borrowed £400 for 9 months and paid £36 interest. What was the equivalent rate of interest per annum?

Also if you could show me how you got the answer it would help greatly.

thanks in advance for any help ;)
• Nov 29th 2007, 08:56 AM
topsquark
Quote:

Originally Posted by Xplicit
16-X
------ = 2+x (thats a fraction lol)
5

Just so you know, there is a difference between the variables X and x.

$\displaystyle \frac{16 - x}{5} = 2 + x$

$\displaystyle \frac{16 - x}{5} \cdot 5 = (2 + x) \cdot 5$

$\displaystyle 16 - x = 10 + 5x$

Can you solve this now?

Quote:

Originally Posted by Xplicit
Find the equation of the line joining the points (0,5) and (2,11)

Given two points $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$, the slope of the line going through them may be calculated as
$\displaystyle m = \frac{y_2 - y_1}{x_2 - x_1}$

To get the rest of the equation for the line you can use the slope-intercept form:
$\displaystyle y = mx + b$

Plug in either point and the slope you just calculated and solve for b.

Quote:

Originally Posted by Xplicit
Calculate
4 and 1 fifth (multiplied by) 1 and 1 seventh

$\displaystyle \left ( 4~\frac{1}{5} \right ) \cdot \left ( 1~\frac{1}{7} \right )$

I am personally against numbers written in this form, though I fully understand why they are useful. But I find that they rather confuse things by the time a student reaches their first Algebra course.

Anyway, I would convert these numbers to improper fractions:
$\displaystyle \left ( 4~\frac{1}{5} \right ) \cdot \left ( 1~\frac{1}{7} \right ) = \frac{21}{5} \cdot \frac{8}{7}$

You should be able to finish this now.

Someone else is going to have to help you with the last problem.

-Dan
• Nov 29th 2007, 09:06 AM
janvdl
Quote:

Originally Posted by Xplicit
Nuno borrowed £400 for 9 months and paid £36 interest. What was the equivalent rate of interest per annum?

This is the formula for simple interest.

$\displaystyle Interest = Capital \ \times \ \frac{Rate}{100} \ \times \frac{Time (Months)}{12 \ Months}$

Thus:

$\displaystyle 36 = 400 \times \frac{r}{100} \times \frac{t}{12}$

$\displaystyle 36 = 400 \times \frac{r}{100} \times \frac{9}{12}$

$\displaystyle 36 = 400 \times \frac{r}{100} \times \frac{3}{4}$

$\displaystyle 36 = 300 \times \frac{r}{100}$

$\displaystyle 36 = 3r$

$\displaystyle r = 12$