How to find Simple Interest in Algebra

In Algebra tutoring ,Interest is a fee paid on borrowed assets. It is the price paid for the use of borrowed money, or, money earned by deposited funds. Assets that are sometimes lent with interest include money, shares, consumer goods through hire purchase, major assets such as aircraft, and even entire factories in finance lease arrangements. The interest is calculated upon the value of the assets in the same manner as upon money. Interest can be thought of as "rent of money". When money is deposited in a bank, interest is typically paid to the depositor as a percentage( how to find percentage) of the amount deposited; when money is borrowed, interest is typically paid to the lender as a percentage of the amount owed. The percentage of the principal that is paid as a fee over a certain period of time (typically one month or year), is called the interest rate.Let's see an example from algebra 1 word problems



Question:-

A city government built a $60 million sports arena.Some of the money was raised by selling bonds that pay simple interest at a rate of 11% annually.The remaining amount was obtained by borrowing money from an insurance company at a simple interest rate of 10% .How much was financed through the insurance company if the annual interest is $6.19 million?


Answer:-

Let $x was financed through the insurance company.

Then $(60-x) was raised through selling bonds.

Interest on $(60-x) for 1 year at 11% is (60-x).11/100

And interest on $x for 1 year at 10% is x.10/100

By the problem

[11(60-x)/100] + [10x/100] = 6.19

multiplying both sides by 100 , we get


11(60-x)+10x = 619

660-11x+10x = 619

660 - x = 619

x = 41

So,$41 was financed through the insurance company.

problem on relations and functions

Domain of a given function is the set of "input" values for which the relations and functions is defined. For instance, the domain of cosine would be all real numbers, while the domain of the square root would be only numbers greater than or equal to 0 (ignoring complex numbers in both cases). In a representation of a function in a xy Cartesian coordinate system, the domain is represented on the x axis (or abscissa).

The x-intercept of a line is the point at which the line crosses the x-axis.The y-coordinate of the
point slope form where a line intersects the x-axis is 0. So, the x-intercept of a line can also be found by substituting y = 0 in the equation of the line.
Let's see a example on this from help solving slope of a line problems in algebra ii
Question:-
Find the domain,x-intercept and vertical Asymp for h(x)=log(4)(x-3)
Answer:-
consider h(x)=log(4)(x-3)
h(x)=log(x-3)/log4
domain : x-3 > 0
x > 3
x-intercept : h(x)=log(x-3)/log4
0 = log(x-3)/log4
0 = log(x-3)
log1 = log(x-3)
1 = x-3
x = 4
vertical Asymp : 3

Sphere Equation

Question : How to find the equation of minimum sphere whose two tangent equation are
(x-1)/1=(y-4)/2=(z-5)/-3 and (x+12)/4=(y-8)/-1=(z-17)

Answer :

The equations (x-1)/1 = (y-4)/2 = (z-5)/-3
and (x+12)/4 = (y-8)/-1 = (z-17)/1

are not coplanar lines nor these are parallel lines ,
these are infact skew lines in space.

The minimum sphere between these lines is the sphere whose diameter is the shortest distance between the given lines .
And it can be found by finding the end points of the shortest distance line
as follows
the coordinates of the one end P of the shortest distance line , which lies on the first line is ( a+1 , 2a+4 , -3a+5) where a is a constant to be found

the coordinates of the other end Q of the shortest distance line , which lies on the second line is ( 4b-12 , -b+8 , b+17) where a is a constant to be found.

hence the direction ratios of the shortest distance line PQ are
4b-12-a-1 , -b+8-2a-4 , b+17+3a-5
that is
4b-a-13 , -b-2a+4 , b+3a+12


now since the shortest distance line PQ is perpendicular to both of the given lines
hence
1(4b-a-13)+2(-b-2a+4)-3(b+3a+12) = 0 .......... (1)
and
4(4b-a-13)-1(-b-2a+4)+1(b+3a+12) = 0 .......... (2)

solving (1) and (2) simultaneously we can find a and b
and hence we can find the end points P( a+1 , 2a+4 , -3a+5) and
Q( 4b-12 , -b+8 , b+17) of the shortest distance line

which are also the end points of the diameter of the required sphere, whose center is the mid point of PQ.