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3 years ago

How are they getting the numerator and denominator for the fractions for the coordinates? PLEASE HELP ASAP!!!!! 20 POINTS!!!!

Mathematics

1 answer:

ycow [4]3 years ago

5 0

Answer:

All except from point E's y-coordinate (which is -1.5 (found it through the equation y=-1.5 and trying to see if these coordinates are solutions to the above equation) All the others are integers, which you can find through aligning the point on the axis of the chosen (y or x) coordinate.

Step-by-step explanation:

Point C has -2 x-coordinate since it is on the x=-2 line

Similarly, point D has -2 y coordinate,

point E has 2 x-coordinate and -1,5 y-coordinate

and point F , since it's the two axis' common point has coordinates of (0,0).

Hope I helped! Further explanation can be given on request on your behalf.

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3 years ago

At an effective annual interest rate of i > 0, each of the following two sets of payments has present value K: (i) A payment

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Answer:

The present value of K is, $K=251.35$

Step-by-step explanation:

First of all, we need to construct an equation system, so

$(1)K=\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}$

$(2)K=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

Then we equalize both of them so we can find $i$

$(3)\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}}$

To solve it we can multiply $(3)*(1+i)^{4}$ to obtain $(1+i)^{4}*(\frac{169}{(1+i)} +\frac{169}{(1+i)^{2}}=\frac{225}{(1+i)^{2}} +\frac{225}{(1+i)^{4}})$ , then we have $225(1+i)^{2}+225=169(1+i)^{3}+169(1+i)^{2}$ .

This leads to a third-grade polynomial $169i^{3}+451i^{2}+395i-112=0$ , after computing this expression, we find only one real root $i=0.2224$ .

Finally, we replace it in (1) or (2), let's do it in (1) $K=\frac{169}{(1+0.2224)} +\frac{169}{(1+0.2224)^{2}}\\\\K=251.35$

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3 years ago