Answer:
23,24,26,27 sorry if you get it wrong
Answer:
1) 12/18
2) 15/25
3) 32/40
4) 6/14
5) 8/15
6) 16/24
7. 12/27
Step-by-step explanation:
Answer:
C. 1,035
Step-by-step explanation:
First step is to simplify the equation into a form that we are familiar with graphing, in this case the simplest way i believe is to factorise the denominator. There are many methods of factorising quadratics but in this case my preferred method would be as follows:
x^2-3x-10, because it is a quadratic equation we know it can be factorised into the form (x+a)(x+b), we know that -10 is the product of a and b, therefore a and b can only be +/-1 and -/+10, or +/-5 and -/+2, the next step is to see which of these factors sum to equal -3, in this case it will be +2 and -5, therefore x^2-3x-10 can be factorised into (x+2)(x-5).
We have now simplified the function into f(x)= (x+2)/(x+2)(x-5), we can still simplify the function further though:
(x+2)/(x+2)(x-5), the numerator and denominator have a common factor therefore we can divide the numerator and the denominator by that factor, ((x+2)/(x+2))/((x+2)(x-5)/(x+2))= 1/(x-5)
The function is now in it's simplest form, so we will now graph it. As a side note you can't perfectly "graph" any function by hand so instead we will sketch the function, to sketch a function you need to know the general shape of the function and any x/y intercepts or asymptotes.
Functions of the form a/(bx+c) are reciprocal functions, you need to be able to remember what reciprocal function looks like to be a able to sketch them by hand, they look something like this https://www.desmos.com/calculator/df2wiewhsz. To work out the x-intercept we will assign f(x) as 0, 0=1/(x-5), multiply both sides by (x-5), (x-5)0=1, 0=1, 0 doesn't equal 1 therefore there is no x-intercept but instead there is an asymptote at y=0. To work out the y-intercept we will assign x as 0, f(x)=1/(x-5), f(x)=1/-5, therefore the only y-intercept is at coordinates (0.-0.2). All reciprocal functions have 2 asymptotes, we have already found one of them, by making f(x)=0, to find the other asymptote we will make 1/(x-5)=∞, 1/0=∞, therefore x-5 has to equal 0, therefore the second asymptote will be x=5.
after putting all of this together the graph should look something like this https://www.desmos.com/calculator/bibeeu7xtv
It would be False, i hope this helps
