Only need question 2
1 answer:
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Check the picture below.
for "b", well, recall that the sum of all interior angles in a triangle is 180°.
for "a", "c" and "d", well, all those are just <u>intercepted arcs</u>.
one way would be to factor
I can't factor it so we will have to use Descartes' Rule of Signs which is helpful for finding how many real roots you have
it goes like this:
for a polynomial with real coefients, consider .
after arranging the terms in decending order in terms of degree, count how many times the signs of the coeffients change direction and minus 2 from that number until you get to 1 or 0. that will be the number of even roots the function can have
We have (-, -, +, +). the signs changed 1 times, so it has 1 real positive root
to get the negative roots, we evaluate f(-x) and see how many times the root changes
signs are (-, -, -, +). there was 1 change in sign
so the function has 1 real negative root
a total of 2 real roots
a function of degree can have at most,
roots
our function is degree 6 so it has 6 roots
if 2 are real, then the others must be complex
6-2=4 so there are 4 complex roots
you can also show that there are only 2 real roots by using a graphing utility to see that there are only 2 real roots
Answer:
or
5.32cm correct to 2 decimal places
Step-by-step explanation:
This is a right angled triangle from what I know so we have to apply pythagoras theorem.
we need to rearrange this to find a²
This can either be written as a fraction or decimal.
OR
correct to 2 decimal places