How many pairs of supplementray angles ate there? PLEASE HELP!!

Why is it important to learn statistics?<br><br>how can you apply statistics in your everyday life?

Setler [38]

Answer:

Statistics is a mathematical technique and science that attempts to extract valid knowledge using empirical data of observation or experiment. The main object of research and study of Statistics is the collection, processing and interpretation of various data with the ultimate goal of drawing safe conclusions for making correct decisions. It is an important science whose applications have a wide field in administration, business, as well as in positive and behavioral or social sciences.

5 0

3 years ago

HELPPPPPPPLLL

love history [14]

Answer:

16

Step-by-step explanation:

4:2

2x4 = 8

4x4=16

7 0

3 years ago

Prove that a cubic equation x 3 + ax 2 + bx+ c = 0 has 3 roots by finding the roots.

evablogger [386]

That's a pretty tall order for Brainly homework. Let's start with the depressed cubic, which is simpler.

Solve

$y^3 + 3py = 2q$

We'll put coefficients on the coefficients to avoid fractions down the road.

The key idea is called a split, which let's us turn the cubic equation in to a quadratic. We split unknown y into two pieces:

$y = s + t$

Substituting,

$(s+t)^3 + 3p(s+t) = 2q$

Expanding it out,

$s^3+3 s^2 t + 3 s t^2 + t^3 + 3p(s+t) = 2q$

$s^3+t^3 + 3 s t(s+t) + 3p(s+t) = 2q$

$s^3+t^3 + 3( s t + p)(s+t) = 2q$

There a few moves we could make from here. The easiest is probably to try to solve the simultaneous equations:

$s^3+t^3=2q, \qquad st+p=0$

which would give us a solution to the cubic.

$p=-st$

$t = -\dfrac p s$

Substituting,

$s^3 - \dfrac{p^3}{s^3} = 2q$

$(s^3)^2 - 2 q s^3 - p^3 = 0$

By the quadratic formula (note the shortcut from the even linear term):

$s^3 = q \pm \sqrt{p^3 + q^2}$

By the symmetry of the problem (we can interchange s and t without changing anything) when s is one solution t is the other:

$s^3 = q + \sqrt{p^3+q^2}$

$t^3 = q - \sqrt{p^3+q^2}$

We've arrived at the solution for the depressed cubic:

$y = s+t = \sqrt[3]{q + \sqrt{p^3+q^2}} + \sqrt[3]{ q - \sqrt{p^3+q^2} }$

This is all three roots of the equation, given by the three cube roots (at least two complex), say for the left radical. The two cubes aren't really independent, we need their product to be $-p=st$ .

That's the three roots of the depressed cubic; let's solve the general cubic by reducing it to the depressed cubic.

$x^3 + ax^2 + bx + c=0$

We want to eliminate the squared term. If substitute x = y + k we'll get a 3ky² from the cubic term and ay² from the squared term; we want these to cancel so 3k=-a.

Substitute x = y - a/3

$(y - a/3)^3 + a(y - a/3)^2 + b(y - a/3) + c = 0$