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Oduvanchick [21]
3 years ago
14

If ∠T and 60˚ form a linear pair what is the measure of ∠T *​

Mathematics
1 answer:
xenn [34]3 years ago
8 0

Answer:

120°

Step-by-step explanation:

angle T + 60°=180° [:. being linear pair ]

or , angle T = 180°-60°

or,angle T = 120°

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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

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

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

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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

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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}

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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

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which we can substitute in to the depressed cubic solution and subtract a/3  to get the three roots.  I won't write that out; it's a little ugly.

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