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3 years ago

18,12,8,.. Find the 10th term

Mathematics

1 answer:

Leya [2.2K]3 years ago

7 0

Answer:

$0.46822130$

Explanation:

The sequence is: $a_{n} =\frac{18 *2^{n-1} }{3^{n-1} }$ , where n = term wanted

$a_{10} =0.46822130$

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Find an equation parallel to x = -9 and passing through (-1, -1).<br> Select one:

BARSIC [14]

Here find slope of x=-9 ie m=Tana and then use double intercept formy-y1=m(x,-x1)

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3 years ago

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If xy=0, what must be true about either x or y

evablogger [386]

Answer:

either x or y must equal 0

Step-by-step explanation:

ts given that xy = 0

Remember that product of two numbers can be zero only if:

Both of them are zero or Either of them is zero as zero multiplied to any non-zero number will always be equal to zero. This is known as Zero Product Property.

So, if the product of x and y is equal to 0 there are two possibilities:

Both x and y are equal to 0

Either x or y must be equal to 0

Note that the condition both x and y are equal to zero is not a must condition, because even if one of them is equal to zero, the entire expression will be equal to zero.

Hence, the condition which has to be true in all cases for xy = 0 is:

either x or y must equal 0

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3 years ago

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HELP HELP HELP PLS If DF = 6x - 2 and FE = 10x - 18, then which the correct equation to solve for x? 2(6x - 2) = 10x - 18 6x - 2

choli [55]

Answer: 6x-2=10x-18

From what u have given this should be the correct answer since they are most likely equal

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3 years ago

7. Which of the following is not a

romanna [79]

A. Rational

B.Not Rational

C.Rational

D.Rational

The Answer is B

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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

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

$y^3 - ay^2 + a^2/3 y - a^3/27 + ay^2-2a^2y/3 + a^3/9 + by - ab/3 + c =0$

$y^3 + (b - a^2/3) y = -(2a^3+9ab) /27$

Comparing that to

$y^3 + 3py = 2q$

we have $p = (3b - a^2) /9, q =-(a^3+9ab)/54$

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.

8 0

4 years ago