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3 years ago

I don't get this one too can someone explain it please

Mathematics

1 answer:

lukranit [14]3 years ago

5 0

Answer:

15ab

Step-by-step explanation:

When you multiply out the square (p+q)^2, you get the trinomial ...

(p +q)^2 = p^2 + 2pq + q^2

Your binomial is (3a+2.5b), so you have ...

p = 3a

q = 2.5b

and the trinomial you get when you multiply it out is ...

(3a)^2 + 2(3a)(2.5b) + (2.5b)^2

= 9a^2 + 15ab + 6.25b^2

The first and last of these terms are shown on the right side of the given expression, so to finish out the identity, you need to replace * with 15ab.

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By graphing the points, The best mode of the data will be the quadratic function

There is 5 steps to find the quadratic function that models the data.

First step           ⇒ Plot the data pairs (x, y) in a coordinate plane.
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Second step     ⇒ Draw the parabola you think best fits the data.
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Third step         ⇒ Estimate the coordinates of three points on the
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Fourth step       ⇒ Substitute the coordinates of the points into the model
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                            a - b + c = 0           ---------(1)
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Fifth step   ⇒ Solve the linear system.
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So, the quadratic model for the data is ⇒ y = x² - 2x - 3

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9514 1404 393

Answer:

possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}

actual roots: -1, (2 ±4i√2)/3

no turning points; no local extrema

end behavior is same-sign as x-value end-behavior

Step-by-step explanation:

The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.

The Rational Root Theorem tells us any rational roots will be of the form ...

±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}

= ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots

Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.

The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.

The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):

f(x) = (x +1)(3x^2 -4x +12)

The complex roots of the quadratic can be found using the quadratic formula:

x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6

x = (2 ± 4i√2)/3 . . . . complex roots

The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.

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