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AnnyKZ [126]
3 years ago
15

Sometimes one form is more beneficial than the other. Identify which form would be more helpful if you needed to do each task li

sted below and explain why. a. Factor the equation. b. Graph the parabola. c. Identify the vertex, minimum, or maximum of the parabola. d. Solve the equation using the quadratic formula.
Mathematics
1 answer:
tankabanditka [31]3 years ago
6 0
The surest way to get many of the points needed to plot a quadratic is to use the quadratic formula. This will give the roots (real or imaginary). It will give you the completed square form also called the vertex form (if you know how to use the discriminant). It can easily give you the y intercept (which you can find before you use the quadratic formula).  It gives the max or min upon solution.

The easiest one to use if it is available to you, is factoring. The quadratic may not be factorable. But if it is and you can see it, then this gives you 2 points immediately (the roots) and a third without much trouble (the y intercept). Factoring will also give you the x value of the vertex. (Find the average between the 2 roots)

This needs an example
Suppose you have y = (x - 5)(x - 9) The roots are 5 and 9, correct? So the x value of the vertex is (5 + 9)/2 = 7 It always works.

Completing the square always gives you the minimum or maximum right away. For example if you have y = (x - 2)^2 - 5 it means you have the vertex at (2,-5) You can get the roots easily enough.  So this form is useful, but not as sure as the quadratic equation or as simple as factoring.

Graphing is the most certain way to check your answer. I find it the most useful thing to do with modern computers. There are all sorts of things that a graph will reveal that algebra by itself might be laborious and prone to leading you to mistakes. Graphing tends to correct that problem.
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Please help me <br><br>If 180°&lt;α&lt;270°, cos⁡ α=−8/17, what is sin -α?
rewona [7]

Starting from the fundamental trigonometric equation, we have

\cos^2(\alpha)+\sin^2(\alpha)=1 \iff \sin(\alpha)=\pm\sqrt{1-\cos^2(\alpha)}

Since 180, we know that the angle lies in the third quadrant, where both sine and cosine are negative. So, in this specific case, we have

\sin(\alpha)=-\sqrt{1-\cos^2(\alpha)}

Plugging the numbers, we have

\sin(\alpha)=-\sqrt{1-\dfrac{64}{289}}=-\sqrt{\dfrac{225}{289}}=-\dfrac{15}{17}

Now, just recall that

\sin(-\alpha)=-\sin(\alpha)

to deduce

\sin(-\alpha)=-\sin(\alpha)=-\left(-\dfrac{15}{17}\right)=\dfrac{15}{17}

6 0
3 years ago
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(x^4)^3=(x^3)^4 true or false
telo118 [61]

Step-by-step explanation:

(x^4)^3=(x^3)^4 , true

=> x^(4×3) = x^(3×4) = x^12

13^4 x 13^7= (13^4)^7, false

13^(4+7) = 13^11

(13^4)^7 = 13^(4×7) = 13^28

y^5 x y^0/y^3=(y^2)^1 , true

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(y^2)^1 = y^(2×1) = y^2

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q^0 x q^5/q^2= q^(0+5-2)= q^3

(q^3)^2/q^3 = q^(3×2-3) = q^3

6 0
3 years ago
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