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PSYCHO15rus [73]
3 years ago
7

Write the equations in graphing form, then state the vertex of the parabola or the center and radius of the circle.

Mathematics
1 answer:
Sati [7]3 years ago
6 0
All we need is to put this form in the vertex form f(x) = (ax+b)^2 + c

So we have <span>f (x)= 3x^2+12x+11 ....

Let's complete the square (if you aware of it)
</span><span>
f(x)= 3x^2+12x+11 = 3(x^2+4x)+11 = 3(x^2+4x+4-4)+11
=</span><span> 3([x^2+4x+4]-4)+11 = 3[(x+2)^2-4]+11 =3</span><span>(x+2)^2 - 12 +11 = 3</span><span><span>(x+2)^2 -1

so our form would be:

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


Here is a parabola with vertex of (-2,-1) and with positive </span> slope (concave up)





</span>

I hope that helps!



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Each set of numbers below represents the lengths of three line segments. Which set represent line segments that could be connect
LenKa [72]

Answer:

B is the triangle

A and E can be thought of as triangles as well with angles 0°, 0° and 180°, but it will be just a line with 3 points. So it depends whether this is a question with multiple answers or not.

Step-by-step explanation:

  • Thinking Process:

We need to find a formula for a triangle that uses all the sides

Cosine rule uses all sides:

c^2 = a^2 + b^2 -2ab(cosA)

we can rearrage this formula and make the angle A the subject:

A = cos^{-1}(\frac{a^2+b^2-c^2}{2ab})

  • now you can just put the values from each option in the question into this formula and see if it gives an answer or not.

<u>Try A. (1,2,3)</u>

put a = 1, b = 2, c = 3 (keep in mind that you can take 'a' 'b' and 'c' as 3,2 and 1 or any other arrangement within the option A(1,2,3). All we are doing here is checking whether the answer exists or not)

A = cos^{-1}(\frac{1^2+2^2-3^2}{2(1)(2)})

A = 180, now graphically it seems that this is just a line with 3 points, or technically you can think of as the sides of the triangle are overlapping each other, and the angle between the sides is 0 0 and 180. It depends on whether the question is asking you to be technical or not. Since the question already stated there are 3 line segments, so we can think of it as a triangle.

<u>Try B.(3,4,5)</u>

A = cos^{-1}(\frac{3^2+4^2-5^2}{2(3)(4)})

A = 90, this is a triangle (specifically, this is a right angled triangle)

<u>Try C.(1,10,100)</u>

A = cos^{-1}(\frac{1^2+10^2-100^2}{2(1)(10)})

from the calculator you'll see a 'math error', this shows that there is no answer, because cos^{-1} can only accept values within -1 and 1 inclusive. Hence no such triangle exists with lengths (1,10,100).

Similarly you can try the formula with all other options, and find whose answers exists or not.

Finally, to complete the answer, let's

Try E.(1,3,4)

A = cos^{-1}(\frac{1^2+3^2-4^2}{2(1)(3)})

A = 180, hence this is also one of those "3 points on a straight line  triangles" similar to A(1,2,3)

so the answer is either A,B and E OR just B

Hope this helps :)

5 0
3 years ago
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