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Sergio039 [100]
3 years ago
13

Which dimensions can create only one unique triangle?

Mathematics
2 answers:
Gnom [1K]3 years ago
5 0

Answer:

B three sides measuring 5 in, 12 in, and 14 in

Step-by-step explanation:

the three angles measuring 25, 65, and 90 is 180 and can make more triangle.

the three angles measuring 50, 50, and 50, doesn't even add up to 180 degrees.

the three sides measuring 4 ft, 8 ft, and 14 ft can make more triangles.

I got this right in my test and I got it right.I hope this helps you

ArbitrLikvidat [17]3 years ago
4 0

When only angles are specified, the sides can be any length, so the triangle is not unique. In any event, the sum of angles must be 180°, so three angles measuring 50° cannot be closed to form a triangle.

Three sides measuring 4, 8, and 14 ft cannot be closed, either, as the long dimension exceeds the sum of the other two. The ends can't meet.

The appropriate choice is the 3rd one ...

... three sides measuring 5 in, 12 in, and 14 in.

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Find+the+positive+value+for+α+if+the+radius+of+the+circle+3x^2+3y-6αx+12y-3α=0+is+4
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<u>Equation of a Circle</u>

A circle of radius r and centered on the point (h,k) can be expressed by the equation

(x-h)^2+(y-k)^2=r^2

We are given the equation of a circle as

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Complete squares and rearrange:

x^2-2\alpha x+y^2+4y=\alpha

x^2-2\alpha x+\alpha^2+y^2+4y+4=\alpha+\alpha^2+4

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We can see that, if r=4, then

\alpha+\alpha^2+4=16

Or, equivalently

\alpha^2+\alpha-12=0

There are two solutions for \alpha:

\alpha=-4,\ \alpha=3

Keeping the positive solution, as required:

\boxed{\alpha=3}

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