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White raven [17]
3 years ago
5

Order the angles in each triangle from smallest to largest. NO LINKS!!!!​

Mathematics
2 answers:
lakkis [162]3 years ago
8 0

Answers:

  • Problem 13)  M, N, L
  • Problem 14)  N, L, M

For each answer above, the angles are sorted from smallest to largest.

=================================================

Explanation:

The general rule used here is: the smallest side is always opposite the smallest angle. Similarly, the largest side is always opposite the largest angle. This trick only works for triangles.

For problem 13, the smallest angle is M because the shortest side is opposite this angle (side NL = 12). The largest side is MN = 21, making the angle opposite this (angle L) to be the largest angle.

We do not need to compute the actual angle values, though you could if you wanted. To find the angle values, you would use the law of cosines. The steps for this are fairly lengthy, so I'll just use the trick mentioned above.

------

Problem 14 is the same idea. Here LM = 7 is the shortest side this time, leading to angle N as the opposite angle that's the smallest of the three angles. Angle M is the largest angle because NL = 14 is the longest side.

Ray Of Light [21]3 years ago
8 0

Answer:

problem 13 MLN

problem 14 NLM

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Answer:

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Step-by-step explanation:

Given the function

f\left(x\right)=x^3-6x^2+3x+10

As the highest power of the x-variable is 3 with the leading coefficients of 1.

  • So, it is clear that the polynomial function of the least degree has the real coefficients and the leading coefficients of 1.

solving to get the zeros

f\left(x\right)=x^3-6x^2+3x+10

0=x^3-6x^2+3x+10              ∵  f(x)=0

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Factor\:x^3-6x^2+3x+10\::\:\left(x+1\right)\left(x-2\right)\left(x-5\right)=0

so

\left(x+1\right)\left(x-2\right)\left(x-5\right)=0    

Using the zero factor principle

if  ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

x+1=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x=-1,\:x=2,\:x=5

Therefore, the zeros of the function are:

x=-1,\:x=2,\:x=5

f\left(x\right)=x^3-6x^2+3x+10 is the function of the least degree has the real coefficients and the leading coefficients of 1 and with the zeros -1, 5, and 2.

Therefore, the last option is true.    

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3 years ago
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