Solve the triangle. A = 51°, b = 14, c = 6, I'll give 20 points

Step-by-step explanation:

We have that

A = 51°, b = 14, c = 6

step 1

find the value of a

Applying the law of cosines

a²=c²+b²-2*c*b*cos A

a²=6²+14²-2*6*14*cos 51> 126.27

a=11.2

we know that

The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

we have

a=11.2

b=14

c=6

so

(a+b) > c> (11.2+14)=25.2

25.2 > 6> is not correct

therefore

the answer is the option

a. No triangles possible

I don't know if i am right sorry if this is wrong

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.

options a,b,c are the right answer.

step-by-step explanation:

a linear parent function is always represented as y = x

if we write this function in the form of y=mx+c

⇒

or  

now if we plot a graph we will find these characteristics:

1) y intercept of the graph is zero so its a straight line passing through origin.

2) as we can see in both the forms of function written, values of x and y may be either positive or negative for the positive slope means points on the graph are either in first quadrant or in third one.

3) it's slope is one.