Answer:
Step-by-step explanation:
You have to use the discriminant for this. If the quadratic is
, then
a = -4, b = -3, and c = 7. The formula for finding the discriminant is
which comes from the quadratic formula, but without the square root sign. Filling in:
which simplifies down to
D = 9 + 112 so
D = 121. This is a perfect square, so the solutions will be 2 real. Just so you know, you will NEVER have a solution like the one offered in the third choice down. If you have one imaginary root, you will ALWAYS have a second by the conjugate rule.
Not sure sorry edit : oh sorry I was suppose to add this in the comment section
Starting from the fundamental trigonometric equation, we have

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

Plugging the numbers, we have

Now, just recall that

to deduce

Answer: 8.954 x 10^-15
Step-by-step explanation:
i took a test with this question and this was the answer
Answer:
Step-by-step explanation:
<em>L</em> is 2* <em>W</em> so L=2W
perimeter= 2w+2l
plug in whats given to get
perimeter=2w+4w 4w is because L is *2 of W so that makes L=2w
36=6w
36/6=6
6w/6=w
w=6
6*2=12=L
now we have 6*12=72=Area