Answer:
well that is a pretty good deal i might have to go there
Step-by-step explanation:
If X > 0 then X must be a positive number. If Y < 0 then Y must be negative. So the quadrant that contains Positive X values and negative Y values will contain the point that you are looking for. This would be the fourth quadrant.
3x^2 = 7x - 3
3x^2 - 7x = -3
x^2 - 7/3x = -1
x^2 - 7/3x + 49/36 = -1 + 49/36 = 13/36
(x - 7/6)^2 = 13/36
x - 7/6 = + or - sqrt(13)/6
x = 7/6 + or - sqrt(13)/6
x = (7 + or - sqrt(13))/6
Answer: You apply the trigonometric ratios of Sine, Cosine and Tangent.
Step-by-step explanation: In a question such as this one of the angles must be given, which is labeled the reference angle. We already know one angle is 90 degrees in a right angled triangle, so the other two angles whatever their size is, must both add up to 90, which means they are both acute angles. If one side is given, for instance 10 units, and one of the two acute angles is given, for example 30 degrees, then the line facing the reference angle (30 degrees) would be the OPPOSITE. The line that lies between the angles 30 and 90 would be the ADJACENT while the third side which is the longest would be the HYPOTENUSE.
So if for instance, the opposite line is 10 units, and you are required to calculate the other line, the adjacent, AS LONG AS YOU HAVE BEEN GIVEN A REFERENCE ANGLE, you can simply calculate as follows;
TanX = opposite/adjacent (and that becomes)
Tan 30 = 10/adj
Adj = 10/Tan 30
Adj = 10/0.5774
Adj = 17.32
**Note that the use of angle 30 was just for the sake of explanation. Your reference angle whatever it is, would be provided by your examiner**
You do parenthesies
exponents
multiply
divide
add
subtract
