Ok, so here your being asked to solve 6x2<span> + 5x = -7
The procedure that I did was using this formula it led me to get the following:
</span>Using the formula:
x = -(-5) ± √(-5)² - 4(6)(-6)/ 2(6)
x = 5 ± √ 25 + 144 / 12
x = 5 ± √ 169 / 12
x = 5 ± 13/12
x1 = 5 + 13/12
x1 = 18/12
x1 = 3/2
x2 = 5 - 13/12
x2 = -8/12
<span>
x2 = - 2/3
Hope this helped :)</span>
First is false, second true, third is true , last is false. Wording is messed up on some gives the hint! Page 1<<<
13 is D no here << second page
ANSWER
See explanation
EXPLANATION
The point (3,-6) is the fourth quadrant.
In this quadrant only the cosine ratio and the secant ratio are positive.
The remaining four trigonometric ratios are negative.
The diagram is shown in the attachment.
We use the Pythagoras Theorem to find the hypotenuse, h.




















