If a secant<span> and a </span><span>tangent of a circle </span><span>are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment.
</span><span>
y</span>² = 7(15+7)
<span>y</span>² = 7*22
<span>y</span>² = 154
<span>y = </span>√154
<span>y = 12.4 </span>← to the nearest tenth<span>
</span><span>
</span>
X^2 + y^2 = 8
X-y=0 so x = y
replace x = y into X^2 + y^2 = 8
y^2 + y^2 = 8
2y^2 = 8
y^2 = 8/2
y^2 = 4
y = - 2 and y = 2
because x = y
so x = - 2 and x = 2
solutions:
x= - 2 and x = + 2
y= - 2 and y = + 2
Your answer is -5.
This is because, if you expand the single bracket, you get -6x -3a, and since the other side of the equals sign is -6x + 15, then you need to do 15 ÷ -3 = -5.
I hope this helps!
