Answer:
14/-11
Step-by-step explanation:
use the slope formula ( y2-y1/x2-x1)
The value of x in the secants intersection is 1 units
The value of NM in the tangent and secant intersection is 51 units
<h3>How to find length when secant and tangent intersect?</h3>
The first question, two secant intersect outside the circle.
Therefore,
(6x + 8x)8x = (9 + 7)7
14x(8x) = 16(7)
112x² = 112
x² = 112 / 112
x = √1
x = 1
The second question, tangent and secant intersect,
Therefore,
(x + 3)² = (x - 3)(16 + x - 3)
(x + 3)² = (x - 3)(x + 13)
(x + 3)(x + 3) = (x - 3)(x + 13)
x² + 3x + 3x + 9 = x² + 13x - 3x - 39
x² + 9x + 9 = x² + 10x - 39
x² - x² + 9x - 10x = -39 - 9
-x = - 48
x = 48
NM = 48 + 3 = 51 units
learn more on secant and tangent here: brainly.com/question/12477905
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Answer:
74
Step-by-step explanation:
Say that arc JL going through M is arc E and JL going the other way is arc D
For the angle formed by two tangents, K=(1/2)(E-D)
64=E-D
Furthermore, angle K and central angle JCL (facing toward K) are supplementary, so 180-K=JCL=180-32=148
Thus, as the angles around angle C add up to 360, angle JCL (facing toward M) is 360-148=32+180=212
E is then 212
64=212-D
212-64=D=148
Thus, as JML is an inscribed angle, M=1/2(D)=1/2(148)=74
The answer is shown above
Answer:
To do this, you need to multiply out the expressions. This is a bit tedious, but remember like FOIL for binomials, for these trinomials you must multiply each term. If you need a step-by-step, I'd be happy to provide it. Let me know.
Once you have simplified the expression, you get
-x-9/2x-4
But, the problem stipulates that a must equal 1. We can equivalently factor out the negative sign and put it on the denominator with no change to write
x+9/-(2x-4) = x+9/-2x+4
So, seeing where each coefficient corresponds between the two expressions, you get a = 1, b = 9, c = –2, and d = 4.
