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:
StartRoot StartFraction 1 Over 41 EndFraction EndRoot
Negative StartRoot 41 EndRoot
Negative StartRoot StartFraction 1 Over 41 EndFraction EndRoot
StartRoot 41 EndRoot
The first expression can be simplified by combining like terms.
First, the terms "4b" and "-3b" can be combined to form "b"
Then, the terms "-7" and "9" can be combined to form "2"
Finally, we simply put these two final terms together, to form "b+2"
Hence, 4b-7-3b+9 is equal to be + 2.
Hope this helps!
2x3=6
3x2=6
6 divided by 2=3
6 divided by 3=2
Answer:
d. 20
Step-by-step explanation:
To answer the question given, we will follow the steps below:
we need to first find p(3)
p(x) = x+ 7/ x-1
we will replace all x by 3 in the equation above
p(3) = 3+7 / 3-1
p(3) = 10/2
p(3) = 5
Similarly to find q(2)
q (x) = x^2 + x - 2,
we will replace x by 2 in the equation above
q (2) = 2^2 + 2 - 2
q (2) = 4 + 0
q (2) = 4
The product of p(3) and q(2) = 5 × 4 = 20
