The answer to this question is: 15
Answer:
d. 6
Step-by-step explanation:
The product of lengths to the near and far intersection points is the same for both secants:
PB×PA = PD×PC
We can use this relation to solve for x, then use the value of x to find PD.
7×12 = (3x)(7x)
4 = x² . . . . . . . . . divide by 21
2 = x . . . . . . . . . . take the square root
__
PD = 3x = 3·2
PD = 6
Answer:
9.9
use the special triangle formula or use cosine to find BC
Answer:
If you mean "How do I write an inequality?", the answer is COEFFICIENTxVARIABLE {any symbol} COEFFICIENTxVARIABLE >/< INTEGER
In other words, you should have something like this: -3x + 4v < 63
Or this: 51x - 16v > 11
Or anything along those lines.
Answer:
x = 28
Step-by-step explanation:
Given that lines AB and CD are straight lines that intersects at O, it follows that the pair of opposite vertical angles formed are congruent.
Thus,
<AOD = <BOC
<AOD = 152°
<BOC = 3x + x + (x + 12) (angle addition postulate)
<BOC = 5x + 12
Since <AOD = <BOC, therefore,
152° = 5x + 12 (substitution)
152 - 12 = 5x (subtraction property of equality)
140 = 5x
140/5 = x (division property of equality)
28 = x
x = 28
