Answer:
x = 7
Step-by-step explanation:
See the attached figure.
As shown: A C D F is a trapezoid and BE is a mid-segment.
The mid-segment of the trapezoid is the average of its bases.
BE = 0.5 ( A F + C D)
Given: CD = 18 , BE = 2x + 10 and A F = 6x - 12
So,
2x + 10 = 0.5 ( 18 + 6x - 12)
Solve for x
2x + 10 = 0.5 (6 + 6x)
2x + 10 = 3 + 3x
10 - 3 = 3x - 2x
∴ x = 7
<u>So, the value of x is 7</u>
<h2>
Answer:</h2>
LP = 8 because LR + PR = LP according to the Segment Addition Postulate, and 8 + 4 = 12 using substitution
<h2>
Step-by-step explanation:</h2>
From this problem, we know that:
LR = 12
PR = 4
So here we have a Line segment. Recall that a line segment has two endpoints, places where they end or stop and they are named after their endpoints, so the line segment here is LR whose measure is 12. Then, according to Segment Addition Postulate it is true that:
LP + PR = LR
By substituting LR = 12 and PR = 4, we have:
LP + 4 = 12
Subtracting 4 from both sides:
LP + 4 - 4 = 12 - 4
LP + 0 = 8
Finally:
LP = 8
90 x 0.4 = 36
90 - 36 = 54
The sale price is $54
Formula :
Base²= Hypotenuse² - Perpendicular ²
Remember the a² in formula has nothing to do with the a we have to find. :)
Answer:
both kinds of tickets are $5 each
Step-by-step explanation:
Let s and c represent the dollar costs of a senior ticket and child ticket, respectively. The problem statement describes two relationships:
12s + 5c = 85 . . . . . revenue from the first day of sales
6s + 9c = 75 . . . . . . revenue from the second day of sales
Double the second equation and subtract the first to eliminate the s variable.
2(6s +9c) -(12s +5c) = 2(75) -(85)
13c = 65 . . . . . simplify
65/13 = c = 5 . . . . . divide by the coefficient of c
Substitute this value into either equation. Let's use the second one.
6s + 9·5 = 75
6s = 30 . . . . . . . subtract 45
30/6 = s = 5 . . . divide by the coefficient of s
The price of a senior ticket is $5; the price of a child ticket is $5.
