.h.i. i think i know the answer
Answer:
33+87+x=180
x=69
x+y=180
by linear pair
69-180
=111
y =111
Step-by-step explanation:
<h3>I hope it's helpful for you</h3>
So basically, these numbers have to be consecutive because they're facing pages.
So you can just divide this by 2, and then add one to the second number.
629/2 = 314 R1
So the first page number is 314, and for the second one, just add one, and you get 315.
Answer:
(3x-2)(4x-1)
Step-by-step explanation:
1. take the a and the c value and multiply them together to isolate x^2. it should be x^2 -11 +24.
2. find what multiplies to 24 and adds up to -11. its -8 and -3.
3. factor like normal and get (x-8) (x-3)
4. divide the -8 and -3 in the factors by 12 (the original a value)
-8/12 = -2/3 and -3/12 = -1/4.
5. the denominator gets moved to the x value to get (3x-2)(4x-1)
Let "a" and "s" represent the costs of advance and same-day tickets, respectively. Your problem statement gives you two relations.
.. a + s = 35 . . . . . the combined cost of one of each is 35
.. 15a +40s = 900 . . total paid for this combination of tickets was 900
There are many ways to solve these equations. You've probably been introduced to "substitution" and "elimination" (or "addition"). Using substitution for "a", we have
.. a = 35 -s
.. 15(35 -s) +40s = 900 . . substitute for "a"
.. 25s +525 = 900 . . . . . . . simplify
.. 25s = 375 . . . . . . . . . . . .subtract 525
.. s = 15 . . . . . . . . . . . . . . .divide by 25
Then
.. a = 35 -15 = 20
The price of an advance ticket was 20.
The price of a same-day ticket was 15.
