Tens: 59
Hundredths: 78
Please give brainliest
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.
The answer is <span>b.3x + 15 = 55 - x
</span>
x - Jill's present age
y - dad's present age
<span>In 5 years, Dad will be three times as old as his daughter Jill will be then:
y + 5 = 3(x + 5)
</span><span>The sum of their present ages is 50:
x + y = 50
We have the system of two equations now:
</span>y + 5 = 3(x + 5)
x + y = 50
Let's rearrange the second equation:
If: x + y = 50
Then: y = 50 - x
Now, substitute y from the second equation (y = 50 - x) into the first one:
y + 5 = 3(x + 5)
50 - x + 5 = 3(x + 5)
50 + 5 - x = 3*x + 3*5
55 - x = 3x + 15
Rearrange it a bit:
3x + 15 = 55 - x
Therefore, the correct choice is b.
Remember
(ab)(cd)=(a)(b)(c)(d)
and
(√a)(√b)=√(ab)
so
(4√7)(2√10)=(4)(√7)(2)(√10)=(4)(2)(√7)(√10)=(8)(√70)=8√70
