first let's name a couples of variable
• the number of adults tickets sold: a
• the number of children tickets sold: c
From the problem we know
a + c = 128
and
$5.40c + $9.20a = $976.20
1) solve the equation to alpha
a+c-c = 128 -c
a+0=128-c
a=128-c
2) substitute (128 - c) for a in the second equation and solve to c
$5.40c + $9.20a = $976.20 become
$5.40c + $9.20(128 - c) = $976.20
$5.40c + ($9.20 × 128) - ($9.20 - c) = $976.20
$5.40c - $9.20c + $ 1177.6 = $976.20
($5.40 - $9.20)c +$1177.6 = $976.20
-$3.80c + $1177.6 = $9.76.20
-$3.80c + $1177.60 - $1177.60 = $976.20 - $1177.60
-$8.30c + 0 = $201.40
-$3.80c = - $201.40
-$3.80c. -$201.40
________. = _________
-$3.80. -$3.80
-$3.80c. -$201.40
________. = _________. - they are 4 cut the no
-$3.80. -$3.80
c = $201.40
________
3.80
c = 53
Answer: Heyaa!! :) ^^
Your Answer Is... y=−6−2x
Step-by-step explanation:
Move all terms that don't contain y to the right side and solve.
Add 2x to both sides of the equation.
Divide each term in −y=6+2x by −1 and simplify.
<h3>^ y=−6−2x ^</h3>
Hopefully this helps <em>you !</em>
<em />
<em> </em>
Answer:
x < 20/3
(negative infinity, 20/3)
Answer:
<u>f(f¹(14)) = 134</u>
Step-by-step explanation:
Given :
Find the value of f(14) :
- f(14) = 3(14) + 2
- f(14) = 42 + 2
- f(14) = 44
Therefore, f(f¹(14)) = f(44).
Solving :
- f(44) = 3(44) + 2
- f(44) = 132 + 2
- <u>f(f¹(14)) = 134</u>
C is false because 30 is both divisible by 5 (30/5=6) and 10 (30/10=3) and is also divisible by 6 since 6*5 is 30