Answer:
Step-by-step explanation:
This problem is similar to many others in which the sum of two quantities and their difference are given. The solution can be found easily when the equations for the relations are written in standard form.
<h3>Setup</h3>
Let s and h represent numbers of sodas and hot dogs sold, respectively. The given relations are ...
- s +h = 235 . . . . . combined total
- s -h = 59 . . . . . . difference in the quantities
<h3>Solution</h3>
Adding the two equations eliminates one variable.
(s +h) +(s -h) = (235) +(59)
2s = 294 . . . . simplify
s = 147 . . . . . .divide by 2
h = 147 -59 = 88 . . . . h is 59 less
147 sodas and 88 hot dogs were sold.
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<em>Additional comment</em>
The solution to a "sum and difference" problem is always the same. One of the numbers is half the sum of those given, and the other is half their difference. ((235-59)/2 = 88)
In order to solve this mathematical problem we can first
consolidate and observe the given values and the values that are not known in
the stated problem.
Annie traveled 5 times the sum of the number of hours Brian
traveled and 2. Together they traveled 20 hours. Find the numbers.
Equation,
<span><span>
1. </span><span> 5(y) + 2 = 20</span></span>
<span><span>2. </span><span> 5y = 20 – 2</span></span>
<span><span>
3. </span><span> 5y = 18</span></span>
<span><span>4. </span><span> Y = 18 / 5</span></span>
<span><span>
5. </span><span> Y = 3.6</span></span>
<span><span>
6. </span><span> 5y = 5(3.6)</span></span>
<span><span>
7. </span><span> Annie = 18</span></span>
Hence, Annie traveled 18 hours while Brian traveled 3.6
hours.
since 10 students is the most amount of kids on each bus, you can fit less than 10 people on it. so i don't know if the question wants each bus to have to same amount of students or each bus having different amounts.
Answer:
Step-by-step explanation:
Because CB = 6 cm, we can find CD
Use Triangle CDB.
<BCD = <BCA - <ACD
<BCD = ?
<BCA = 90
<ACD = 60
<BCD = 90 - 60
<BCD = 30
Cos 30 = CD / CB
CD = Cos(30) * BC
CD = 5.196 cm
<A = 90 - ACD
<ACD = 60
<A = 90 - 60
<A = 30
Sin(<A) = CB / AB
AB = CB / sin(<A)
AB = 6 / 0.5
AB = 12
Area =1/2 CD * AB
Area = 1/2 * 5.196 * 12
Area = 31.18
