We want to translate the given statement into an algebraic expression.
The expression is:
d = 0.45*L
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When we have a discount of an X% on a given value V, that discount will be:
D = (X%/100%)*V.
Here we have:
<em>"A </em><em>discount</em><em> of a 45% on the list price L"</em>
The discount will be given by:
d = (45%/100%)*L = 0.45*L
d = 0.45*L
This is the expression we wanted to get.
If you want to learn more, you can read:
brainly.com/question/2736271
9514 1404 393
Answer:
- 75 adult tickets
- 125 child tickets
Step-by-step explanation:
Let 'a' represent the number of adult tickets sold. Then (200-a) is the number of child tickets sold, and the revenue is ...
8a +5(200 -a) = 1225
3a = 225 . . . . . . . . . . subtract 1000, simplify
a = 75 . . . . . . . . . . . . .divide by 3
200 -a = 125
75 adult ($8) and 125 child ($5) tickets were sold.
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<em>Additional comment</em>
The question asked here is "how many tickets did Kay sell?" The second line of your problem statement tells you the answer: "Kay sold 200 tickets ...". We have assumed that you are interested in the breakdown of tickets sold, even though that is not the question that is asked here.
1.5 would be the answer I think
Answer:
C) 0 ≤ x ≤ 25
Step-by-step explanation:
We are supposed to find a reasonable constraint so that the function is at least 300 i.e. the value of x at which f(x) is greater or equal to 300
A)x ≥ 0
Refer the graph
At x = 0
f(x)=300
On increasing the value of x , f(x) increases but at x = 12 it starts decreasing
So, x ≥ 0 can also have f(x)<300
So, Option A is wrong
B)−5 ≤ x ≤ 30
At x = -5
f(x) = 100
So, Option B is wrong since we require f(x) is greater or equal to 300
c)0 ≤ x ≤ 25
At x = 0
f(x)=300
At x = 12 , it starts decreasing
At x = 25
f(x)=300
So, The value of f(x) is at least 300 when 0 ≤ x ≤ 25
D)All real numbers
At x = 30
f(x)=0
But we require f(x) greater or equal to 300
Hence Option C is true
