Jillian is selling boxes of cookies to raise money for her basketball team. the 10 oz. box costs $3.50, while the 16 oz. box costs $5.00. at the end of one week, she collected $97.50, selling a total of 24 boxes. the system of equations that models her sales is below.x y= 24
3.50x 5.00y = 97.50
solve the system of equations. how many 10 oz. boxes were sold?
6
9
12
15
Answers
Answer:
15
Step-by-step explanation:
If 10 oz. box costs $3.50 and a 16 oz. box costs $5.00 and she collected $97.50 we can represent an expression that relates for how many money Jillian made:
(1)
where x is the number of 10 oz. boxes and y is the number of 16 oz. boxes.
She sold 24 boxes therefore:
(2)
We can right equation in terms of y and substitute it into equation 1:
(2)
Solve for x:
Jillian sold 15 10oz. boxes
3.50x + 5.00y = 97.50
3.50x + 5(24 - x) = 97.50
3.50x + 120 - 5x = 97.50
3.50x - 5x = 97.50 - 120
-1.5x = -22.50
x = -22.50/-1.5
x = 15....there were 15 ten oz boxes (x) sold
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solve this question by taking the value of pi 22/7
Answers
r+h=U/π
r=U/π - h
Data:
U=16 1/2=16 + 1/2=(16*2+1)/2=33/2
h=2
π=22/7
Then:
r=U/π - h
r=(33/2)/(22/7) - 2
r=(33*7)/(2*22) - 2
r=231 /44 - 2
r=(231-2*44)/ 44
r=(231-88)/44
r=143/44 = (3*44 + 11)/44=3 11/44
Answer: r=143/44 or 3 11/44
c) y=1/2sinx
d) -1/2xcosx
Answers
Answer:
Option (d)
Step-by-step explanation:
Given,
y" +y=sin x ...........(1)
The particular solution
Putting the value of y" and y in equation (1)
Therefore 2A =0 -2B=1
⇒A=0
Therefore
Final answer:
The solutions of the differential equation y'' + y = sin(x) are y = cos(x), y = (1/2)sin(x), and y = -(1/2)xcos(x).
Explanation:
To determine which of the given functions are solutions of the differential equation y'' + y = sin(x), we can substitute each function into the equation and check if it satisfies the equation. Let's go through each option:
- Substituting y = sin(x) into the equation, we get -sin(x) + sin(x) = sin(x), which is not true. So, y = sin(x) is not a solution.
- Substituting y = cos(x) into the equation, we get -cos(x) + cos(x) = sin(x), which is true. So, y = cos(x) is a solution.
- Substituting y = (1/2)sin(x) into the equation, we get -(1/2)sin(x) + (1/2)sin(x) = sin(x), which is true. So, y = (1/2)sin(x) is a solution.
- Substituting y = -(1/2)xcos(x) into the equation, we get (-1/2)xcos(x) + (1/2)xcos(x) = sin(x), which is true. So, y = -(1/2)xcos(x) is a solution.
Therefore, the solutions of the differential equation y'' + y = sin(x) are y = cos(x), y = (1/2)sin(x), and y = -(1/2)xcos(x).
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600!!
Step-by-step explanation:
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Step-by-step explanation:
In every triangle, all the measures of the angles combined is 180 degrees.
So x + y + z = 180 degrees
Answer:
180 dg
Step-by-step explanation:
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Answer:
Step-by-step explanation:
Class limits are the minimum data value(lower) and maximum data value (upper) that a class can contain. They usually have the same numerical accuracy as the original data values.
Class boundaries are boundary lines that mark or separate where one class stops and the other begins. The lower class boundary of a given class is got by finding the average of the previous upper class limit and the given lower class limit while the upper class boundary is got by finding the average of the given upper class limit and the next lower class limit.
Final answer:
Class limits and class boundaries are statistical terms used in frequency distributions. Class limits are the smallest and largest values of a class, while class boundaries are the points separating one class from another.
Explanation:
The terms class limits and class boundaries are used in the field of statistics, particularly in the context of frequency distributions. Class limits are the smallest and largest values that can fall within each class in a frequency distribution, whereas class boundaries are the points that separate one class from another, and each boundary forms the end of one class and the start of the next.
For example, imagine you are analyzing the frequency of test scores and you have a class with limits of 80 and 89. These limits are the smallest and largest scores that fit into that class. However, the class boundaries are 79.5 and 89.5, serving as the dividing lines between this class and the next ones.
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