Write a recursive formula for
Answer:
= 5
Step-by-step explanation:
The sequence has a common ratio r between consecutive terms, that is
r = 10 ÷ 2 = 50 ÷ 10 = 250 ÷ 50 = 5
Thus the sequence is geometric with recursive formula
= 5
Answer:
12/40 non simplified and 3/10 simplified.
Step-by-step explanation:
b. 18.2 feet
c. 23.4 feet
d. 39.7 feet
Answer:
A
Step-by-step explanation:
i just did it
B. 115
C. 75
D. 66
E. 52
Write an equation to represent the possible numbers of cats and dogs that could have been at the
shelter on Wednesday.
Pat said that there might have been 8 cats and 14 dogs at the shelter on Wednesday. Are Pat’s
numbers possible? Use your equation to justify your answer.
Later, Pat found a record showing that there were a total of 22 cats and dogs at the shelter on
Wednesday. How many cats were at the shelter on Wednesday?
The equation representing the costs to care for the animals is 2.35c + 5.5d = $89.50. Pat's proposal of 8 cats and 14 dogs does not match the costs. Upon using additional information of the total number of animals being 22, we can solve the system of equations to find that there were 16 cats and 6 dogs at the shelter on Wednesday.
Let's denote the number of cats as c and the number of dogs as d. We know from the cost of care per day per animal that 2.35c + 5.5d = $89.50. This is the equation representing the possible number of cats and dogs at the shelter in terms of costs. For Pat's numbers, we can substitute c=8 and d=14 into the equation and check if they are possible. That is, substitute and check if 2.35(8) + 5.5(14) equals to 89.5. In this case, it does not since the total here comes to more than $89.50. So, Pat's numbers were not possible.
Regarding the total number of animals, we know that c + d = 22. With this equation, we can solve the system of equations to find out the actual number of cats and dogs at the shelter on Wednesday. Solving these two equations, we will find that 16 cats and 6 dogs were present at the shelter on Wednesday.
#SPJ3
Answer the following questions.
How many points was each part worth?
points
How many questions did Part A have?
questions
How many questions did Part B have?
questions
Answer:
1. How many points was each part worth?
- 12 points
2. How many questions did part A have?
- 2 questions
3. How many questions did Part B have?
- 3 questions
Step-by-step explanation:
We can set up our equation like this:
6x = 4y
In the above equation, x is representing the number of true/false questions and y is representing the nymber of multiple choice questions.
Now, the problem tells us that they want the least number of points possible so we know we need to use low numbers.
Since 6 is higher than 4, it's easier to go off of there.
6 x 1 = 6 4 is too big to go into 6 so we will move on.
6 x 2 = 12 4 goes into 12 3 times so we can use this.
Now that we've figured this out, we can put it in our equation:
6(2) = 4(3)
In the above equation, we can see that I've put 2 in for x because we multiplied 6 by 2 to get 12. I also put 3 in for y because we multiplied 4 by 3.
Now we can start with the questions:
1. How many points was each part worth?
Each part was worth 12 points because we can multiply 6 by 2 and get 12 or 4 by 3 and get the same thing
2. How many questions did part A have?
Part A had 2 questions because this is what x was when we multiplied by 6
3. How many questions did Part B have?
Part B had 3 questions because this is what y was when we multiplied by 4
Hope this helps!!
Each part is worth 12 points. Part A has 2 questions. Part B has 3 questions.
The problem states that the number of points for Part A is equal to the number of points for Part B, and we need to find the least number of points for which this is possible. Let's represent the number of questions in Part A as x. Since each true/false question is worth 6 points, the total points for Part A will be 6x. Similarly, let's represent the number of questions in Part B as y. Since each multiple choice question is worth 4 points, the total points for Part B will be 4y. To find the least number of points for which the two parts are equal, we need to find the smallest common multiple of 6 and 4.
The prime factorization of 6 is 2 x 3.
The prime factorization of 4 is 2 x 2.
From the prime factorization, we can see that the least common multiple (LCM) of 6 and 4 is 2 x 2 x 3 = 12.
Therefore, each part is worth 12 points.
To find the number of questions in Part A and Part B, we can substitute 12 for the total points in each part and solve for x and y:
6x = 12
x = 2
4y = 12
y = 3
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