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timurjin [86]
3 years ago
7

Janice wants to create a test containing 20 questions worth 50 points. If Janice creates questions worth either two points or fo

ur points, she can include (blank)
two-point questions and (blank) four-point questions.
Mathematics
1 answer:
Elena L [17]3 years ago
3 0

Answer:

15 two-point questions and 5 four-point questions.

Step-by-step explanation:

Let x represent number of two-points questions and y represent number of four-points questions.

We have been given that Janice wants to create a test containing 20 questions. We can represent this information in an equation as:

x+y=20...(1)

Since all questions are worth 50 points. We can represent this information in an equation as:

2x+4y=50...(2)

From equation (1), we will get:

x=20-y

Upon substituting this value in equation (2), we will get:

2(20-y)+4y=50

40-2y+4y=50

40+2y=50

40-40+2y=50-40

2y=10

\frac{2y}{2}=\frac{10}{2}

y=5

Therefore, there are 5 questions that are worth 4 points each.

Now, we will substitute y=5 in equation (1) to solve for x.

x+5=20

x+5-5=20-5

x=15

Therefore, there are 15 questions that are worth 2 points each.

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Please make sure to check with classmates for answers, to help each other.




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Step-by-step explanation:

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6 0
3 years ago
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Use the given graph to determine the limit, if it exists. A coordinate graph is shown with a horizontal line crossing the y axis
Lesechka [4]

Answer:

The limit of the function does not exists.

Step-by-step explanation:

From the graph it is noticed that the value of the function is 6 from all values of x which are less than 2. At x=2, the line y=6 has open circle. It means x=2 is not included.

For x<2

f(x)=6

The value of the function is -3 from all values of x which are greater than 2. At x=2, the line y=-3 has open circle. It means x=2 is not included.

For x>2

f(x)=-3

The value of y is 1 at x=2, because of he close circles on (2,1).

For x=2

f(x)=1

Therefore the graph represents a piecewise function, which is defined as

f(x)=\begin{cases}6& \text{ if } x2 \end{cases}

The limit of a function exist at a point a if the left hand limit and right hand limit are equal.

lim_{x\rightarrow a^-}f(x)=lim_{x\rightarrow a^+}f(x)

The function is broken at x=2, therefore we have to find the left and right hand limit at x=2.

lim_{x\rightarrow 2^-}f(x)=6

lim_{x\rightarrow 2^+}f(x)=-3

6\neq-3

Since the left hand limit and right hand limit are not equal therefore the limit of the function does not exists.

6 0
3 years ago
Read 2 more answers
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