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marysya [2.9K]
3 years ago
14

If f(x)=3x+8/2x-Aand f(0)=2.what is the value of a?

Mathematics
1 answer:
skad [1K]3 years ago
8 0

Answer:

a=-4

Step-by-step explanation:

Since f(x)=\frac{3x+8}{2x-a}, we have

2=f(0)=\frac{3(0)+8}{2(0)-a}=-\frac8a

Taking the reciprocal gives

-\frac a8=\frac12

Then multiplying by -8 gives

a=\frac12(-8)=-4

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If the expression is 2y+9x the answer is: 10x+12y

If the expression is 2y-9x the answer is -8x+12y

Step-by-step explanation:

You would combine all your x values:-4,9,9,-4 and that equals 10 then you would add your variable, so your x value is 10x.

You would then combine all your y values: 8,-2,8,-2, and that equals 12, then you would add your variable, so your y value is 12y.

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When solving a linear system by substitution, how do you decide which variable to solve first?
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Step-by-step explanation:

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Lizzie rolls two dice. What is the probability that the sum of the dice is:
zhenek [66]

Answer:

A.\ \dfrac{1}{3}\\B.\ \dfrac{5}{12}\\C.\ \dfrac{7}{36}\\

Step-by-step explanation:

Total outcomes possible: 36

A. Divisible by 3

Possible options are:

3, 6, 9 and 12.

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Possible outcomes for 9 are: {(3,6), (4,5), (5,4),(6,3)} Count 4

Possible outcomes for 12 are: {(6,6)} Count 1

Total count = 2 + 5 + 4 + 1 = 12

Probability of an event E can be formulated as:

P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}

P(A)  = \dfrac{12}{36} = \dfrac{1}{3}

B. Less than 7:

Possible sum can be 2, 3, 4, 5, 6

Possible cases for sum 2: {(1,1)}  Count 1

Possible cases for sum 3: {(1,2), (2,1)}  Count 2

Possible cases for sum 4: {(1,3), (3,1), (2,2)}  Count 3

Possible cases for sum 5: {(1,4), (2,3), (3,2),(4,1)}  Count 4

Possible cases for sum 6: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

Total count = 1 + 2 + 3 + 4 + 5 = 15

P(B)  = \dfrac{15}{36} = \dfrac{5}{12}

C. Divisible by 3 and less than 7:

P(A \cap B) = \dfrac{n(A\cap B)}{\text{Total Possible outcomes}}

Here, common cases are:

Possible outcomes for 3 are: {(1,2), (2,1)} Count 2

Possible outcomes for 6 are: {(1,5), (2,4), (3,3), (5,1),(4,2)} Count 5

P(A \cap B) = \dfrac{7}{\text{36}}

5 0
3 years ago
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