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Dmitry [639]
3 years ago
8

Which equation represents a line that passes through the point (−2, 6) and is parallel to the line whose equation is 3x − 4y = 6

? 1. 3x + 4y = 18 2. 4x + 3y = 10 3. −3x + 4y = 30 4. −4x + 3y = 26?
Mathematics
1 answer:
mart [117]3 years ago
8 0
Method 1:  Realize that the desired equation will look exactly the same as the given equation 3x − 4y = 6, except that the constant will be different.

Thus, the equation of the new line is   3(-2) − 4(6) = C, where C comes out to 30.

Then the new line is 3x - 4y = 30.


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Vikki [24]

Answer:the answer is NO.

Step-by-step explanation:

Because the number 9, repeats itself in the domain.

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3 years ago
Help me with the right answer
Nadya [2.5K]

Answer:

x = 3

y = 1

Step-by-step explanation:

9x - y = 26  ---------------(I)

9x + y = 28 ---------------(II)

9x - y = 26

9x = 26 + y

Substitute 9x = 26 + y in equation (II)

26 + y + y = 28               {add like terms}

26 + 2y = 28             {subtract 26 from both sides}

         2y = 28 - 26

        2y = 2    {divide both sides by 2}

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y = 1

Substitute y =1 in equation (I)

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9x = 27

x = 27/9

x = 3

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6 0
3 years ago
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What is the equation of g?<br><br><br> PLEASE HELP!!! I NEED TO PASS THIS CLASS PLEASE
Ugo [173]

Answer:

g(x)= 2|x+3|-3

Step-by-step explanation:

hope it helps

8 0
3 years ago
An instructor who taught two sections of engineering probability last term, the first with 20 students and thesecond with 30, de
Oliga [24]

Answer:

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Step-by-step explanation:

This is an hypergeometric distribution problem

An hypergeometric distribution has the same sense as the discrete probabilities of binomial distribution, but unlike binomial distribution, hypergeometric distribution does not allow replacement.

Binomial distribution expresses the probability of picking k objects from n with replacement, but hypergeometric distribution expresses picking k objects from n without replacement, with the finite total population, N, containing K objects.

It is expressed mathematically as

h(k: n, K, N) = (ᴷCₖ)(ᴺ⁻ᴷCₙ₋ₖ)/(ᴺCₙ)

where

k = number of students in the 2nd section required to be in the first 15 graded projects (number of successes) = 10

n = total number of first graded projects (number of trials) = 15

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N = total number of students = 50

h(10: 15, 30, 50) = (³⁰C₁₀)(⁵⁰⁻³⁰C₁₅₋₁₀)/(⁵⁰C₁₅)

h(10: 15, 30, 50) = (³⁰C₁₀)(²⁰C₅)/(⁵⁰C₁₅)

= (30,045,015)(15,504)/(2,250,829,575,120)

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mr_godi [17]
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