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Nikitich [7]
3 years ago
14

Which equations represent the line that is parallel to 3x − 4y = 7 and passes through the point (−4, −2)? Check all that apply.

Mathematics
1 answer:
DIA [1.3K]3 years ago
8 0
Reta a:

3x-4y=7 \\ -4y=-3x+7 \\ 4y=3x-7 \\ y= \frac{3}{4}x- \frac{7}{4}

reta x:

y=ax+b

Retas paralelas possuem coeficientes angulares iguais ⇒ mₐ=mₓ

y= \frac{3}{4}x+b

(-4,-2)

\frac{3}{4}.(-4)+b=-2 \\ -3+b=-2 \\ b=-2+3 \\ b=1

∴ y=3/4x+1
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salantis [7]

Answer:

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8 0
2 years ago
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The volume of a where is 38808 cm cube .find the radius​
Leokris [45]
638272 is the radius I think
4 0
3 years ago
Eliminating by multiplying <br> X+3y=1<br> -5x+4y=-24
gavmur [86]

Answer:

y = -1         x = 4

Step-by-step explanation:

(X+3y=1) 5            Multiply this equation by 5 to cancel out the x

-5x+4y=-24

-5x + 4y= -24

+ 5x + 15y = 5       Add both equations together

19y = -19                Divide both sides by 19

y = -1

Plug -1 for y into one of the equations

x + 3(-1) = 1             Multiply 3(-1)

x - 3 = 1

 + 3 + 3                  Add 3 to both sides

x = 4  

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3 0
3 years ago
If lin runs 24 laps at the same rate how long does it take her
sveta [45]

Answer:

is that all

Step-by-step explanation:

if she runs 24 laps a minute then the next minute she would of ran 48

5 0
2 years ago
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Fifty-three percent of employees make judgements about their co-workers based on the cleanliness of their desk. You randomly sel
GarryVolchara [31]

Answer:

The unusual X values ​​for this model are: X = 0, 1, 2, 7, 8

Step-by-step explanation:

A binomial random variable X represents the number of successes obtained in a repetition of n Bernoulli-type trials with probability of success p. In this particular case, n = 8, and p = 0.53, therefore, the model is {8 \choose x} (0.53) ^ {x} (0.47)^{(8-x)}. So, you have:

P (X = 0) = {8 \choose 0} (0.53) ^ {0} (0.47) ^ {8} = 0.0024

P (X = 1) = {8 \choose 1} (0.53) ^ {1} (0.47) ^ {7} = 0.0215

P (X = 2) = {8 \choose 2} (0.53)^2 (0.47)^6 = 0.0848

P (X = 3) = {8 \choose 3} (0.53) ^ {3} (0.47)^5 = 0.1912

P (X = 4) = {8 \choose 4} (0.53) ^ {4} (0.47)^4} = 0.2695

P (X = 5) = {8 \choose 5} (0.53) ^ {5} (0.47)^3 = 0.2431

P (X = 6) = {8 \choose 6} (0.53) ^ {6} (0.47)^2 = 0.1371

P (X = 7) = {8 \choose 7} (0.53) ^ {7} (0.47)^ {1} = 0.0442

P (X = 8) = {8 \choose 8} (0.53)^{8} (0.47)^{0} = 0.0062

The unusual X values ​​for this model are: X = 0, 1, 7, 8

6 0
3 years ago
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