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Aleksandr-060686 [28]
3 years ago
14

What is the equation of the line in slope intercept form of slope 8 and point (0,-6)

Mathematics
1 answer:
kipiarov [429]3 years ago
8 0

Answer:

y = 8x - 6

Step-by-step explanation:

Point-slope form is y - y₁ = m(x - x₁) where (x₁, y₁) is a point on the line and m is a slope.

Slope-intercept form is y = mx + b where m is the slope and b is the y-intercept.

Normally, when you are given a slope and a point on the line and asked to find the equation of the line in slope-intercept form, you would first write a line in point-slope form and simplify it into slope-intercept form (there is an easier way to do it). You would do it like this:

y - (-6) = 8(x - 0)

Simplify

y + 6 = 8x

Subtract 6 from both sides.

y = 8x - 6

You can do this here, but if you look at the point carefully, you can see that the point we are given lies on the y-axis, and is therefore the y-intercept of the line. So we can put it into slope-intercept form with 8 as the slope and -6 as the y-intercept:

y = 8x + (-6)

Simplify

y = 8x - 6

As you can see, both ways of solving this problem yield the same answer, but the second way that I showed is, in mu opinion, easier.

I hope you find my answer helpful.

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A ↔ B ↔ C ↔ D ↔ E ↔ F

 \frac{1}{4} AD      8      7      \frac{4}{5} AD   ???

AB + BC + CD = AD   <em>segment addition postulate</em>

\frac{1}{4} AD   +   8   +   7   = AD

\frac{1}{4} AD   +       15        = AD

AD         + 60    = 4AD

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AB =  \frac{1}{4} AD    =  \frac{1}{4}(20)    = 5

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7   +   4   + EF  = CF

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Answer: 11 + EF

Note: You did not provide any info about EF.  If you have additional information that you did not type in, calculate EF and add it to 11 to find the length of CF.

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Part I:

1. If you have to circle each outcome in the table that at least 1 die is a 4, then you have to circle row corresponding to the number 4 and column corresponding to the number 4.

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Here you can see 36 possible outcomes.

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Pr(A)=\dfrac{11}{36} - (favorable outcomes are 4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9, 4+6=10, 1+4=5, 2+4=6, 3+4=7, 5+4=9, 6+4=10).

2. The probability that the sum of the dice is greater than 5 is:

Pr(B)=\dfrac{26}{36}=\dfrac{13}{18}.

3. The probability that at least 1 die is a 4 and that the sum of the dice is greater than 5 is

Pr(A\cap B)=\dfrac{9}{36}=\dfrac{1}{4}.

4. The probability that at least 1 die is a 4 or that the sum of the dice is greater than 5 is:

Pr(A\cup B)=\dfrac{28}{36}=\dfrac{7}{9}.

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