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3 years ago

Which equation represents the line that passes through the points (2,4)and (1,-3)

Mathematics

1 answer:

garik1379 [7]3 years ago

8 0

Answer:

y=7x-10

Step-by-step explanation:

m=(-3-4)/(1-2)

m=-7/-1

m=7

y-y1=m(x-x1)

y-4=7(x-2)

y=7x-14+4

y=7x-10

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3 years ago

8x-6+2x=44 what does x equal

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Answer:

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Step-by-step explanation:

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What is the equation of this line in standard form? (-4, -1) & (1/2, 3)

BabaBlast [244]

bearing in mind that standard form for a linear equation means

• all coefficients must be integers, no fractions

• only the constant on the right-hand-side

• all variables on the left-hand-side, sorted

• "x" must not have a negative coefficient

$\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-1})\qquad (\stackrel{x_2}{\frac{1}{2}}~,~\stackrel{y_2}{3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{\frac{1}{2}}-\underset{x_1}{(-4)}}}\implies \cfrac{3+1}{\frac{1}{2}+4}\implies \cfrac{4}{~~\frac{9}{2}~~}\implies \cfrac{4}{1}\cdot \cfrac{2}{9}\implies \cfrac{8}{9}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-1)}=\stackrel{m}{\cfrac{8}{9}}[x-\stackrel{x_1}{(-4)}]\implies y+1=\cfrac{8}{9}(x+4) \\\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{9}}{9(y+1)=9\left( \cfrac{8}{9}(x+4) \right)}\implies 9y+9=8(x+4)\implies 9y+9=8x+32 \\\\\\ 9y=8x+23\implies -8x+9y=23\implies 8x-9y=-23$

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3 years ago

Factor 6y + 8y 6y² +8y?

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Answer:uiuyiuytghjv 312

7iuty

Step-by-step explanation:

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3 years ago

On Texas Avenue between University Drive and George Bush Drive, accidents occur according to a Poisson process at a rate of thre

Zarrin [17]

Answer:

(a) The probability is 0.6514

(b) The probability is 0.7769

Step-by-step explanation:

If the number of accidents occur according to a poisson process, the probability that x accidents occurs on a given day is:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is the mean number of accidents per day and t is the number of days.

So, for part (a), a is equal to 3/7 and t is equal to 1 day, because there is a rate of 3 accidents every 7 days.

Then, the probability that a given day has no accidents is calculated as:

$P(x)=\frac{e^{-3/7}*(3/7)^{x}}{x!}$

$P(0)=\frac{e^{-3/7}*(3/7)^{0}}{0!}=0.6514$

On the other hand the probability that February has at least one accident with a personal injury is calculated as:

P(x≥1)=1 - P(0)

Where P(0) is calculated as:

$P(x)=\frac{e^{-at}*(at)^{x} }{x!}$

Where a is equivalent to (3/7)(1/8) because that is the mean number of accidents with personal injury per day, and t is equal to 28 because 4 weeks has 28 days, so:

$P(x)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{x}}{x!}$

$P(0)=\frac{e^{-(3/7)(1/8)(28)}*((3/7)(1/8)(28))^{0}}{0!}=0.2231$

Finally, P(x≥1) is:

P(x≥1) = 1 - 0.2231 = 0.7769

3 0

3 years ago