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3 years ago

Y + 5 = 2(x + 1) order pairs

Mathematics

1 answer:

kolbaska11 [484]3 years ago

8 0

Y = 2x - 3, xeR
X = 3/2

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Given the midpoint and one endpoint of a line segment, find the other endpoint.

lozanna [386]

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{\frac{5}{8}}~,~\stackrel{y_1}{\frac{27}{8}})\qquad (\stackrel{x_2}{x}~,~\stackrel{y_2}{y}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{x+\frac{5}{8}}{2}~~,~~\cfrac{y+\frac{27}{8}}{2} \right)=\stackrel{midpoint}{\left( -\frac{7}{6}~~,~~-\frac{10}{3} \right)}$

$\bf -------------------------------\\\\ \cfrac{x+\frac{5}{8}}{2}=-\cfrac{7}{6}\implies x+\cfrac{5}{8}=-\cfrac{14}{6}\implies x=-\cfrac{14}{6}-\cfrac{5}{8} \\\\\\ x=\cfrac{-14(4)-5(3)}{24}\implies x=\cfrac{-56-15}{24}\implies \boxed{x=-\cfrac{71}{24}}\\\\ -------------------------------\\\\ \cfrac{y+\frac{27}{8}}{2}=-\cfrac{10}{3}\implies y+\cfrac{27}{8}=-\cfrac{20}{3}\implies y=-\cfrac{20}{3}-\cfrac{27}{8} \\\\\\ y=\cfrac{-20(8)-27(3)}{24}\implies y=\cfrac{-160-81}{24}\implies \boxed{y=-\cfrac{241}{24}}$

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3 years ago

What is the distance between point G(−4, 1) and point H(−12, 1) ?

Naya [18.7K]

Answer:

8 units

Step-by-step explanation:

because they have the same y axis all we need to do is subtract the distance between the x axis to get the distance between the points

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3 years ago

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Is this a polygon? If it is not a polygon please say why.

Eva8 [605]

Answer:

if you where trying to draw it straight then yes it is a polygon but if it is supposed to be rounded then no it is not.

Step-by-step explanation:

a plane figure with at least three straight sides and angles, and typically five or more.

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3 years ago

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Mary ate 1 4/7 pies, while fred ate 1 4/7 pies. in total . how much pie did these two eat?

nydimaria [60]

Hi there, 1+1=2 and 4/7+4/7=8/7. Therefore, Mary and Fred ate 2 8/7 of the pie.

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3 years ago

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Suppose p" must approximate p with relative error at most 10-3 . Find the largest interval in which p* must lie for each value o

goblinko [34]

Answer:

$[p-|p|*10^{-3} \, , \, p+|p|* 10^-3]$

Step-by-step explanation

The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is

r = |p*-p|/|p|

we want r to be at most 10⁻³, thus

|p*-p|/|p| ≤ 10⁻³

|p*-p| ≤ |p|* 10⁻³

therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]

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3 years ago