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3 years ago

Find the equation of the line that is perpendicular to the line x = −5 and passes through the point (1, π).

1 answer:

5 0

The line x = -5 is a vertical line, therefore the line perpendicular to this line is the horizontal line of the form y = a.

We know that it passes through the point (1, π) → x = 1, y = π.

Therefore we have answer: A) y = π.

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Sarah lives 3 9/10 miles from school. Her classmate, Roger, lives 1 mile from school. How much farther from school does Sarah li

Veseljchak [2.6K]

Answer:

2 9/10

Step-by-step explanation:

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2 years ago

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omeli [17]

Answer:

Step-by-step explanation:

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2 years ago

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ollegr [7]

Answer:

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Step-by-step explanation:

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3 years ago

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Bezzdna [24]

Answer:

25 2/7

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6 0

2 years ago

The points A BC :(2, 2,1), :(1,1,3), :(2,0,5) − are the vertices of a right triangle. The radius of the sphere with center at th

ElenaW [278]

Answer:

r=1

Step-by-step explanation:

First we need to know the length of each side of the triangle, so we use the formula of the vector modulus:

$|AB|= \sqrt{(b_{1}-a_{1})^{2}+(b_{2}-a_{2})^{2}+(b_{3}-a_{3})^{2}$

By doing so, we find:

$|AB|=\sqrt {6}\\|BC|=\sqrt {6}\\|AC|=2\sqrt {5}$

With this we know that the triangle is not right, but, we assume the longest side as the hypotenuse of the problem.

As we have two equal sides, we know that the line between point |AB| and the center of the hypotenuse is perpendicular, therefore, we can calculate it using Pythagoras theorem:

$|BC|^{2}=r^{2}+(\frac{|AC|}{2})^{2}\\\\r^{2}=|BC|^{2}-(\frac{|AC|}{2})^{2}\\\\r^{2}=(\sqrt{6})^{2}- (\frac{2\sqrt{5}}{2})^{2}\\\\r^{2}=6-5=1\\r=1$

4 0

2 years ago