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2 years ago

Segment BD is a median. Solve for x. Round to the nearest tenth, if necessary. (Image not necessarily to scale.)

Mathematics

1 answer:

nata0808 [166]2 years ago

6 0

Answer: 10

Step-by-step explanation:

By definition, the point at which a median intersects the side of the triangle opposite from the vertex from which it is drawn is the midpoint of that side.

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If you're on a car trip, and you've already done 13/17 of the trip, but there are still 80km to go, what's the total length in K

Radda [10]

Answer:

total length of trip = 340km

u have done 260km

Step-by-step explanation:

you've already done 13/17 of the trip so 1-13/17=4/17 of the trip left

4/17 of the trip = 80km

total length of trip = 80*17/4 = 340km

u have done 340 - 80 = 260km

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olga2289 [7]

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3 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:

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

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3 years ago