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3 years ago

Point E is on line segment DF. Given DE EF. 6 and DF =9, determine the length EF

Mathematics

1 answer:

nlexa [21]3 years ago

5 0

Answer:

Step-by-step explanation:

IF point E lies on the line segment DF, this means that all the points DEF are collinear and DE+EF = DF.

Given parameter

DE = 6

DF = 9

Required

Substituting the given parameter into the expression above to get the required will be;

DE+EF = DF.

EF = DF-DE

EF = 9-6

EF = 3

Hence the length of EF is equivalent to 3

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alukav5142 [94]

Answer

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Step-by-step explanation:

because we don't know how big the room is and what the shape of the room is we have no idea what the answer will be it could be a little less than the distance around the world or it could be no distance cause it is the same size as the room

(also the rug is in the middle of the the room)

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3 years ago

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3 years ago

Find the value of x for which the figure is the given special parallelogram

Aleksandr-060686 [28]

All the corners of a rectangle is equal to 90 degrees. So you need to set (x-1) and (2x+10) equal to 90 and then solve for x.

(x - 1) + (2x + 10) = 90
<em>Combine the like terms...
</em>3x + 9 = 90
<em>Subtract both sides by 9...
</em>3x + 9 - 9 = 90 - 9
3x = 81
<em>Divide both sides by 3...
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</em>(3x) / 3 = 81 / 3
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3 years ago

Assume that lines that appear to be tangent are tangent o is the center of the circle

Greeley [361]

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4 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