In trapezoid $abcd$, $\overline{ab}$ is parallel to $\overline{cd}$, $ab = 7$ units, and $cd = 10$ units. segment $ef$ is drawn parallel to $\overline{ab}$ with $e$ lying on $\overline{ad}$ and $f$ lying on $\overline{bc}$. if $bf: fc = 3: 4$, what is $ef$? express your answer as a common fraction.

EF = 58/7

Step-by-step explanation:

If you extend both BC and AD to meet at P you get 3 similar triangles PAB, PEF, and PCD. Let's call BF = 3x and FC = 4x. Therefore the following proportions can be expressed:

BP/7 = (BP+7x)/10

10BP = 7*(BP+7x)

10BP = 7BP+49x

3BP = 49x

BP = (49/3)x

BP/7 = (BP+3x)/EF

EF = (BP+3x)*7/BP

EF = ((49/3)x + 3x)*7/((49/3)x)

EF = ((58/3)x*7*3)/(49x)

EF = 58/7

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)