Triangles JFE and JLK are similar provided that line segments FE and LK are parallel. This means that corresponding parts of these triangles occur proportionally to one another. In particular,

$\dfrac{JF}{JL}=\dfrac{JE}{JK}=\dfrac{FE}{LK}$

From this relation we get

$\dfrac{JF}{JL}=\dfrac{JE}{JK}\implies\dfrac{(3x+21)-12}{12}=\dfrac{56}{70}$

Now we can solve for $x$ :

$\dfrac{3x+9}{12}=\dfrac45$

$3x+9=12\cdot\dfrac45=\dfrac{48}5$

$3x=\dfrac{48}5-9=\dfrac35$

$x=\dfrac{\frac35}3=\dfrac15$

5 0

4 years ago

Please help with question

nikklg [1K]

48%
Take 7 divided by all the rest added up

7 0

3 years ago

HelppppppppppppppppppppppppppppppppppppppppppPp

Zigmanuir [339]

Answer:

Step-by-step explanation:

4) You just have to find the lcm of the denominators

a) lcm(4,3,6)=4*3=12

$\dfrac{3}{4} x-\dfrac{2}{3} =\dfrac{1}{6}x\\\\12*(\dfrac{3}{4} x-\dfrac{2}{3}) =12*(\dfrac{1}{6}x)\\\\12*\dfrac{3}{4} x-12*\dfrac{2}{3} =12*\dfrac{1}{6}x\\\\9*x-8 =2*x\\\\9*x-2*x =8\\\\7*x=8\\\\x=\dfrac{8}{7} \\$

b) I don't know the name of the property but it is :

"We can multiply the two members of an equality by the same non-zero number, we obtain an equality. "

$-6x+10+8x=12-4x\\\\(-6x+10)+8x=12-4x\\\\8x+(-6x+10)=12-4x\\\\8x-6x+10=12-4x\\\\2x+10=12-4x\\....$

7 0

3 years ago