Answer:
<h2><DEF = 40</h2><h2><EBF = <EDF = 56</h2><h2><DCF = <DEF =40</h2><h2><CAB = 84</h2>
Step-by-step explanation:
In triangle DEF, we have:
<u>Given</u>:
<EDF=56
<EFD=84
So, <DEF =180 - 56 - 84 =40 (sum of triangle angles is 180)
____________
DE is a midsegment of triangle ACB
( since CD=DA(given)=>D is midpoint of [CD]
and BE = EA => E midpoint of [BA] )
According to midsegment Theorem,
(DE) // (CB) "//"means parallel
and DE = CB/2 = FB =CF
___________
DEBF is a parm /parallelogram.
<u>Proof</u>: (DE) // (FB) ( (DE) // (CB))
AND DE = FB
Then, <EBF = <EDF = 56
___________
DEFC is parm.
<u>Proof</u>: (DE) // (CF) ((DE) // (CB))
And DE = CF
Therefore, <DCF = <DEF =40
___________
In triangle ACB, we have:
<CAB =180 - <ACB - <ABC =180 - 40 - 56 =84(sum of triangle angles is 180)

4.
-3/4 - 2/3 ÷ (-4/5)
= -3/4 - [2/3 ÷ (-4/5)]
{switch the numerator and denominator when changing from ÷ to ×, or × to ÷}
= -3/4 - [2/3 × (-5/4)]
{numerator × numerator, and denominator × denominator}
= -3/4 - [-10/12]
= -3/4 + 10/12
= -9/12 + 10/12
= 1/12
5.
-2 2/5 + (-2 4/5) × 4/7
= -2 2/5 - (2 4/5 × 4/7)
= -12/5 - (14/5 × 4/7)
= -12/5 - (56/35)
{Simplify fraction to make denominators same}
= -12/5 - (8/5)
= -20/5
= -4/1
= -4
6.
-(-3/5) - 1 3/7 ÷ (-5/14)
{Negative signs cancelled out, fractions compounded}
= 3/5 - 10/7 ÷ (-5/14)
{Switch numerator and denominator when switch from ÷ to ×}
= 3/5 - [10/7 × (-14/5)]
= 3/5 - [-140/35]
= 3/5 + 20/5
= 23/5
{Simplify fraction}
= 4 3/5
Hope this helps! :) You may ask me for more help if you need
Answer:
y = -x + 2
Step-by-step explanation:
The desired equation must have the form y = mx + b.
Start with -2Y - 2X = -6 and solve for y: y + x = 2
Now solve this result for y:
y = -x + 2
You divide 134 by 8 and the answer would be how many classes. The answer is 16.75, I would think anyways.