<h3><em>1st quadrant,</em></h3><h3><em>(+,+) in 1st quadrants </em></h3><h3><em>HOPE IT HELPS !!!</em></h3>
Answer:
72
Step-by-step explanation:
I did the math
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Yes, this identity is true. Those two operations are equivalent.
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)

Answer:
x > 9/5
Step-by-step explanation:
Step 1: Write inequality
2/3x - 1/5 > 1
Step 2: Solve for <em>x</em>
- Add 1/5 to both sides: 2/3x > 6/5
- Divide both sides by 2/3: x > 9/5
Here we see that any <em>x</em> value greater than 9/5 will work.