<em>Greetings from Brasil...</em>
As we have a line, the function will be given by the expression:
F(X) = AX + B
where
<em>A = ΔY/ΔX</em>
<em>B = where the line intersects the Y axis</em>
Looking at the graph we have already concluded that
B = - 3
A = ΔY/ΔX
A = (5 - 3)/(4 - 3) see attached picture
A = 2
So,
F(X) = AX + B
<h2>F(X) = 2X - 3</h2>
Step-by-step explanation:
Hey there!
The equation is; y= -3x
or, 3x+y = 0.......... (I)
From equation (I)
Therefore, m= -3
Now, As per the condition of perpendicular lines;
M1*M2 = -1
-3*M2 = -1
M2= -1/-3
Therefore, the slope of a line which is perpendicular to the equation is; 1/3.
<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
Answer:
x=4 and y=2
Step-by-step explanation:
Name the triangles as ABC and DEF,
Now, since both the triangles are congruent by HL rule, therefore
(1)
and (2)
Substituting the value of in equation (1), we get
Therefore, Putting y=2 in equation (2),
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)
