Draw an equilateral triangle with side lengths of 1. Each angle here is 60 degrees, which is true of any equilateral triangle.
Now draw another equilateral triangle that has side lengths of 2 units. Clearly this triangle is not the same size as the previous one, but the angles are all still 60 degrees.
We have an example in which there are 2 triangles with the same angles, but the triangles are not congruent. Therefore, having info about congruent angles only isn't sufficient to prove triangles to be congruent.
The graphs of the given equations are parallel.
<h3>What is the slope-intercept form of a line?</h3>
The slope-intercept form of a line is y = mx +c where m is the slope of the line and c is the y-intercept.
The given equations are y = x +10 and y = x+5.
The equations are in the slope-intercept form of y = mx +b.
Therefore, the slopes of both lines are 1.
Two lines are parallel if they have the same slope.
Therefore, the graphs of the given equations are parallel.
To learn more about the slope-intercept form of a line, click here:
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Answer:
3 players
Step-by-step explanation:
So, all you have to do is multiply 27 by 1/9 to get 3. Or, you could try figure out what number needs to be multiplied by 9 to get 27, which is also 3.
3) In Δ BDC
|DC|/|BC| = cos C
cos C= 16/17.89
C= cos⁻¹( 16/17.89)=26.57⁰
In triangle ABC:
x=180-(90+26.57)=63.43
x=63.43
2)
AB/BD= tan(70⁰), AB=BD*tan(70⁰)
AB/BC=tan(40⁰), AB=BC*tan(40⁰)
BD*tan(70⁰)=BC*tan(40⁰)
BD=BC-CD=BC - 15
(BC -15)*tan(70⁰)=BC*tan(40⁰)
BC*tan(70⁰) -15*tan(70⁰)= BC*tan(40⁰)
BC*tan(70⁰) - BC*tan(40⁰) = 15*tan(70⁰)
BC(tan(70⁰)-tan(40⁰))= 15*tan(70⁰)
BC = 15*tan(70⁰)/(tan(70⁰)-tan(40⁰)) = 21.60
BC=21.60
1) In a quadrilateral sum of angles =360⁰.
PQS=SQR=50⁰, because SQ bisects PQR.
Using Law of sine in ΔSRQ
SR/sin ∠SQR = SQ/sinR, SQ = SR * sinR/sin SQR = 3*sin30/sin50 =3.26
SQ=3.26 cm
ΔPQS:
cos PQS= PQ/SQ
PQ=SQ*cosPQS =3.26*cos 50⁰=2.09=2.1
PQ=2.1 cm
