Let's see
In ∆ABE and ∆CBE
- BE=BE(Common side)
- AE=EC[Diagonals of a parallelogram bisect each other]
- <AEB=<BEC[90°]
So by
SAS congruence the triangles are congruent
AB=BC
Fact:-
It's already given AC is perpendicular to BD
- It means diagonals are perpendicular to each other
According to general property of rhombus this parallelogram is also a rhombus.
So sides are equal hence AB =BC
10 different groups can ski.
This is a combination of 5 items taken 3 at a time:
Translate it 8units to the right then reflect it over the line y=-3
Why?
- We can see the Quadrilateral is in Quadrant 3.
- If we translate it by 8units right it come to Quadrant 4.
- Now reflect it over line y=-3
- we will get Quadrilateral 2
<span>4. Simplify the expression.
sine of x to the second power minus one divided by cosine of negative x</span>
<span>(1−sin2(x))/(sin(x)−csc(x))<span>
</span>sin2x+cos2x=1</span>
<span>1−sin2x=cos2x<span>
</span>cos2(x)/(sin(x)−csc(x))</span>
<span>csc(x)=1/sin(x)</span>
<span>cos2(x)/(sin(x)− 1/sin(x))= cos2(x)/((sin2(x)− 1)/sin(x))</span>
<span>sin2(x)− 1=-cos2(x)</span>
<span>cos2(x)/(( -cos2(x))/sin(x))
=-sin(x)</span>
<span>
the answer is the letter a)
-sin x
</span><span>
5. Find all solutions in the interval [0, 2π). (6 points)sin2x + sin x = 0</span> using a graphical tool
the solutions
x1=0
x2=pi
<span>x3=3pi/2
the answer is the letter </span><span>
D) x = 0, π, three pi divided by two</span>
Answer:
y = (1/4)x+3
Step-by-step explanation: