∆ABC ~ ∆WXY. What is the value of x? (show work)
1 answer:
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Answer:
D. Triangle QRS is an isosceles triangle because QR = RS.
Step-by-step explanation:
Find the length of each side of the triangle using the formula for calculating the distance between two points.
D = √(x2-x1)²+(y2-y1)²
For side RS
R(0,0) and S(5, -3.322)
RS = √(5-0)²+(-3.322-0)²
RS = √25+11.035684
RS = √36.035684
RS = 6.0029
For side RQ
R(0,0) and Q(-3, -5.2)
RQ = √(-3-0)²+(-5.2-0)²
RQ = √9+27.04
RQ = √36.04
RQ = 6.0033
For side QS
Q(-3,-5.2) and S(5, -3.322)
QS = √(5+3)²+(-3.322+5.2)²
QS = √64+3.526884
QS = √67.526884
QS = 8.22
From the calculation it can be seen that RS=QR
Since the two sides f the triangle are equal, hence the triangle is an isosceles triangle. An isosceles triangle is a triangle that has two of its sides equal
You could complete the square to state the vertex.
You could use the quadratic equation to find the roots (which are complex).
Try an example that will require both.
y = x^2 + 2x + 5
Step One
Get the graph. That's included below.
Step Two
Provide the steps for completing the square.
Note: we should get (-1,4)
y= (x^2 +2x ) + 5
y = (x^2 +2x + 1) + 5 - 1
y = (x +1)^2 + 4
The vertex is at (-1,4)
Step Three
Find the roots. Use the quadratic equation. Note that the graph shows us that the equation never crosses or touches the x axis. The roots are complex.
a = 1
b = 2
c = 5
x = -1 +/- 2i
x1 = -1 + 2i
x2 = -1 - 2i And we are done.
You could use the quadratic equation to find the roots (which are complex).
Try an example that will require both.
y = x^2 + 2x + 5
Step One
Get the graph. That's included below.
Step Two
Provide the steps for completing the square.
Note: we should get (-1,4)
y= (x^2 +2x ) + 5
y = (x^2 +2x + 1) + 5 - 1
y = (x +1)^2 + 4
The vertex is at (-1,4)
Step Three
Find the roots. Use the quadratic equation. Note that the graph shows us that the equation never crosses or touches the x axis. The roots are complex.
a = 1
b = 2
c = 5
x = -1 +/- 2i
x1 = -1 + 2i
x2 = -1 - 2i And we are done.