Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle QNP has vertices Q(-4,4), N(-3,0), and P(-7,1). Is (triangle)JKL
Tems11 [23]
Answer:
Yes they are
Step-by-step explanation:
In the triangle JKL, the sides can be calculated as following:
=> JK = 
=> JL = 
=> KL = 
In the triangle QNP, the sides can be calculate as following:
=> QN = ![\sqrt{[-3-(-4)]^{2} + (0-4)^{2} } = \sqrt{1^{2}+(-4)^{2} } = \sqrt{1+16}=\sqrt{17}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-3-%28-4%29%5D%5E%7B2%7D%20%2B%20%280-4%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B1%5E%7B2%7D%2B%28-4%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B1%2B16%7D%3D%5Csqrt%7B17%7D)
=> QP = ![\sqrt{[-7-(-4)]^{2} + (1-4)^{2} } = \sqrt{(-3)^{2}+(-3)^{2} } = \sqrt{9+9}=\sqrt{18} = 3\sqrt{2}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-7-%28-4%29%5D%5E%7B2%7D%20%2B%20%281-4%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B%28-3%29%5E%7B2%7D%2B%28-3%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B9%2B9%7D%3D%5Csqrt%7B18%7D%20%3D%203%5Csqrt%7B2%7D)
=> NP = ![\sqrt{[-7-(-3)]^{2} + (1-0)^{2} } = \sqrt{(-4)^{2}+1^{2} } = \sqrt{16+1}=\sqrt{17}](https://tex.z-dn.net/?f=%5Csqrt%7B%5B-7-%28-3%29%5D%5E%7B2%7D%20%2B%20%281-0%29%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B%28-4%29%5E%7B2%7D%2B1%5E%7B2%7D%20%20%7D%20%3D%20%5Csqrt%7B16%2B1%7D%3D%5Csqrt%7B17%7D)
It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP
=> They are congruent triangles
Finding y intercept and x intercept is easy:
X intercept will be of the form (x,0) and y intercept will be of the form (0,y)
● If you put x=0 in the equation, you will get y-intercept.
● If you put y=0 in the equation, you will get x-intercept.
______________________________
Given equation: 2x - 4y = 10
◆ Put x = 0
2×0 - 4y = 10
=> -4y = 10
=> y = 10/(-4)
=> y = -5/2
Thus y intercept is (0, -5/2)
◆Put y = 0
2x - 4×0 = 10
=> 2x = 10
=> x = 10/2
=> x = 5
Thus the x intercept is (5,0)
Answer:
<h3>Q1</h3>
The graph of y = f(x), has vertex at (1, -2)
<u>The vertex of a function f(x - 3) is going to be:</u>
<h3>Q2</h3>
- <em>The graph of y = f(x) has the line x = 5 as an axis of symmetry. The graph also passes through the point (8,-7). Find another point that must lie on the graph of y = f(x).</em>
The axis of symmetry is at the same distance from the symmetric points.
x = 5 is a vertical line. The point (8, -7) is 3 units to the right. So the mirror point will be 3 units to the left and have same y-coordinate: x = 5 - 3 = 2
The point is (2, -7)
<h3>Q3</h3>
The graph in blue is the translation of the red to the left by 2 units.
<u>So the equation is:</u>
<h3>Q4</h3>
y = h(x) is graphed
- h(7) = 5
- h(h(7)) = h(5) = -1
<h3>Q5</h3>
The graph of the function y = u(x) given
This is a odd function.
The coordinates of u(x) and u(-x) add to zero because u(-x) = -u(x)
<u>Therefore:</u>
- u(-2.72) + u(-0.81) + u(0.81) + u(2.72) =
- [u(-2.72) + u(2.72)] + [u(-0.81) + u(0.81)] =
- 0 + 0 = 0
Answer: 120 ft
<u>Step-by-step explanation:</u>
The Ellipse equation is: 
where
- (h, k) is the center
- "a" is the horizontal radius
- "b" is the vertical radius
- c is the distance from the center to the foci: |a² - b²| = c²
Given: horizontal diameter = 122 → horizontal radius = 61 = a
vertical diameter = 22 → vertical radius = 11 = b
Foci (c): 61² - 11² = c²
3721 - 121 = c²
3600 = c²
±60 = c
Let (h, k) = (0, 0)
Then the foci are located at -60 and +60.
The distance between them is 60 + 60 = 120
