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
-7 - 3x = 9x(x + 2)
Standard Form:
-9x^2 - 21x - 7 = 0
Answer:
1)-
How to solve your question
Your question is
4(4−72)−9(5+2)
4(4y-7y^{2})-9(5y+2)4(4y−7y2)−9(5y+2)
Simplify
1
Rearrange terms
4(4−72)−9(5+2)
4({\color{#c92786}{4y-7y^{2}}})-9(5y+2)4(4y−7y2)−9(5y+2)
4(−72+4)−9(5+2)
4({\color{#c92786}{-7y^{2}+4y}})-9(5y+2)4(−7y2+4y)−9(5y+2)
2
Distribute
4(−72+4)−9(5+2)
{\color{#c92786}{4(-7y^{2}+4y)}}-9(5y+2)4(−7y2+4y)−9(5y+2)
−282+16−9(5+2)
{\color{#c92786}{-28y^{2}+16y}}-9(5y+2)−28y2+16y−9(5y+2)
3
Distribute
−282+16−9(5+2)
-28y^{2}+16y{\color{#c92786}{-9(5y+2)}}−28y2+16y−9(5y+2)
−282+16−45−18
-28y^{2}+16y{\color{#c92786}{-45y-18}}−28y2+16y−45y−18
4
Combine like terms
2)
−17y+17z+24
See steps
Step by Step Solution:

STEP1:Equation at the end of step 1
((24 - 4 • (5y - 6z)) + 3y) - 7z
STEP2:
Final result :
-17y + 17z + 24
−282+16−45−18
-28y^{2}+{\color{#c92786}{16y}}{\color{#c92786}{-45y}}-18−28y2+16y−45y−18
−282−29−18
-28y^{2}{\color{#c92786}{-29y}}-18−28y2−29y−18
Solution
−282−29−18
Answer:
I think it's a or b the picture is fuzzy
