You need to add up all the sides of the thing thank you
Your answers are a and d hope this helps
GCF(24, 30, 42) = 6 I think it is.
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
Answer:
AB=29; BC=27
Step-by-step explanation:
So they told us AB=4x+9 and that BC=5x+2, and AC=56 , now to help with the question you can draw this information on a number line. Now on a number you can see that basically AC=AB+BC.
So you would write it as such,,
4x+9+5x+2=56
Combine like terms
9x+11=56
Now you have to isolate the x by itself but first get rid of the 11.
9x+11-11=56-11
You would get
9x=45
Here you can divide 9 by both sides to isolate x.
9x/9=45/9
{x=5}
Now to find the value for both substitue x in the equations for both
1. AB=4x+9 where x is 5
4(5)+9 =AB
20+9 =AB
29=AB
You would do the same with BC
2. BC= 5x+2 where x is 5
5(5)+2= BC
25+2= BC
27=BC
If you want to check your answers you can just substitute x for 5 in the first equation we did where AC=AB+BC
