I was never sure of what the "additive inverse" is.
So, just now, just for you, I went and looked it up.
The additive inverse of any number ' A ' is the number
that you need to ADD to A to get zero. That's all !
So now, let's check out the choices:
a), 6, -(-6)
That second number, -(-6), is the same as +6 .
So the two numbers are the same.
Do you get zero when you add them up ? No.
b). -7, 7
What do you get when you add -7 and 7 ?
You get zero.
So these ARE additive inverses.
c). -7, -7
What do you get when you add -7 to -7 ?
You get -14 . That's not zero, so these
are NOT additive inverses.
d). 7, 7
What do you get when you add 7 to 7 ?
You get 14. That's NOT zero, so these
are NOT additive inverses.
e). 6, -6
What do you get when you add 6 to -6 ?
You get zero.
So these ARE additive inverses.
What do we end up with from the list of choices:
a)., c)., and d). are NOT additive inverses.
b). and e). ARE additive inverses.
Answer:Although the Quadratic Formula always works as a strategy to solve quadratic equations, for many problems it is not the most efficient method. Sometimes it is faster to factor or complete the square or even just "out-think" the problem. For each equation below, choose the method you think is most efficient to solve the equation and explain your reason. Note that you do not actually need to solve the equation. a. x2+7x−8=0x
2
+7x−8=0, b. (x+2)2=49(x+2)
2
=49, c. 5x2−x−7=05x
2
−x−7=0, d. x2+4x=−1x
2
+4x=−1.
I drew the segment and used Pythagorean theorem to solve for its measure. The line formed is the hypotenuse of the imaginary right triangle.
Among the choices only -1.33 and -1.25 is a feasible choice. But I am leaning towards -1.25 as the y-value of point F based on my diagram. Please see attachment.
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:
None of the numbers are Perfect Square.
Step-by-step explanation:
6, 10, 12, and 14 are not <u>Perfect Square</u>, because each number are multiplied by two different numbers:



