Answer:
The distance between A and D to the nearest tenth is;
Explanation:
Given the two points;
Applying the distance between two points formula;
![d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}](https://tex.z-dn.net/?f=d%3D%5Csqrt%5B%5D%7B%28x_2-x_1%29%5E2%2B%28y_2-y_1%29%5E2%7D)
substituting the given coordinates we have;
![\begin{gathered} AD=\sqrt[]{(-3-6)^2+(-2-2)^2} \\ AD=\sqrt[]{(-9)^2+(-4)^2} \\ AD=\sqrt[]{81+16} \\ AD=\sqrt[]{97} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20AD%3D%5Csqrt%5B%5D%7B%28-3-6%29%5E2%2B%28-2-2%29%5E2%7D%20%5C%5C%20AD%3D%5Csqrt%5B%5D%7B%28-9%29%5E2%2B%28-4%29%5E2%7D%20%5C%5C%20AD%3D%5Csqrt%5B%5D%7B81%2B16%7D%20%5C%5C%20AD%3D%5Csqrt%5B%5D%7B97%7D%20%5Cend%7Bgathered%7D)
Simplifying;
Therefore, the distance between A and D to the nearest tenth is;
Answer:
Draw an open circle on 2 and continue the arrow left.
Step-by-step explanation:
So for this, I will be using the substitution method. Firstly, subtract both sides in the second equation by 2x: 
Next, substitute y in the first equation with 7-2x, then solve for x from there:
Now that we got x, substitute it into either equation to solve for y:
In short, x = 1/2 and y = 6.
Answer:
CLASS FREQUENCIES RELATIVE FREQUENCIES
A 60 0.5
B 12 0.1
C 48 0.4
TOTAL 120 1
Step-by-step explanation:
Given that;
the frequencies of there alternatives are;
Frequency A = 60
Frequency B = 12
Frequency C = 48
Total = 60 + 12 + 48 = 120
Now to determine our relative frequency, we divide each frequency by the total sum of the given frequencies;
Relative Frequency A = Frequency A / total = 60 / 120 = 0.5
Relative Frequency B = Frequency B / total = 12 / 120 = 0.1
Relative Frequency C = Frequency C / total = 48 / 120 = 0.4
therefore;
CLASS FREQUENCIES RELATIVE FREQUENCIES
A 60 0.5
B 12 0.1
C 48 0.4
TOTAL 120 1
Answer: (1,3) or x = 1 and y = 3
Step-by-step explanation:
The solution is the intersection thus the answer is (1,3) or x = 1 and y = 3
