Answer:
(4,-3)
Step-by-step explanation:
Easiest way to do this without any advanced methods is to use the answer choices to your advantage.
For a.) we have (1,1) meaning x = 1 and y = 1 if we get a 7 for the first equation and a -2 for the second equation then that is the correct answer.
Let x=1, y =1:
4(1)+3(1)=4+3=7 Correct so far.
1+2(1)=1+2=3 Incorrect since we should have got a -2 if this was the solution
Let (0,-1) x = 0, y = -1:
4(0) + 3(-1) = 0 -3 = - 3 Incorrect so we can stop there next answer choice.
Let (4,-3) x = 4 y = -3:
4(4)+3(-3)=16-9=7 Correct so far.
4+2(-3)=4-6=-2 Both are correct!
Therefore the solution (where the lines intersect) is (4,-3).
BEDMAS
Brackets
exponents
division & multiplication }
addition & subtraction} -these 2 last ones can be done interchangably or from left to right
False because it’s not correct it’s a false statement because (again)…
Answer:
1st one: -24a-6b-36c
2nd one: 14mn-12m
Step-by-step explanation:
1st one: Multiply '-6' w/ '4a', 'b', and '6c'
2nd one: Multiply '2m' w/ '7n' and '-6'
Answer:
0.01863, yes preference
Step-by-step explanation:
given that a class consists of 12 boys and 12 girls. The teacher picks five students to present their work to the rest of the class and says that the five students are being selected at random
Here selecting any 5 students from the group of total 24 students is combination because order does not matter.
Total no of ways of selecting any 5 from total 24 = 
No of ways of selecting only 5 girls = No of ways of selecting random 5 girls from total 12 girls
= 
a) the probability be that all five students
selected are girls=
b) Since the probability for selecting all girls is very small and near to 0, it is unusual to select all girls if done at random. Hence the teacher had a preference for girls.