I believe it's the first option. Hope this helps.
Answer:
Step-by-step explanation:
because both equations do not have matching numbers for the x or y variable, and both equations are positive you are going to have to multiply each equation by a number so that there will be at least one variable with the same number but with opposite signs.
it does not matter which variable you choose.
lets use y because 2 and 3 are smaller then 2 and 5.
so lets multiply the first equation by 2 in order to get y equal to 6.
2(2x)+2(3y)=(2)6
(do not forget to multiply what the equation is equal to also)
4x+6y=12
now for the second equation we need y to equal negative 6
-3(5x)+-3(2y)=-3(4)
-15x-6y=-12
now lets put the 2 new equations next to each other and see what we can cancel out
4x+6y=12
-15x-6y=-12
-11x=0
x=0
now plug 0 in for x and solve for y (it does not matter which of the 4 equations you choose to solve.
2(0)+3y=6
3y=6
y=2
so your answer is x=0, y=2
The last terms must multiply to the last terms (confusing)
example
if ax^3+bx^2+c+d=(ex+f)(gx+h)(jx+k) then
d=fhk
so
70 is last term
we got 2 and 5
2*5*?=70
10*?=70
divide by 10
?=7
the missing number is 7
Step-by-step explanation:
According to the question a(n) = -5 for all real values of n.
So the 12th term in the sequence is -5.
Answer: 143
Step-by-step explanation:
Thanks for the free points
