Step-by-step explanation:
For quadratic equation ax^2 + bx + c = 0 to have two distinct real roots,
b^2 - 4ac must be positive.
b^2 - 4ac > 0
(k - 3)^2 - 4(3 - 2k) > 0
k^2 - 6k + 9 - 12 + 8k > 0
k^2 + 2k - 3 > 0
<span>let the angles be x and y.
as they are compliment
x+y=90
also given
x=7y
substituting value of x in first equation
7y+y=90
8y=90
therefore
y=11.25
x=7*y=78.75</span>
Rational numbers.
Please give brainliest and have a great day!
A - 3b = 4
a = b -2
(b - 2) - 3b = 4
b - 2 - 3b = 4
-2b = 4 + 2
-2b = 6
b = 6/-2
b = -3
a = b - 2
a = -3 -2
a = -5
to check: a = -5 ; b = -3 ⇒ (-5,-3)
a - 3b = 4
-5 - 3(-3) = 4
-5 + 9 = 4
4 = 4
Answer:
I am not sure how your teacher wanted you to estimate the answer but I solved it for you. Hopefully this helps.
Step-by-step explanation:
7x-y=7
x+2y=6
14x-2y=14
x+2y=6
add both equations
15x=20
x=20/15
x=4/3
x+2y=6
4/3+2y=6
2y=6-4/3
2y=18/3 -4/3
2y=14/3
divide both sides by 2
y= 14/3 divided by 2
y=14/3(1/2)
y=14/6
2y=
