Answer:
the answer is 3
Step-by-step explanation:
because if you do the methoes method you know that it rules out 1 and 2 after you do that its just between 3 and 4, for the caunususicon there has to be one, that is why i think its 3. hope this helps!
Hello,
Answer D
ax-b < -c or ax-b >c
If ax-b >0 then |ax-b| = ax-b ==> ax-b > c
if ax-b <0 then |ax-b|=-(ax-b) > c ==> ax-b <-c
If you mean "factor over the rational numbers", then this cannot be factored.
Here's why:
The given expression is in the form ax^2+bx+c. We have
a = 3
b = 19
c = 15
Computing the discriminant gives us
d = b^2 - 4ac
d = 19^2 - 4*3*15
d = 181
Note how this discriminant d value is not a perfect square
This directly leads to the original expression not factorable
We can say that the quadratic is prime
If you were to use the quadratic formula, then you should find that the equation 3x^2+19x+15 = 0 leads to two different roots such that each root is not a rational number. This is another path to show that the original quadratic cannot be factored over the rational numbers.
Answer:
10.2
How I found this answer:
4+6=10
3-1=2
10^2+2^2=104
The square root of 104 is 10.1980390272, which can be rounded to 10.2
Answer:
Step-by-step explanation:
vector u=<(11-(-7),(-5-2)>=<18,-7>
as direction is opposite to u
so vector v=-3(18,-7)=(-54,21)
