Q1=3 Q2[median]=12 Q3=14 and the IQR is 11
The first 3 are examples of the difference of 2 squares so you use the identity
a^2 - b^2 = (a + b)(a - b)
x^2 - 49 = 0
so (x + 7)(x - 7) = 0
so either x + 7 = 0 or x - 7 = 0
giving x = -7 and 7.
Number 7 reduces to 3x^2 =12, x^2 = 4 so x = +/- 2
Number 8 take out GCf (d) to give
d(d - 2) = 0 so d = 0 , 2
9 and 10 are more difficult to factor
you use the 'ac' method Google it to get more details
2x^2 - 5x + 2
multiply first coefficient by the constant at the end
that is 2 * 2 = 4
Now we want 2 numbers which when multiplied give + 4 and when added give - 5:- -1 and -4 seem promising so we write the equation as:-
2x^2 - 4x - x + 2 = 0
now factor by grouping
2x(x - 2) - 1(x - 2) = 0
(x - 2) is common so
(2x - 1)(x - 2) = 0
and 2x - 1 = 0 or x - 2 = 0 and now you can find x.
The last example is solved in the same way.
Answer:
I'll help if you need it.
Step-by-step explanation:
With the curve
parameterized by
with
, and given the vector field
the work done by
on a particle moving on along
is given by the line integral
where
The integral is then
Answer:
3 pints = 6 cups
Step-by-step explanation:
Formula: multiply the volume value by 2
