To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
Answer:
= - 21
+ 6x² + 1
Step-by-step explanation:
Differentiate each term using the power rule
(a
) = na
Given
y = - 3
+ 2x³ + x , then
= (7 × - 3 ) + (3 × 2)x² + (1 × 1 )
= - 21 + 6x² + 1
Answer: x = 37.8
Step-by-step explanation: We start with triangle ABC with two sides given as 15 and 18. We shall make angle C the reference angle and thereby calculate the third side, line BC.
Since we have the opposite side as 15, and the adjacent side (which lies between the reference angle and the right angle) as 18, we can use the tangent of the angle C
Tan C = Opp/Adj
Tan C = 15/18
Tan C = 0.8333
From our table of values/use of the calculator
Tan C = 39.8
Angle C in triangle ACB = Angle C in triangle ECD (Opposite angles are equal).
That takes us to triangle ECD, since the reference angle is known (39.8) and the opposite side is also given (31.5), we can now calculate the adjacent which is side x.
Tan C = Opp/Adj
Tan 39.8 = 31.5/x
when you cross multiply, x moves to the left hand side, while Tan 39.8 moves to the right hand side
x = 31.5/Tan 39.8
x = 31.5/0.8333
<u>x = 37.8</u>
First convert the terms to fractional exponents
u = t^2/3 - 3t^3/2
differentiating
u' = 2/3 t^ (2/3 - 1) - 3* 3/2 t^(3/2 - 1)
= 2/3 t ^(-1/3) - 9/2 t ^(1/2)
= 2 / (3∛t) - 9 √ t / 2 in radical form
Answer:
is this a question .....write it correctly
