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:
176cm³
Step-by-step explanation:
Multiply area of base times height
22cm² × 8cm = 176cm³
Answer:
x+5=5
Step-by-step explanation:
This is an equation where you are actually solving it. The others are expressions.
Answer:
dV/dt = 100 cm³/min
Step-by-step explanation:
Given
V = 728 cm³
P = 182 kPa
dP/dt = - 25 kPa/min
dV/dt = ?
If we apply the ideal gas equation
P*V = n*R*T
where n*R*T is constant
we have
d(P*V)/dt = d(n*R*T)/dt
⇒ d(P*V)/dt = 0
⇒ V*(dP/dt) + P*(dV/dt) = 0
⇒ dV/dt = - (V/P)*(dP/dt)
Plugging the known values we obtain
⇒ dV/dt = - (728 cm³/182 kPa)*(- 25 kPa/min)
⇒ dV/dt = 100 cm³/min
Last choice 1 9/50
118/100 =
(2 × 59)/(22 × 52) =
((2 × 59) ÷ 2) / ((22 × 52) ÷ 2) =
59/(2 × 52) =
59/50
Then
59 ÷ 50 = 1 and remainder = 9 =>
59 = 1 × 50 + 9 =>
59/50 =
(1 × 50 + 9) / 50 =
1 + 9/50 =
1 9/50
