Answer:
Step-by-step explanation:
f(x) = 9x³ + 2x² - 5x + 4; g(x)=5x³ -7x + 4
Step 1. Calculate the difference between the functions
(a) Write the two functions, one above the other, in decreasing order of exponents.
ƒ(x) = 9x³ + 2x² - 5x + 4
g(x) = 5x³ - 7x + 4
(b) Create a subtraction problem using the two functions
ƒ(x) = 9x³ + 2x² - 5x + 4
-g(x) = <u>-(5x³ - 7x + 4)
</u>
ƒ(x) -g(x)=
(c). Subtract terms with the same exponent of x
ƒ(x) = 9x³ + 2x² - 5x + 4
-g(x) = <u>-(5x³ - 7x + 4)
</u>
ƒ(x) -g(x) = 4x³ + 2x² + 2x
Step 2. Factor the expression
y = 4x³ + 2x² + 2x
Factor 2x from each term
y = 2x(2x² + x + 1)
Answer:
47 and 49
Step-by-step explanation:
This would not be between two consecutive whole numbers. Any number from 47.5 and less than 48 would round up, and any number from 48 to less than 48.5 would round down. It might be consecutive odd numbers that you meant, or your question is not clear enough.
Answer:
θ is decreasing at the rate of 
units/sec
or 
(θ) = 
Step-by-step explanation:
Given :
Length of side opposite to angle θ is y
Length of side adjacent to angle θ is x
θ is part of a right angle triangle
At this instant,
x = 8 , 
= 7
( denotes the rate of change of x with respect to time)
y = 8 , 
= -14
( The negative sign denotes the decreasing rate of change )
Here because it is a right angle triangle,
tanθ = 
-------------------------------------------------------------------1
At this instant,
tanθ = 
= 1
Therefore θ = π/4
We differentiate equation (1) with respect to time in order to obtain the rate of change of θ or (θ)
(tanθ) = (y/x)
( Applying chain rule of differentiation for R.H.S as y*1/x)
θ(θ) = 
- 
-----------------------2
Substituting the values of x , y , , , θ at that instant in equation (2)
2(θ) = 
*(-14)- 
*7
(θ) =
Therefore θ is decreasing at the rate of units/sec
or (θ) =
Answer:
(a) 2ab - a/b
(b) c² + d + 2cd
(c) c + c²
(d) Let x be the number that is 1 greater than b, then x = b + 1
Let y be the number twice as large as b, then y = 2b
Then the quotient of x and y is what we want. This is (b + 1)/2b
(e) pq - 3(p + q)
(f) The quotient of n² and n³ + 5, this is
n²/(n³ + 5)
