Answer:
- <u><em>The lamp burns 6 ounces of oil per min.</em></u>
Explanation:
That the oil of a lamp burns at a <em>linear rate</em> means that the rate of burn is constant.
Hence, you can calcualte the rate with which the lamp burns the oil using any two ordered pairs.
- Rate of burn = [ change in oil ] / [change in time]
- Rate of burn = [Oil at time 2- Oil at time 1 ] / [time 2 - time 1]
Your two ordered pairs are (13 ounces, 10 min) and (7 ounces, 38 min).
Substituting those pairs in the equation for the rate, you get:
- Rate = [ 7 ounces - 13 ounces ] / [38 min - 10 min]
- Rate = ( - 6 ounces) / (18 min) = - 1/3 oz / min ≈ - 0.33 oz / min
The negative sign just shows that the amount of oil is decreasing (the oil is being burned).
Hence, the lamp burns 6 ounces of oil per min.
The solutions associated with each case are listed below:
- 3 · x + 2
- 2 - x
- 2 · x² + 4 · x
- 1 / 2 + 1 / x
- 2 · x + 2
- 2 · x + 4
- x + 4
- 4 · x
- - 8
- 2
<h3>How to use operations between functions and evaluate resulting expressions</h3>
According to the statement, we find that the two functions are f(x) = x + 2 and g(x) = 2 · x and we are asked to perform on the functions to obtain all resulting expressions and, if possible, to evaluate on each case:
Case 1
(f + g) (x) = f (x) + g (x) = (x + 2) + 2 · x = 3 · x + 2
Case 2
(f - g) (x) = f (x) - g (x) = (x + 2) - 2 · x = 2 - x
Case 3
(f · g) (x) = f (x) · g (x) = (x + 2) · (2 · x) = 2 · x² + 4 · x
Case 4
(f / g) (x) = f (x) / g (x) = (x + 2) / (2 · x) = 1 / 2 + 1 / x
Case 5
(f ° g) (x) = f [g (x)] = 2 · x + 2
Case 6
(g ° f) (x) = g [f (x)] = 2 · (x + 2) = 2 · x + 4
Case 7
(f ° f) (x) = f [f (x)] = (x + 2) + 2 = x + 4
Case 8
(g ° g) (x) = g [g (x)] = 2 · (2 · x) = 4 · x
Case 9
(g ° g) (- 2) = 4 · (- 2) = - 8
Case 10
(f ° f) (- 2) = - 2 + 4 = 2
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Answer:
D
Step-by-step explanation:
17 and 11 are numbers while all 3 other equations have either x or n.
Answer:
A) A = X^2 + 11X + 24
B) X = 18 ft
C) 414 ft^2
D) 132 ft^2
Step-by-step explanation:
A) Width W of yard and path = X + 3
Lenght L of path = X + 5 + 3 = X + 8 ft, therefore,
Area of yard and path = L x W = (X +3) x (X +8)
A = X^2 + 11X + 24
B) if area of the yard and path is 546 ft^2,
546 = X^2 + 11X + 24
X^2 +115X - 522 = 0 (quadratic equation)
Solving the quadratic equation gives
X = 18 and X = -29
Our answer can only be positive so we choose X = 18 ft
C) lenght of yard = 5 + 18 = 23 ft
Width = 18 ft
Therefore area = L x W = 23 x 18 = 414 ft^2
D) area of path = area of path and yard minus area of yard
= 546 - 414 = 132 ft^2
The other coordinate B is = (-4,-4)