Answer:
$61
Step-by-step explanation:
Given
Number of days, x = 5
Expenditure per day, E'(x) = 10x + 11
Required
Determine the expenditure per day (E'(x))
From the question, we have that

Substitute 5 for x



Hence, the expenditure for 5 days is $61
Answer:
Step-by-step explanation:
So first, lets solve for the intercepts.
Let’s start by finding the x intercept.
Plug in 0 for y.
10x+9(0)=-90
10x=-90
X=-9, so your first point on your line will be (-9,0).
Now the y intercept.
Plug in 0 for x.
10(0)+9y=-90
9y=-90
Y=-10, so your second point on your line will be (0,-10).
Use those points to create your line.
Answer:
11.
Step-by-step explanation:
I got the answer of 11, as I:
-First, I worked out how many rides Margaret would be able to go on, so I did $35 -/- $3, which was a decimal, but the most amount of times would be 11 rides that she could go on, therefore, that's the amount of rides she would take, and have $2 left over.
Hope this helps!
--Emilie Xx
Answer:
There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.
Step-by-step explanation:
Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.
The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is
As a result, we have 165 ways to distribute the blackboards.
If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is
. Thus, there are only 35 ways to distribute the blackboards in this case.
Answer:
Γ = 15
Step-by-step explanation:
Given
f(x) = x² - 8x + Γ
with a = 1, b = - 8 and c = Γ , then
sum of zeros α + β = -
= -
= 8
product of zeros = αβ =
= Γ
Given α - β = 2 , then
(α - β)² = 2²
α² - 2αβ + β² = 4 → (1)
and
(α + β)² = 8²
α² + 2αβ + β² = 64 → (2)
Add (1) and (2) term by term
2α² + 2β² = 68 ( divide through by 2 )
α² +β² = 34
Substitute α² + β² = 34 into (1)
34 - 2αβ = 4 ( subtract 34 from both sides )
- 2αβ = - 30 ( divide both sides by - 2 )
αβ = 15
Now
αβ = Γ = 15
Thus
f(x) = x² - 8x + 15