Answer:
Option A.
Step-by-step explanation:
The given question is incomplete. Here is the complete question.
P(n) models the price (in dollars) of a pack of n bulbs at a certain store.
When does the price of a pack increase faster ?
n 4 10 12
P(n) 12 25 28
When does the price of a pack increase faster ?
A. Between 4 and 10 bulbs
B. Between 10 and 12 bulbs
C. The price increases at the same rat over both the intervals.
To solve this question we will find the rate of increase in the prices per pack in the given intervals.
From n = 4 to n = 10
Rate of increase in price = 
= 
= 2.166 ≈ $2.17 per pack
From n = 10 to n = 12
Rate of increase in price = 
=
= $1.5 per pack
Therefore, price per pack increases faster between n = 4 and n = 10 as compared to n = 10 to n = 12.
Option A is the answer.
-1/2=3/2k+3/2
Rewrite the equation to have k on left side
3/2k+3/2=-1/2
Simplify 3/2k
3k/2+3/2=12
Move all terms not containing k to right side
3k/2=1/2-3/2
3k/2=-2
Multiply both sifmdes by 2
2(3k/2)=-2(2)
3k=-4
Divide both sides by 3 to get k alone
3k/3=-4/3
K= -4/3
Answer:
FIFTY - FIVE THOUSAND FIVE HUNDRED AND FIFTY-FIVE
Step-by-step explanation:
Step-by-step explanation:
The outlier is excluded, so the minimum is 59.
Since ∠E and ∠F are vertical angles, they are congruent, meaning that m∠E = m∠F
Plugging in the equations that were originally given, we can form the equation 9x + 12 = 3x + 24
Subtract both sides of the equation by 3x
6x + 12 = 24
Subtract 12 from both sides
6x = 12
Divide both sides by 6
x = 2
This should be your answer. Have an awesome day! :)