Answer:
Do you want to be extremely boring?
Since the value is 2 at both 0 and 1, why not make it so the value is 2 everywhere else?
is a valid solution.
Want something more fun? Why not a parabola? 
.
At this point you have three parameters to play with, and from the fact that 
we can already fix one of them, in particular 
. At this point I would recommend picking an easy value for one of the two, let's say 
(or even 
, it will just flip everything upside down) and find out b accordingly:
Our function becomes
Notice that it works even by switching sign in the first two terms: 
Want something even more creative? Try playing with a cosine tweaking it's amplitude and frequency so that it's period goes to 1 and it's amplitude gets to 2: 
Since cosine is bound between -1 and 1, in order to reach the maximum at 2 we need 
, and at that point the first condition is guaranteed; using the second to find k we get 
Or how about a sine wave that oscillates around 2? with a similar reasoning you get
Sky is the limit.
Answer:
A) answer c ; B) X+3.7=15.3 ; C) 15.3 - 3.7 = 11.6 ---> 11.6+3.7=15.3
Step-by-step explanation:
Solution for A: First you have to look for the line underneath the bar says 15.3 as this will represent the total amount of liters in the bucket altogether. This would be bar c since we also know that we have 3.7 liters added but are missing the amount that was previously inside the bucket.
Solution for B: Let X resemble the unknown amount of water. To solve B, we need to write an equation. We know that there was already an amount of water in the bucket prior to us pouring more in so X is our first addend. We then added 3.7 liters of water to the unknown amount so that is our next addend giving you X+3.7=15.3
Solution for C: Now that we know our equation, we do the inverse operation (subtraction) in order to figure out the missing amount of water. This means we do
15.3 - 3.7 = 11.6
Answer:
can you explain yourself more please
B is the right thing 15 squints
Answer:
Step-by-step explanation:
Given that the figure is made up of portion of a square and a semicircle, we have;
BC ≅ AB = 6 cm
The area of semicircle BC with radius BC/2 = 3 is 1/2×π×r² = 1/2×π×3² = 4.5·π cm²
Triangle ABC = 1/2 × Area of square from which ABC is cut
The area of triangle ABC = 1/2×Base ×Height = 1/2×AB×BC = 1/2×6×6 = 18 cm²
The area of the figure = The area of semicircle BC + The area of triangle ABC
The area of the figure = 4.5·π cm² + 18 cm² = 
.
