Answer:
(-∞, 4]
Step-by-step explanation:
The range of the function is the values that the output can take
Y goes from negative infinity to and including 4 ( assuming the scale is 1)
(-∞, 4]
Answer: 16
Step-by-step explanation:
multiply 6 by the answers below until you get 1/3 of 384.
You can predict that it's odd because the end of the 2 numbers is 1 and 1 x 1 is 1 and one is an odd number. (answer is odd)
Answer:
It would take 5.9 years to the nearest tenth of a year
Step-by-step explanation:
The formula of the compound continuously interest is A = P , where
- A is the value of the account in t years
- P is the principal initially invested
- e is the base of a natural logarithm
- r is the rate of interest in decimal
∵ Serenity invested $2,400 in an account
∴ P = 2400
∵ The account paying an interest rate of 3.4%, compounded continuously
∴ r = 3.4% ⇒ divide it by 100 to change it to decimal
∴ r = 3.4 ÷ 100 = 0.034
∵ The value of the account reached to $2,930
∴ A = 2930
→ Substitute these values in the formula above to find t
∵ 2930 = 2400
→ Divide both sides by 2400
∴ =
→ Insert ㏑ in both sides
∴ ㏑() = ㏑()
→ Remember ㏑() = n
∴ ㏑() = 0.034t
→ Divide both sides by 0.034 to find t
∴ 5.868637814 = t
→ Round it to the nearest tenth of a year
∴ t = 5.9 years
∴ It would take 5.9 years to the nearest tenth of a year
Answer:
The data that we have is:
"Adrian's backyard pool contains 6.4 gallons of water. Adrian will begin filling his pool at a rate of 4.1 gallons per second."
Then we can write the amount of water in Adrian's pool as a linear function:
A(t) = 6.4gal + (4.1gal/s)*t
Where t is our variable and represents time in seconds.
We also know that:
"Dale's backyard pool contains 66.4 gallons of water. Dale will begin draining his pool at a rate of 0.9 gallons per second. "
We can also model this with a linear function:
D(t) = 66.4 gal + (0.9gal/s)*t
Both pools will have the same amount of water when:
D(t) = A(t)
So we can find the value of t:
6.4gal + (4.1gal/s)*t = 66.4 gal + (0.9gal/s)*t
(4.1gal/s)*t - (0.9gal/s)*t = 66.4gal - 6.4gal
(3.2gal/s)*t = 60gal
t = 60gal/(3.2gal/s) = 18.75s
In 18.75 seconds both pools will have the same amount of water.