Serenity invested $2,400 in an account paying an interest rate of 3.4% compounded continuously. Assuming no deposits or withdrawals are made, how long would it
take, to the nearest tenth of a year, for the value of the account to reach $2,930?

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

∵ 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

£56

step-by-step explanation:

u divide 24 by 3 and u get 8 so if u do 8 time 7 or 8+8+8+8+8+8+8 u get 56 and 56 divided by 7 is 8.

c. y = 1/4 x + 5

step-by-step explanation:

first, put the equation of the line on the graph into slope-intercept form

(y = mx + b):   y = -4x + 8

second, a perpendicular line has to have "an opposite/reciprocal slope" to the original slope. so, find the "m" value in your original equation (it's -4, right? )

original slope: m = -4   perpendicular slope: change the sign to its opposite and find the reciprocal ("flip" the number)

new slope is now positive (b/c the original slope was negative 4), and it's 1/4 (not the original slope of 4 or 4/1)