An amount was invested at r% per quarter. What value of r will ensure that accumulated amount at the end of one year is 1.5 times more than amount invested? Correct to 2 decimal places

25.75 %  interest rate

Step-by-step explanation:

Given:

Amount was invested = r% per quarter  

Amount invested = P

Rate of interest = r %  per quarter

Time (n) = 4  Quarters

Computation:

A = P(1 + r/100)ⁿ

According to question.

⇒ A = P + 1.5P  = 2.5P

⇒ 2.5P = P(1 + r/100)⁴

⇒ 2.5  = (1  + r/100)⁴

⇒ 1 + r/100  =  1.2575

⇒ r/100 = 0.2575

⇒ r = 25.75

25.75 %  interest rate

25 questions , because they start with 100.  

350-100=250 .  

. 250/10=25

step-by-step explanation:

first take off 100, then what x 10 = 250? that is your answer. (25)

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)