Y=a(1+r)^t 135272-Y=a(1+r)^t
Y = a(1 r)^t Example of Exponential Growth Equation y = 2500(105)^4 Example of Exponential Decay Equation y = 1(085)^6 Exponential Growth Graph Exponential Decay Graph Initial Value (yintercept) "a" in the function y = ab^x Compound Interest Formula A = P(1 r/n)^nt HalfLife Formula A = P(05)^t YOU MIGHT ALSO LIKEThe general equation for depreciation is given by y = a (1 – r)t, where y = current value, a = original cost, r = rate of depreciation, and t = time, in years the original value of a car is $24,000 it depreciates 15% annually what is its value in 4 years?I think you use the following formula y=a(1r)^t where y is the amount after t years, a is the initial amount, r is the annual growth rate, and t is the time in years I will appreciate everyones
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Y=a(1+r)^t
Y=a(1+r)^t-Solve for t A=P(1r/n)^(nt) Rewrite the equation as Divide each term by and simplify Tap for more steps Divide each term in by Cancel the common factor of Tap for more steps Cancel the common factor Divide by Take the natural logarithm of both sides of the equation to remove the variable from the exponentExponential decay equation #1 – y = a (1 – r) t y = what's leftover a = what you start with r = rate t = time ex Timmy drank hot chocolate which has 110 milligrams of sugar If the sugar was eliminated from the body at a rate of 12% per hour
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Remaining grams a = 16 Starting value r = 50% = 05 Decimal form b = 1 05 Decay Factor x = 500 5730 No of Half livesOf the polynomial are r = −1 and −4 The general solution is then y = C1 e −t C 2 e −4t Suppose there are initial conditions y(0) = 1, y′(0) = −7 A unique particular solution can be found by solving for C1 and C2 using the initial conditions First we need to calculate y′ = −C1 e −t − 4 C 2 e −4 t, then apply theThe general equation for depreciation is given by {eq}\displaystyle{ y = A(1 r)t }{/eq}, where y = current value, A = original cost, r = rate of depreciation, and t = time, in years A car was
Answer 3 📌📌📌 question Rewrite the function in the form y=a(1r)^t of y=a(1r)^t Then state the growth of or decay rate Y=a(8)^t/2 the answers to estudyassistantcomAn equation for the depreciation of a car is given by y = A(1 – r)t, where y = current value of the car, A = original cost, r = rate of depreciation, and t = time, in years The current value of a car is $12,250 The car originally cost $,000 and depreciates at a rate of 15% per year How old is the car?Hence, r = 1 = , or, rounding the value of "r" to the nearest tenthousandth, r = 0630 Therefore, your answer is y = 5*()^t It is EXACTLY the form you requested, with the value of r = 0630, rounded as it is assigned by the problem
Rewrite y=(14)^t8 in the form y=a(1r)^t I keep getting the wrong answer not sure if it goes to 4%?Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreAn equation for the depreciation of a car is given by y = A (1 – r)t, where y = current value of the car, A = original cost, r = rate of depreciation, and t = time, in years The value of a car is half what it originally cost The rate of depreciation is 10% Approximately how old is the car?
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The general equation for depreciation is given by y = A(1 r)t, where y = current value, A = original cost, r = rate of depreciation, and t = time, in years The original value of a car is $24,000 It depreciates 15% annually What is its value in 4 years?Other time period) The amount y of such a quantity after t years can be modeled by one of these equations Exponential Growth Model Exponential Decay Model y = a(1 r)t y = a(1 − r)t Note that a is the initial amount and r is the percent increase or decrease written as a decimal The quantity 1 r is the growth factor, and 1 − r is theX(t) = x 0 × (1 r) t x(t) is the value at time t x 0 is the initial value at time t=0 r is the growth rate when r>0 or decay rate when r
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👍 Correct answer to the question The general equation for depreciation is given by y = a(1 – r)t, where y = current value, a = original cost, r = rate of depreciation, and t = time, in years the original value of a car is $24,000 it deprecia eeduanswerscom(1 r ) is the growth factor, r is the growth rate The percent of increase is 100 r y = C (1 r ) t 21 E XPONENTIAL D ECAY M ODEL C is the initial amount t is the time period (1 – r ) is the decay factor, r is the decay rate The percent of decrease is 100 r y = C (1 – r ) tQuestion An equation for the depreciation of a car is given by y = A(1 – r)t , where y = current value of the car, A = original cost, r = rate of depreciation, and t = time, in years The value of a car is half what it originally cost The rate of depreciation is 10% Approximately how old is the car?
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Solve for t A=P(1r/n)^(nt) Rewrite the equation as Divide each term by and simplify Tap for more steps Divide each term in by Cancel the common factor of Tap for more steps Cancel the common factor Divide by Take the natural logarithm of both sides of the equation to remove the variable from the exponentX(t) = x 0 × (1 r) t x(t) is the value at time t x 0 is the initial value at time t=0 r is the growth rate when r>0 or decay rate when rWarmup y = a(1r)t 10) 00(1 003)1 = $60 11) 0(1 003)10 = $ 12) 600(1 007)4 =$ 13) 1500(1 004)8 = $5285
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Rewrite y=a (2)^1/3 in the form y=a (1r)^t or y=a (1r)^t then state the growth or decay rate Mathematics, 0500 NewKidnewlessonsThe decay factor is b = 1 r In this situation x is the number of halflives If one halflife is 5730 years then the number of halflives after 500 years is x = 500 5730 y = ?Using the growth formula we have y = a(1 r) x where a = 1 (we start with 1 bacteria), and r = 100%, since the amount doubles y = 1(1 100) x = 2 x (same result) Notice that the graph is a scatter plot You cannot have a fractional part of a bacteria The dotted line is the exponential function which contains the scatter plots (the model)
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