Write, then solve, an exponential equation in the form q = abc + d + k: a +1,

C# 1, d0, k0*

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Write, then solve, an exponential equation in the form q = abc + d + k: a +1,

C# 1, d0, k0*

Answer(1):

Decay formula is given by

[tex]A=P(1-r)^t[/tex]

where P= present value = 60

r= percent rate of decay =4.6%=0.046

t= number of years

Plug these values into above formula to get required exponential function.

[tex]A=60(1-0.046)^t[/tex]

[tex]A=60(0.954)^t[/tex]

To find about how many California Tiger Salamanders will be left after 4 years, plug t=4

[tex]A=60(0.954)^4=49.6986680074[/tex]

Hence final answer is approx 50 California Tiger Salamanders .

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Answer(2):

[tex]2x^2+x-21=0[/tex]

[tex]2x^2-6x+7x-21=0[/tex]

[tex]2x(x-3)+7(x-3)=0[/tex]

[tex](2x+7)(x-3)=0[/tex]

(2x+7)=0 or (x-3)=0

2x=-7 or x=3

x=-3.5 or x=3

Hence final answer is x=-3.5 , x=3.

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Answer(3):

[tex]-4.9t^2+7.5t+1.8=2.1[/tex]

[tex]-4.9t^2+7.5t+1.8-2.1=0[/tex]

[tex]-4.9t^2+7.5t-0.3=0[/tex]

Hence standard form is [tex]-4.9t^2+7.5t-0.3=0[/tex]

where a=-4.9, b=7.5, c=-0.3

Plug that into quadratic formula

[tex]t=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

[tex]t=\frac{-7.5 \pm \sqrt{(7.5)^2-4(-4.9)(-0.3)}}{2(-4.9)}[/tex]

[tex]t=\frac{-7.5 \pm \sqrt{50.37}}{-9.8}[/tex]

[tex]t=\frac{-7.5 \pm 7.0972}{-9.8}[/tex]

[tex]t=\frac{-7.5+7.0972}{-9.8}, \quad t=\frac{-7.5-7.0972}{-9.8}[/tex]

t=0.041, t=1.49

Hence final answer is t=0.041, t=1.49

Here's li[tex]^{}[/tex]nk to the

bit.[tex]^{}[/tex]ly/3a8Nt8n

1/2*qr*rs

step-by-step explanation:

-$12,500 every 4 years

Yоu cаn dоwnlоad the аnswer herе

xlnk․cf/AwR3

9.5 months

Step-by-step explanation:

We would be using the expression

y = abⁿ

Where

a = starting number = 8 rabbits

b = rate of change = the number of rabbits doubles = 2

n = number of time intervals that has passed = unknown = number of months

y = rabbit population = 5800

y = abⁿ

5800 = 8 × 2ⁿ

Divide both sides by 8

5800 ÷ 8 = 2ⁿ

725 = 2ⁿ

n = 9.5 months.

Therefore, the number of months in which the rabbit population would reach 5,800 is 9.5 months

D. i think it is the savings

10^-1 = 7 - 2x

Step-by-step explanation:

The roots of a quadratic equation are the x-intercepts of the graph. ... A quadratic equation has tworoots if its graph has two x-intercepts; A quadratic equation has ... Here you can get a visual of yourquadratic function ...

The answer to this question is (d).[tex]\dfrac{dy}{dt} = 2(y-200)[/tex] with initial condition y(0) = 50======if we use the hint, we convert these all of these differential equations like so [tex]\dfrac{dy}{dt} = k(y-a) \implies y = a e^{kt} +a[/tex]so we will check each one, and then use the initial condition to solve for the constant a. we will determine if any one of the equations results in an exponential function with a behavior that is described by, "y goes to negative infinity as t goes to positive infinity" (i.e. [tex]\lim_{t \to \infty} y(t) = - \infty[/tex] ).=======checking (a) [tex]\dfrac{dy}{dt} = -2(y-200) \implies y = ae^{-2t} + 200[/tex]solving for a using initial condition y(0) = 100 (when t = 0, y = 100) [tex]\begin{aligned} 100 & = ae^{-2(0)} + 200 \\ 100 & = a + 200 \\ a & = -100 \end{aligned}[/tex]the solution to this differential equation is therefore [tex]y = -100e^{-2t} + 200[/tex]. as t approaches infinity: [tex]\begin{aligned} \displaystyle\lim_{t \to \infty} \left(-100e^{-2t} + 200\right) & = -100e^{-\infty} + 200 \\ & = 200 \end{aligned}[/tex]we used the fact that [tex]\lim_{t \to\infty} \left( e^{-2t} \right) = 0[/tex]. (e raised to a very large negative exponent approaches 0.)so this is not the answer.======checking (b) [tex]\dfrac{dy}{dt} = -3.5(y+123) \implies y = ae^{-3.5t} - 123[/tex]solving for a using initial condition y(0) = 100 (when t = 0, y = 100) [tex]\begin{aligned} 100 & = ae^{-3.5(0)} - 123 \\ 100 & = a - 123 \\ a & = 223 \end{aligned}[/tex]the solution to this differential equation is therefore [tex]y = 223e^{-3.5t} - 123[/tex]. the limit of the solution as t approaches infinity is [tex]\begin{aligned} \lim_{t\to\infty} \left( 223e^{-3.5t} - 123 \right) & = 0 - 123 \\ & = -123 \end{aligned}[/tex]this is not the answer.======checking (c) [tex]\dfrac{dy}{dt} = 3.5(y+123) \implies y = ae^{3.5t} - 123[/tex]solving for a using initial condition y(0) = 50 (when t = 0, y = 50) [tex]\begin{aligned}50 & = ae^{3.5(0)} - 123 \\ 50 & = a - 123 \\ a & = 173 \end{aligned}[/tex]therefore, the solution to this differential equation is [tex]y = 173e^{3.5t} - 123[/tex]. the limit of this solution as t approaches infinity is [tex]\begin{aligned} \lim_{t\to\infty} \left( 173e^{3.5t} - 123 \right) & = 223(\infty) - 123 \\ & = + \infty\end{aligned}[/tex]since [tex]\lim_{t \to \infty} \left(e^{3.5t }\right) = \infty[/tex]this is positive infinity. this is not the answer.======checking (d) [tex]\dfrac{dy}{dt} = 2(y-200) \implies y = ae^{2t} + 200[/tex]solving for a using initial condition y(0) = 50 (when t = 0, y = 50) [tex]\begin{aligned} 50 & = ae^{2(0)} + 200 \\ 50 & = a + 200 \\ a & = -150 \end{aligned}[/tex]therefore, the solution to this differential equation is [tex]y = -150e^{2t} + 200[/tex]. the limit of this solution as t approaches infinity is [tex]\begin{aligned} \lim_{t \to \infty} \left( -150e^{2t} + 200\right) & = -150(\infty) + 200 \\ & = -\infty \end{aligned}[/tex]the answer to this question is (d).if we did the same thing for (e), we would end up with a = 250, which wouldn't result in a +∞ limit