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Elena L [17]
3 years ago
7

What is the value of y in this exponential growth function when x equals 3?

Mathematics
2 answers:
Reika [66]3 years ago
8 0
Y = 2(1.06)^x

y = 2(1.06)^3

y = 2(1.19)

y = 2.38
docker41 [41]3 years ago
7 0
Its hard to do the problem without the function or graph for the function but if you can get that i would love to help.
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Identify the horizontal asymptote of f(x) = quantity 2 x minus 1 over quantity x squared minus 7 x plus 3. y = 0 y = 1 over 2 y
Ratling [72]
The horizontal asymptote is the value at the y-axis where the graph approaches the line but not necessarily touching it. Hence, the asymptotic characteristic of the graph. The standard form of a function in fraction form is y = (ax^n +...)/(bx^m+...). There are rules to follow to determine the horizontal asymptote of a function.
1) if n = m , then the horizontal equation is y = a/b
2) if n>m, then there is no horizontal equation
3) if n<m, then the horizontal equation is the x axis ; y = 0.

The function given falls on the third rule hence the horizontal asymptote of the function is at y = 0.
8 0
3 years ago
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As part of a screening process, computer chips must be operated in an oven at 145 °C. Ten minutes after starting, the temperatur
Novosadov [1.4K]

Answer:

Step-by-step explanation:

I solved this using initial conditions and calculus, so I hope that's what you are doing in math.  It's actually NOT calculus, just a concept that is taught in calculus.

The initial condition formula we need is

y=Ce^{kt}

Filling in our formula with the 2 conditions we are given:

65=Ce^{10k}   and   85=Ce^{15k}

With those 2 equations, we have 2 unknowns, the C (initial value) and the k (the constant). We know that the initial value (or starting temp) for both conditions is the same, so we solve for C in one equation, sub it into the other equation and solve for k.  If

65=Ce^{10k} then

\frac{65}{e^{10k}}=C which, by exponential rules is the same as

C=65e^{-10k}

Since that value of C is the same as the value of C in the other equation, we sub it in:

85=65e^{-10k}(e^{15k})

Divide both sides by 65 and use the rules of exponents again to get

\frac{85}{65}=e^{-10k+15k} which simplifies down to

\frac{85}{65}=e^{5k}

Take the natural log of both sides to get

ln(\frac{85}{65})=5k

Do the log thing on your calculator to get

.2682639866 = 5k and divide both sides by 5 to find k:

k = .0536527973

Now that we have k, we sub THAT value in to one of the original equations to find C:

65=Ce^{10(.0536527973)}

which simplifies down to

65=Ce^{.536527973}

Raise e to that power on your calculator to get

65 = C(1.710059171) and divide to solve for C:

C = 38.01038064

Now sub in k and C to the final problem when t = 23:

y=38.01038064e^{(.0536527973)(23)} which simplifies a bit to

y=38.01038064e^{1.234014338}

Raise e to that power on your calculator to get

y = 38.01038064(3.434991111) and

finally, the temp at 23 minutes is

130.565

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3 years ago
Using the eliminated method joe solved the system of equations below for y using the following steps which statements correctly
elixir [45]

Answer:

c

Step-by-step explanation:

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Answer:

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Consecuative whole numbers lie n the square root of 59
Art [367]

I think it would be 7 and 8. 7 squared is 49, and 8 squared is 64. The closest number is 7 tho

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