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algol13
3 years ago
10

1. Based on the construction below, which conclusion

Mathematics
1 answer:
trapecia [35]3 years ago
6 0

Answer: B ,I just completed the assignment

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Which statement describes the end behavior of the exponential function f(x) = 2x – 3?
charle [14.2K]
The <u>correct answer</u> is:

As x→-∞, y→-3.
As x→∞, y→∞.

Explanation:

As our values of x get further into the negative numbers, the value of 2ˣ will approach 0.  This is because raising a number to a negative exponent "flips" the number below the denominator and raises it to a power; we end up with smaller and smaller fractions, eventually so small that they nearly equal 0.  

This will make the value of the function 0-3=-3.

As x gets larger and larger (towards ∞), the value of y, 2ˣ, continues to grow as well.  Since it continues to grow exponentially, we say the value approaches ∞.
3 0
2 years ago
Read 2 more answers
Determining When a Function is increasing or Decreasing
Juliette [100K]

Answer:

B

Step-by-step explanation:

As x tends to infinity and - infinity, the function will tend to +infinity for both

7 0
3 years ago
Determine the value of x?
Alexxx [7]

hope it will help you..........

6 0
2 years ago
Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt
ser-zykov [4K]

Answer:

I(t)=\frac{1}{3}(1-e^{-30t})

Step-by-step explanation:

We are given that

\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}

R=150 ohm

L=5 H

V(t)=10 V

P=\frac{R}{L}=\frac{150}{5}=30

I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}

I(t)\times I.F=\int e^{30 t}\times 10 dt+C

I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C

I(t)=\frac{1}{3}+Ce^{-30 t}

I(0)=0

Substitute t=0

0=\frac{1}{3}+C

C=-\frac{1}{3}

Substitute the values

I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}

I(t)=\frac{1}{3}(1-e^{-30t})

5 0
3 years ago
I need help please...​
Ostrovityanka [42]

Answer:

B

Step-by-step explanation:

I think it's hitting a golf ball in the air, as it is a transformation, but the general shape of it is not affected.

Opening a locker, you are still affecting the general shape.

6 0
3 years ago
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