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jonny [76]
3 years ago
11

Easy math problems!!!!!

Mathematics
2 answers:
Alexandra [31]3 years ago
7 0

Answer: 36

Step-by-step explanation:

Karolina [17]3 years ago
6 0

Answer:

36

Step-by-step explanation:

isnt this science.

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Hey dhsidvxhejfbeifbshdjdbd
8 0
3 years ago
On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and increases from quadrant 2 into quadrant 1. It
dusya [7]

Answer:

f(x) = 2^{(3x)}

Step-by-step explanation:

The given options are,

a) f(x) = 2^{(3x)}

b) f(x) = ({\frac {1}{2}})^{3x}

c) f(x) = 2 \times ({\frac {1}{3}})^{x}

d) f(x) = \frac {1}{2} \times ({\frac {1}{3}})^{x}

Now, among the given options only option (a) satisfies the criteria stated in the question i. e.,

     1. f(x) → 0 as x → - ∞  and

     2 f(0) = 1

so, the correct answer is,

f(x) = 2^{(3x)}

7 0
3 years ago
Read 2 more answers
Can someone please help me out with this
Sliva [168]

Answer:

hope it helps

4 0
3 years ago
Pleas help me ASAP thank you :)
RoseWind [281]

Answer:

11

Step-by-step explanation:

it says 11

8 0
2 years ago
Identify whether the equation represents an exponential growth or exponential decay function.1. y = 1/4 (1/e)^-2x2. y = (1/e)^4x
zvonat [6]

The general formula for exponential growth and decays is:

y=y_0e^{kx}

if k>0 then then it is an exponential growth function. If k<0 then the function represents an exponential decay.

Now we need to classify each of the functions:

1.

The function

y=\frac{1}{4}(\frac{1}{e})^{-2x}

can be wrtten as:

\begin{gathered} y=\frac{1}{4}(e^{-1})^{-2x}^{} \\ =\frac{1}{4}e^{2x} \end{gathered}

comparing with the general formula we notice that k=2, therefore this is an exponential growth.

2.

The function

y=(\frac{1}{e})^{4x}

can be written as:

\begin{gathered} y=(\frac{1}{e})^{4x} \\ y=(e^{-1})^{4x} \\ y=e^{-4x} \end{gathered}

comparing with the general formula we notice that k=-4, therefore this is an exponential decay.

3.

The function

y=2e^{-x}+1

comparing with the general formula we notice that k=-1, therefore this is an exponential decay.

5 0
1 year ago
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