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Paladinen [302]
2 years ago
5

9. P(no more than 16 | prime

Mathematics
1 answer:
koban [17]2 years ago
7 0
17 prom is the answer
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The inequality x S-8 would represent which of the following values?
irinina [24]
S refers to side so it’s 64 = 8x8
8 0
3 years ago
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A 48 gram sample of a substance that’s used for drug research has a k value of 0.1325
Lyrx [107]

Using an exponential function, it is found that the half-life of the substance is of 5.23 units of time.

<h3>What is the exponential function for the amount of a substance?</h3>

The function is given by:

A(t) = A(0)e^{-kt}

In which:

  • A(0) is the initial value.
  • k is the exponential decay rate.

The parameters are given by:

A(0) = 48,  k = 0.1325.

Hence the equation is:

A(t) = 48e^{-0.1325t}

The half-life is the amount of time for which A(t) = 0.5A(0) = 24, hence:

A(t) = 48e^{-0.1325t}

24 = 48e^{-0.1325t}

e^{-0.1325t} = 0.5

\ln{e^{-0.1325t}} = \ln{0.5}

0.1325t = -\ln{0.5}

t = -\frac{\ln{0.5}}{0.1325}

t = 5.23.

More can be learned about exponential functions at brainly.com/question/25537936

#SPJ1

3 0
2 years ago
the piece wise defined function f(x) is graphed below. What is the value of f(2)? PLS HELP THIS IS TIMED
Margarita [4]

Answer:

B

Step-by-step explanation:

We have to look at the full dot

5 0
3 years ago
Please help :3 I really need help with this
Gelneren [198K]

Answer:

0

Step-by-step explanation:

It would not change at all. Mode is the most often occuring number, and adding one 9 will not change the fact that there are threes 6s. Therefore, the mode will stay 6

6 0
3 years ago
Find an explicit solution to the Bernoulli equation. y'-1/3 y = 1/3 xe^xln(x)y^-2
NNADVOKAT [17]

y'-\dfrac13y=\dfrac13xe^x\ln x\,y^{-2}

Divide both sides by \dfrac13y^{-2}(x):

3y^2y'-y^3=xe^x\ln x

Substitute v(x)=y(x)^3, so that v'(x)=3y(x)^2y'(x).

v'-v=xe^x\ln x

Multiply both sides by e^{-x}:

e^{-x}v'-e^{-x}v=x\ln x

The left side can be condensed into the derivative of a product.

(e^{-x}v)'=x\ln x

Integrate both sides to get

e^{-x}v=\dfrac12x^2\ln x-\dfrac14x^2+C

Solve for v(x):

v=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

Solve for y(x):

y^3=\dfrac12x^2e^x\ln x-\dfrac14x^2e^x+Ce^x

\implies\boxed{y(x)=\sqrt[3]{\dfrac14x^2e^x(2\ln x-1)+Ce^x}}

4 0
3 years ago
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