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Paladinen [302]

2 years ago

5

9. P(no more than 16 | prime

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

Read 2 more answers

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

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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