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3 years ago

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1 answer:

3241004551 [841]3 years ago

3 0

The correct answer to your question is d

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Solve for x<br> 3^x+4=9<br> 3 to the power of x plus 4 is equal to 9

Bingel [31]

$3^{x} + 4 = 9 \\3^{x} = 5 \\ln(3^{x}) = ln(5) \\xln(3) = ln(5) \\\frac{xln(3)}{ln(3)} = \frac{ln(5)}{ln(3)} \\x = \frac{ln(5)}{ln(3)}$

or

$3^{x + 4} = 9 \\ln(3^{x + 4}) = ln(9) \\(x + 4)ln(3) = ln(9) \\\frac{(x + 4)ln(3)}{ln(3)} = \frac{ln(9)}{ln(3)} \\x + 4 = \frac{ln(3^{2})}{ln(3)} \\x + 4 = \frac{2ln(3)}{ln(3)} \\x + 4 = 2 \\x = -2$

6 0

3 years ago

Read 2 more answers

What two numbers multiply to get -36 and add to get 5

attashe74 [19]

-4 x 9= -36 and the when you add 5 to -36 you get -31. Hope this helps

8 0

2 years ago

-10=-15+5x heeeeeeeeeeeeeeeeeeeeeeelp me

Allisa [31]

The answer is -10=-15+5x x = 1

6 0

3 years ago

You can eliminate one of the variable terms in an ___ by adding or subtracting another equation

MakcuM [25]

Answer:

system of equations

Step-by-step explanation:

You can eliminate one of the variable terms in a <u>system of equations</u> by adding or subtracting another equation.

8 0

2 years ago

A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

GuDViN [60]

Answer:

1. $\frac{dy}{dt}=ky$

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

$\frac{dy}{dt}$ =Number of bacteria per unit time

$\frac{dy}{dt}\proportional y$

$\frac{dy}{dt}=ky$

Where k=Proportionality constant

2. $\frac{dy}{y}=kdt$ ,y'(0)=100

Integrating on both sides then, we get

$lny=kt+C$

We have y(0)=200

Substitute the values then , we get

$ln 200=k(0)+C$

$C=ln 200$

Substitute the value of C then we get

$ln y=kt+ln 200$

$ln y-ln200=kt$

$ln\frac{y}{200}=kt$

$\frac{y}{200}=e^{kt}$

$y=200e^{kt}$

Differentiate w.r.t

$y'=200ke^{kt}$

Substitute the given condition then, we get

$100=200ke^{0}=200 \;because \;e^0=1$

$k=\frac{100}{200}=\frac{1}{2}$

$y=200e^{\frac{t}{2}}$

Substitute t=2

Then, we get $y=200e^{\frac{2}{2}}=200e$

$y=200(2.718)=543.6=543.6$

e=2.718

Hence, the number of bacteria after 2 hours=543.6

4 0

3 years ago