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3 years ago

What is the range of f(x) = 3x + 9?

Mathematics

2 answers:

maw [93]3 years ago

7 0

We are given function f(x) = 3x+9.

We need to find the range of the given function.

We are given options

A) {y | y < 9}

B) {y | y > 9}

C) {y | y > 3}

D) {y | y < 3}

From the given options we can think that x is in the exponent of 3.

So, the given function is actually f(x) = 3^x +9

Now, we need to find the range of given function.

We can see that first term is and exponential term 3^x and second term is 9.

We know that 3^x will always be greater than 0.

<em>Therefore, 3^x +9 would always be greater than 9.</em>

<h3>Therefore, range would be y>9.</h3><h3>So, the correct option is B) {y | y > 9}.</h3>

aivan3 [116]3 years ago

4 0

The answer is b broskie

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I'm studying a new bacteria that doubles in numbers every 25 minutes. If i start with 50 bacteria, how long until i have 5 milli

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415.63 minutes

Step-by-step explanation:

Growth can be represented by the equation $A=A_0e^{rt}$ . We can find the rate at which it grows by using t=25 minutes and $\frac{A_{0}}{A} =2$ or double the amount at that time. The first step we always take is to divide $A_0$ by A.

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To solve for r, we will take the natural log of both sides and use log rules to isolate r.

$ln 2=ln e^{(25)r}\\ln 2=25r (ln e)\\\frac{ln2}{25} =r$

We know $lne=1$ so we were able to cancel it out and divide both sides by 25.

We solve with a calculator $\frac{ln2}{25} =r\\0.0277=r$

We change 0.0277 into a percent by multiplying by 100 to get 2.77% as the rate.

The equation is $A=A_0e^{0.0277t}$ .

We repeat the step above substituting A=5,000,000, $A_0$ =50, and r=0.02777. Then solve for t.

$5000000=50e^{0.0277t}\\\frac{5000000}{50} =e^{0.0277t}\\100000=e^{0.0277t}\\ln100000=lne^{0.0277t}\\ln100000=0.02777t(lne)\\\frac{ln100000}{0.0277} =t$