Ask question

Ask question

All categories

3 years ago

5

The number of bacteria in a Petri dish increases by 18% every hour. If there were initially 200 bacteria placed in the Petri dis

h, which function can be used to determine the number of bacteria in the Petri dish in exactly x hours?

1 answer:

In-s [12.5K]3 years ago

5 0

Answer:

Step-by-step explanation:

an exponential equation has the form y=ar^x where y=final amount, a=initial amount, r=rate, and x=time, in this case

y=200(1.18)^x

You might be interested in

How many more marmosets are there than mandrills?

LekaFEV [45]

Answer:

3

Step-by-step explanation:

5 0

2 years ago

Our scale is that 100 mm on the map is equal to 6371 km in Earth as our unknown conversion factor is to 1km.

fiasKO [112]

Answer:

The distance from sea level in mm is 0.5494 mm.

Step-by-step explanation:

Consider the provided information.

It is given that Our scale is that 100 mm on the map is equal to 6371 km in Earth.

This can be written as:

100 mm = 6371 km

Divide both the sides by 6371.

100/6371 mm = 1 km

We need to find the distance from sea level in mm, if the distance is 35 km.

Multiply both sides by 35.

100/6371 × 35 mm = 1 × 35 km

0.5494 mm = 35 km

Hence, the distance from sea level in mm is 0.5494 mm.

7 0

3 years ago

In the triangle below, what is the measure of angle R?

Dmitriy789 [7]

C. 30 degrees because it mirrors the angle of P.

3 0

2 years ago

Read 2 more answers

Solve for x. 0=65892.56*x^(-1.67)

scoundrel [369]

$0=65892.56\cdot x^{-1.67}$

It would seem the only solution is 0, but $x^{-1.67}=\frac{1}{x^{1.67}}$ and the denominator of the fraction can't be equal zero, therefore there's no solution.

3 0

2 years ago

Uhm mainly 11 and 12 but if u solve all of them I’ll give u brainliest

Dvinal [7]

Question 11)

Answer:

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80$

Step-by-step explanation:

Given the expression

$\log _2\left(63\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(\frac{63}{9}\right)$

$=\log _2\left(\frac{63}{9}\right)$

$\mathrm{Divide\:the\:numbers:}\:\frac{63}{9}=7$

$=\log _2\left(7\right)$

$=2.80$

Therefore,

$\log _2\left(63\right)-\log _2\left(9\right)=\log _2\left(7\right)=2.80$

Question 12)

Answer:

$\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32$

Step-by-step explanation:

Given the expression

$\log _2\left(3\right)+\log _2\left(15\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)+\log _c\left(b\right)=\log _c\left(ab\right)$

$\log _2\left(3\right)+\log _2\left(15\right)=\log _2\left(3\cdot \:15\right)$

$=\log _2\left(3\cdot \:15\right)-\log _2\left(9\right)$

$\mathrm{Multiply\:the\:numbers:}\:3\cdot \:15=45$

$=\log _2\left(45\right)-\log _2\left(9\right)$

$\mathrm{Apply\:log\:rule}:\quad \log _c\left(a\right)-\log _c\left(b\right)=\log _c\left(\frac{a}{b}\right)$

$\log _2\left(45\right)-\log _2\left(9\right)=\log _2\left(\frac{45}{9}\right)$

$=\log _2\left(\frac{45}{9}\right)$

$\mathrm{Divide\:the\:numbers:}\:\frac{45}{9}=5$

$=\log _2\left(5\right)$

$=2.32$

Therefore,

$\log \:_2\left(3\right)+\log \:_2\left(15\right)-\log \:_2\left(9\right)=\log \:_2\left(5\right)=2.32$

3 0

2 years ago