All categories

3 years ago

What Is The Prime Factorization of 212.

1 answer:

6 0

212
/ \
2 106
/ \
2 53

2800
/ \
2 1400
/ \
2 700
/ \
2 350
/ \
2 175
/ \
5 35
/ \
5 7
/ \
7 1

900
/ \
2 450
/ \
2 225
/ \
3 75
/ \
3 25
/ \
5 5
/ \
5 1

This might help if you need it again

You might be interested in

The average discharge at the mouth of the Amazon River is 4,900,000 cubic feet per second. How much water is discharged from the

Misha Larkins [42]

Answer: Average discharge at the mouth of the Amazon River in 1 hour = $1.764\times 10^{10}$ cubic feet

Average discharge at the mouth of the Amazon River in 1 year= $3.66912\times10^{13}$ cubic feet[/tex]

Step-by-step explanation:

Given : The average discharge at the mouth of the Amazon River =4,900,000 cubic feet per second.

i.e. The average discharge at the mouth of the Amazon River in 1 second= 4,900,000 cubic feet

Since 1 hour = 3600 seconds

Then , the average discharge at the mouth of the Amazon River in 1 hour =

$4900000\times3600=17640000000$ cubic feet

Since , there are 2080 hours in a typical year.

Then, the average discharge at the mouth of the Amazon River in 1 year=

$2080\times17640000000\\\\=36,691,200,000,000=3.66912\times10^{13}$ cubic feet

6 0

2 years ago

Use the function below to find f(1). f9t0=4

ExtremeBDS [4]

The value of the F(1) is 1/2 option (C) is correct after plugging the value of t = 1 in the provided function.

<h3>What is a function?</h3>

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

The question is incomplete.

The complete question is attached in the picture please refer to the picture.

We have a function:

$\rm F(t) = 4\dfrac{1}{2^{3t}}$

Plug t = 1

$\rm F(1) = 4\dfrac{1}{2^{3\times1}}$

After solving:

F(1) = 1/2

Thus, the value of the F(1) is 1/2 option (C) is correct after plugging the value of t = 1 in the provided function.

Learn more about the function here:

brainly.com/question/5245372

#SPJ1

6 0

1 year ago

If f and g are differentiable functions for all real values of x such that f(1) = 4, g(1) = 3, f '(3) = −5, f '(1) = −4, g '(1)

belka [17]

Answer:

h'(1)=0

Step-by-step explanation:

We use the definition of the derivative of a quotient:

If $h(x)=\frac{f(x)}{g(x)}$ , then:

$h'(x)=\frac{f'(x)*g(x)-f(x)*g'(x)}{(g(x))^2}$

Since in our case we want the derivative of $h(x)$ at the point x = 1, which is indicated by: h'(1), we need to evaluate the previous expression at x = 1, that is: