How many times greater is the value of the 2 in 204,936 than the value of the 2 in 124,936

Mathematics

2 answers:

iragen [17]2 years ago

8 0

The value of the 2 in 204,936 is 10 times greater than the value of the 2 in 124,936. So the answer is C

Sliva [168]2 years ago

5 0

I believe it would be C.10

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F(x) = 3x + 6 please help quickly!!!!!!!

Georgia [21]

Given the function, f(x) = 3x + 6, we can solve for f(a), f(a + h) and $\frac{(f(a + h) - f(a)) }{h}$ by substituting their values into f(x) = 3x + 6. We will have the following:

$\mathbf{f(a) = 3a + 6}\\\\\mathbf{f(a + h) = 3a + 3h + 6}\\\\\mathbf{\frac{(f(a + h) - f(a)) }{h} = 6}$

Given:

f(x) = 3x + 6

We are told to find:

f(a)
f(a + h), and
$\frac{(f(a + h) - f(a)) }{h}$

1. Find f(a):

Substitute x = a into f(x) = 3x + 6

f(a) = 3(a) + 6

f(a) = 3a + 6

2. Find f(a + h):

Substitute x = a + h into f(x) = 3x + 6

f(a + h) = 3(a + h) + 6

f(a + h) = 3a + 3h + 6

3. Find $\frac{(f(a + h) - f(a)) }{h}$ :

Plug in the values of f(a + h) and f(a) into $\frac{(f(a + h) - f(a)) }{h}$

Thus:

$\frac{((3a + 3h + 6) - (3a + 6)) }{h}\\\\\frac{(3a + 3h + 6 - 3a - 6) }{h}\\\\$

Add like terms

$\frac{(3a - 3a + 3h + 6 - 6) }{h}\\\\= \frac{3h }{h}\\\\\mathbf{= 3}$

Therefore, given the function, f(x) = 3x + 6, we can solve for f(a), f(a + h) and $\frac{(f(a + h) - f(a)) }{h}$ by substituting their values into f(x) = 3x + 6. We will have the following:

$\mathbf{f(a) = 3a + 6}\\\\\mathbf{f(a + h) = 3a + 3h + 6}\\\\\mathbf{\frac{(f(a + h) - f(a)) }{h} = 6}$

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