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noname [10]
3 years ago
10

What is the common difference of the arithmetic sequence 1/5, 6/5 , 11/5 ,16/5 , ...?

Mathematics
2 answers:
stepladder [879]3 years ago
8 0
+ 5/5 each time

hope this helps
snow_lady [41]3 years ago
8 0
The common difference is 1 because if you simplify the improper fractions, each hole number is 1 greater than the one before. So the common difference is 1 or 5/5
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The apple weighs 100 grams 
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3 years ago
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F(x) = 3x + 6 please help quickly!!!!!!!
Georgia [21]

Given the function, <em>f(x) = 3x + 6,</em> 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}

<em><u>Given:</u></em>

  • f(x) = 3x + 6

<em>We are told to find:</em>

  1. f(a)
  2. f(a + h), and
  3. \frac{(f(a + h) - f(a)) }{h}

1. <em><u>Find f(a):</u></em>

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

f(a) = 3(a) + 6

f(a) = 3a + 6

<em>2. Find f(a + h):</em>

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

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

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

<em>3. Find </em>\frac{(f(a + h) - f(a)) }{h}<em>:</em>

  • 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, <em>f(x) = 3x + 6,</em> 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}

Learn more here:

brainly.com/question/8161429

6 0
2 years ago
Please help solve for angles Q X and Z
Iteru [2.4K]

Answer:

125

Step-by-step explanation:

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60 + 70 + 40 + 65 + x = 360

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6 0
3 years ago
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Darina [25.2K]

Answer:

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Step-by-step explanation:

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We know that

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Using the property then, we get

\frac{y-x}{xy}\times \frac{xy}{x+y}

\frac{y-x}{x+y}

This is required expression which is equivalent to given expression.

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