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2 years ago

How to solve 5/a+1/2=6/a

Mathematics

2 answers:

swat322 years ago

6 0

Answer:

a = 2

Step-by-step explanation:

Multiply equation by 2a to get rid of denominators.

10 + a = 12

a = 2

Illusion [34]2 years ago

5 0

A=2 is what the answer would be

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I need the slope of each line.. need the answer

s344n2d4d5 [400]

2 over -1 im pretty sure that's it we just learned this like 2 weeks ago

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2 years ago

One of your friends gives you $10 for a charity walkathon. Another friend gives you an amount per mile. After 5 miles, you have

irinina [24]

One of your friends gives you $10 for a charity walkathon -- this is your beginning amount.
Another friend gives you an amount per mile -- you will to calculate this value.
After 5 miles, you have raised $13.50 total --
-- Since one friend gave you $10.00, subtract this from the $13.50 to give you the amount by walking the 5 miles
-- $10.00 - $13.50 = $3.50
-- Since you walked 5 miles, divide $3.50 by 5 to give you $0.70 per mile earned.

So, an equation that represents the amount of money earned is: y = $0.70x+$10.00.
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3 years ago

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ser-zykov [4K]

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3 years ago

Factorise: 6ax+2abx+3ay+by

Slav-nsk [51]

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Step-by-step explanation:

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2 years ago

Stephon makes the following statements: statement 1 lim x -3 f(x) exists and is equal to 1 statement 2 lim x 1 f(x) exists and i

Mandarinka [93]

Neither statement 1, nor statement 2 are correct

The given Stephon's statements are;

Statement 1; $\mathbf{\lim \limits _{x \to -3} f(x)} \ \mathbf{Exist} \ and \mathbf{\lim \limits _{x \to -3} f(x) = 1}$

Statement 2; $\mathbf{\lim \limits _{x \to 1} f(x)} \ \mathbf{Exist} \ and \mathbf{\lim \limits _{x \to 1} f(x) = 1}$

The analysis of the graph and reason for the answer

From the graphed line on the left of the y-axis, we have an open circle at x = -3, and an arrow at the other end pointing towards negative infinity, (-∞) which indicates that the domain is -∞ ≤ x < -3, therefore, at x = -3, f(x) does not exist, therefore, we can write the following statement

The limits of the domain and range of the graph includes;

$\mathbf{\lim \limits _{x \to -3} f(x)} = \mathbf{Does \ not \ exist}$

f(x) = Defined for -∞ ≤ x < -3

Similarly, from the graphed line on the right of the y-axis, we have an open circle at x = 1 and an arrow at the other end of the line f(x) = 4 pointing towards positive (+∞) infinity, which indicates the domain and the graph of the function is 1 < x ≤ ∞ , therefore, f(x) does not exist at x = 1, and we can write

$\mathbf{\lim \limits _{x \to 1} f(x)} = \mathbf{Does \ not \ exist}$

From we above, we have that neither statement 1, nor statement 2 are correct

Learn more about open and closed circles on graph lines

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2 years ago