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3 years ago

Solve 2/3x-1/5>1 Need help asap

Mathematics

2 answers:

Ksenya-84 [330]3 years ago

4 0

Answer:

Step-by-step explanation:

Step 1: Add 1/5 to both sides

$\frac {2}{3}x - \frac {1}{5} + \frac {1}{5} > 1 + \frac {1}{5}$

$\frac {2}{3}x > 1 + \frac {1}{5}$

Step 2: Combine on the right side

$\frac {2}{3}x > 1 + \frac {1}{5}$

$\frac {2}{3}x > \frac {6}{5}$

Step 3: Multiply both sides by 3/2

$\frac {2}{3}x * \frac {3}{2} > \frac {6}{5} * \frac {3}{2}$

$x > \frac {18}{10}$

Step 4: Simplify

$x > \frac {18/2}{10/2}$

$x > \frac{9}{5}$

Answer: $x > \frac {9}{5}$

nalin [4]3 years ago

3 0

Answer:

x > 9/5

Step-by-step explanation:

might not be correct but ur welcome for the help

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Matt Jones' gross pay for this week is $630.00. He is single and claims 2 allowances. Using the percentage method of withholding

atroni [7]

Answer:

Step-by-step explanation:

I will give you the answer in the following steps. In addition to solving this exercise it is necessary the table found in the attachments -And that was probably in the book where this exercise is taken-.

Here we have:

Percentage Method of Withholding –Single Person

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Step 1:

= ∙

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Step 2:

=−

=$560.00- $190.38

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Step 3: (Here we refer to the table, as the income is less than $ 645, and more than $ 195, the tax is 14.40 plus 15% for being the income above $ 195)

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3 years ago

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beks73 [17]

Answer:

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Step-by-step explanation:

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3 years ago

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Determine whether or not the given set is open, connected, and simply-connected. (Select all that apply.) (x, y) | 16 ≤ x2 + y2

Yuliya22 [10]

Answer:

The set is closed, connected and simyple connected

Step-by-step explanation:

A set is closed if contains all the point in its boundaries. A set is open if it doesn't contain any of the points in its boundaries. In this set, all the points of the boundaries are included because it is using the less than or equal to and greater than or equal to define the set.

The set is connected if you can find a path inside the set to connect any two points of the set. If you make the graph of the set you would see the set covers this condition because the set hasn't any division.

The set is simply connected if you can draw a closed curve inside the set and in the interior of the curve there are only points of the set. In other words, if the set has holes is not simply connected. This set doesn't have holes, it's simply connected.