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Lesechka [4]
3 years ago
11

Which expressions simplify to 0? Check all that apply. 8g – 8g, (–2g) – 2g, 5g + (–5g), One-half g + one-half, g Negative two-th

irds g + two-thirds g
Mathematics
2 answers:
victus00 [196]3 years ago
6 0

We have been given 4 expressions. We are asked to choose the expressions, which simplifies to 0.

Let us check our given choices one by one.

1. 8g-8g

8(g-g)

8(0)=0

When we will subtract 8g from 8g, we will get 0. Therefore, expression 8g-8g simplifies to 0.

2. (-2g)-2g

Let us remove parenthesis.

-2g-2g

Upon combining like terms, we will get:

-4g

Since  (-2g)-2g simplifies to -4g, therefore, expression -2g-2g is not a correct choice.

3. 5g+(-5g)

Let us remove parenthesis.

5g-5g

Upon combining terms, we will get:

5g-5g=0

Since  5g+(-5g) simplifies to 0, therefore, expression  5g+(-5g) is a correct choice.

4. \frac{1}{2}g+\frac{1}{2}g-\frac{2}{3}g+\frac{2}{3}g

Let us combine the terms as:

\frac{1+1}{2}g-\frac{2}{3}g+\frac{2}{3}g

\frac{2}{2}g-\frac{2}{3}g+\frac{2}{3}g

g-g(-\frac{2}{3}+\frac{2}{3})

g-g(0)

g-(0)=g

Since \frac{1}{2}g+\frac{1}{2}g-\frac{2}{3}g+\frac{2}{3}g simplifies to g, therefore, expression \frac{1}{2}g+\frac{1}{2}g-\frac{2}{3}g+\frac{2}{3}g is not a correct choice.

Setler [38]3 years ago
6 0

Answer:

its 1,3,5

1Step-by-step explanation:

trust me

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Answer:

Step-by-step explanation:

Notice that the inputs {5, -8, -1, -5} are all different.  So we can immediately conclude that this is a function.

If the inputs were {5, -8, -1, 5}, this would not be a function, as the same input value (5) would result in two different outputs.

6 0
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An electronics store makes a profit of $72 for every standard CD player sold and $90 for every portable CD sold. the managers ta
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s -> no: of standard CD players sold

p -> no: of portable CD players sold.

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Profit per Portable CD player =90$

Target Amount =360$

<em>I</em><em>n</em><em> </em><em>short</em><em>:</em><em> </em><em>The</em><em> </em><em>combined</em><em> </em><em>sales</em><em> </em><em>of</em><em> </em><em>Both</em><em> </em><em>the</em><em> </em><em>CD</em><em> </em><em>player</em><em>'s</em><em> </em><em>must</em><em> </em><em>exceed</em><em> </em><em>or</em><em> </em><em>atleast</em><em> </em><em>become</em><em> </em><em>equal</em><em> </em><em>to</em><em> </em><em>3</em><em>6</em><em>0</em><em>$</em><em> </em><em>to</em><em> </em><em>a</em><em>c</em><em>q</em><em>u</em><em>i</em><em>r</em><em>e</em><em> </em><em>the</em><em> </em><em>target</em><em>.</em>

<em>Thus</em><em>;</em><em> </em><em>the</em><em> </em><em>inequality</em><em> </em><em>can</em><em> </em><em>be</em><em> </em><em>stated</em><em> </em><em>as</em><em>:</em>

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<em>-</em><em>></em><em> </em><em>3</em><em>6</em><em>(</em><em>s</em><em>)</em><em> </em><em>+</em><em> </em><em>4</em><em>5</em><em>(</em><em>p</em><em>)</em><em> </em><em>≥</em><em>1</em><em>8</em><em>0</em>

<em>-</em><em>></em><em> </em><em>1</em><em>2</em><em>(</em><em>s</em><em>)</em><em> </em><em>+</em><em>1</em><em>5</em><em>(</em><em>p</em><em>)</em><em> </em><em>≥</em><em> </em><em>9</em><em>0</em>

<em> </em><em> </em><u><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em><em>_</em></u>

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<em><u>Hope it helps...</u></em>

<em><u>Hope it helps...Regards;</u></em>

<em><u>Hope it helps...Regards;Leukonov/Olegion.</u></em>

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Mademuasel [1]

Answer:

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sashaice [31]
Heres the decimal <span>0.600</span>
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