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stiks02 [169]
2 years ago
6

Solve the equation 2a+b=2a, where a and b are nonzero, real numbers. Describe the solution to this equation and justify your des

cription.
Mathematics
1 answer:
ss7ja [257]2 years ago
3 0
<span>To solve this equation we can first assume that both a and b are nonzero real numbers. Hence, A = 1 b = 1 <span><span>
1.   </span>2 (1) + 1 = 2(1)</span> <span><span>
2.   </span><span> 2 + 1 = 2: now this a false equation since there is not equality, the equation cannot retain the equal sign but will become 2 + 1 > 2. Leaving the relationship unequal.

</span></span>However, the alternative to this problem is to be b = 0. To oversee the rule in order to solve the equation retaining it as an “equation”. Further, there is no other solution for this equation. A = 1 b = 0
<span>1.   Which becomes 2(1) + 0 = 2(1)</span> <span><span>
2.   </span><span> 2 + 0 = 2 :
3.    2 = 2. Here we can observe the equality. </span></span> </span>
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Step 2: Divide the decrease by the original number

1.15 \div 5.75\\\\= \frac{1.15}{5.75}\\\\= 0.2

Step 3: Multiply the result by 100 to get a percent decrease

0.2 \times 100\\\\= 20

Therefore, the percent decrease of the price of paper per package is 20%

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3/8 was the swimming segment!

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2 years ago
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2 years ago
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Which statements are true about the graph of the system of linear inequalities? Check all that apply. y &gt; 3x – 4 y &lt; x + 1
romanna [79]

Answer:

  • The graph of y > 3x − 4 has shading above a dashed line.
  • The graph of y < x + 1 has shading below a dashed line.
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y > ... means the shading will be above the corresponding line.

y < ... means the shading will be below the corresponding line.

These lines have different slopes, so the solution spaces <em>must</em> overlap, hence <em>there must be solutions</em> to the system.

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<em>Comment on last choice</em>

It isn't clear exactly what is intended by the last offered statement. Both of the listed points are in the solution space of y > 3x-4. Neither point is in the solution space of y < x+1.

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The y-intercepts of the two boundary lines are (0, 1) and (0, -4). Neither of these points is in the solution space of the system of inequalities.

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