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pashok25 [27]
3 years ago
9

Solve 0.05p-0.02(5-2p)=0.05(p-2)-0.12

Mathematics
1 answer:
Montano1993 [528]3 years ago
8 0
0.05p-0.02(5-2p)=0.05(p-2)-0.12 

0.05p-0.02(5-2p)=0.05(p-2)-0.12:p 


0.05p-0.1+0.04p=0.05p-0.1-0.12
 
  |Expand \ the \ equation| 

0.09p-0.1=0.05p-0.1-0.12
 
 

0.09p-0.1=0.05p-0.22 


0.09p-0.1-0.05p=-0.22
 

  |Subtract \ 0.05p \ from \ both \ sides| 


0.04p-0.1=-0.22
 
 


0.04p=-0.22+0.1
 
  |Add \ 0.1 \ to \ both \ sides| 


0.04p=-0.12
 
 

0.04p/0.04=-0.12/0.04
  |Divide \ both \ sides \ by \ 0.04| 

p=-3 

\star \ Hence \ Solved
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Answer:

Yes.

Step-by-step explanation:

Let's write out each problem:

\sqrt{5} = 2.3

2^3 = 8

Now that we have both answers to each expression, we can tell that 2^3, in fact, is larger than the square root of 5.

So your answer is, "Yes, 2^3 is bigger than the square root of 5.

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Which are the solutions of the quadratic equation? x2 = –5x – 3 –5, 0 StartFraction negative 5 minus StartRoot 13 EndRoot Over 2
boyakko [2]

<u>Given</u>:

The quadratic equation is x^{2}=-5 x-3

We need to determine the solutions of the quadratic equation.

<u>Solution</u>:

Let us solve the equation to determine the value of x.

Adding both sides of the equation by 5x and 3, we get;

x^{2}+5 x+3=0

The solution of the equation can be determined using quadratic formula.

Thus, we get;

x=\frac{-5 \pm \sqrt{5^{2}-4 \cdot 1 \cdot 3}}{2 \cdot 1}

x=\frac{-5 \pm \sqrt{25-12}}{2 }

x=\frac{-5 \pm \sqrt{13}}{2 }

Thus, the two roots of the equation are x=\frac{-5 + \sqrt{13}}{2 } and x=\frac{-5- \sqrt{13}}{2 }

Hence, the solutions of the equation are x=\frac{-5 + \sqrt{13}}{2 } and x=\frac{-5- \sqrt{13}}{2 }

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Answer:

52

Step-by-step explanation:

3^3+3(3)^2-7(3)+19

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