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SIZIF [17.4K]
3 years ago
10

Solve for P: 9(p−4)=−18 Please Explain

Mathematics
1 answer:
vekshin13 years ago
3 0

Answer:

<em>p</em> = 2

Step-by-step explanation:

Happy to help.

When we have numbers in parenthesis, we generally want to deal with those first. However, we can hit a rough patch when a variable is in there. Consider this:

2(3 + 4) = 2(7), or 14.

But, the two can also be distributed into both numbers in the parenthesis, like this:

2(3 + 4) = 2*3 + 2*4

That leaves us with the same answer—14! We can apply this to a variable, and that will help us figure out 9(<em>p</em> - 4), or the left side of your equation you presented.

9(<em>p</em> - 4) = -18

9*<em>p</em> - 9*4 = -18

9<em>p</em> - 36 = -18

Add 36 on both sides to isolate the variable (in this case, <em>p)</em>

9<em>p</em> = -18 + 36

You can also write it like this; 9<em>p</em> = 36 - 18

9<em>p</em> = 18

Divide 9 to isolate <em>p</em>

<em>p</em> = 2

So, we would get (<em>p</em> = 2). Make sure to practice a few more questions like these to really get the hang of it—you'll be using this a lot in the future!

Good luck!

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Answer:

2.-2m^4-6m^2+4m+9

3.-2m^4-6m^2-4m+9

Step-by-step explanation:

<em>In general we can write a polynomial in standard form as </em>

ax^n+bx^n^-^1+cx^n^-^2+...+px+q

<em>Given</em> 4m-2m^4-6m^2-4m+9

<em>Combine the like terms: 4m and -4m</em>

<em>4m-4m=0</em>

<em>We have 4m-4m=0</em>

<em>So, write the remaining terms</em>

-2m^4-6m^2+9+0

= -2m^4-6m^2+9

<em>This is in decreasing order of powers.</em>

<em>Hence the answer is the standard form is</em>

-2m^3-6m^2+9

<em>But in the given options, you can choose option 2 and option 3 are in standard form.</em>

<em>Because they are in decreasing order of powers.</em>

<em>In other two options, the constants term is first and the highest power term is at the last. So, they are not in standard form.</em>

<em>-2m^4-6m^2+4m+9</em>

<em>-2m^4-6m^2-4m+9</em>

<em>I hope this helps you.</em>

<em>And please comment if I need to do corrections.</em>

<em>Please let me know if you have any questions.</em>

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