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vodomira [7]
3 years ago
11

PLEASEEEEEEEEEE HELP ASAP PLEASEEEEEEEEEE

Mathematics
2 answers:
icang [17]3 years ago
7 0

Answer:

3rd choice

Step-by-step explanation:

In division for variables with same base, you do subtract top exponent minus bottom exponent. She did that correctly since -3-(-1)=-3+1=-2 and -2-1=-3.  

The problem said m=-2 and n=4 and she replace m with (-2) and n with (4). She did this correctly.

You can multiply base numbers unless the exponents are the same 4 doesn't have the exponent -2 on it so you can't do (4(-2))^(-2)

The error is the 3rd option.

svlad2 [7]3 years ago
4 0

We start with

\dfrac{4m^{-3}n^{-2}}{m^{-1}n}

Simplifying the exponents, we have

\dfrac{4m^{-3}n^{-2}}{m^{-1}n} = 4m^{-3}n^{-2} (mn^{-1}) = 4m^{-3+1}n^{-2-1}=4m^{-2}n^{-3}

So, the exponents are ok.

If we plug the values, we have

4m^{-2}n^{-3} \mapsto 4(-2)^{-2}(4)^{-3} = 4\cdot \dfrac{1}{(-2)^2}\cdot\dfrac{1}{4^3} = 4\cdot \dfrac{1}{4}\cdot \dfrac{1}{64} = \dfrac{1}{64}

So, she didn't apply the exponent -2 correctly.

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Solving simultaneous equations with one quadratic, x^2+y^2=16 and y=x+3
Svetlanka [38]
If y=x+3, then in the first equation you have

x^2+(x+3)^2=x^2+(x^2+6x+9)=2x^2+6x+9=16

Rewriting a bit, you get

x^2+3x+\dfrac92=8

x^2+3x+\dfrac94+\dfrac94=8

\left(x+\dfrac32\right)^2=\dfrac{23}4

x=-\dfrac32\pm\dfrac{\sqrt{23}}2

Now, since y=x+3, you also get

y=-\dfrac32\pm\dfrac{\sqrt{23}}2+3=\dfrac32\pm\dfrac{\sqrt{23}}2

So, there are two solutions here:

(x,y)=\left(-\dfrac32+\dfrac{\sqrt{23}}2,\dfrac32+\dfrac{\sqrt{23}}2\right)
(x,y)=\left(-\dfrac32-\dfrac{\sqrt{23}}2,\dfrac32-\dfrac{\sqrt{23}}2\right)
7 0
3 years ago
Help plsss(+Brainliest too.)
gavmur [86]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
1 3/4 and 1/8 divided as a simplify form
Sati [7]

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3 0
4 years ago
Which of these contexts describes a situation that is unlikely?
attashe74 [19]

Answer:

The first one

Step-by-step explanation:

This is because all the other ones there is a likely chance that you will get them. While the first one it is less of a chance that you will role a die and land on a one. It is a 1 out of 6 chance

4 0
3 years ago
Multiply using partial products
Alex
I don't see what we need to multiply but  here is an example: the numbers of the equation that I'm doing is  12 times 13 and this is how I do it .

  (1)(2)                           Step 1: multiply 3 times 2 then when you get the      
x 1 (3)                             answer you would need to multiply then 2 times 3.
  -------                                     
       (36) ( when multiplying 1 times 3 and then multiply 2 times 3)  
     
       
     Step 3: know that in the third step you would need to multiply 1 times 2 then 1 times 1.

    (1)(2)                                    
  x(1) 3
-------------                                        
       36
   + 120  (when multiplying 1 times 2 then 1 times 1 but I added a zero )
-----------
      156 ( I add them all up and that's the answer )

If need any questions just message me and I will answer back .




5 0
3 years ago
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