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AleksandrR [38]
3 years ago
14

Please help me with this question

Mathematics
2 answers:
Lorico [155]3 years ago
8 0
This is what I got I hope it helps

tamaranim1 [39]3 years ago
7 0

Step-by-step explanation:

We need to simply like terms. Let's break this out into four different problems that are easy to solve, then we can put it all back together for the final answer.

There are four like terms in both the numerator (top) and denominator (bottom) of the fraction. Numbers, m, p, and v. We will look at each of these seperately.

What is 45/15? Easy right, just 3.

Let's move on to the harder parts.

\frac{m^{-6}}{m^{-2}}

Remember that negavite exponents mean to divide.

For example, 2^{-2} = \frac{1}{2^2} = 1/4

So we just need to apply this rule to both of these exponents to get,

\frac{m^2}{m^6}

Now remember that when dividing by exponents, the exponents subtract from each other\frac{x^y}{x^z} = x^{y-z}

So we get, m^{2-6} = m^{-4} = \frac{1}{m^4}

So far we have

\frac{3}{m^4}

Keep applying the same rules for the other two terms. Reply back if you need more help and I can keep working through with you.

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HELLO PLEASE GIVE ME THE RIGHT ANSWER ASAP THANK YOU !
seropon [69]

\large\underline{\sf{Solution-}}

\textsf{Given expression;}\\

\rm{23 =  \frac{7}{9}(6x - 36) + 9 } \\

\rm{\longmapsto} \: 23 =  \frac{(42x - 256)}{9}  + 9 \\

\rm{\longmapsto} \: 23 =  \frac{(42x - 256 + 81)}{9}  \\

\rm{\longmapsto} \: 23 =  \frac{(42x - 175)}{9}  \\

\rm{\longmapsto} \:  \frac{23}{1}  =  \frac{(42x - 175)}{9}  \\

\textsf{By doing cross multiplication, we get}\\

\rm{\longmapsto}23(9) = 1(42x - 175) \\

\rm{\longmapsto}207 =42x - 175\\

\rm{\longmapsto} \: 42x =  - 175 - 207 \\

\rm{\longmapsto} \: 42x =  - 379 \\

\rm \therefore \: x =  -  \frac{375}{42}  \\

\textsf{Hence, the value of x will be -375/42 respectively.}\\

\textbf{Read more:}

\textsf{Similar Questions}

Solve the expression and find the value of x. [tex] \frac{x⁠…

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5 0
3 years ago
Which value for the constant ccc makes w=5w=5w, equals, 5 an extraneous solution in the following equation? \sqrt{29+4w}=23-cw 2
miv72 [106K]

Answer:

c=6

Step-by-step explanation:

In order to solve the original equation, we would have to square both sides of the equation:

\begin{aligned}\sqrt{29+4w}&=23-cw\\\\ \left(\sqrt{29+4w}\right)^2&=(23-cw)^2\\\\ 29+4w&=(23-cw)^2\end{aligned}  

29+4w

​  

 

(  

29+4w

​  

)  

2

 

29+4w

​  

 

=23−cw

=(23−cw)  

2

 

=(23−cw)  

2

 

​  

 

However, squaring both sides of an equation can create extraneous solutions! [Why?]

Hint #22 / 4

Let's plug \blueD w=\blueD{5}w=5start color #11accd, w, end color #11accd, equals, start color #11accd, 5, end color #11accd into the last equation we obtained:

\begin{aligned}29+4\blueD w&=(23-c\blueD{w})^2\\\\ 29+4(\blueD 5)&=(23-c(\blueD{5}))^2\\\\ 49&=(23-5c)^2\end{aligned}  

29+4w

29+4(5)

49

​  

 

=(23−cw)  

2

 

=(23−c(5))  

2

 

=(23−5c)  

2

 

​  

 

This equation is correct, both when 23-5c=723−5c=723, minus, 5, c, equals, 7 and when 23-5c=-723−5c=−723, minus, 5, c, equals, minus, 7.

However, the original equation is not correct for 23-5c=-723−5c=−723, minus, 5, c, equals, minus, 7, since this way we obtain \sqrt{49}=-7  

49

​  

=−7square root of, 49, end square root, equals, minus, 7.

Hint #33 / 4

Therefore, an extraneous solution is obtained for the ccc-value that makes 23-5c23−5c23, minus, 5, c equal -7−7minus, 7, which is c=6c=6c, equals, 6.

Substituting this back into the original equation gives \sqrt{29+4w}=23-6w  

29+4w

​  

=23−6wsquare root of, 29, plus, 4, w, end square root, equals, 23, minus, 6, w. You can now solve this for www and see for yourselves that w=5w=5w, equals, 5 is indeed extraneous.

Hint #44 / 4

The answer is:

c=6c=6

4 0
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200/5
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The painter must work AT LEAST 13.33 hours to make AT LEAST 200 dollars.
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Y = mx + b looking for x
Tanya [424]

Use the formula y=1 and so forth

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3 years ago
Convert 7.2 ⋅ 10–3 to standard form.
DENIUS [597]
0.0072. Hope this helps.
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3 years ago
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