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kipiarov [429]
3 years ago
6

Solve

%7Bx%7D%20%3D%20%5Cdfrac%7B1%7D%7Bp%20%2B%20q%20%2B%20x%7D" id="TexFormula1" title="\sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x}" alt="\sf \dfrac{1}{p} + \dfrac{1}{q} + \dfrac{1}{x} = \dfrac{1}{p + q + x}" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Nostrana [21]3 years ago
3 0

Answer:

\displaystyle   \begin{cases} \displaystyle  {x} _{1} =  - p \\   \displaystyle x _{2}   =  -  q \end{cases}

Step-by-step explanation:

we would like to solve the following equation for x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}  +  \frac{1}{x}  =  \frac{1}{p  + q + x}

to do so isolate \frac{1}{x} to right hand side and change its sign which yields:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{1}{p  + q + x}  -  \frac{1}{x}

simplify Substraction:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{x - (q + p +  x)}{x(p  + q + x)}

get rid of only x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{  - (q + p )}{x(p  + q + x)}

simplify addition of the left hand side:

\displaystyle  \frac{q + p}{pq}     =  \frac{  - (q + p )}{x(p  + q + x)}

divide both sides by q+p Which yields:

\displaystyle  \frac{1}{pq}     =  \frac{  -1}{x(p  + q + x)}

cross multiplication:

\displaystyle    x(p  + q + x)  =   - pq

distribute:

\displaystyle    xp  + xq +  {x}^{2} =   - pq

isolate -pq to the left hand side and change its sign:

\displaystyle    xp  + xq +  {x}^{2} + pq =  0

rearrange it to standard form:

\displaystyle   {x}^{2} +    xp  + xq  + pq =  0

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so

factor out x:

\displaystyle  x( {x}^{} +   p ) + xq  + pq =  0

factor out q:

\displaystyle  x( {x}^{} +   p ) +q (x + p)=  0

group:

\displaystyle  ( {x}^{} +   p ) (x + q)=  0

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

\displaystyle   \begin{cases} \displaystyle  {x}^{} +   p  = 0 \\   \displaystyle x + q=  0 \end{cases}

cancel out p from the first equation and q from the second equation which yields:

\displaystyle   \begin{cases} \displaystyle  {x}^{}   =  - p \\   \displaystyle x  =  -  q \end{cases}

and we are done!

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The degree measure of an angle and its supplement are consecutive odd integers. Find the measure of each angle.
Daniel [21]

Answer:

89

91

Step-by-step explanation:

Supplementary means the pair of angles that have this relationship add up to 180 degrees.

If you assume n is odd, then the very next odd number is n+2.

Example 5 is odd and the next odd number after 5 is 7. This fits the pattern of n and n+2 when n=5.

So we are given the degree measurement of an angle and it's supplement are consecutive odd integers.

So let's pretend the supplement of A is B. This means A+B=180.

We are also given A and B are consecutive odd integers, so this means we are going to let A=n and B=n+2.

Let's put this into our equation for A+B=180:

(n)+(n+2)=180

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2n     +2 =180

Subtract 2 on both sides:

2n          =178

Divide both sides by 2:

 n           =178/2

Simplify right hand side:

 n           =89

If n=89, then n+2=89+2=91.

Does 89+91=180? Yep.

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Here are some consecutive numbers including those: 89,90,91

Yes 91 is the next odd number after 89.

3 0
2 years ago
Read 2 more answers
A test of abstract reasoning is given to a random sample of students before and after they completed a formal logic course. The
Natasha_Volkova [10]

Answer:

The right answer is "0.2".

Step-by-step explanation:

The given values are:

\bar{x_0} = 3.7

s_o=4.95

t_{\frac{0.05}{2} }=2.2621

As we know,

95% confidence for \mu_0 will be:

= \bar{x_0} \pm t_{\frac{0.05}{2} },n-1\times \frac{s_o}{\sqrt{n} }

The lower bound will be:

= 3.7-2.2621\times \frac{4.95}{\sqrt{10} }

= 0.16\simeq 0.2

The upper bound will be:

= 3.7+2.2621\times \frac{4.95}{\sqrt{10} }

= 7.23\simeq7.2

Thus the right answer is "0.2"

8 0
3 years ago
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