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3 years ago

What is x divided by x-1 - 1 divided by 2-2x expessed as a single fraction ?

Mathematics

1 answer:

Kay [80]3 years ago

8 0

To simplify into 1 fraction, remember, we must make the denominator the same
x/ (x-1 ) - 1/ (2-2x)
= -x/ -(1-x) - 1/(2-2x) [see the similarity now? ] [whatever we multiply for the denominator, as long as we multiply to the numerator, it will be ok]
= -2x / -(2-2x) + 1/-(2-2x)
= (-2x+1 )/ -(2-2x)
=(2x-1)/(2-2x)

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How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

3 years ago

A healthcare industry is reviewing the number of liabilities each location has. Location A has a z-score of −1.12 and Location N

elixir [45]

Answer:

0.8686 or 86.86 %

0.2148 or 21.48 %

Step-by-step explanation:

In z table the value of z > - 1,12 is 0.1314 (value from the z point to the left of the curve ) then 1 - 01314 will be value from z point to the right

Again from z table we get for z = - 0.79 the value 0.2148 s the vale from the point up to the left tail

6 0

3 years ago

Solve |4| + |11 - 6| - |-2|

Pie

Answer:

...........the answer is 7?!

5 0

3 years ago

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Janine inflated a ball to a radius of 18cm and another ball to a radius of 12cm. How much greater was the volume of the air in t

bagirrra123 [75]

Answer:

17,182.08 cm³

Step-by-step explanation:

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3 years ago

Find so that the point (-1,2) is on the graph f(x)=ax^2+4. What does A equal?

natali 33 [55]

We have to find the parameter a so that (-1,2) is part of the function f(x) = ax²+4.

To check if a point is part of a function, we can replace the values of x and y = f(x) with the coordinates of the point and then, if the equation stays true, then the point is part of the function.

So for (x,y) = (-1,2) to be part of the function y = f(x), this equation has to stands true:

$\begin{gathered} y=f(x) \\ y=ax^2+4 \\ 2=a(-1)^2+4 \\ 2=a+4 \\ a=2-4 \\ a=-2 \end{gathered}$

Then, the function would have to be f(x) = -2x² + 4.

We can check with a graph as:

Answer: a = -2

8 0

1 year ago