True or false fraction are not rational #’s

Which equation has the solutions x=1+/- 5?

nikitadnepr [17]

Answer:

$x^{2} -2x-4=0$

Solving Steps

Use the quadratic formula

$x=\frac{-(-2)(+/-)\sqrt{(-2)^2-4*1*(-4)} }{2*1}$

Multiply / Remove the parentheses / Evaluate

$x=\frac{2(+/-)\sqrt{4+16} }{2}$

Calculate

$x=\frac{2(+/-)\sqrt{20} }{2}$

Separate the solutions

$x=\frac{2(+/-)2\sqrt{5} }{2} \\x=\frac{2-2\sqrt{5} }2}$

Simplify

$x=1+\sqrt{5} \\x=1-\sqrt{5}$

4 0

2 years ago

Please help me with this problem please!!!!!!!!!!!!!!

777dan777 [17]

Answer:

Brain

Step-by-step explanation:

7 0

2 years ago

One angle measure in an acute triangle is 38°. What could the measure of one of the other angles be?

aivan3 [116]

Acute, Obtuse, Acute, Acute

6 0

3 years ago

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441 is 63% of what number

blagie [28]

Question: 441 is 63% of what number?
=> 63% = 63% / 100% = 0.63
Since 441 is the 63% let’s find the value of 37% to get the toal of 100%
Solution, follow the formula
=> 0.63x = 441 – where x will be the value of the missing 100% value.
=> x = 441 / 0.63
=> x = 700 – the 100% value that have 441 as 63%
let’s check if we have the correct answer:
=> 700 * .63 = 441

6 0

3 years ago

Find all possible values of α+

const2013 [10]

Answer:

$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

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we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

recall angle sum indentity of tan:

$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore when k is 1 we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

when is 0:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

likewise by Opposite Angle Identity we obtain:

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

$\rm\displaystyle \gamma = - \alpha - \beta \pm 0$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }$

and we're done!

8 0

3 years ago

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