Need help desperately T~T

Mathematics

1 answer:

babunello [35]3 years ago

4 0

As you can see, angle 5 and angle 6 are supplementary. And angle 5 and angle 3 are congruent because they are alternate interior angles.

So it will be
x+2 = 180- x+3
move x over from the right to the left
2x+2 = 183
move 2 over from the left to the right
2x = 181
divide by 2
x= 90.5

and angle 3 and angle 1 are vertical angles so they are congruent. Using the angle 3 formula to solve for the answer:

90.5+3 =93.5

When angles are congruent, their measures are congruent, therefore, measure of angle 1 is 93.5

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Find two additional roots of P(x)=0.<br><br> 13+sqrt 3 and 7i

vova2212 [387]

So, x = 13, x = √3 and x =7i.

now, recall that for an EVEN radical, there are two possible roots, namely is say √3 is say hmmm some value "a", that means that a*a = √3, however, -a*-a is also √3, therefore, ±√3 are two valid values, and therefore -√3 is another one.

now.... keep in mind that, complex solutions or roots, never come all by their lonesome, their sister is always with them, the conjugate, so, for 7i or namely 0 + 7i, her sister is always around, 0 - 7i, which is the other root.

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3 years ago

If x^2+1/x^2=3 find the value of x^2/(x^2+1)^2<br> Express answer as a common fraction.<br> Thanks!!

timama [110]

$x^2+\dfrac1{x^2}=3\implies x^4+1=3x^2\implies x^4-3x^2+1=0$

By the quadratic formula,

$x^2=\dfrac{3\pm\sqrt5}2\implies x^2+1=\dfrac{5\pm\sqrt5}2$

Then

$(x^2+1)^2=\dfrac{25\pm10\sqrt5+5}4=\dfrac{15\pm5\sqrt5}2$

$\implies\dfrac{x^2}{(x^2+1)^2}=\dfrac{\frac{3\pm\sqrt5}2}{\frac{15\pm5\sqrt5}2}=\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}$

Multiply numerator and denominator by the denominator's conjugate:

$\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}\cdot\dfrac{15\mp5\sqrt5}{15\mp5\sqrt5}=\dfrac{45\pm15\sqrt5\mp15\sqrt5-25}{15^2-(5\sqrt5)^2}=\dfrac{20}{100}=\dfrac15$