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1 year ago

I need help on this one question, anyone know the answer?

Mathematics

1 answer:

JulsSmile [24]1 year ago

6 0

X = 43

4(43)+22= 194°

194 + 24 + 142 = 360

adjacent angles on a point

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Use the quadratic formula to determine the exact solutions to the equation.

Ilia_Sergeevich [38]

ANSWER

$x = \frac{5}{4} - \frac{ \sqrt{ 17} }{4}$

$x = \frac{5}{4} + \frac{ \sqrt{ 17} }{4}$

EXPLANATION

The given equation is:

$2 {x}^{2} - 5x + 1= 0$

Comparing this to

$a {x}^{2} + bx + c = 0$

we have a=2, b=-5, c=1

The quadratic formula is given by

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

We substitute the values to get,

$x = \frac{ - - 5 \pm \sqrt{ {( - 5)}^{2} - 4(1)(2)} }{2(2)}$

$x = \frac{ 5 \pm \sqrt{ 17} }{4}$

$x = \frac{5}{4} - \frac{ \sqrt{ 17} }{4}$

$x = \frac{5}{4} + \frac{ \sqrt{ 17} }{4}$

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Answer:

Step-by-step explanation:

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Troyanec [42]

Answer:

Proved

Step-by-step explanation:

The options are not given. So, I will solve from scratch

Given

$\frac{secx-1}{tan x}= \frac{tanx}{secx+1}$

Required

Prove

Multiply the right-hand side by $\frac{secx + 1}{secx + 1}$

$\frac{secx-1}{tan x} * \frac{secx + 1}{secx + 1}= \frac{tanx}{secx+1}$

Apply difference of two squares on the numerator

$\frac{sec^2 x - 1}{(tanx)(secx + 1)} =\frac{tanx}{secx+1}$

In trigonometry:

$tan^2x = sec^2x - 1$

So, we have:

$\frac{tan^2 x}{(tanx)(secx + 1)} =\frac{tanx}{secx+1}$

$\frac{tan x * tan x}{(tanx)(secx + 1)} =\frac{tanx}{secx+1}$

tan x cancels out

$\frac{tan x}{secx + 1} =\frac{tanx}{secx+1}$

Proved

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its answer 1.

Step-by-step explanation:

please mark me brainliest thank you for your time and have a nice rest of your day :)

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