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2 years ago

Gail is competeting in a checkers tournament. She won 12 out of a total of 20 games. What is her ratio of wins to loses

Mathematics

1 answer:

Dmitrij [34]2 years ago

8 0

The first answer is a and the second answer is d

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Solve and graph the inequality. 4/5x + 5 < -3

kiruha [24]

For the Solving Part, Answer is: −110<x<0

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2 years ago

Chad owns twice as many cds as carlin owns. if you add 6 to the number of cds chad owns and then divide by 7, you get the number

Luba_88 [7]

35 cds does carlin have

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3 years ago

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2 years ago

SinA=√1-x÷√1+X Find the value of cosA

Irina-Kira [14]

The value of cos A is √(1 + x²)/ (1 - x²) /√1 + x

<h3>Trigonometric ratios</h3>

It is important to note that

sin A = opposite/ hypotenuse

cos A = adjacent/ hypotenuse

Then,

opposite = $\sqrt{1} - x$

Hypotenuse = $\sqrt{1} + x$

Let's find the adjacent side using the Pythagorean theroem

$(\sqrt{1} + x)^2 = (\sqrt{1 -x } )^2 + x^2$

$1 + x^2 = 1 - x^2 + x^2$

$x = \sqrt{\frac{1 + x^2}{1 -x^2} }$

cos A = x/hypotenuse

$cos A = \frac{\sqrt{\frac{1+x^2}{1 -x^2} } }{\sqrt{1} +x}$

cos A = √(1 + x²)/ (1 - x²) /√1 + x

Thus, the value of cos A is √(1 + x²)/ (1 - x²) /√1 + x

Learn more about trigonometric identities here:

brainly.com/question/7331447

#SPJ1

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1 year ago

What are the roots to the equation 2x² + 6x-1=0?

lys-0071 [83]

Answer:

The roots of equations are as m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$ And n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$

Step-by-step explanation:

The given quadratic equation is 2 x² + 6 x - 1 = 0

This equation is in form of a x² + b x + c = 0

Let the roots of the equation are ( m , n )

Now , sum of roots = $\frac{ - b}{a}$

And products of roots = $\frac{c}{a}$

So, m + n = $\frac{ - 6}{2}$ = - 3

And m × n = $\frac{ - 1}{2}$

Or, (m - n)² = (m + n)² - 4mn

Or, (m - n)² = (-3)² - 4 ( $\frac{ - 1}{2}$ )

Or, (m - n)² = 9 + 2 = 11

I.e m - n = $\sqrt{11}$

Again m + n = - 3 And m - n = $\sqrt{11}$

Solving this two equation

(m + n) + ( m - n) = - 3 + $\sqrt{11}$

I.e 2 m = - 3 + $\sqrt{11}$

Or, m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$

Similarly n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$

Hence the roots of equations are as m = $\frac{-3}{2} + \frac{\sqrt{11} }{2}$ And n = $\frac{-3}{2} - \frac{\sqrt{11} }{2}$ Answer

6 0

3 years ago