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prisoha [69]
4 years ago
6

How do you write a ratio as a percent

Mathematics
1 answer:
Vikentia [17]4 years ago
3 0
Take the ratio
4:8
Convert it to a fraction
4/8
Divide
.5
Multiply by 100 and add a % sign
50%
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Which value of m satisfies the inequality 170-7m 99?<br><br> 10<br> 12<br> 14<br> 15
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Hi, help with question 18 please. thanks​
Vadim26 [7]

Answer:

See Below.

Step-by-step explanation:

We are given the equation:

\displaystyle y^2 = 1 + \sin x

And we want to prove that:

\displaystyle 2y\frac{d^2y}{dx^2} + 2\left(\frac{dy}{dx}\right) ^2 + y^2 = 1

Find the first derivative by taking the derivative of both sides with respect to <em>x: </em>

<em />\displaystyle 2y \frac{dy}{dx}  = \cos x<em />

Divide both sides by 2<em>y: </em>

<h3><em />\displaystyle \frac{dy}{dx} = \frac{\cos x}{2y}<em /></h3>

<em />

Find the second derivative using the quotient rule:

\displaystyle \begin{aligned} \frac{d^2y}{dx^2} &= \frac{(\cos x)'(2y) - (\cos x)(2y)'}{(2y)^2}\\ \\  &= \frac{-2y\sin x-2\cos x \dfrac{dy}{dx}}{4y^2} \\ \\ &= -\frac{y\sin x + \cos x\left(\dfrac{\cos x}{2y}\right)}{2y^2} \\ \\ &= -\frac{2y^2\sin x+\cos ^2 x}{4y^3}\end{aligned}

Substitute:

\displaystyle 2y\left(-\frac{2y^2\sin x+\cos ^2 x}{4y^3}\right)  + 2\left(\frac{\cos x}{2y}\right)^2 +y^2 = 1

Simplify:

\displaystyle \frac{-2y^2\sin x-\cos ^2x}{2y^2} + \frac{\cos ^2 x}{2y^2} + y^2 = 1

Combine fractions:

\displaystyle \frac{\left(-2y^2\sin x -\cos^2 x\right)+\left(\cos ^2 x\right)}{2y^2} + y^2 = 1

Simplify:

\displaystyle \frac{-2y^2\sin x }{2y^2} + y^2 = 1

Cancel:

\displaystyle -\sin x + y^2 = 1

Substitute:

-\sin x + \left( 1 + \sin x\right) =1

Simplify. Hence:

1\stackrel{\checkmark}{=}1

Q.E.D.

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A large bowl of Halloween candy is going to
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