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3 years ago

May someone assist me?

Mathematics

1 answer:

raketka [301]3 years ago

3 0

Answer:

<h2><u><em>15 is the answer.</em></u></h2>

Step-by-step explanation:

The side lengths are equal - supposedly

30 25

? ?

25 + ? = 45

-25 - 25

? = 20

45 - 30 = 15

The question mark should be equal to 15.

Hope this helps,

Kavitha

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Is 1:2:3 equivalent to 2:4:6

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Answer:

Yes it is.

Step-by-step explanation:

Just multiply everything by 2.

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3 years ago

Write each expression as an algebraic (nontrigonometric) expression in u, u > 0.

max2010maxim [7]

Answer:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

Step-by-step explanation:

We want to write the trignometric expression:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0$

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

$\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)$

Take the secant of both sides:

$\displaystyle \sec(\theta)=\frac{u}{10}$

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

$\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}$

By substitutition:

$\displaystyle= \sin(2\theta)$

Using an double-angle identity:

$=2\sin(\theta)\cos(\theta)$

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

$\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)$

Simplify. Therefore:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

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3 years ago

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sveticcg [70]

Answer:

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Step-by-step explanation:

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4 years ago

Solve the equation on the interval [0,2π]

WARRIOR [948]

$\bf 16sin^5(x)+2sin(x)=12sin^3(x)
\\\\\\
16sin^5(x)+2sin(x)-12sin^3(x)=0
\\\\\\
\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\
-------------------------------\\\\
2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=
\begin{cases}
0\\
\pi \\
2\pi 
\end{cases}\\\\
-------------------------------$

$\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0$

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

$\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0
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~~~~~~~~~~~~\textit{quadratic formula}
\\\\
\begin{array}{lcccl}
& 8 sin^4& -6 sin^2(x)& +1\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
sin(x)= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}
\\\\\\
sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}
\\\\\\
sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=
\begin{cases}
\frac{1}{2}\\\\
\frac{1}{4}
\end{cases}$

$\bf \measuredangle x=
\begin{cases}
sin^{-1}\left( \frac{1}{2} \right)
sin^{-1}\left( \frac{1}{4} \right)
\end{cases}\implies \measuredangle x=
\begin{cases}
\frac{\pi }{6}~,~\frac{5\pi }{6}\\
----------\\
\approx~0.252680~radians\\
\qquad or\\
\approx~14.47751~de grees\\
----------\\
\pi -0.252680\\
\approx 2.88891~radians\\
\qquad or\\
180-14.47751\\
\approx 165.52249~de grees
\end{cases}$

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3 years ago

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lianna [129]

The slope would be 0.

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