Can someone help with this?

Mathematics

2 answers:

Zina [86]3 years ago

7 0

Answer:

Step-by-step explanation:

Natasha2012 [34]3 years ago

5 0

Answer:

Step-by-step explanation:

With $25+ and 25 -

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What is <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B5%7D%20" id="TexFormula1" title=" \frac{3}{5} " alt=" \frac{3}

Ilia_Sergeevich [38]

Answer:

Step-by-step explanation:

3/5 * 15

Rewriting as

3 * 15/5

3 * 3

3 0

2 years ago

Read 2 more answers

What is the solution to the equation?

UkoKoshka [18]

$\frac{1}{3}x+\frac{1}{3}(2x-15)=3\frac{1}{2}\\\\
\frac{1}{3}x+\frac{1}{3}(2x-15)=\frac{7}{2}\ \ \ |Multiply\ by\ 6\\\\
2x+2(2x-15)=21\\\\
2x+4x-30=21\\\\
6x-30=21\ \ |Add\ 30\\\\
6x=51\ \ \ |Divide\ by\ 6\\\\
\boxed{x=\frac{51}{6}=8\frac{1}{2}}$

7 0

3 years ago

Solve using algebraic equation: 5sin2x=3cosx (No exponents)

DaniilM [7]

$5\sin2x=3\cos x\iff10\sin x\cos x=3\cos x$

by the double angle identity for sine. Move everything to one side and factor out the cosine term.

$10\sin x\cos x=3\cos x\iff10\sin x\cos x-3\cos x=\cos x(10\sin x-3)=0$

Now the zero product property tells us that there are two cases where this is true,

$\begin{cases}\cos x=0\\10\sin x-3=0\end{cases}$

In the first equation, cosine becomes zero whenever its argument is an odd integer multiple of $\dfrac\pi2$ , so $x=\dfrac{(2n+1)\pi}2$ where $n[/tex ]is any integer.\\Meanwhile,\\[tex]10\sin x-3=0\implies\sin x=\dfrac3{10}$

which occurs twice in the interval $[0,2\pi)$ for $x=\arcsin\dfrac3{10}$ and $x=\pi-\arcsin\dfrac3{10}$ . More generally, if you think of $x$ as a point on the unit circle, this occurs whenever $x$ also completes a full revolution about the origin. This means for any integer $n$ , the general solution in this case would be $x=\arcsin\dfrac3{10}+2n\pi$ and $x=\pi-\arcsin\dfrac3{10}+2n\pi$ .