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3 years ago

6

Pls help me!!!!!! Its due tomorrow

1 answer:

prisoha [69]3 years ago

3 0

Pls help its due tomorrow

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3x ^2 - 9 x y ^2+ 12 X ^3Y^2

shepuryov [24]

$\huge \boxed{\mathfrak{Question} \downarrow}$

$3 x ^ { 2 } y - 9 x y ^ { 2 } + 12 x ^ { 3 } y ^ { 2 }$

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

$3 x ^ { 2 } y - 9 x y ^ { 2 } + 12 x ^ { 3 } y ^ { 2 }$

Factor out $3$ .

$3\left(x^{2}y-3xy^{2}+4x^{3}y^{2}\right)$

Consider $x^{2}y-3xy^{2}+4x^{3}y^{2}$ . Factor out $xy$ .

$xy\left(x-3y+4x^{2}y\right)$

Rewrite the complete factored expression.

$\boxed{ \boxed{ \bf \: 3xy\left(x-3y+4x^{2}y\right) }}$

4 0

2 years ago

Please help, very confused!

My name is Ann [436]

$\bf cot(\theta)=\cfrac{1}{tan(\theta)}
\qquad \qquad 
csc(\theta)=\cfrac{1}{sin(\theta)}\qquad \qquad sin(-\theta )=-sin(\theta )
\\\\\\
\textit{also recall }sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)\\\\
-------------------------------$

$\bf \cfrac{csc^2(x)-cot^2(x)}{sin(-x)cot(x)}\implies \cfrac{\frac{1}{sin^2(x)}-\frac{cos^2(x)}{sin^2(x)}}{-sin(x)\frac{cos(x)}{sin(x)}}\implies \cfrac{\frac{1-cos^2(x)}{sin^2(x)}}{-cos(x)}
\\\\\\
\cfrac{\frac{sin^2(x)}{sin^2(x)}}{-cos(x)}\implies \cfrac{1}{-cos(x)}\implies -sec(x)$

7 0

3 years ago

Can someone help me with these math problems

LiRa [457]

The first one is 40 pie

8 0

3 years ago

Read 2 more answers

Write an equation for the description.<br> The difference between a number x and 19 is 27.

ra1l [238]

Answer

x-19=27

Step-by-step explanation:

4 0

3 years ago

Given <br><img src="https://tex.z-dn.net/?f=%20log_%7B2%7D%28x%29%20%20%3D%20%20%5Cfrac%7B3%7D%7B%20log_%7Bxy%7D%282%29%20%7D%20

Naily [24]

Answer:

$\displaystyle y = x^{-\frac{2}{3}}$

Step-by-step explanation:

<u>Logarithms</u>

Some properties of logarithms will be useful to solve this problem:

1. $\log(pq)=\log p+\log q$

2. $\displaystyle \log_pq=\frac{1}{\log_qp}$

3. $\displaystyle \log p^q=q\log p$

We are given the equation:

$\displaystyle \log_{2}(x) = \frac{3}{ \log_{xy}(2) }$

Applying the second property:

$\displaystyle \log_{xy}(2)=\frac{1}{ \log_{2}(xy)}$

Substituting:

$\displaystyle \log_{2}(x) = 3\log_{2}(xy)$

Applying the first property:

$\displaystyle \log_{2}(x) = 3(\log_{2}(x)+\log_{2}(y))$

Operating:

$\displaystyle \log_{2}(x) = 3\log_{2}(x)+3\log_{2}(y)$

Rearranging:

$\displaystyle \log_{2}(x) - 3\log_{2}(x)=3\log_{2}(y)$

Simplifying:

$\displaystyle -2\log_{2}(x) =3\log_{2}(y)$

Dividing by 3:

$\displaystyle \log_{2}(y)=\frac{-2\log_{2}(x)}{3}$

Applying the third property:

$\displaystyle \log_{2}(y)=\log_{2}\left(x^{-\frac{2}{3}}\right)$

Applying inverse logs:

$\boxed{y = x^{-\frac{2}{3}}}$

7 0

3 years ago