All categories

2 years ago

How will you compare your experience in factoring by grouping with factoring the difference of two squares? Describe

Mathematics

1 answer:

Sergio [31]2 years ago

8 0

Answer:

Factoring in groups is like breaking it apart piece by piece and takes more steps.

Factoring the difference of two squares takes few steps and is easier.

You might be interested in

Rewrite 9cos 4x in terms of cos x.

rosijanka [135]

$\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]
\\\\\\
72cos^4(x)-72cos^2(x)+9$

---------------------------------------------------------------------------

as far as the previous one on the 2tan(3x)

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\$

$\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[ \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[ \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
\\\\\\$

$\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
\\\\\\
2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}$

4 0

3 years ago

Negative 7 bigger or smaller than negative 8

dalvyx [7]

-7 is bigger than -8

4 0

3 years ago

Read 2 more answers

A rectangular auditorium seats 1564 people. the number of seats in each row exceeds the number of rows by 1212. find the number

yanalaym [24]

<span>We can safely assume that 1212 is a misprint and the number of seats in a row exceeds the number of rows by 12. Let r = # of rows and s = # of seats in a row. Then, the total # of seats is T = r x s = r x ( r + 12), since s is 12 more than the # of rows. Then r x (r + 12) = 1564 or r**2 + 12*r - 1564 = 0, which is a quadratic equation. The general solution of a quadratic equation is: x = (-b +or- square-root( b**2 - 4ac))/2a In our case, a = 1, b = +12 and c = -1564, so x = (-12 +or- square-root( 12*12 - 4*1*(-1564) ) ) / 2*1 = (-12 +or- square-root( 144 + 6256 ) ) / 2 = (-12 +or- square-root( 6400 ) ) / 2 = (-12 +or- 80) / 2 = 34 or - 46 We ignore -46 since negative rows are not possible, and have: rows = 34 and seats per row = 34 + 12 = 46 as a check 34 x 46 = 1564 = total seats</span>

4 0

3 years ago

Can someone help me with this x+5=13

kkurt [141]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Please help ASAP!

Kruka [31]

First, recall that

$\sin(-\theta)=-\sin\theta$

So if $\sin(-\theta)=\dfrac15$ , then $\sin\theta=-\dfrac15$ .

Second,

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$

We know that $\tan\theta>0$ and $\sin\theta$ , which means we should also have $\cos\theta$ .

Third,

$\sin^2\theta+\cos^2\theta=1\implies\cos\theta=\pm\sqrt{1-\sin^2\theta}$

but as we've already shown, we need to have $\cos\theta$ , so we pick the negative root.

Finally,

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}\iff\dfrac6{\sqrt{12}}=\dfrac{-\frac15}{\cos\theta}\implies\cos\theta=-\dfrac1{5\sqrt3}$

Unfortunately, none of the given answers match, so perhaps I've misunderstood one of the given conditions... In any case, this answer should tell you everything you need to know to find the right solution from the given options.

6 0

3 years ago