Which expression is a factor of 3xy + 2x − 18y − 12?

Mathematics

2 answers:

trapecia [35]3 years ago

6 0

X(3y+2)-6(3y+2)
(x-6)(3y+2)
so the answer is 2+3y

cluponka [151]3 years ago

3 0

Answer: The correct option is (C) 2 + 3y.

Step-by-step explanation: We are given to select the expression that is a factor of the following expression :

$E=3xy+2x-18y-12.$

To select the correct factor, first we need to factorize the given expression.

The factorization is as follows :

$E\\\\=3xy+2x-18y-12\\\\=x(3y+2)-6(3y+2)\\\\=(x-6)(3y+2)\\\\=(x-6)(2+3y).$

Thus, the required factors of the given expression are (x - 6) and (3y + 2).

The correct option is (C) 2 + 3y.

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Read 2 more answers

Find cos(2*ABC) 100POINTS

juin [17]

Answer:

$-\dfrac{7}{25}$

Step-by-step explanation:

Trigonometric Identities

$\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B$

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

$\cos(\angle ABC)=\dfrac{3}{5}$

$\sin(\angle ABC)=\dfrac{4}{5}$

Therefore, using the trig identities and ratios:

$\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}$