$\displaystyle\frac{n^2-10n+24}{n^2-13n+42}-\frac{9}{n-7}=\frac{(n-6)(n-4)}{(n-6)(n-7)}-\frac{9}{n-7}\\\\=\frac{n-4}{n-7}-\frac{9}{n-7}=\frac{n-4-9}{n-7}\\\\=\frac{n-13}{n-7}$

The last choice is appropriate.

3 0

3 years ago

Please help ASAP. Need it for homework

Slav-nsk [51]

Speedy Taxis
you will put 30 instead of X in each equation and see which one is the cheapest
you put 30 beacuse it is said 30 miles and also said That X is the mile
Do you understand?

6 0

3 years ago

Find cos(2*ABC) 100POINTS

juin [17]

Answer:

$-\dfrac{7}{25}$

Step-by-step explanation:

<u>Trigonometric Identities</u>

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

<u>Trigonometric ratios</u>

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

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:

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

7 0

2 years ago

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