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2 years ago

What is -5(4x + 3) = -5x - 5 - 15x - 10?

Mathematics

2 answers:

Ber [7]2 years ago

7 0

Answer:

Step-by-step explanation:

-5(4x + 3) = -5x - 5 - 15x - 10

-5*4x - 5*3 = -5x - 5 - 15x - 10

-20x - 15 = -5x - 5 - 15x - 10

-20x + 5x + 15x = -5 -10 - 15

0= 0

inysia [295]2 years ago

7 0

Answer:

Infinite Solutions

Step-by-step explanation:

Firstly we distribute

-20x - 15 = -5x - 5 - 15x - 10

Collect like terms

-20x - 15 = -20x - 15

This represents infinite solution because they will result in 0.

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Westkost [7]

Answer:

$\displaystyle \cos\left(-\frac{7\,\pi}{12}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}$ .

Step-by-step explanation:

Convert the angle $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ to degrees:

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ .

Note, that $\left(-105^\circ\right)$ is the sum of two common angles: $\left(-45^\circ\right)$ and $\left(-60^\circ\right)$ .

$\displaystyle \cos\left(-45^\circ\right) = \cos\left(45^\circ\right) = \frac{\sqrt{2}}{2}$ .
$\displaystyle \cos\left(-60^\circ\right) = \cos\left(60^\circ\right) = \frac{1}{2}$ .

$\displaystyle \sin\left(-45^\circ\right) = -\sin\left(45^\circ\right) = -\frac{\sqrt{2}}{2}$ .
$\displaystyle \sin\left(-60^\circ\right) = -\sin\left(60^\circ\right) = -\frac{\sqrt{3}}{2}$ .

By the sum-angle identity of cosine:

$\cos(A + B) = \cos(A)\cdot \cos(B) - \sin(A) \cdot \sin(B)$ .

Apply the sum formula for cosine to find the exact value of $\cos\left(-105^\circ \right)$ .

$\begin{aligned}\cos\left(-105^\circ \right) &= \cos\left(\left(-45^\circ\right) + \left(-60^\circ\right)\right) \\ &= \cos\left(-45^\circ\right) \cdot \cos\left(-60^\circ\right)\right) - \sin\left(-45^\circ\right) \cdot \sin\left(-60^\circ\right)\right) \\ &= \frac{\sqrt{2}}{2} \times \frac{1}{2} - \left(-\frac{\sqrt{2}}{2}\right)\times \left(-\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{2} - \sqrt{6}}{4}\end{aligned}$ .

$\displaystyle \left(-\frac{7\, \pi}{12}\right) = \left(-\frac{7\, \pi}{12}\right) \times \frac{180^\circ}{\pi} = -105^\circ$ . In other words, $\displaystyle \left(-\frac{7\, \pi}{12}\right)$ and $\left(-105^\circ\right)$ correspond to the same angle. Therefore, the cosine of $\displaystyle \left(-\frac{7\, \pi}{12}\right)\!$ would be equal to the cosine of $\left(-105^\circ\right)\!$ .