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3 years ago

Solve the equation

Mathematics

1 answer:

erastova [34]3 years ago

6 0

Look at the picture.

$|a|=\left\{\begin{array}{ccc}a&for&a\geq0\\-a&for&a < 0\end{array}\right\\\\|9-8x|=\left\{\begin{array}{ccc}9-8x&for&x\in\left(-\infty,\ \dfrac{9}{8}\right]\\-(9-8x)&for&x\in\left(\dfrac{9}{8},\ \infty\right)\end{array}\right$

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P= -24 q= 12 r= -6 a. (pxq):(q+r) b. (pxr):(r-q) what the answer?

kolezko [41]

Step-by-step explanation:

a. ( p × q ) ( q + r )

= ( -24 × 12 ) ( 12 + -6 )

= 288 × 18

= 5184

b. ( p × r ) ( r - q )

= ( -24 × -6 ) ( -6 - 12 )

= ( 144 ) ( -18 )

= -2592

5 0

3 years ago

Question is in the image

storchak [24]

Answer:

Step-by-step explanation:

y = -2x + 3

-1 = -2(2) + 3

-1 = -4 + 3

-1 = -1

y = x - 3

-1 = 2 - 3

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8 0

3 years ago

9. A box contains three disks painted gold (81,82,83), three disks painted silver (s1, 32, 33) and one

Illusion [34]

Answer:

Step-by-step explanation:

7 0

3 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .