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3 years ago

Sec^2x+2sec=0. solve the equation for 0

1 answer:

8 0

U = sec x ( substitution )
u² + 2 u = 0
u ( u + 2 ) = 0
u = 0, or u = - 2
1/cos x = 0 1/cos x = - 2
(not accepted ) cos x = -1/2
x 1 = 2π/3, x 2 = 4π/3

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The x value is the domain

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3 years ago

1. Write the standard form of the line that passes through the given points. (7, -3) and (4, -8)

Sveta_85 [38]

Answer:

1. $-5x+3y+44=0$

2. $2x+y-2=0$

3. $2x+y-4=0$

Step-by-step explanation:

Standard form of a line is $Ax+By+C=0$ .

If a line passing through two points then the equation of line is

$y-y_1=m(x-x_1)$

where, m is slope, i.e., $m=\dfrac{y_2-y_1}{x_2-x_1}$ .

The line passes through the points (7,-3) and (4,-8). So, the equation of line is

$y-(-3)=\dfrac{-8-(-3)}{4-7}(x-7)$

$y+3=\dfrac{-5}{-3}(x-7)$

$y+3=\dfrac{5}{3}(x-7)$

$3(y+3)=5(x-7)$

$3y+9=5x-35$

$-5x+3y+9+35=0$

$-5x+3y+44=0$

Therefore, the required equation is $-5x+3y+44=0$ .

We need to find the equation of the line, in standard form, that has a y-intercept of 2 and is parallel to $2 x + y =-5$ .

Slope of the line : $m=\dfrac{-\text{Coefficient of x}}{\text{Coefficient of y}}=\dfrac{-2}{1}=-2$

Slope of parallel lines are equal. So, the slope of required line is -2 and it passes through the point (0,2).

Equation of line is

$y-2=-2(x-0)$

$y-2=-2x$

$2x+y-2=0$

Therefore, the required equation is $2x+y-2=0$ .

We need to find the equation of the line, in standard form, that has an x-intercept of 2 and is parallel to $2x + y =-5$ .

From part 2, the slope of this line is -2. So, slope of required line is -2 and it passes through the point (2,0).

Equation of line is

$y-0=-2(x-2)$

$y=-2x+4$

$2x+y-4=0$

Therefore, the required equation is $2x+y-4=0$ .

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3 years ago

Solve 3x ≤ -6. Graph the solution.

Alinara [238K]

$as \: given \: \: 3x \leqslant - 6 \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: so \: \frac{3x}{3} \leqslant \frac{ - 6}{3} \: \ \\ results \: in \: x \leqslant - 2$