Answer pls! No spam!

If anyone needs someone to talk to its ok and im here if you need me :)

umka2103 [35]

Answer:

hi

Step-by-step explanation:

um sometimes i feel like i always have to be perfect because of my parents high expectations for me,but sometimes i can live up to those expectations and i feel like i failed them and i feel like i always have to be perfect my hair my clothes my grades i hate it . i try to be me but i end up doing something wrong and i just feel sad.

4 0

2 years ago

Line m has the equation y = 1/2x - 4

Mandarinka [93]

Answer:

As the slopes of both lines 'm' and 'n' are the same.

Therefore, we conclude that the equation x-2y=4 represents the equation of the line 'n' if lines m and n are parallel to each other.

Step-by-step explanation:

We know that the slope-intercept of line equation is

$y=mx+b$

Where m is the slope and b is the y-intercept

Given the equation of the line m

y = 1/2x - 4

comparing with the slope-intercept form of the line equation

y = mx + b

Therefore,

The slope of line 'm' will be = 1/2

We know that parallel lines have the 'same slopes, thus the slope of the line 'n' must be also the same i.e. 1/2

Checking the equation of the line 'n'

$x-2y=4$

solving for y to writing the equation in the slope-intercept form and determining the slope

$x-2y=4$

Add -x to both sides.

$x - 2y + (-x) = 4+(-x)$

$-2y = 4 - x$

Divide both sides by -2

$\frac{-2y}{-2}=\frac{-x+4}{-2}$

$y=\frac{1}{2}x-2$

comparing ith the slope-intercept form of the line equation

Thus, the slope of the line 'n' will be: 1/2

As the slopes of both lines 'm' and 'n' are the same.

Therefore, we conclude that the equation x-2y=4 represents the equation of the line 'n' if lines m and n are parallel to each other.

3 0

2 years ago

A projectile is launched from the ground at an angle of theta above the horizontal with an initial speed of v in ft/s. The rang

VARVARA [1.3K]

Answer:

$\theta=24.4^\circ$

Step-by-step explanation:

The formula you wrote has the fraction the other way around. You can find that the range of a projectile is in reality approximated by the equation $x=\frac{v^2}{g} sin(2\theta)=\frac{v^2}{32ft/s^2} sin(2\theta)$ , where we will use ft for distances.

From the given equation we have then $\frac{x32ft/s^2}{v^2} =sin(2\theta)$ , which means $\theta=\frac{Arcsin(\frac{x32ft/s^2}{v^2})}{2}$ since the arcsin is the inverse function of the sin.

Since we have x = 170 ft and v = 85 ft/s, we can substitute these values from the equation written, and we will have $\theta=\frac{Arcsin(\frac{(170ft)(32ft/s^2)}{(85ft/s)^2})}{2}$ , and from now on we have just to use a calculator, obtaining $\theta=\frac{Arcsin(0.75294117647)}{2}=\frac{48.84579804^\circ}{2}=24.4^\circ$