All categories

2 years ago

12. Determine the slope of the line having the equation 2x + y = 12

Mathematics

2 answers:

olchik [2.2K]2 years ago

7 0

Answer:

b. -2

Step-by-step explanation:

2x+y=12

add -2x to both sides

y = -2x + 12

this is in slope intercept form y = mx + b

m meaning the slope

in this case the slope is -2

the answer will be b. -2

Virty [35]2 years ago

4 0

Answer:

b. -2

Step-by-step explanation:

Note the slope-intercept form: y = mx + b

Note that:

y = (x , y)

m = slope

x = (x , y)

b = y-intercept

First, isolate the variable, y. Note the equal sign, what you do to one side, you do to the other. Subtract 2x from both sides of the equation:

2x (-2x) + y = (-2x) + 12

y = -2x + 12

Remember, m is the variable for the term that stands for the slope (directly next to the x variable). In this case, your answer will be -2

b. -2 is your answer.

You might be interested in

Doug hits a baseball straight towards a 15 ft high fence that is 400 ft from home plate. The ball is hit 2.5ft above the ground

Scorpion4ik [409]

Answer:

$125.4\ \text{m/s}$

Step-by-step explanation:

u = Initial velocity of baseball

$\theta$ = Angle of hit = $30^{\circ}$

x = Displacement in x direction = 400 ft

y = Displacement in y direction = 15 ft

$y_0$ = Height of hit = 2.5 ft

$a_y$ = g = Acceleration due to gravity = $32.2\ \text{ft/s}^2$

t = Time taken

Displacement in x direction

$x=u_xt\\\Rightarrow x=u\cos\theta t\\\Rightarrow t=\dfrac{x}{u\cos\theta}\\\Rightarrow t=\dfrac{400}{u\cos30^{\circ}}\\\Rightarrow t=\dfrac{400}{u\dfrac{\sqrt{3}}{2}}\\\Rightarrow t=\dfrac{800}{u\sqrt{3}}$

Displacement in y direction

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow y=y_0+u\sin\theta t+\dfrac{1}{2}a_yt^2\\\Rightarrow 15=2.5+u\sin30^{\circ}(\dfrac{800}{u\sqrt{3}})+\dfrac{1}{2}\times -32.2\times (\dfrac{800}{u\sqrt{3}})^2\\\Rightarrow \dfrac{400}{\sqrt{3}}-\dfrac{10304000}{3u^2}-12.5=0\\\Rightarrow 218.44=\dfrac{10304000}{3u^2}\\\Rightarrow u=\sqrt{\dfrac{10304000}{3\times218.44}}\\\Rightarrow u=125.4\ \text{m/s}$