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NikAS [45]
3 years ago
6

Write the equation of a line with a slope of -2 and a y-intercept of 5:

Mathematics
2 answers:
Fiesta28 [93]3 years ago
7 0

The equation of a line with a slope of -2 and a y-intercept of 5 is y = -2x + 5

<u>Solution:</u>

Given that slope is -2 and y-intercept is 5

We need to find the equation of line

The equation of line when slope and y-intercept is given can be calculated using "slope-intercept form"

<em><u>The slope intercept form is given as:</u></em>

y = mx + b

where "m" is the slope of line and "b" is the y-intercept

Plugging the given values in slope intercept form, we get

y = -2x + 5

Thus the required equation of line is found

saveliy_v [14]3 years ago
6 0
Answer: y = -2x +5 , so C (you probably just forgot to put the x)

Explanation: y = mx+b m represents the slope, b represents the y intercept
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Step-by-step explanation:

You just have to plug the numbers they give you into the equation. For the first one, you have 0 + 5y = 10; 5y = 10; y = 2. For the second one, x + 5(0) = 10; x + 0 = 10; x = 10. And lastly, for the third one, x + 5(6) = 10; x + 30 = 10; x = -20.

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3 years ago
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9966 [12]
Point slope formula is y= mx+b. To find the slope, or m, you need to find the "rise over run". rise = y coordinates, run = x coordinates. And the slope equation is y1-y2/x1-x2. So let's say the first point is (x1,y1) and the second is (x2,y2). that would be 35-(-31)/5-(-6)= 66/11 or 6/1 aka up six, across one. That is your slope. So far you have y=6x+b, next plug (5,35) into that equation and solve for b (aka the y intercept). So: 35=6(5)+b. 35-30=b, b=5. So your final equation is y=6x+5.
8 0
3 years ago
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}

The minimum initial velocity needed for the ball to clear the fence is 125.4\ \text{m/s}

5 0
3 years ago
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