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3 years ago

Passing through (-2, 8) and PERPENDICULAR to y = 2x + 5

Mathematics

1 answer:

tangare [24]3 years ago

7 0

Answer:

$y=\frac{-1}{2} x+7$

Step-by-step explanation:

The slope of a perpendicular line will always be the NEGATIVE RECIPROCAL of the slope of the original line.

y=2x+5

Because this equation is organized in slope-intercept form, y=mx+b, we can identify the slope of the original line is 2. The negative reciprocal of 2 is $\frac{-1}{2}$ . Therefore, the slope of the perpendicular line is $\frac{-1}{2}$ . So far, our equation looks like this:

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

Now, we must solve for b, the y-intercept of this line. To do that, we can plug in the given point (-2,8) and solve for b:

$y=\frac{-1}{2}x+b\\8=\frac{-1}{2}(-2)+b\\8=\frac{2}{2} +b\\8=1+b$

Subtract both sides by 1 to isolate b

$8-1=1+b-1\\7=b$

Therefore, the y-intercept of this line is 7.

After plugging both the slope and the y-intercept into y=mx+b, we get:

$y=\frac{-1}{2} x+7$

I hope this helps!

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3 years ago

At time t equals0, a particle is located at the point (2,1,6). It travels in a straight line to the point (9,8,5), has

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Answer:

$r(t)=$

Step-by-step explanation:

We are given that

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$\mid \mid=8$

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There exist a constant s such that

$v_1=7s, v_2=7s, v_3=-s$

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$r(t)=\int dt$

$r(t)= +r_0$

Where $r_0=r(0)=$ =Some constant vector

$r_0=$

Substitute the value

$r(t)=+$

$r(t)=$

This is required equation of the position vector r(t) of the particle at time t.

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3 years ago

PLZZ HELP!

shtirl [24]

Answer:

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