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nordsb [41]
3 years ago
15

Write the equation of the line that passes through (-4, 8) and is perpendicular to 5x - 10y = -12. The equation of the line in s

lope-intercept form is _____.
Mathematics
1 answer:
arlik [135]3 years ago
8 0

Answer:

y= -2x

Step-by-step explanation:

To find the slope, convert the standard form of the equation into slope intercept form, y=mx+b.

5x-10y=-12

-10y=-5x-12

y=1/2 x + 1.2

The slope is 1/2 and the slope of the new line will be -2 because the negative reciprocal create a perpendicular line. Use the point-slope form to write the equation, then simplify and convert into the slope intercept form.

(y-8)=-2(x--4)

y-8=-2(x+4)

y-8=-2x -8

y= -2x +0

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49x - 21y = -1,470<br><br> put in slope intercept form
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Answer:

Slope intercept form:

<em>y = mx + b</em>

What we are given:

49x - 21y = -1,470

we need to keep y on the left side so we subtract 49x from both sides;

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2 years ago
Given: sin (A) = 5/13, pi/2 &lt; A &lt; pi and tan (B) = - sqr root 13, pi/2 &lt; B &lt;. What is tan(A - B)? A. (5 + 12sqr root
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3 years ago
Hey I know this is kind of a lot but can someone pls help.
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100 POINTS
Shkiper50 [21]

Answer:

  6√3 ±3 ≈ {7.392, 13.392}

Step-by-step explanation:

The length of AB is the long side of a right triangle with hypotenuse CD and short side (AC -BD). The desired radius values will be half the length of EF, with AE added or subtracted.

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<h3>length of AB</h3>

Radii AC and BD are perpendicular to the points of tangency at A and B. They differ in length by AC -BD = 12 -9 = 3 units.

A right triangle can be drawn as in the attached figure, where it is shaded and labeled with vertices A, B, C. Its long leg (AB in the attachment) is the long leg of the right triangle with hypotenuse 21 and short leg 3. The length of that leg is found from the Pythagorean theorem to be ...

  AB = √(21² -3²) = √432 = 12√3

<h3>tangent circle radii</h3>

This is the same as the distance EF. Half this length, 6√3, is the distance from the midpoint of EF to E or F. The radii of the tangent circles to circles E and F will be (EF/2 ±3). Those values are ...

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