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Y_Kistochka [10]
3 years ago
12

All of the techniques for finding a line's equation use the definition of slope, which is given the symbol m. slope is defined a

s the difference between the y coordinates of two points, divided by the difference between the x coordinates of those two points. that is, if you have two points (x1,y1) and (x2,y2) on a line, the slope is m=y2−y1x2−x1. you might be able to remember the definition more easily in the form "the difference in the y values over the difference in the x values." the end result that you want is an equation that looks like y=mx+b, where m is the slope and b is the y intercept—the value of y where the line intersects the y axis. for instance, you might have as the final result the equation y=2x+4. in the example given of distance walked, you could easily determine how far you had walked after 2.5 minutes if you had an equation for the graphed line.
Mathematics
1 answer:
a_sh-v [17]3 years ago
5 0

That's a nice story. It's mostly true, but the standard form of a line and the form of a line between two points each avoid a direct expression of the slope.

The standard form for a line is

ax + by = c

Either both not both <em>a</em> and <em>b</em> may be zero. When <em>b</em> is zero there's no slope, but it's a perfectly good line.

The line through (a,b) and (c,d) is

(c-a)(y-b) = (d-b)(x-a)

Again, we have no slope when a=c but a perfectly good equation and line.




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1. if m&lt;b = 60 what is AB<br><br><br>2. if m&lt;b =45 what is the perimeter of the trapezoid.​
nika2105 [10]

Answer:....

Step-by-step explanation:

first we need to find the base of the triangle, so the total is 20, they also give us 12, so we subtract 20-12=8 now divide. 8/2=4. Now we use the 4 and the 60. We use SOH-CAH-TOA, they give us the base now we need to find AB which is the hypotenuse. So we take out the TOA, and the 4 is the adjacent so we take out the SOH, we have CAH now, so we use cos. Now equation time....

cos60=\frac{4}{x}   since x is the bottom we switch \frac{4}{cos60}, which equals 8. AB=8.

Now to find the perimeter. We still use the 4 to find the hypotenuse so we  can find the outside side. For this one they tell us to use 45, so now lets see what method we use. SOH-CAH-TOA. We need the hypotenuse and they give us the adjacent, so we use cos again. Now setting up the equation...

cos45=\frac{4}{x}    x is at the bottom again to we switch.

\frac{4}{cos45}   which equals to 5.657 or you can round it. Now we add everything.

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5 0
3 years ago
The altitude drawn to the hypotenuse of a right triangle divides the hypotenuse into segments such that their lengths are in rat
Zanzabum

In geometry, it would be always helpful to draw a diagram to illustrate the given problem.

This will also help to identify solutions, or discover missing information.

A figure is drawn for right triangle ABC, right-angled at B.

The altitude is drawn from the right-angled vertex B to the hypotenuse AC, dividing AC into two segments of length x and 4x.


We will be using the first two of the three metric relations of right triangles.

(1) BC^2=CD*CA (similarly, AB^2=AD*AC)

(2) BD^2=CD*DA

(3) CB*BA = BD*AC


Part (A)

From relation (2), we know that

BD^2=CD*DA

substitute values

8^2=x*(4x) => 4x^2=64, x^2=16, x=4

so CD=4, DA=4*4=16 (and AC=16+4=20)


Part (B)

Using relation (1)

AB^2=AD*AC

again, substitute values

AB^2=16*20=320=8^2*5

=>

AB

=sqrt(8^2*5)

=8sqrt(5)

=17.89 (approximately)

6 0
4 years ago
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