Answer:
Given all right triangles
<u>Use Pythagorean to find the missing side:</u>
- a² + b² = c², where c- hypotenuse
=1=
- x² = 6² + 3² = 45
- x = √45
- x = 6.7
=2=
- x² = 12² + 5² = 169
- x = √169
- x = 13
=3=
- x² = 2² + 5² = 29
- x = √29
- x = 5.4
Answer:
I attached it.
Step-by-step explanation:
This is the graph.
Answer:
12x +8y +1 = 0
Step-by-step explanation:
You can write the equation by swapping the x- and y-coefficients and negating one of them. Then compute the constant that makes the line go through the given point. You can do that like this:
3(x-(-3/4)) +2(y -1) = 0
3x +9/4 +2y -2 = 0 . . . . eliminate parentheses
3x +2y +1/4 = 0 . . . . . . . simplify
To eliminate the fraction, multiply by 4:
12x +8y +1 = 0
_____
<em>Comment on the equation</em>
The line ax+by=0 passes through the origin. Replacing x and y with x-h and y-k, respectively, makes the line pass through the point (h, k). That's what we did above to make the line pass through the given point.
The business of swapping coefficients and negating one causes the slope of the new line to be the negative reciprocal of the slope of the original line. That is what makes the new line perpendicular to the original.
Answer:
Option C.
The distance from the point to the circle is 6 units
Step-by-step explanation:
we know that
The distance between the circle and the point will be the difference of the distance of the point from the center of circle and the radius of the circle
step 1
Find the center and radius of the circle
we have

Convert to radius center form
Complete the square

Rewrite as perfect squares

so
The center is the point (0,4)
The radius is

step 2
Find the distance of the point from the center of circle
the formula to calculate the distance between two points is equal to

we have
(6,-4) and (0,4)
substitute the values




step 3
Find the difference of the distance of the point from the center of circle and the radius of the circle

therefore
The distance from the point to the circle is 6 units
see the attached figure to better understand the problem