b) parallel to the line 2y-6x=9, and that passes through point (-2,1)
c) parallel to the line y=4x+3, and that has the same y-intercept as the line y=5x-3
d) parallel to the line -1/2x, and that has the same y-intercept as the line 2y=7x+6
e) perpendiclar to the line 2y-4x=8, and that passes through point (6,-6)
f) perpendicular to the line 2y-5x=15, and has an x-intercept of 3
please help me with all the questions it would mean alot
Answer:
25.6 units
Step-by-step explanation: From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).
First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:
d=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2}}
where
(x_1,y_1) are the coordinates of the first point
(x_2,y_2) are the coordinates of the second point
- For AB:
d=\sqrt{[1-(-5)]^{2}+(4-4)^2}
d=\sqrt{(1+5)^{2}+(0)^2}
d=\sqrt{(6)^{2}}
d=6
- For BC:
d=\sqrt{(3-1)^{2} +(-4-4)^{2}}
d=\sqrt{(2)^{2} +(-8)^{2}}
d=\sqrt{4+64}
d=\sqrt{68}
d=8.24
- For AC:
d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}
d=\sqrt{(3+5)^{2} +(-8)^{2}}
d=\sqrt{(8)^{2} +64}
d=\sqrt{64+64}
d=\sqrt{128}
d=11.31
Next, now that we have our lengths, we can add them to find the perimeter of our triangle:
p=AB+BC+AC
p=6+8.24+11.31
p=25.55
p=25.6
We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.
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Answer:
could you give what the problem is
Step-by-step explanation: