Given:
∠JEK + ∠DFH = 90°
Find:
relationship, if any, of ∠JEK to angles CFL, HFG, LFE
Solution:
We note that ∠DFG = 90° = ∠DFH + ∠HFG, so ∠JEK ≅∠HFG.
We also note that ∠HFG and ∠CFL are vertical angles, so ∠JEK ≅∠CFL.
And ∠LFE and ∠DFH are vertical angles, so ∠JEK + ∠LFE = 90°.
Evaluating the choices, we find
A) false. The angles are not complementary, they are equal.
B) false. The angles are not equal, they are complementary.
C) false. The angles are not supplementary, they are equal.
D) TRUE. The angles are equal.
The appropriate choice is ...
D) ∠JEK ≅∠CFL
Insert the point, (5,-2) with the slope of the line to figure out the equation. You can either use point-slope form, or plug it for for slope-intercept form.
Slope-intercept form:
-2 = -2(5) + b
-2 = -10 + b
Add 10 on both sides
8 = b
Thus,
y = -2x + 8
Answer is B
Answer:
y= 2x -3
Step-by-step explanation:
Let's rewrite the given equation into the form of y=mx+c, so that we can find the gradient of the line. In this form, m (coefficient of x) is the gradient.
4x -2y= 3
2y= 4x -3
<em>Divide</em><em> </em><em>by</em><em> </em><em>2</em><em> </em><em>throughout</em><em>:</em>

Thus the gradient is 2.
Parallel lines have the same gradient thus the line would also have a gradient of 2.
Substitute m=2 into the equation:
y= 2x +c
To find the value of c, substitute a pair of coordinates.
When x=2, y=1,
1= 2(2) +c
1= 4 +c
c= 1 -4
c= -3
Thus, the equation of the line is y= 2x -3.
Answer: 16.97
<u>Step-by-step explanation:</u>
To find the distance from W to X, you can use the distance formula ...or... draw a right triangle using W and X as the vertices and use Pythagorean Theorem to find the hypotenuse. I will solve using the latter.
The y-distance from W to X is 12.
The x-distance from W to X is 12.
hypotenuse² = 12² + 12²
= 144 + 144
= 288
hypotenuse = 
= 16.97