Answer:
an isosceles right triangle
Step-by-step explanation:
The square of the length of a side can be found from the distance formula:
d^2 = (x2-x1)^2 +(y2-y1)^2
The square of the length of WX is ...
WX^2 = (-3-(-10))^2 +(-1-4)^2 = 49+25 = 74
The square of the length of XY is ...
XY^2 = (-5-(-3))^2 +(11-(-1))^2 = 4 +144 = 148
The square of the length of YW is ...
YW^2 = (-10-(-5))^2 +(4 -11)^2 = 25 +49 = 74
The sum of the squares of the short sides is equal to the square of the long side, so this is a right triangle. The squares of the short sides are equal, so this is an isosceles right triangle.
Cos(A-B) = cosAcosB + sinAsinB
<span>
cos(</span>π/2 - θ) = cos(π/2)cosθ + sin(π/2)sinθ
π/2 = 90°
cos(π/2) = cos90° = 0. sin(π/2) = sin90° = 1
cos(π/2 - θ) = cos(π/2)cosθ + sin(π/2)sin<span>θ
</span>
= 0*cosθ + 1*sin<span>θ = </span>sin<span>θ
Therefore </span>cos(π/2 - θ) = sin<span>θ
QED </span>
<h3>
Answer: -20</h3>
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Work Shown:
Let x be the location of E on the number line.
Since C is the midpoint of E and F, this means we can find C's location by adding E and F together and dividing that sum by 2
midpoint = (endpoint1 + endpoint2)/2
C = (E+F)/2
Plug in E = x, C = -8 and F = 4. Then solve for x
C = (E+F)/2
-8 = (x+4)/2
(x+4)/2 = -8
x+4 = 2(-8) .... multiplying both sides by 2
x+4 = -16
x = -16-4 .... subtract 4 from both sides
x = -20
The location of point E on the number line is -20
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As a check, lets add E and F to get E+F = -20+4 = -16
Then cut this in half to get -16/2 = -8, which is the proper location of point C
This confirms our answer.
Answer:
b
Step-by-step explanation:
because it passes the vertical line test and A does not if you look at x = 1 there are 2 points along that point (1,3) (1,-1)
Answer:
Infinite number of solutions
Step-by-step explanation:
when you solve for V, you notice that the term in "v" goes away, and you end up with a true statement:
2 = 2
This is a true statement no matter what values the variable V has, so it is true for all possible (infinite) values of "v".
