Answer:
(x+y)² =(x+y)(x+y) Then you FOIL (First, outer, inner, last)
(x+y)² =(x+y)(x+y) = xx + xy + xy + yy [and when you combine like terms] = x² + 2xy + y²
(x+y)3 = (x² + 2xy + y²)(x+y) Then you FOIL (First, outer, inner, last)
(x+y)3 = (x² + 2xy + y²)(x+y) = x2x +2xxy + xy2 + x2y + 2xyy + y2y [and when you combine like terms] = x3 + 3x2y+ 3xy2 + y3
Step-by-step explanation:
The line drops 3 squares for each one it moves to the right, so the slope is
.. -3/1 = -3
The 1st selection is appropriate.
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Given the choices, you can tell at a glance that -3 is the correct one. When it goes down to the right, the slope is negative. when the angle is steep, the slope has a magnitude greater than 1. Of the negative choices offered, {-3, -1/3}, the obvious choice is -3.
A line has a slope of ±1 when it makes a 45° angle with the x- and y-axes. Steeper than that is a slope whose magnitude is greater than 1. Shallower than that, and the slope has a magnitude less than 1.
Answer:
- 5.8206 cm
- 10.528 cm
- 23.056 cm^2
Step-by-step explanation:
(a) The Law of Sines can be used to find BD.
BD/sin(48°) = BD/sin(50°)
BD = (6 cm)(sin(48°)/sin(60°)) ≈ 5.82064 cm
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(b) We can use the Law of Cosines to find AD.
AD^2 = AB^2 +BD^2 -2·AB·BD·cos(98°) . . . . . angle ABD = 48°+50°
AD^2 ≈ 110.841
AD ≈ √110.841 ≈ 10.5281 . . . cm
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(c) The area of ∆ABD can be found using the formula ...
A = ab·sin(θ)/2 . . . . . where a=AB, b=BD, θ = 98°
A = (8 cm)(5.82064 cm)sin(98°)/2 ≈ 23.0560 cm^2
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Angle ABD is the external angle of ∆BCD that is the sum of the remote interior angles BCD and BDC. Hence ∠ABD = 48° +50° = 98°.
The answer would be B 21/4
Look at the attached picture
Hope it will help you
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