If the whole angle (ABC) is equal to 133, we can add angles ABD and DBC to find x.
10x+13=133
Subtract 13 from both sides to get
10x=120
Divide both sides by 10 to get that x=12
We can now plug the value of x into ABD to find its measure.
7(12)+4
Simplifies to 84+4, which is 88
So angle ABD is equal to 88 degrees.
:)
A triangle<span> can't have more than one </span>right angle<span> to begin with. If it had 2 </span>rightangles<span>, then it is no longer a </span>triangle<span>. The </span>angles<span> of the </span>triangle<span> ha…ve to add up to 180 degrees and a </span>right angle<span> is 90 degrees. So if you had two </span>right angles<span> you would already be at 180 degrees. hope </span>that helped
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Answer:
1. B, E
2. A, E
16. B'(1, 5)
Step-by-step explanation:
Lines p and r are parallel, as are lines a and b. Either of the lines in one parallel pair is perpendicular to the lines in the other parallel pair. Line q has no necessary relation to anything, so can be ignored. (No answer containing q can be correct.)
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1. p ⊥ b
r ⊥ a
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2. r║p
a║b
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16. Point B(3, 2) translated (right, up) = (-2, 3) will be ...
B' = B + (-2, 3) = (3, 2) +(-2, 3) = (3 -2, 2 +3)
B' = (1, 5)
Note: (x, y) coordinates are (<em>right</em>, <em>up</em>). So, in translation problems it is important to <em>pay attention.</em> Often, the up/down translation is listed before the left/right translation. If you're not paying attention, you may inadvertently reverse the numbers that need to be added to the coordinates to do the translation.
Answer:
critical point of the given function f(x,y) = x²+y²+2xy is from line y = -x is the critical point of the function f(x0,y0) = 0
and it local minimum.
Step-by-step explanation:
Let the given function be;
f(x,y) = x²+y²+2xy
From above function, we can locate relative minima, maxima and the saddle point
f(x,y) = x²+y²+2xy = (x+y)²
df/dx = 2x+2y = 0 ---- (1)
df/dy =2y+2x = 0 ---- (2)
From eqn 1 and 2 above,
The arbitrary point (x0,y0) from line y = -x is the critical point of the function f(x0,y0) = 0
Then, from f(x,y) >= 0 for arbitrary (x,y) € R^n, the arbitrary point from the line x = -y is local minima of the function f.
Answer:
It's c :) Hope this helps
