The angle between the segment labled x and the chord labeled 30 is not specified. The measure of x cannot be determined, except to say that it is somewhere between 16 and the radius of the circle, √(16²+15²) = √481 ≈ 21.93.
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If the angle between x and the chord were marked as a right angle, one could say x=16, because all chords of the same length are the same distance from the center of the circle.
D
Divide the whole thing by 2
2(X^2+8x-5)
X^2+8x. =5
Half of 8 is 4 sq it so you add 16
X^2+8x+16= 5+16
(X+4)^2=21
Answer:
2kg
Step-by-step explanation:
Let the sum of the other 2 bricks be R. Given that the heaviest brick weighs 2/3 times as much as the other 2 bricks in total, then the weight of the heaviest brick H in terms of the weight of the other 2 will be
H = 2/3 * R
= 2R/3
Given that the 3 bricks weigh 5kg in total then,
2R/3 + R = 5
Multiply through the equation by 3
2R + 3R = 15
5R = 15
Divide both sides by 5
R = 15/5
= 3
The weight of the heaviest brick is 2R/3. Since R = 3, the heaviest block weighs
= 2 * 3/3
= 2kg
Step-by-step explanation:
From this figure TUW+WUV=TUV ; (5x+3)+(10x-5)=17x-16 so 15x-2=17x-16 ;14=2x therefore x=7
then substitute x on it.
TUW=5x+3=5(7)+3=35+3=38
WUV=10x-5=10(7)-5=70-5=65
TUV=17x-16=17(7)-16=119-16=103
