Answer:
$79 is the answer.
The answer is hard to explain, but I asked someone if it was right, and it is.
Hope this helps, - Good Luck!
It is a binomial. Two terms: 3x, 1
Let the weight of triangle be x units. You are given that the weight of square is 1 unit.
1) On the left side of the balanced beam you can see 3 triangles and 5 squares. The weight on left side is 3·x+5·1=3x+5 units.
2) On the right side of the balanced beam you can see 2 triangles and 7 squares. The weight on right side is 2·x+7·1=2x+7 units.
3) If the whole system is balanced, then the weights on left and right sides are equal:
3x+5=2x+7.
Solve this equation:
3x-2x=7-5,
x=2 units.
Answer: option A
The answer is o there is no value of x to make the equation true.
Answer:
46. ΔQRP ≅ ΔUTV
47. y = 42
48. ∠KMN = 101°
Step-by-step explanation:
46. It is convenient to name the vertices in the order of the number of arc identifying the angle. Corresponding vertices need to be named in the same order in the congruence statement.
47. From the sum of angles of a triangle, you know that ...
x +62 +65 = 180
x = 180 -127 = 53
x - y = 11 since the corresponding sides are the same length
53 - y = 11 . . . . . .substitute the value of x
y = 53 - 11 . . . . . . add y-11
y = 42
48. The exterior angle has the same measure as the sum of the opposite interior angles:
81 +4x = 13x +36
45 = 9x . . . . . . . . . add -4x-36
5 = x . . . . . . . . . . . divide by 9
∠KMN = (13x +36)° = (13·5 +36)° = 101°