1.43 x (-19/37) please answer
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Answers

Answer 1

Answer: 1.43 X (-19/37)=1.43 X -0.51=-0.73

Answer 2

Answer: 1.43 × (-19/37) = -0.734324

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Solve these systems of equations Using the elimination method.a). 3x + y = 13 b) 6x - y = 14
x - 5y = 15 2x + y = 12

c) 3x - 4y = 22
2x + 5y = 7

Answers

a. 3x + y = 13 → 15x + 5y = 65
x - 5y = 15 →   x - 5y = 15
16x = 80
16 16
x = 5
x - 5y = 15
5 - 5y = 15
- 5 - 5
-5y = 10
-5 -5
y = -2
(x, y) = (5, -2)

b. 6x - y = 14
  2x + y = 12
  8x = 26
8 8
  x = 3¹/₄
6x - y = 14
6(3¹/₄) - y = 14
19¹/₂ - y = 14
- 19¹/₂   - 19¹/₂
-y = -5¹/₂
-1 -1
y = 5¹/₂
  (x, y) = (3¹/₄, 5¹/₂)

c. 3x - 4y = 22 → 15x - 20y = 110
2x + 5y =   7 → 8x + 20y =   28
  23x = 138
23   23
x = 6
3x - 4y = 22
  3(6) - 4y = 22
18 - 4y = 22
- 18 - 18
-4y = 4
-4 -4
y = -1
(x, y) = (6, -1)

a). 3x + y = 13
x - 5y = 15
multiplier equation 1 by 5
15x+5y=65
DSA
15x+5y=65
x - 5y = 15
remaining
16x=80
x=5
5-5y=15
-5y=15-5
-5y=10
y=10 ÷ -5
y=-2
x,y
5,-2

Two forces of 30 pounds and 40 pounds act upon a body, forming an acute angle with each other. The angle between the resultant and the 30-pound force is 35degree 10'. Find, to the nearest ten minutes, the angle between the two given forces

Answers

angles in the triangle are
35°10'
180° - A (with A = angle between the 2 forces)
A - 35°10'

40/sin35°10' = 30/sin(A-35°10')
sin(A-35.17) = 30sin35.17°/40 = 0.432

A-35.17= 25.6
A = 60.77° = 60°46'
60°50'

In the inverse variation function, what happens to the output when the function's input value is divided by 5?