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crimeas [40]
3 years ago
10

How do i solve 4g + 2 ≥ - 14

Mathematics
1 answer:
Kryger [21]3 years ago
5 0

Answer:

g ≥ -4

Step-by-step explanation:

4g + 2 ≥ - 14

      -2      -2   Subtract 2 from both sides

4g ≥ -16

g ≥ -4      Divide both sides by 4

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If the sum of two numbers is 15 and their product is 50 what are they
Westkost [7]
Xy = 50
x + y = 15

x = 15 - y

xy = 50

(5 - y)y = 50

5y - y^2 = 50

y^2 - 5y + 50 = 0

(y - 5)(y - 10) = 0

y - 5 = 0 or y - 10 = 0

y = 5 or y = 10

x + y = 15

For y = 5: 

x + 5 = 15

x = 10

For y = 10

x + 10 = 15

x = 5

Answer: the numbers are 5 and 10.

5 0
3 years ago
Solve for y.<br> 3X+2y=12
Montano1993 [528]
3x + 2y = 12

2y = -3x + 12
  y = -1.5x + 6
3 0
3 years ago
Convert 65:36 in to a smaller number
VMariaS [17]
65:36 cannot be simplified. It would turn into a decimal number and we don’t put decimals in ratios
8 0
3 years ago
Answer with proper units too!
EastWind [94]

Answer:

Step-by-step explanation:

1. Area of a parallelogram = base * height

 area = 10 * 4 = 40 in squared

perimeter = 2(a + b)

perimeter = 2 * (5 + 10) = 2 * 15 = 30in

2. Area of triangle = 1/2 * base * height

area = 1/2 * 21 * 8 = 84 cm squared

perimeter = sum of outer perimeters

perimeter= 10cm + 17cm + 21cm = 48cm

3.  area of trapezium = 1/2 * (a+b) * h

 area = 1/2 * (6+12) * 4 = 36ft squared

 perimeter = sum of outer perimeters

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3 0
3 years ago
An angle bisector of a triangle divides the opposite side of the triangle into segments 5 cm and 3 cm long. A second side of the
attashe74 [19]
There is a little-known theorem to solve this problem.

The theorem says that
In a triangle, the angle bisector cuts the opposite side into two segments in the ratio of the respective sides lengths.

See the attached triangles for cases 1 and 2.  Let x be the length of the third side.

Case 1:
Segment 5cm is adjacent to the 7.6cm side, then
x/7.6=3/5  => x=7.6*3/5=4.56 cm

Case 2:
Segment 3cm is adjacent to the 7.6 cm side, then
x/7.6=5/3 => x=7.6*5/3=12.67 cm

The theorem can be proved by considering the sine rule on the adjacent triangles ADC and BDC with the common side CD and equal angles ACD and DCB.

6 0
3 years ago
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