Let AB extended intersect DC extended at point E
<span>We now have right triangle BEC with E = 90 degrees </span>
<span>For triangle BEC: </span>
<span>Exterior angle at E = 90 </span>
<span>Exterior angle at C = 148 (given) </span>
<span>Exterior angle of all polygons add up to 360 degrees </span>
<span>Exterior angle at B = 360−148−90 = 122 </span>
<span>So in quadrilateral ABCD </span>
<span>B = 122 </span>
<span>D = 360−44−148−122 = 46</span>
Jamal because he has a higher percentage of shots made 78%>62%
<h3>Answer:</h3>
Equation of the ellipse = 3x² + 5y² = 32
<h3>Step-by-step explanation:</h3>
<h2>Given:</h2>
- The centre of the ellipse is at the origin and the X axis is the major axis
- It passes through the points (-3, 1) and (2, -2)
<h2>To Find:</h2>
- The equation of the ellipse
<h2>Solution:</h2>
The equation of an ellipse is given by,

Given that the ellipse passes through the point (-3, 1)
Hence,

Cross multiplying we get,
- 9b² + a² = 1 ²× a²b²
- a²b² = 9b² + a²
Multiply by 4 on both sides,
- 4a²b² = 36b² + 4a²------(1)
Also by given the ellipse passes through the point (2, -2)
Substituting this,

Cross multiply,
- 4b² + 4a² = 1 × a²b²
- a²b² = 4b² + 4a²-------(2)
Subtracting equations 2 and 1,
- 3a²b² = 32b²
- 3a² = 32
- a² = 32/3----(3)
Substituting in 2,
- 32/3 × b² = 4b² + 4 × 32/3
- 32/3 b² = 4b² + 128/3
- 32/3 b² = (12b² + 128)/3
- 32b² = 12b² + 128
- 20b² = 128
- b² = 128/20 = 32/5
Substituting the values in the equation for ellipse,


Multiplying whole equation by 32 we get,
3x² + 5y² = 32
<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
6 and 8 I believe is the answer