Please look at the question in the picture it’s a statistics problem
1 answer:
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Answer: The hook would be 2.2 inches (approximately) above the top of the frame
Step-by-step explanation: Please refer to the picture attached for further details.
The top of the picture frame has been given as 9 inches and a 10 inch ribbon has been attached in order to hang it on a wall. The ribbon at the point of being hung up would be divided into 5 inches on either side (as shown in the picture). The line from the tip/hook down to the frame would divide the length of the frame into two equal lengths, that is 4.5 inches on either side of the hook. This would effectively give us two similar right angled triangles with sides 5 inches, 4.5 inches and a third side yet unknown. That third side is the distance from the hook to the top of the frame. The distance is calculated by using the Pythagoras theorem which states as follows;
AC^2 = AB^2 + BC^2
Where AC is the hypotenuse (longest side) and AB and BC are the other two sides
5^2 = 4.5^2 + BC^2
25 = 20.25 + BC^2
Subtract 20.25 from both sides of the equation
4.75 = BC^2
Add the square root sign to both sides of the equation
2.1794 = BC
Rounded up to the nearest tenth, the distance from the hook to the top of the frame will be 2.2 inches
<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)
- The equation of the ellipse
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>