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
Answer:
The length of the lights is 2088 inches
Step-by-step explanation:
<em>The question is mixed up with another (See comment for correct question)</em>
Given
Required
The perimeter of the deck (this is what the question implies)
The perimeter (P) is:
We are given the following quadratic equation
The vertex is the maximum/minimum point of the quadratic equation.
The x-coordinate of the vertex is given by
Comparing the given equation with the general form of the quadratic equation, the coefficients are
a = 2
b = 7
c = -10
The y-coordinate of the vertex is given by
This means that we have a minimum point.
Therefore, the minimum point of the given quadratic equation is
Note: Consider we need to find the vertices of the triangle A'B'C'
Given:
Triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C'.
Triangle A,B,C with vertices at A(-3, 6), B(2, 9), and C(1, 1).
To find:
The vertices of the triangle A'B'C'.
Solution:
If triangle ABC is rotated 90 degrees clockwise about the origin to create triangle A'B'C', then
Using this rule, we get
Therefore, the vertices of A'B'C' are A'(6,3), B'(9,-2) and C'(1,-1).