A figure with length and width but no depth is called
2 answers:
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Part 1
We are given
. This can be rewritten as 
.
Therefore, a=1, b=-18, c=0.
Using the quadratic formula
The values of x are
Part 2
Since the values of y change drastically for every equal interval of x, the function cannot be linear. Therefore, the kind of function that best suits the given pairs is a quadratic function.
Part 3.
The first equation is
.
The second equation is
.
We have
Factoring, we have
Equating both factors to zero.
When the value of x is 6, the value of y is
When the value of x is -3, the value of y is
Therefore, the solutions are (6,38) or (-3,11)
We are given
Therefore, a=1, b=-18, c=0.
Using the quadratic formula
The values of x are
Part 2
Since the values of y change drastically for every equal interval of x, the function cannot be linear. Therefore, the kind of function that best suits the given pairs is a quadratic function.
Part 3.
The first equation is
The second equation is
We have
Factoring, we have
Equating both factors to zero.
When the value of x is 6, the value of y is
When the value of x is -3, the value of y is
Therefore, the solutions are (6,38) or (-3,11)
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
