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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
Answer:
a) The estimates for the solutions of are
and
.
b) The estimates for the solutions of are
and
Step-by-step explanation:
From image we get a graphical representation of the second-order polynomial , where
is related to the horizontal axis of the Cartesian plane, whereas
is related to the vertical axis of this plane. Now we proceed to estimate the solutions for each case:
a)
There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:
,
b)
There are two approximate solutions according to the graph, which are marked by red circles in the image attached below:
,
