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
1.25 is the slope 5/4 rise over run
Line 1 and line 4 are parallel
Step-by-step explanation:
Two lines are said to be parallel if they have same slope.
In order to compare the lines given, we have to write them all in the same form, and then compare their slopes.
Line 1:
Applying distributive property,
Line 2:
Line 3:
Applying distributive property,
Line 4:
re-arranging,
Now we have rewritten all the lines in the form , where m is the slope. By comparing the values of m, we see:
Therefore, the lines which are parallel are line 1 and line 4.
Learn more about parallel and perpendicular lines:
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Answer:
77
Step-by-step explanation:
Answer:
The correct answer is 9 m per
Step-by-step explanation:
Force is defined by mass of an object times it's acceleration.
Let the mass of the object be m kilograms.
When a force of 30 N acts on a certain object the acceleration of the object is 5 m per
∴ 30 = m × 5
⇒ m = 6.
If the force is changed to 54 N, let the acceleration of the object be x m per .
∴ 54 = m × x
⇒ x = = 9
Thus when the force is 54 N, the acceleration of the object is 9 m per .