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
It is 780.99999 sorry if i'm incorrect
Answer:
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Step-by-step explanation:
you have to write about the remaining people if they had gas problems and write what happened next maybe he was arrested for the sauce made Thanks
Answer:
The volume of the cone is;
100 π inches^3
Step-by-step explanation:
To calculate the volume, we need the height
We can get the height from the slant height and the radius
Mathematically, we use Pythagoras’ theorem for this as the slant height, the radius and the height forms a right-angled triangle
The right-angled triangle has the slant height as its hypotenuse
By the theorem, the square of the hypotenuse equals the sum of the squares of the two other sides
Let the height be h
Thus;
12^2 = 5^2 + h^2
169= 25 + h^2
h^2 = 169-25
h^2 = 144
h = √1144 inches = 12 inches
The volume of the cone is;
1/3 * π * r^2 * h
= 1/3 * π * 5^2 * 12
= 100 π inches^3
Answer:
∠BAD=20°20'
∠ADB=34°90'
Step-by-step explanation:
AB is tangent to the circle k(O), then AB⊥BO. If the measure of arc BD is 110°20', then central angle ∠BOD=110°20'.
Consider isosceles triangle BOD (BO=OD=radius of the circle). Angles adjacent to the base BD are equal, so ∠DBO=∠BDO. The sum of all triangle's angles is 180°, thus
∠BOD+∠BDO+∠DBO=180°
∠BDO+∠DBO=180°-110°20'=69°80'
∠BDO=∠DBO=34°90'
So ∠ADB=34°90'
Angles BOD and BOA are supplementary (add up to 180°), so
∠BOA=180°-110°20'=69°80'
In right triangle ABO,
∠ABO+∠BOA+∠OAB=180°
90°+69°80'+∠OAB=180°
∠OAB=180°-90°-69°80'
∠OAB=20°20'
So, ∠BAD=20°20'
