Answer:
The first choice is the one you want
Step-by-step explanation:
First thing you need to know about this greatest integer graph is that it is aptly called a step graph. It literally looks like stair steps on your calculator: short horizontal lines that are not connected vertically. Really cool graph.
Second thing you need to know is about transformations of functions. ANY side-to-side movement in ANY function will be in a set of parenthesis (or absolute value symbols, or under a radical sign, or inside the greatest integer brackets, etc.) and ANY up or down movement will be either added or subtracted. Added means you move the function up from its starting position, subtracting means you move the function down from its starting position. Since we have no numbers inside the greatest integer brackets, there is no side-to-side movement. Since there is a "-2" after the brackets, we are moving the whole function down.
If you do not know how to graph these without a calculator and you have no idea what this graph looks like, I recommend going to your calculator to see it. First, call up your "y = " window. Next, hit 2nd-->0 (catalog), then hit the x^2 button (this will take you to the letter I in the catalog). Scroll down til you see "int( " and hit that button. It will take you back to the "y = " window. Enter an x after that set of parenthesis and then close it, then hit " - 2 " and then "graph". Your steps should begin to appear. Notice that the horizontal line between x = 0 and x = 1 is at y = -2. The parent graph has this line between x = 0 and x = 1 on y = 0. The -2 in ours moved the graph down from y = 0 to y = -2
Summing up, the first choice is the one you want as your answer.
Answer:
14.4in so your radius is your third answer
Answer:
The Area of Δ TOP is 43.55 units².
Step-by-step explanation:
Given:
P ≡ ( x₁ ,y₁ ) ≡ ( -5 , -7)
T ≡ ( x₂ ,y₂ ) ≡ ( 1 , 8)
O ≡ ( x₃ ,y₃ ) ≡ ( 6 , 6)
To Find :
Area of Δ TOP = ?
Solution :
We have
![\textrm{Area of Triangle TOP} = \frac{1}{2}\times Base\times Height\\\textrm{Area of Triangle TOP} = \frac{1}{2}\times OT\times PT](https://tex.z-dn.net/?f=%5Ctextrm%7BArea%20of%20Triangle%20TOP%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20Base%5Ctimes%20Height%5C%5C%5Ctextrm%7BArea%20of%20Triangle%20TOP%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20OT%5Ctimes%20PT)
Now Distance formula we have
![l(PT) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}](https://tex.z-dn.net/?f=l%28PT%29%20%3D%20%5Csqrt%7B%28%28x_%7B2%7D-x_%7B1%7D%29%5E%7B2%7D%2B%28y_%7B2%7D-y_%7B1%7D%29%5E%7B2%7D%20%29%7D)
![l(OT) = \sqrt{((x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2} )}](https://tex.z-dn.net/?f=l%28OT%29%20%3D%20%5Csqrt%7B%28%28x_%7B3%7D-x_%7B2%7D%29%5E%7B2%7D%2B%28y_%7B3%7D-y_%7B2%7D%29%5E%7B2%7D%20%29%7D)
Substituting the given values we get
![l(PT) = \sqrt{((1--5)^{2}+(8--7)^{2} )}\\l(PT) = \sqrt{((1+5)^{2}+(8+7)^{2} )}\\l(PT) = \sqrt{((6)^{2}+(15)^{2} )}\\l(PT) = \sqrt{261}\\l(PT) = 16.16\ units](https://tex.z-dn.net/?f=l%28PT%29%20%3D%20%5Csqrt%7B%28%281--5%29%5E%7B2%7D%2B%288--7%29%5E%7B2%7D%20%29%7D%5C%5Cl%28PT%29%20%3D%20%5Csqrt%7B%28%281%2B5%29%5E%7B2%7D%2B%288%2B7%29%5E%7B2%7D%20%29%7D%5C%5Cl%28PT%29%20%3D%20%5Csqrt%7B%28%286%29%5E%7B2%7D%2B%2815%29%5E%7B2%7D%20%29%7D%5C%5Cl%28PT%29%20%3D%20%5Csqrt%7B261%7D%5C%5Cl%28PT%29%20%3D%2016.16%5C%20units)
And
![l(OT) = \sqrt{((x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2} )}\\l(OT) = \sqrt{((6-1)^{2}+(6-8)^{2} )}\\l(OT) = \sqrt{((5)^{2}+(-2)^{2} )}\\l(OT) = \sqrt{29}\\l(OT) = 5.39\ units](https://tex.z-dn.net/?f=l%28OT%29%20%3D%20%5Csqrt%7B%28%28x_%7B3%7D-x_%7B2%7D%29%5E%7B2%7D%2B%28y_%7B3%7D-y_%7B2%7D%29%5E%7B2%7D%20%29%7D%5C%5Cl%28OT%29%20%3D%20%5Csqrt%7B%28%286-1%29%5E%7B2%7D%2B%286-8%29%5E%7B2%7D%20%29%7D%5C%5Cl%28OT%29%20%3D%20%5Csqrt%7B%28%285%29%5E%7B2%7D%2B%28-2%29%5E%7B2%7D%20%29%7D%5C%5Cl%28OT%29%20%3D%20%5Csqrt%7B29%7D%5C%5Cl%28OT%29%20%3D%205.39%5C%20units)
Now substituting OT and PT in area formula we get
![\textrm{Area of Triangle TOP} = \frac{1}{2}\times 5.39\times 16.16\\\textrm{Area of Triangle TOP} = 43.55\ units^{2}](https://tex.z-dn.net/?f=%5Ctextrm%7BArea%20of%20Triangle%20TOP%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%205.39%5Ctimes%2016.16%5C%5C%5Ctextrm%7BArea%20of%20Triangle%20TOP%7D%20%3D%2043.55%5C%20units%5E%7B2%7D)
Therefore, Area of Δ TOP is 43.55 units².
Answer:
Therefore the magnitude of the applied force is
N.
Step-by-step explanation:
Torque: Torque is the cross product of force and the distance of applied force from the rotational axis.
![\tau=\vec F\times \vec r= |F||r| sin\theta](https://tex.z-dn.net/?f=%5Ctau%3D%5Cvec%20%20F%5Ctimes%20%5Cvec%20r%3D%20%7CF%7C%7Cr%7C%20sin%5Ctheta)
Where
is angle between F and r.
Newton-meter is the S.I unit of torque.
Along the positive y axis, the wrench lies .
The length of wrench is 0.1 m.
[ since the vector
lies on y axis]
The direction of applied force is along the vector <0,2,-4>.
Then the angle between
and <0,2,-4> = the angle between F and
.
We know that,
![=0+(0.1).2+0.(-4)=0.2](https://tex.z-dn.net/?f=%3D0%2B%280.1%29.2%2B0.%28-4%29%3D0.2)
![\therefore\sqrt{(0.1)^2}.\sqrt{0^2+2^2+(-4)^2} \ cos \theta= 0.2](https://tex.z-dn.net/?f=%5Ctherefore%5Csqrt%7B%280.1%29%5E2%7D.%5Csqrt%7B0%5E2%2B2%5E2%2B%28-4%29%5E2%7D%20%5C%20%20cos%20%5Ctheta%3D%200.2)
![\Rightarrow (0.1).2\sqrt5 cos \theta =0.2](https://tex.z-dn.net/?f=%5CRightarrow%20%280.1%29.2%5Csqrt5%20cos%20%5Ctheta%20%3D0.2)
![\Rightarrow cos \theta =\frac{0.2}{0.2\sqrt 5}](https://tex.z-dn.net/?f=%5CRightarrow%20%20cos%20%5Ctheta%20%3D%5Cfrac%7B0.2%7D%7B0.2%5Csqrt%205%7D)
![\Rightarrow cos \theta =\frac{1}{\sqrt 5}](https://tex.z-dn.net/?f=%5CRightarrow%20%20cos%20%5Ctheta%20%3D%5Cfrac%7B1%7D%7B%5Csqrt%205%7D)
We know that,
![sin \theta =\sqrt{1-cos^2\theta}=\sqrt{1-(\frac{1}{\sqrt5})^2}](https://tex.z-dn.net/?f=sin%20%5Ctheta%20%3D%5Csqrt%7B1-cos%5E2%5Ctheta%7D%3D%5Csqrt%7B1-%28%5Cfrac%7B1%7D%7B%5Csqrt5%7D%29%5E2%7D)
![=\sqrt {1-\frac15}](https://tex.z-dn.net/?f=%3D%5Csqrt%20%7B1-%5Cfrac15%7D)
![=\sqrt {\frac{5-1}{5}](https://tex.z-dn.net/?f=%3D%5Csqrt%20%7B%5Cfrac%7B5-1%7D%7B5%7D)
![=\frac{2}{\sqrt 5}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%7D%7B%5Csqrt%205%7D)
Here
= 100 N-m,
and ![sin \theta = \frac{2}{\sqrt 5}](https://tex.z-dn.net/?f=sin%20%5Ctheta%20%3D%20%5Cfrac%7B2%7D%7B%5Csqrt%205%7D)
Now
![\tau=|F||r| sin\theta](https://tex.z-dn.net/?f=%5Ctau%3D%7CF%7C%7Cr%7C%20sin%5Ctheta)
![\Rightarrow 100= |F|(0.1) \frac{2}{\sqrt 5}](https://tex.z-dn.net/?f=%5CRightarrow%20100%3D%20%7CF%7C%280.1%29%20%5Cfrac%7B2%7D%7B%5Csqrt%205%7D)
![\Rightarrow |F|=\frac{100\times \sqrt5}{0.2}](https://tex.z-dn.net/?f=%5CRightarrow%20%7CF%7C%3D%5Cfrac%7B100%5Ctimes%20%5Csqrt5%7D%7B0.2%7D)
![\Rightarrow |F|=500\sqrt 5](https://tex.z-dn.net/?f=%5CRightarrow%20%7CF%7C%3D500%5Csqrt%205)
Therefore the magnitude of the force is
N.