Answer:
.3 feet
Step-by-step explanation:
3.8 - 3.5 = .3
Answer: 98 millimeters
Explanation: Since angle VTY is congruent to angle VTK, segment TY bisects angle VTK. Since Y is on segment VK, between V and K, we can use the Angle Bisector Theorem, which states that
![\frac{VY}{YK} = \frac{VT}{TK}](https://tex.z-dn.net/?f=%20%5Cfrac%7BVY%7D%7BYK%7D%20%3D%20%20%5Cfrac%7BVT%7D%7BTK%7D%20%20%20)
(1)
Since x= VK = VY + YK, we need to obtain VY since YK = 68.
VY is obtained by multiplying the denominator YK on both sides of equation (1). So,
![VY = \frac{(VY)(VT)}{TK} = \frac{(68)(57)}{129.2} \newline VY = 30](https://tex.z-dn.net/?f=VY%20%3D%20%5Cfrac%7B%28VY%29%28VT%29%7D%7BTK%7D%20%3D%20%5Cfrac%7B%2868%29%2857%29%7D%7B129.2%7D%20%5Cnewline%20VY%20%3D%2030)
Hence,
x = VK = VY + YK
x = 30 + 68
x = 98 millimeters
11+5=
3+4=
done you need to add them together and then those to answers you add them and done
Answer:
f(x)=0.05x + 10
Step-by-step explanation:
.50 is the price per ticket, and 10 is the entrance fee.
Answer:
1. )![\vec {v}.\vec{u}= -11](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D.%5Cvec%7Bu%7D%3D%20-11)
2.) The angle between v and u is 147.52°.
Step-by-step explanation:
Given:
![\vec {v}= ( -2 , 1 )\\\vec {u}= ( 3 , -5 )](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D%3D%20%28%20-2%20%2C%201%20%29%5C%5C%5Cvec%20%7Bu%7D%3D%20%28%203%20%2C%20-5%20%29)
To Find:
1. ![\vec {v}.\vec{u}= ?](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D.%5Cvec%7Bu%7D%3D%20%3F)
2.![\theta = ?](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%3F)
Solution:
is scalar product given as,
![\vec {v}.\vec{u}=|\vec {v}||\vec {u}|\cos \theta](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D.%5Cvec%7Bu%7D%3D%7C%5Cvec%20%7Bv%7D%7C%7C%5Cvec%20%7Bu%7D%7C%5Ccos%20%5Ctheta)
![|\vec {v}|=|(-2, 1)|=\sqrt{(-2)^{2} +1^{2}}=\sqrt{5}\\|\vec {u}|=|(3, 5)|=\sqrt{(3)^{2} +(-5)^{2}}=\sqrt{34}](https://tex.z-dn.net/?f=%7C%5Cvec%20%7Bv%7D%7C%3D%7C%28-2%2C%201%29%7C%3D%5Csqrt%7B%28-2%29%5E%7B2%7D%20%2B1%5E%7B2%7D%7D%3D%5Csqrt%7B5%7D%5C%5C%7C%5Cvec%20%7Bu%7D%7C%3D%7C%283%2C%205%29%7C%3D%5Csqrt%7B%283%29%5E%7B2%7D%20%2B%28-5%29%5E%7B2%7D%7D%3D%5Csqrt%7B34%7D)
![\vec {v}.\vec{u}=(-2i +j).(3i-5j)\\](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D.%5Cvec%7Bu%7D%3D%28-2i%20%2Bj%29.%283i-5j%29%5C%5C)
Here only i.i = j.j =1 and i.j = j.i = 0
∴ ![\vec {v}.\vec{u}=(-2\times 3 +1\times -5)\\\vec {v}.\vec{u}=-6-5=-11 \\](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D.%5Cvec%7Bu%7D%3D%28-2%5Ctimes%203%20%2B1%5Ctimes%20-5%29%5C%5C%5Cvec%20%7Bv%7D.%5Cvec%7Bu%7D%3D-6-5%3D-11%20%5C%5C)
Now, Substituting the above values we get
![-11=\sqrt{5}\times \sqrt{34}\cos \theta\\ \cos \theta=\frac{-11}{\sqrt{170}} \\ \cos \theta =-0.84366\\\therefore \theta =cos^{-1}(-0.84366)\\\therefore \theta =147.52\°](https://tex.z-dn.net/?f=-11%3D%5Csqrt%7B5%7D%5Ctimes%20%5Csqrt%7B34%7D%5Ccos%20%5Ctheta%5C%5C%20%5Ccos%20%5Ctheta%3D%5Cfrac%7B-11%7D%7B%5Csqrt%7B170%7D%7D%20%5C%5C%20%5Ccos%20%5Ctheta%20%3D-0.84366%5C%5C%5Ctherefore%20%5Ctheta%20%3Dcos%5E%7B-1%7D%28-0.84366%29%5C%5C%5Ctherefore%20%5Ctheta%20%3D147.52%5C%C2%B0)
As it is negative mean
is in Second Quadrant Because Cosine is negative in Second Quadrant.
1. )![\vec {v}.\vec{u}= -11](https://tex.z-dn.net/?f=%5Cvec%20%7Bv%7D.%5Cvec%7Bu%7D%3D%20-11)
2.) The angle between v and u is 147.52°.