Find the angle between u =(8,-2) and v =(9,3) round to the nearest tenth of a degree. a.32,5b.6.3c.42.5d.16.3

Find magnitude of each vector.

||u|| = sqrt{8^2 + (-2)^2}

||u|| = sqrt{64 + 4}

||u|| = sqrt{68}

Now the other vector.

||v|| = sqrt{9^2 + 3^3}

||v|| = sqrt{81 + 9}

||v|| = sqrt{90}

Now find u • v.

u•v = <8, -2> <9, 3>

u•v = 8(9) + (-2)(3)

u•v = 72 - 6

u•v = 66

Now use

cos x = (u•v)/||u||•||v||

Remaining Steps:

1. Let u•v = 66

2. Let ||u||•||v|| = sqrt{68}•sqrt{90}

3. Simplify your substitution.

4. Take the arccos on both sides to find x.

Note: x = angle between vectors u and v.

Take it from here.

the angle between u=(8, -2) and v=(9,3) is 32.5°

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

alright well the answers to your question is

2xy/9 hope this have a nice day : )

step-by-step explanation: