Timedd find the angle between u = sqr5i-8j and v = sqr5i+j. round to the nearest tenth of a degree. (the square root is only on the 5i in both) a.65.9 b.98.5 c.90.4 d33.3 use the dot product to find [v] when v =(-2,-1) a.-1 b.-3 c; sqr5 dsqr3

98.5 degrees

Step-by-step explanation:

To find the angle measurement between two vectors use: cos(theta)=(u dot v)/(|u|*|v|)

The vectors are u=sqrt(5)i-8j and v=sqrt(5)i+1j .

So find the dot product of u and v.  u dot v=sqrt5*sqrt(5)+(-8)*(1)=5-8=-3.

Second step: Find the magnitude of both vectors.  So to find the magnitude of a vector, let's call it t when t=ai+bj, you just do |t|=sqrt(a^2+b^2).

|u|=sqrt(5+64)=sqrt(69) and |v|=sqrt(5+1)=sqrt(6)

Now multiply your magnitudes of your vectors: sqrt(69)*sqrt(6)=sqrt(414).

So now we have:

cos(theta)=-3/sqrt(414)

Now arccos( ) or cos^(-1) to find the angle theta.

theta=cos^(-1)[-3/sqrt(414)] which is approximately 98.5 degrees.

answer: a.   one is a factor of every whole number since every number is divisible by itself.

1 x=22.52

2: 65/39=604+39x

65x-39x=604

26x=604

x=604/26

×=23.23