What is the simplest form of 8/28?

2 answers:

7 0

Both 8 and 28 are divisible by 4.

8/4=2
28/4=7

2 and 7 don't have any other common factors, so the simplest form of 8/28 is 2/7

Marizza181 [45]3 years ago

5 0

2/7 because both 8 and 28 are divisible by 4

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Find an equation of the line that satisfies the given conditions.

Makovka662 [10]

Well, the very first thing you need to do is to <span> find the slope of the line
According to the fact that we already know that the slope is parallel </span>to the line, you can easily find out the slope, to be more exact : $m=(Y2-Y1)/X2-X1)=4/4=1$
Straight after, you have to use <span>point slope formula</span> $Y-Y1=m(X-X1)

Y-7=X-6$

6 0

4 years ago

Vector question. Let v=<-2,1> u=<3,-5> , find:

lyudmila [28]

Answer:

1. ) $\vec {v}.\vec{u}= -11$

2.) The angle between v and u is 147.52°.

Step-by-step explanation:

Given:

$\vec {v}= ( -2 , 1 )\\\vec {u}= ( 3 , -5 )$

To Find:

1. $\vec {v}.\vec{u}= ?$

2. $\theta = ?$

Solution:

$\vec {v}.\vec{u}$ is scalar product given as,

$\vec {v}.\vec{u}=|\vec {v}||\vec {u}|\cos \theta$

$|\vec {v}|=|(-2, 1)|=\sqrt{(-2)^{2} +1^{2}}=\sqrt{5}\\|\vec {u}|=|(3, 5)|=\sqrt{(3)^{2} +(-5)^{2}}=\sqrt{34}$

$\vec {v}.\vec{u}=(-2i +j).(3i-5j)\\$

Here only i.i = j.j =1 and i.j = j.i = 0

∴ $\vec {v}.\vec{u}=(-2\times 3 +1\times -5)\\\vec {v}.\vec{u}=-6-5=-11 \\$

Now, Substituting the above values we get

$-11=\sqrt{5}\times \sqrt{34}\cos \theta\\ \cos \theta=\frac{-11}{\sqrt{170}} \\ \cos \theta =-0.84366\\\therefore \theta =cos^{-1}(-0.84366)\\\therefore \theta =147.52\°$