This is the concept of algebra, to solve the expression we proceed as follows;
cos 2x-cosx=0
cos 2x=cosx
but:
cos 2x+1=2(cos^2x)
thereore;
from:
cos 2x=cos x
adding 1 on both sides we get:
cos 2x+1=cos x+1
2(cos^2x)=cosx+1
suppose;
cos x=a
thus;
2a^2=a+1
a^2-1/2a-1/2=0
solving the above quadratic we get:
a=-0.5 and a=1
when a=-0.5
cosx=-0.5
x=120=2/3π
when x=1
cos x=1
x=0
the answer is:
x=0 or x=2/3π
Answer:
2(x-9)(x+5)
Step-by-step explanation:
<h2>
Answer:</h2>
a. <-13/2,-13/2>
<h2>
Step-by-step explanation:</h2>
The projection of a vector u onto another vector v is given by;
=
----------------(i)
Where;
u.v is the dot product of vectors u and v
|v| is the magnitude of vector v
Given:
u = <-6, -7>
v = <1, 1>
These can be re-written in unit vector notation as;
u = -6i -7j
v = i + j
<em>Now;</em>
<em>Let's find the following</em>
(i) u . v
u . v = (-6i - 7j) . (i + j)
u . v = (-6i) (1i) + (-7j)(1j) [Remember that, i.i = j.j = 1]
u . v = -6 -7 = -13
(ii) |v|
|v| = 
|v| = 
<em>Substitute these values into equation (i) as follows;</em>
= ![[\frac{-13}{(\sqrt{2}) ^2}][i + j]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B-13%7D%7B%28%5Csqrt%7B2%7D%29%20%5E2%7D%5D%5Bi%20%2B%20j%5D)
= ![\frac{-13}{2} [i + j]](https://tex.z-dn.net/?f=%5Cfrac%7B-13%7D%7B2%7D%20%5Bi%20%2B%20j%5D)
This can be re-written as;
= 
= 
The answer is B
Step-by-step explanation:
the table B represent an inverse variation, the constant of variation k is 30