Questions on Algebra: Introduction to vectors, addition and scaling answered by real tutors!

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Since , this means that


So for any vector the magnitude or norm of that vector is:





So in this case





So


Geometrically, this means that the length of is 13 units


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To find the unit vector that has the same direction as , simply divide EACH component of the vector by the length of the vector .


So...




So the unit vector that has the same direction as is

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Since v = i + j and w = - i + j, this means that v=<1,1> and w=<-1,1>


So their dot product is

v · w = 1*1+1*(-1) = 1 - 1 = 0


Since their dot product is equal to zero, this means that the two vectors are orthogonal (perpendicular)


Here's a picture to help visualize the problem and verify the answer:

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The vectors are orthogonal if their dot product is 0. So in this case v=<1,1> and w=<1,c>

Now take the dot product:

v · w = 1*1+1*c = 1+c


Now set the dot product equal to zero


1+c=0


Now solve for c

c=-1

So if c=-1, then the dot product will be zero. This means that if c=-1, then v and w are orthogonal


If this is hard to grasp, draw a picture of the vectors and you'll see that the two vectors <1,1> and <1,-1> are orthogonal (perpendicular)

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please graph and classify each system as independent, dependent, and inconsistent
-2x-y=9
3x-4y=-8
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Solve each for "y":
y = -2x-9
y = (3/4)x+2
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Since the slopes are different the functions are independent.
The lines intersect at one point.
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Cheers,
Stan H.

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To find the unit vector in the direction of the given vector, divide the vector by its magnitude,M.
This way it has the same direction and magnitude of 1.
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i)


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ii)

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iii)


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iv)

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Start with the given equation


Multiply both sides by 5


Distribute


Multiply


So the answer is c) -15i+10j

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Let . Since ||v||=2, this means that


Now let's compute :


Start with the given vector


Multiply both sides by -4


Distribute


So this means:

||-4v||==4||v||


So this shows us that the length of ||-4v|| is simply 4 times that of ||v||. So the length of ||-4v|| is 8 units


In other words, ||-4v||=8

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Find the unit vector having the same direction as v if v = 2i - j
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Find the magnitude of v
||v|| = sqrt(4+1) = sqrt(5)
unit vector = v/sqrt(5)
=(2i-j)/sqrt(5)
=

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Find the magnitude of v, // v // if v = -5i + 12j
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=
=
= 13

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Find // v // if v = - i - j
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Assuming the // v // means the length of a vector,
it's
=

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If the vector v has and initial point P = (0, 0) and a terminal point
Q = (-3,-5) find its positional vector ai + bj.
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It's -3i -5j. It's easy when one end is the Origin.

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If v = 3i-5j and w = -2i + 3j find
3v - 2w
3(3i-5j)-2(-2i+3j)
9i-15j+4i-6j
13i-21j answer.

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dot product means multiply each component of one vector with another vector, and add them.
For instance, <1,1,1>*<0,2,3>=2+3=5
So for <1,1>*<-1,1>=0

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For v and w to be orthogonal, their dot product must be 0.
(1,1)*(1,b)=1+b=0
b=-1

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Being with the first equation solving for Y so:
Y = 6 + 2X
Then, plug in what you got for Y into the second equation:
4X - 2(6 + 2X) = 5
Then, simplify:
4X - 12 - 4X = 5
-12 = 5
This is an inconsistency so the system has no solution.

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The car headed north at 50 km/hr will have traveled a distance of 100km after 2 hours, right? ((2hrs)*(50km/hr) = 100km.
The car headed east at 70 km/hr will have traveled 140km after 2 hours, ok? (2hrs)*(70km/hr) = 140km.
So you now have two legs of a right triangle in which the base is a vector of magnitude 140km and the height is a vector of magnitude 100km.
The distance between the two cars, after 2 hours of travel, is represented by the hypotenuse of this right triangle.
So you can call upon Pythagoras to help find the length (distance vector) of the hypotenuse (d).




km. Approximately.
This would be a vector whose magnitude is 172.

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(7-2)x(5-3)+2=12

5x2+2=12

10+2=12

12=12

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1.
2.


3. (5,-3) Starting at the origin, that 5 right, 3 down.
4th Quadrant.

4.

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The cost C (in dollars) for a company to make x dozen units of a certain product is given by C = 0.1x^2 + 0.6x + 10. How many units can be made for $60?
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60 = 0.1x^2 + 0.6x + 10
0.1x^2 + 0.6x -50 = 0
10x^2 + 60x -500 = 0
x = [-60 +- sqrt(3600 - 4*10*-500)]/20
x = [-60 +- 153.62..]/20
x = 4.6811... dozen units
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Cheers,
Stan H.

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A ball is thrown vertically upward from the ground with an initial speed of 80 ft/sec. Its height is given by h=80t-16t^2. How high does the ball go? When does the ball hit the ground?
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h(t) = 80t-16t^2
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maximum height occurs when t = -b/2a = -80/(2*-16) = 5/2 = 2.5 seconds
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The height when t=2.5 sec is t(2.5) = 80*2.5 - 16*(2.5)^2 = 100 ft.
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The ball hits the ground when the h(t) = 0
80t - 16t^2 = 0
16t(5-t) = 0
t= 0 (it starts on the ground)
or
t= 5 seconds (it is back on the ground in 5 seconds)
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Cheers,
Stan H.

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X^2+2X-3=0
(X+3)(X-1)=0
X+3=0
X=-3 ANSWER.
X-1=0
X=1 ANSWER.

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You need pulleys to help you out.
Check out the link for compound pulleys here to understand how it works.
http://en.wikipedia.org/wiki/Pulley

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first take these values equal =
Now by cross multiplication
4(p+2)=6(p-2)
4p+8=6p-12
6p-4p-12-8=0
2p-20=0
p-10=0 dividing by 2
this is an equation of a line .In order to draw a graph we must have two variables connected by either equality or inequality sign.but if we consider a variable like q and take its values on vertical line then graph of above equation will be a vertical line otherwise a horizontal line be formed.as first value is greater than second value so p-10>0.hence graph contains the set of all points on the left of this line.

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The rectangular form for the posar equation r = -2csc(0 with a line through it) is
a. y = x-2
b. y = 2
c. y = 2x
d. y = -2
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Conversion formulas:
r = sqrt(x^2+y^2)
theta = tan^-1(y/x)
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Note: If tangent is y/x then csc is r/y
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Substituting you get:
r =-2(r/y
Divide both sides by r to get:
1 = -2/y
y = -2
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Cheers,
Stan H.

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Reflexive a=a
Symmetric If a=b, then b=a
Transitive If a=b and b=c, then a=c
Substitution If a=b, a may replace b in any equation

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When looking at a vector, you need to look at the the 'size' or amplitude of the the vector. A vector has an x component and a y component (or in your case, an i and a j).
Look at the vector like a hypotenuse of a right triangle where the two legs are given in terms of i an j.
You know that in a right triangle,
Thus, in this case




Got it?

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To add the vectors and simply add each component like this:

Start with the given expression

Add each component

Combine like terms

So

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tan angle a=side a/side b
sin angle a=side a/side c
cos angle a=side b/side c

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the vertical components of the tensions must sum to 100 lbs (to hold up the sign)

the horizontal components of the tensions cancel each other (equal but opposite) __ otherwise, the sign would move

T1(sin(43))+T2(sin(62))=100

T1(cos(43))=T2(cos(62))

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The Richter scale is a logarithmic scale.
You subtract the two values.
The value is then raised to the power of ten.




A magnitude 8.2 earthquake is 100 times stronger than a magnitude 6.2 earthquake.

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I tried to solve this and I have had problems.
I have got 3 equations so far. There is one more needed.
f >=0
r >=0
f>2r
A toy manufacturer is introducing two new dolls, My First Baby and My Real Baby. In one hour, the company produces at least twice as many First Babies as Real Babies. The company spends no more than 48 hours a week making these dolls. The profit on each First Baby is $3.00 and the profit on each Real Baby is $7.50. Find the number an type of dolls that should be produced to maximize profit.
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If in an hour 2 first Babies are produced for 1 Real Baby, the
time to produce one first baby is 1/2 hr.; the time for a Real
Baby is 1 hr
So, if f is the number of first babies and r is the # of real,
(1/2)f+r <= 48
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Also I think you should have f>=2r, not just f>2r
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Cheers,
Stan H.

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THESE PROBLEMS DON'T NEED A CALCULATOR IF YOU KNOW HOW TO SOLVE SQUARE ROOTS, MULTIPLY & DIVIDE (LONG METHOD).
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3x^2-2x+7=0
using the quadratic equation we get:
x=(2+-sqrt[-2^2+4*3*7])/2*3
x=(2+-sqrt[4-84])/6
x=(2+-sqrt-80)/6
x=(2+-8.944i)/6
x=2/6-8.944i/6
x=1/3+1.49i answer.
x=1/3-1.49i answer.
then:
(x-1/3)^2=(1/3+1.49i-1/3)^2=(1.49i)^2=-2.22 answer.
or (x-1/3)^2=(1/3-1.49i-1/3)^2=(-1.49i)^2=-2.22 answer.

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the balls meet when they are at the same height ... you can solve for the time and then find the height

(second ball height)=(first ball height) ... -.5gt^2+6=-.5g(t+1.5)^2+8(t+1.5)+6

once you find t, use h(t)=-.5gt^2+6

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A) constant acceleration is (change in velocity)/(change in time) ... (53-18)/(7-0)

B) velocity=(initial velocity)+(acceleration*time) ... Vb=18+3a

C) distance=(average velocity)*(time) ... d=((53+Vb)/2)*(7-3)

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the general equation is h(t)=-.5gt^2+Vot+Ho
... where h(t) is the height at time t, g is gravitational acceleration, Vo is initial velocity, and Ho is initial height

plug in the values and solve for t when h(t)=3 ... you will get two answers ... you want their difference

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Do you need to graph?
If so, you choose a value for x and solve for y;
y=4x-3
x=1
y=4(1)-3
y=1
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x=0
y=4(0)-3
y=-3
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x=2
y=4(2)-3
y=5
:)
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you'll notice that the displacement you think should happen (9i-12j) is half of the correct one (18i-24j)

the direction vector (3i-4j) has a resultant of magnitude 5 (3, 4, 5 triangle)
...when you apply these proportions to the known resultant of 30 you get 18, 24, 30
...the 6 is the ratio between 5 and 30
...your answer was half the correct value because you multiplied by 3 (the 3 sec), but not the 2 ((10m/sec)/5)

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You start at (2,1)
In one second you move 10 meters or directionally (6,-8) since

6,8,10 are the Pythagorean triple (3,4,5)x2.
So in one second you are at (using ordered pair(vector) addition),
(2,1)+(6,-8)=(8,-7)
After another second you move another (6,-8)
(8,-7)+(6,-8)=(14,-15)
and after the third second, you move another (6,-8)
(14,15)+(6,-8)=(20,-23)
If you went on another second, you would move to
(20,-23)+(6,-8)=(26,-31)
and so on.

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Given:
The speed is ; let assume there is no acceleration, and the particle travels in .
Since we need to find a vector with magnitude and parallel to , to do this first find a unit vector parallel to and multiply it by .
A unit vector parallel to is:





So the vector we want is

=
=
Now, add that to the particle’s initial position:

That’s its position vector after seconds

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A particle has position 2i + j initially and is moving with speed 10ms-1 in the direction 3i - 4j. Find it's position vector when t = 3 and the distance it has travelled in those 3 seconds.
I know the distance travelled is Speed * time, so 10 * 3 = 30metres.
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Draw the figure:
A vector from (2,1) thru (3,-4) and extended to (x,y)
A vertical segment from (2,1) down to (2,-4)
A horizontal segment form (2,-4) to (3,-4)
These form a triangle with hypotenuse = sqrt(26),
vertical = 5, horizontal = 1.
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Need to find x and y.
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The diagonal distance is 30 meters as you pointed out.
So x^2 + y^2 = 900
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Draw a vertical line from (2,1) down to (2,y)
Draw a horizontal line from (2,y) to (x,y)
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You now have two proportional triangles.
Vertical Proportion: 30/sqrt(26) = v/5
v = 150/sqrt26 = 29.417
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Horizontal Proportion:
30/sqrt(26)= h/1
h = 5.88
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Therefore:
x = 2+h = 7.88
y = 1-29.417 = -28.417
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These are the position coordinates after 3 seconds.
====================
Cheers,
Stan H.

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Yes, 30 m is the magnitude of the position change vector (AB) but the direction is (3,-4) in (i,j) coordinates.
You start at A (2,1) and go to B (2+i component of AB,1+j component of AB).
Find the magnitudes of the i component and the j component using the angle with vertex at A.

Hint : 3,4,5 make up a Pythagorean triple.
Sine and cosine of an angle using these sides are either 3/5 or 4/5.

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5x^2 + 5x – 30
Suppose that two numbers: a and b satisfy conditions as such:


Therefore a = 15 and b = -10.
To factorize the quadratic equation above,



To find the solution,

so either or
Therefore the solutions for x is either or .
Cheers.

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You lose $126 in the stock market.
A submarine dives 126 ft below the surface of theocean.
Your checking acount is $126 in debt.
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All these relate to -126
Cheers,
Stan H.

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I'm not sure how far you are in your course, but to find the angle between any two vectors, you would use this formula:

where is the magnitude (ie length) of vector and is the magnitude (ie length) of vector




So let's assume that is true. That means we can replace with 3





Now plug in the given magnitudes and




Now multiply the values in the denominator




So this equation is implying that there is some value of that will make true. However, there is no such value. Remember, the range of cosine is . Since equals 1.5 (which is greater than 1), it is outside the range of cosine. So if you try to take arccosine of both sides, you will not get a real answer for


So that means there are no two vectors that have magnitudes of 1 and 2 while having a dot product of 3.

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If you have vectors (2,3) and (-1,6)
Add the corresponding x and y values to get:
(1,9)
Then the angle is tan^-1(9/1) = 83.66 degrees
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Be careful when you take the tan^-1; you must
remember tan^-1 has a restricted range.
Visualize where your resultant is so you get
the proper angle.
For example:
If the resultant is (-3,-5), tan^-1(-5/-3) = 59.036.. degrees.
But that resultant is in the 3rd quadrant. The angle you
want is 180+59.036 = 239.036 degrees.
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Cheers,
Stan H.