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From
y - 2x = 0
y = 4x - 1
I can show you the solution but graphing it here is impossible...
Normally you can set up an x-and-y chart and plug in values to get the points you plot...or you can put each line in slope-intercept form and graph...
In either case, the place where the lines intersect is your solution...
In this case, this is x = 1/2 and y = 1 or (1/2, 1)

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Let x be the number of miles.
Then we convert the facts of the problem into algebra and get:
21.95 + .19x = 18.95 + .21x
Let us multiply everything by 100 to get rid of decimals...
2195 + 19x = 1895 + 21x
Now subtract 19 x from both sides
2195 = 1895 + 2x
Now subtract 1895 from both sides and then divide by two...
300 = 2x
x = 150 miles

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From
2Y - 8 = X
2Y + 2X = 2
let us substitute 2y - 8 for x in the second equation and get
2y + 2(2y - 8) = 2
2y + 4y - 16 = 2
6y - 16 = 2
6y = 18
y = 3
and then
x = 2y - 8
x = 2(3) - 8
x = -2
Choice C.

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From
2Y=X+3
X=Y
we can substitute x for y in the first equation and get
2x = x + 3
x = 3
then
y = 3
Choice B.

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Answered this already...see below...y = 2x + 8

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3sqrt(5) - 5sqrt(3)
cannot be combined.
It is its own answer.
If you want an approximate number, well
3*2.236 - 5*1.732 =
-1.952

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Please scroll down and find my earlier solution...y = 2x + 8

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From -3x + 3/4 = 1/8
we need to to subtract 3/4 from both sides
-3x = 1/8 - 3/4
-3x = -5/8
then divide both sides by -3
x = 5/24

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Okay from
2sqrt(3x - 1) = 4
we divide both sides by two first
sqrt(3x - 1) = 2
then square both sides
3x - 1 = 4
then solve and check
3x = 5
x = 5/3
It checks.

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Okay, from x = 1 + 6sqrt(x-9), move the one over first and get
x - 1 = 6sqrt(x-9)
Now square both sides
x^2 - 2x + 1 = 36(x - 9)
Now expand, combine, factor, solve and check...
x^2 - 2x + 1 = 36x - 324
x^2 - 38x + 325 = 0
(x - 13)(x - 25) = 0
x = 13 or x = 25
Both check out, so we have two solutions...

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Okay, from 6e^(-x) = 3
we'll divide by 6 first and get
e^(-x) = 1/2
then take the ln of both sides
ln[e^(-x)] = ln(1/2)
-x = ln(1/2)
x = -ln(1/2)
x = ln 2

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5 lb/ft X 3 ft =
15 lb
You can cancel units like feet just like you cancel x's or y's...

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From sqrt(8y^3 / 49x), we will need to rationalize the denominator by multiplying by sqrt(x) / sqrt(x)...this will give us a perfect square in the denominator...
sqrt(8y^3 / 49x) =
sqrt(8y^3 / 49x) * [sqrt(x) / sqrt(x)] =
sqrt(8xy^3 / 49x^2)
We can pull perfect squares out of the radical from the numerator and denominator, giving us the answer...
sqrt(8xy^3 / 49x^2) =
(2y / 7x) * sqrt(2xy)

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Okay, from
-4 = -4(f - 7)
we can divide both sides by -4 first...
1 = f - 7
then add seven to both sides, and
f = 8

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Okay from 6sqrt(x) = x + 5, let us square both sides...
36x = x^2 + 10x + 25
Subtract the 36x, factor, solve and check...
x^2 - 26x + 25 = 0
(x - 25)(x - 1) = 0
x = 25 or x = 1
They both check out okay...two solutions here...

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Parallel means same slope...so we need a line going thru (-2, 4) with a slope of 2...we use the point-slope formula...
y - y1 = m(x - x1)
y - 4 = 2(x -(-2))
y - 4 = 2(x + 2)
y - 4 = 2x + 4
y = 2x + 8

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Gauss showed the shortcut for this hundreds of years ago, when he was but a child...
Notice that if you add the first and last terms you get 4002.
If you add the second and second-to-last terms you get 4002.
If you add the third and third-to-last terms you get 4002.
And so on, and so on...
Now how many pairs making 4002's are there?
Well, half the number of numbers present.
There are 2000 even numbers from 2 to 4000.
That means 1000 pairs, each adding to 4002.
So the sum is 1000 * 4002 = 4002000.
Cool!

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From 5x^2 - 20y^2, we first factor out the 5 and get
5(x^2 - 4y^2)
But x^2 - 4y^2 is the difference of squares, and can be factored again...
5(x^2 - 4y^2) =
5(x + 2y)(x - 2y)

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From
sqrt27(a^2)b / sqrt7(a^3)(b^3)
we can make it all one root and then cancel as possible, thus we get
sqrt[27(a^2)b / 7(a^3)(b^3)] =
sqrt[27 / 7ab^2]
From here, we need to rationalize the denominator by multiplying top and bottom by sqrt(7a)/sqrt(7a)...so we have
sqrt[27 / 7ab^2] * sqrt(7a)/sqrt(7a) =
sqrt[189a / 49(a^2)(b^2)]
Now 49(a^2)(b^2) is a perfect square so we can remove it from the root...our final answer is
sqrt(189a) / 7ab

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First we factor the trinomial
x^2 - 4x - 21 = 0
into its binomial factors:
(x - 7)(x + 3) = 0
Now if two quantities multiply together to make zero, one (or both) has to be zero...
Thus x - 7 = 0 and x + 3 = 0 which leads to
x = 7 or x = -3

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Okay, from the equation
A = P(1 + r/n)^nt
we plug in what we know and solve for what we don't know...
10000 = 5000(1 + (.07/4))^4t
now divide by 5000
2 = (1 + (.07/4))^4t
take the log of both sides
log 2 = 4t*log(1 + (.07/4))
now solve for t...
t = log 2 / 4(log 1.0175)
t = 9.988 years or roughly 10 years

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From
5x + 9y = -12
x - 2y = 10
let us multiply the bottom equation by (-5) and add it to the top one...
5x + 9y = -12
-5x + 10y = -50
which yields
19y = -62
and the not-so-nice
y = -62/19
Now plug that into the second equation to find x...
x - 2(-62/19) = 10
x + 124/19 = 10
x + 124/19 = 190/19
x = 66/19

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3*sqrt(25) / 3*sqrt(5) can be figured by direct division.
The 3's cancel and sqrt(25) / sqrt(5) = sqrt(5).

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Those nmbers which can be factored into numbers other than itself and one are composite.
Thus, 15, 49, 55, 87, 91 and 105 are composite.
Aside from 0 and 1, the rest are prime.

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Perpendicular lines have slopes which are negative reciprocals of each other. thus we need a line with a slope of -1/4.
Choice B is correct.

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Perpendicular

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A positive quantity multiplied by a negative quantity equals
==> a negative quantity.
Remember ==> same sign is positive, different sign is negative.

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Yes. Because the opposite angles would have to be equal to 90 degrees. Since you are given it is a rhombus, and not merely an arbitrary rectangle, it would have to be a square. Of course both squares and rectangles have congruent diagonals.

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From
x - y = 1 and
x = (1/2)y + 2
Let us subtract 1/2y in the second one and then subtract equations..
x - y = 1
x - (1/2)y = 2
which leads to
-(1/2)y = -1
and
y = 2
plug that in the first eqn and get
x = 3

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Parallel lines have equal slopes.
Thus which of the choices has a slope of .5?
Choice C is best. Choice is the exact same line and not technically parallel.

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If the tail was 9 lbs, the head was 9 lbs too. If that was half the body, the body must have been 18 lbs, and the sum of head and tail, as stated...
Thus the entire fish, head, body and tail was 36 pounds. That's some fish!