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An airplane flew for 4 hours with a 20-km/h tail wind. The return flight against the same wind took 5 hours. Find the speed of the plane in still air.
.
Let x = speed of plane in still air
then
from distance formula:
4(x+20) = 5(x-20)
4x+80 = 5x-100
80 = x-100
180 km/hr = x

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The hypotenuse of a right triangle is 5 feet long. One leg is 3 feet longer then the other leg. Find the perimeter of the triangle.
.
Let x = length of "other leg"
then
x+3 = length of "one leg"
.
applying Pythagorean theorem:
x^2 + (x+3)^2 = 5^2
x^2 + (x+3)(x+3) = 25
x^2 + x^2+6x+9 = 25
2x^2+6x+9 = 25
2x^2+6x-16 = 0
x^2+3x-8 = 0
applying the "quadratic formula" results in:
x = {1.7, -4.7}
Throw out the negative solution leaving:
x = 1.7 feet (other leg)
"one leg":
x+3 = 1.7+3 = 4.7 feet
Perimeter:
4.7 + 1.7 + 5 = 11.4 feet
.
Details of quadratic to follow:

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Quadratic equation (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=41 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 1.70156211871642, -4.70156211871642. Here's your graph:

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If an angle has a measure of 45°12' 10", the measure of its complement is what?
The sum of two angles that are complementary is 90 degrees:
90° - 45°12' 10"
89°60' - 45°12' 10"
89°59'60" - 45°12' 10"
(89°-45°)(59'-12')(60"-10")
(44°)(47')(50")
44°47'50"

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give the equation for a line with a slope of 5 that passes through the point (4,1)
.
Plug the given information into the "point-slope" form:
y - y1 = m(x - x1)
y - 1 = 5(x - 4)
y - 1 = 5x - 20
y = 5x - 21 (this is what they're looking for - in "slope-intercept" form)

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-.25 = logx - log(.0375) ?
-.25 = logx/(.0375)
10^(-.25) = x/(.0375)
.0375*10^(-0.25) = x
0.0211 = x

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1.43=1.05(1+g)^4
1.43/1.05 = (1+g)^4
(1.43/1.05)^(1/4) = 1+g
(1.43/1.05)^(1/4) - 1 = g
0.08 = g
or
8% = g

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x^2-5x-14=0
factoring the left side:
(x-7)(x+2) = 0
x = {-2, 7}

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domain for log3(2x-7)
.
Since, you can't take a log of a negative number:
2x-7 >= 0
2x >= 7
x >= 7/2
x >= 3.5 (domain)
'x' must be "greater than or equal" to 3.5

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50% of 40
50/100 * 40
1/2 * 40
20

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two numbers total 57 and have a difference of 9
Let x,y be the two numbers
x+y = 57
x-y = 9
.
Add both equations together:
x+y = 57
x-y = 9
------------
2x = 66
x = 33
.
substitute above into:
x+y = 57
33+y = 57
y = 24
.
solution: 24 and 33

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f(x)=1/(4x+3) Domain
.
Domain specifies all possible values of x:
The only value you must exclude is when the denominator is zero:
4x+3 = 0
4x = -3
x = -3/4
x = -0.75
.
In set-notation:
(-oo, -0.75) U (-.75, +oo)
where
oo stands for infinity

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Yes, you are correct.

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LOGa9+LOGa70
LOGa(9*70)
LOGa630

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√(16Tt^2)
√(4*4Tt^2)
4*√(Tt^2)
4*√(Ttt)
4t*√(T)

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What is the slope and y- intercept of -3y = 4(6x - 3)?
Begin by rearranging given equation inot "slope-intercept" form of:
y = mx + b
.
-3y = 4(6x - 3)
-3y = 24x - 12
3y = -24x + 12
y = -8x + 4
.
From inspection then:
slope is -8
y-intercept is at (0,4)

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What is the vertex of (x-3)^2?
Since the "vertex form" of a parabola is:
y = a(x-h)^2 + k
where
(h,k) is the vertex
.
From:
(x-3)^2
we can rewriet
as:
(x-3)^2 + 0
So, the vertex is at (3,0)

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THe height in feet a ball is t seconds after being throan in the air is given by H(t)=-16t^2+64t+5
A)How long did it take the ball to reach its maximum height above the ground
Vertex of parabola is the max:
t = -b/(2a)
t = -64/(2(-16))
t = -64/(-32)
t = 2 seconds
.
B)What was the maximum height above the ground the ball reached
plug t=2 into:
H(t)=-16t^2+64t+5
H(2)=-16(2)^2+64(2)+5
H(2)=-16(4)+128+5
H(2)=-64+128+5
H(2)= 69 feet

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A certain radioactive isotope has a half-life of approximately 1300 milliseconds. How much time would be required for a given amount of this isotope to decay to 35% of that amount? Use the formula A=Ie^kt where A is the amount at the time, I is the initial amount and k is the decay constant unique to the substance.
.
Given:
A=Ie^kt
Use:
"A certain radioactive isotope has a half-life of approximately 1300 milliseconds."
to find the constant 'k':
Let x = initial amount
x/2 = half the initial amount
x/2 = xe^(k*1300)
dividing both sides by x:
1/2 = e^(k*1300)
ln(.5) = k*1300
ln(.5)/1300 = k
.
Now we can answer:
How much time would be required for a given amount of this isotope to decay to 35% of that amount?
Let x = initial amount
then
.35x = 35% of initial amount
our formula:
A=Ie^kt
.35x = xe^(t*ln(.5)/1300)
dividing both sides by x:
.35 = e^(t*ln(.5)/1300)
ln(.35) = t*ln(.5)/1300
1300*ln(.35) = t*ln(.5)
1300*ln(.35)/ln(.5) = t
1968.95 milliseconds = t
OR,
1.96 seconds = t

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Since:
0.001
is equivalent to:
0.001 = 10^(-3)
then
log base 10 of 0.001 = -3
.
-3 is your answer

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at 8 am your car's odometer reads 2475 km at noon it reads 2895 km what is the average speed
.
time traveled: 12-8 = 4 hours
distance traveled: 2895 - 2475 = 420 km
.
speed = 420/4
speed = 105 km/hr

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A line has a rise of 6 and a slope of 1/20. What is the run?
.
Since slope = rise/run
.
6/run = 1/20
cross-multiplying
6*20 = run
120 = run

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what is the y-intercept of the equation, y=1.7x+40
.
Because the equation is already in "slope-intercept" form of
y = mx + b
where
m is the slope
b is the y-intercept at (0,b)
.
From inspection then,
y-intercept is at (0, 40)

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find the domain of the function
f(x)=log base 2 (x+1)
.
since you can't take a log of a negative number:
x+1 >= 0
x >= -1
x has to be greater than or equal to -1

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log base 2 (x^2-7x+14)=1
(x^2-7x+14)=2^1
x^2-7x+14 = 2
x^2-7x+12 = 0
(x-4)(x-3) = 0
x = {3, 4}

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Does the system of equations have no solution, one solution, or infinitely many solutions?
y = 5x – 4
y = 5x – 5
.
Since slope (5) is the same for both but the y-intercepts are diffferent
we conclude that
there is "no solution"
.
See the site below, it describes the three cases:
http://www.mathwarehouse.com/algebra/linear_equation/systems-of-equation/index.php

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find the x and y intercepts of: y = -2x - 6
.
to find the x-intercepts, set y=0 solve for x:
y = -2x - 6
0 = -2x - 6
6 = -2x
-3 = x
so, x-intercept is at (-3,0)
.
to find the y-intercepts, set x=0 solve for y:
y = -2x - 6
y = -2(0) - 6
y = -6
so, y-intercept is at (0,-6)

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From the top of the building in the illustration, a ball is thrown straight up with an initial velocity of 40 feet per second. The equation s = -16t2 + 40t + 48 gives the height s of the ball t seconds after it is thrown. Find the maximum height reached by the ball and the time it takes for the ball to hit the ground. (Round your answer to one decimal place.)
s = -16t2 + 40t + 48
The "vertex" defines the max height.
It reaches this height when
t = -b/(2a)
t = -40/(2(-16))
t = -40/(-32)
t = 1.25 seconds
.
Max height:
s = -16(1.25)^2 + 40(1.25) + 48
s = 73.0 feet

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The height (in feet) of the water level in a reservoir over a 1 year period is modeled by the function H(t) = 2.4(t - 10)2 + 15, where t = 1 represents January, t = 2 represents February, and so on. How low did the water level get that year, and when did it reach the low mark?
The vertex of H(t) = 2.4(t - 10)2 + 15 gives you the low point:
equation given is in the "vertex" form:
y= a(x-h)^2+k
vertex is at (10,15)
when did it reach the low mark?
t = 10 (October)
How low did the water level get:
h(t) = 15 feet

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How do you change 4x-2y=10 into slope-intercept form?
.
The "slope-intercept" form of a line is:
y = mx + b
all you need to do, is to rearrange the terms of the given equation to make it look like the form above...
You started with:
4x-2y=10
begin by subtracting 4x from both sides:
-4x+4x-2y = -4x+10
-2y = -4x+10
multiply both sides by -1:
2y = 4x-10
divide both sides by 2:
y = 2x-5 (this is what they're looking for)

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A 18-foot ladder leans against a wall at a 74 degree angle of elevation. Find the distance from the ladder's base to the wall.
.
Let x = distance (base to wall)
.
Applying trigonometry:
cos x = adjacent/hypotenuse
cos(74) = x/18
18*cos(74) = x
4.96 feet = x
or approximately:
5 feet = x