3 (n+6) do I do 3xN=3n than 3x6=18 n=6?
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Helllllp >_< Compare: 1.2057 O 1.1421OA) >OB) <OC)=OD) none of the above
The two points (-2,4) and (4,2) are the endpoints of the diameter of a circle. What is the equation of this circle in standard form?
Wedgewood is looking to buy new computers from one of two companies. ComputersRus will charge $650 per computer. SchoolzToolz will charge the school $200 per computer as long as the school gives them a $75,000 contract. Which expression represents SchoolzTools? a) $650x + $0 b) $650 + $0x c) $200x + $75,000 d) $200 + $75,000x
-2x - 35 = 9 - 4x answers
B) Plug in 0 for x into both equations and solve for y. Then plug that answer back into the other equation to find the corresponding x-coordinate.
C) Solve both equations for x and set them equal to each other. This will give you the y-coordinates of the points of intersection. Then plug back into one of the equations to find the corresponding x-coordinates.
D) Solve both equations for y and set them equal to each other. This will give you the x-coordinates of the points of intersection. Then plug back into one of the equations to find the corresponding y-coordinates.
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OptionC) Solve both equations for x and set them equal to each other. This will give you the y-coordinates of the points of intersection. Then plug back into one of the equations to find the corresponding x-coordinates.
x = 2y^2 + 3y + 1 x = - 3y^2 / 2
2y^2 + 3y + 1 = - 3y^2 / 2
4y^2 + 6y +2 = -3y^2
4y^2 + 3y^2 + 6y +2 = 0
7y^2 +6y + 2 = 0
Now you apply the quadratic formula and find y-values. Then use those values to find the x -values.
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A=93.5 sq ft
h=6 ft
V=93.5*6*1/3=93.5*2=187 cubic ft
So: 1/3 * 93.5 * 6 = 187.
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Answer:
To compute the length of the curve f(x)=47(4−x2) over the interval 0≤x≤2, we need to use the formula for the arc length of a function:
L=∫ab1+(f′(x))2dx
where a and b are the endpoints of the interval. First, we need to find the derivative of f(x), which we can do by using the chain rule and the power rule:
f′(x)=4dxd7(4−x2)
f′(x)=427(4−x2)1dxd(7(4−x2))
f′(x)=427(4−x2)1(−14x)
f′(x)=−7(4−x2)28x
Next, we need to plug in f′(x) into the formula and simplify:
L=∫021+(−7(4−x2)28x)2dx
L=∫021+7(4−x2)784x2dx
L=∫027(4−x2)7(4−x2)+784x2dx
L=∫024−x228−21x2dx
Now, we need to evaluate the integral, which we can do by using a trigonometric substitution. Let x=2sinu, then dx=2cosudu and u=arcsin(x/2). The limits of integration change as follows:
x=0⟹u=0
x=2⟹u=2π
The integral becomes:
L=∫02π4−(2sinu)228−21(2sinu)2(2cosu)du
L=∫02π4−4sin2u28−84sin2u(2cosu)du
L=∫02π1−sin2u7−21sin2u(2cosu)du
L=∫02πcos2u7−21sin2u(2cosu)du
L=∫02π27−21sin2udu
Using a trigonometric identity, we can write:
L=∫02π4127−1221cos(2u)du
Using another trigonometric substitution, let v=2u, then dv=2du and u=v/2. The limits of integration change as follows:
u=0⟹v=0
u=2π⟹v=π
The integral becomes:
L=∫0π4127−1221cosv(21)dv
L=6∫0π
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Answer:
x = 1
Step-by-step explanation:
Plan A: 20 + 10x
Where, x = fee per lesson
Plan B: $3 × 10 lessons = $30
Equate plan A and plan B
Plan A = Plan B
20 + 10x = 30
10x = 30 - 20
10x = 10
x = 10/10
= 1
x = 1
70°
12x + 14
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Answer:
Step-by-step explanation:
12x=70-14=56
x=
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OK. Relax, settle down, don't worry.
We're in luck here. The shape hasn't been rotated, or swelled up
or shrunk. It's still exactly the same size and shape, and in the same
position. It's only been moved, from here to there. All you really
have to do is look at one point, and figure out how that point moved,
and all the other points moved exactly the same ... same distance,
in the same direction.
When I looked at this, I worked with point-D. But now I think
point-B may be a little easier to work with.
-- What's the 'x'-value (left/right) of point-B before it moves ?
It's right on the y-axis, so its x-value is zero.
-- What's the 'x' value of point-B', after the move ?
Look at the light light line that B' is on, follow it down,
and you see that the x-value of point-B' is 6 .
-- The x-value of point-B moved from x=0 to x=6.
If you look at just the x-value of any other point on the shape,
you'll see that every point moved 6 units to the right.
Every 'x'-value became 6 units bigger after the move.
x --> x+6
===================
-- What's the 'y'-value (up/down) of point-B before it moves ?
It's in the middle between the 2 and the 4, so its y-value is 3.
-- What's the 'y' value of point-B', after the move ?
Look at the light light line that B' is sitting on, follow it left,
and you see that the y-value of point-B' is 8 .
-- The y-value of point-B moved from y=3 to y=8.
If you look at just the y-value of any other point on the shape,
you'll see that every point moved 5 units UP.
Every 'y'-value became 5 units bigger after the move.
y --> y+5
Look for a choice that says what we found.
The second choice says it.