Actually, y=4x2 is narrower than y=x2
If graphed, the 4, as the coefficient of x2 multiplies the y value by 4, resulting in the following points:
y=4x2: (0,0), (1,4), (-1, 4), (2, 8), (-2, 8);as opposed to the points for y=x2: (0,0), (1,1),(-1,1),(2,4),...
Remember that 1-sin2x = cos2x
So if you multiply this fraction (cosx)/(1-sinx) by (1+sinx)/(1+sinx) you will get:
(cosx)(1+sinx)/(1-sin2x) = (cosx)(1+sinx)/(cos2x) or (1+sinx)/(cosx)
or: 1/cosx + sinx/cosx =...
Therefore, if f'(x) = 2x and Δx = 1, Δy = 2(3)(1) = 6.
I hope this helps.
9in = .75 ft.
(.75 ft)3 = .421875 ft3
110 ft3/(.421875 ft3/balloon) = 260.7 balloons/ft3
Find the perpendicular bisectors of segments AQ, QB, BT and TA. If a point lies on the perpendicular bisector of a segment, all points on that line are equidistant from the endpoints. Therefore, the point at which these lines meet is the center of a circle.
That is a nice,...
AB is a chord bisected by PO, the radius since OT is an angle bisector. The arcs PA and PB are also congruent since the arc AB is bisected also.
Tangent TA and chord AB intercept arc PA and is half its measure.
Chord PA and chord AB intercept arc PB and...
(n+3)/(n2-5n+6) + (6)/(n2-7n+12)
Factor the denominators: (n+3)/((n-3)(n-2)) + (6)/((n-3)(n-4))
Note that the LCD is (n-3)(n-2)(n-4).
Now, multiply the first fraction by the missing factor: (n-4) and the second fraction...
Use the log and the equation becomes cos2x + tanx = 0
So, cos2x - sin2x + tanx = 0
Does this help?? Does using the identities to put this into sin and cos help?
csc2x = cot2x + 1, from the Pythagorean identities.
Therefore, csc2x - 1 = cot2x.
Substitute: (csc2x - 1)/cot2x = cot2x/cot2x = 1
I hope this helps.
Use the double angle formula: cos2x = 1 - 2sin2x.
Therefore, if sinx = -3/5, and sin2x = 9/25,
cos2x = 1 - 2(9/25) = 1 - 18/25 = 25/25 - 18/25 = 7/25.
BUT: Since the angle lies in quadrant 3, (π < x < 3π/2), the double angle, 2x lies between...
... or using de Moivre's Theorem: wk = z1/n = (rcisθ)1/n = r1/n (cis(θ+360k)/n), where k = 0 and 1, and n = 2,
z = 0 + 2i, so r = 2, and θ = 90°.
we get w0 = √2 (cis(90/2)), and w1 = √2 (cis(90 + 360)/2);
So, w0 = √2 cos 45 + √2 i sin...
The half angle formula: cos(θ/2) = ±√((1 + cosθ)/2).
Since sinθ = 3/5, cosθ = 4/5.
Therefore, (1+cosθ)/2 = 9/10, and cos(θ/2) = ±√.9.
Since 0 < θ < π, we know that 0 < θ/2 < π/2, and cos(θ/2)...
The half angle formula: sin(θ/2) =± √((1-cosθ)/2).
Therefore, (1-cosθ)/2 = (1+.6)/2 = .8.
sin(θ/2) = ±√(.8) and since π/2 < θ < 3π/2, Π/4 < θ/2 < 3π/4, sin(θ/2) is ±√(.8).
The arithmetic mean is the average of 2 or more numbers.
The arithmetic mean of x and y is: (x + y)/2.
The geometric mean is the square root of the product of two numbers.
For example: the geometric mean of 4 and 9 is 6, the square root...
y = cos-1(1/2) = π/3 and, since the cos is also positive for (3π/2, 2π), the answer needs to include 5π/3.
The value of arccos0 has two values: at π/2, and 3π/2.
Arccos(-15/17): x = -15, r = 17, and y = 8. Therefore, the cot must = x/y, or -15/8.
Using the product formula, taking the derivative of: x cos y = 2,
We get: 1 cos y + x (-sin y) dy/dx = 0
So -x sin y (dy/dx) = -cos y, Divide by -xsiny, dy/dx = (cos y)/(x sin y) = (cot y)/x.
First: multiply all equations by 10.
5a + 3b = 22
12c + 85b = -244
33c + 13a = 290
Eliminate b from the first 2 equations, getting an equation in a and c.
Use that and the 3rd equation to solve for a and c.
Bus: r = 45 m/h, t = 2 h, d = rt, therefore, the bus travels 90 miles in 2 hours.
Car: let its rate = r, time = 2 h, d = 20 miles more than bus (90 miles + 20 miles) = 110 miles.
So: r = 110/2 = 55 mph.
Q1. Do you mean hextagon or hexagon? The last three are parallelograms, and therefore, by definition, are polygons with parallel sides.
Also, all hexagons don't have any parallel sides unless they are regular or they may have only 2 parallel sides.
If you take the equation and square both sides, here is what happens:
(√x - √5)2 = (√x-5)2 becomes: x -2√(5x) + 5 = x-5
So, subtract x and subtract 5: -2√(5x) = -10
Square both sides: 4(5x) = 100, ...