Trigonometry

x= 30 degrees

first, subtract 3 from 4 and you get 2sinx=1

then, divide both sides by 2 to get sinx=1/2

by using a 30, 60, 90 triangle you can see that 1 is the side opposite theta and

2 is the hypotenuse

therefore, your answer is 30

By statistical analysis.

It is very difficult to calculate these using mechanics. Calculations of the trajectory of a projectile assume that the mass of the projectile is such that air resistance has a negligible effect. This is not the case when the projectile is confetti - even if it is packed densely to start with.

0.75

Consider the Complex Plane, with Real numbers along the horizontal axis, and Pure Imaginary numbers on the vertical axis. Any Complex number (a + ib) can be plotted as a point (a,b) on this plane. The point can be represented as a vector from the 'origin' (0,0) to the point (a1,b1). If the second 'complex vector' (a2,b2) is added to the first, this can be shown as a translated vector with it's 'tail' starting at the arrowhead of the first vector, and then the arrowhead of the second vector will terminate at the sum of: a1 + ib1 + a2+ ib2 [coordinate point: (a1+a2,b1+b2)

The TI-83 does not have the cot button, however, if you type 1/tan( then this will work the same as the cot since cot=1/tan.

The other way to do this is to type (cos(x))/(sin(x)) where x is the angle you're looking for. This works because cot=cos/sin

An example of a real life exponential function in electronics is the voltage across a capacitor or inductor when excited through a resistor.

Another example is the amplitude as a function of frequency of a signal passing through a filter, when past the -3db point.

You can choose either or but tangent which is sin/cos seems to be the most common way.

x + y = 13 . . . . . . . . . . (A)2*x + 3*y = 1 . . . . . . . (B)

(B) - 2*(A) : y = -25

Then, by (A) x - 25 = 13 so that x = 38.

There are only three kinds of triangle - equilateral, where all three sides are the same - isoceles, where only two sides are the same - and scalene, where no sides are the same.

The right triangle is a special case of the isoceles or scalene. Even if you consider the right triangle different, which is is not, that only make five kinds of triangle.

The sine of an angle of a right triangle - which is a triangle containing one 90o angle - is calculated as the length of the side opposite the angle divided by the length of the hypotenuse.

For very small values of x, sin(x) is approximately equal to x.

There are many ways to find the area. We restrict the answer to a regular pentagon with is one with equal sides and angles.

One method uses the apothem which is a line segment from the center to the midpoint of any of the sides. We use the length of the apothem. In this case, the area is 1/2 x apothem x perimeter.

This method works for any regular polygon.

Here is a slightly more complicated method and explanationl

We have n sides, all equal length a

We have n interior angles, all equal measure beta

Alpha is the central angle subtending one side

P is the Perimeter

K is the Area

Let R be the radius of circumscribed circle and

r be the radius of inscribed circle.

The for any regular polygon we have:

beta = Pi(n-2)/n radians = 180o(n-2)/n

alpha = 2 Pi/n radians = 360o/n

alpha + beta = Pi radians = 180o

P = na = 2nR sin(alpha/2)

K = na2 cot(alpha/2)/4

= nR2 sin(alpha)/2

= nr2 tan(alpha/2)

= na sqrt(4R2-a2)/4

R = a csc(alpha/2)/2

r = a cot(alpha/2)/2

a = 2r tan(alpha/2) = 2R sin(alpha/2)

For a regular pentagon

Number of sides n = 5

Internal angles beta = 3/5 radians = 108 degrees

Central angles alpha = 2/5 radians = 72 degrees

Perimeter P = 5a = 5R sqrt(10-2 sqrt[5])/2

Area K = 5a2 sqrt(1+2/sqrt[5])/4

= 5R2 sqrt(10+2 sqrt[5])/8

= 5r2 sqrt(5-2 sqrt[5])

= 5a sqrt(4R2-a2)/4

Circumradius R = a sqrt(2+2/sqrt[5])/2

Apothem r = a sqrt(1+2/sqrt[5])/2 = R(1+sqrt[5])/4

Side a = 2r sqrt(5-2 sqrt[5]) = R sqrt(10-2 sqrt[5])/2

Yes, Anuita her older sister and a brother

Isosceles, right, acute, obtuse, scalene, equilateral, (anyone else feel free to tag on)

equiangular

-14 + 3p = -9p - 21

add 14 and 9p to both sides, then 12p = -7

p = -7/12

To solve equations with absolute values in them, square the absolute value and then take the square root. This works because the square of a negative number is positive, and the square root of that square is the abosolute value of the original number.

Yes, the law of sines can be used in a right triangle. The law applies to any arbitrary triangle.

The slope is a negative number.

A sine wave is a simple vertical line in the frequency domain because the horizontal axis of the frequency domain is frequency, and there is only one frequency, i.e. no harmonics, in a pure sine wave.

False.

The angles can be formed by two skew lines intersecting a third line.

Farmer John stores grain in a large silo located at the edge of his farm. The cylinder-shaped silo has one flat, rectangular face that rests against the side of his barn. The height of the silo is 30 feet and the face resting against the barn is 10 feet wide. If the barn is approximately 5 feet from the center of the silo, determine the capacity of Farmer John's silo in cubic feet of grain.

Hint: Look for connections between this problem and the problem you solved in Part 1 of this week's Discussion (a square inscribed in a circle).

Is this correct?

The volume of any prism is the area of the face multiplied by the height. The height is given, so all you need to find is the area of the base. Start off by drawing a circle. One side has a chord of length 10 feet cut across it. Its given that the chord is 5 feet from the center of the circle. Therefore, you draw a line from the center bisecting the chord, as well as a line from the center to each end of the chord (the radius), and you have two equal isosceles triangles with two sides of length 5. Pythagorean's relation tells us that the third side has a length of 5*sqroot(2) ft. Now that you have the radius, you can find the area of the circle. However, its not a true circle you're concerned with. You want to exclude the area on the outside of the chord. Using trig, you can find that the total angle subtended by the chord is 90°. Therefore, the area that you're concerned with is the circular area for the outside 270° plus the area of the triangles. 270° is (3/4) of the circle, so its area is (3/4) the area of the entire circle. A = (3/4)*π*r² = (3/4)*π*(5*sqroot(2)ft)² = 118ft². The area of each of the smaller triangles is given by (1/2)*b*h. Since there's two triangles, the total area of the two is 2*(1/2)*b*h = b*h = 5ft*5ft = 25ft². The total area of the base is 118ft² + 25ft² = 143ft². Multiplied by the height will give the total volume of the prism, 143ft²*30ft = 4290ft³.

If you know the length of the sides of a triangle you can find all the angles of the triangle using the Law of cosines such as:

Step 1.

cos A = (b^2 + c^2 - a^2)/(2bc)

cos B = (a^2 + c^2 - b^2)/(2ac)

cos C = (a^2 + b^2 - c^2)/(2ab)

Step 2.

Find the arc cosine A, arc cosine B, and arc cosine C in order to find the angles A, B, and C.

Zero.

The simplest geometric concept is an item that has no dimensions = a point. The next simplest is an item that has only one dimension = a straight line. The next simplest is an item that covers area but only needs one number to completely describe it . . . a circle.