proyaop

If you graph it on desmos, you'll see that x = 10.9

Let the load that the small trucks be equal to a variable, small_load. Let the load that the large truck can carry be large_load.

By the statements:

large_load >= small_load + 600

large_load <= (small_load)*4/2 = 2(small_load)

Thus, 2(small_load) >= large_load >= small_load + 600, so 2(small_load) >= small_load + 600

Thereafter, small_load >= 600.

Plugging it into the upper-bound inequality:

large_load <= 2(small_load) = 2(600) = 1200 pounds.

By special right triangls, triangle XYZ is a 45-45-90 with legs XY and XZ = 10. 

Let the radius of the semicircle = r. Also, let the point of tangency between the semicircle and XY, XZ be P and Q respectively. Let the center of the semicircle be O (which also happens to be the midpoint of YZ). 

OP = OQ = r, and since P and Q are points of tangency, they are perpendicular to the tangent line XY and XZ.

[XYZ] = 10*10/2 = 50 units squared

[XYZ] = XY*PO/2 + YZ*QO/2 (because you can divide the triangle in two, but still obtain the same area). 

[XYZ] = 50 = 10r/2 + 10r/2 = 10r, so r = 5.

If r = 5, then the area of the semicirlce is pi*r^2 / 2 = 25pi/2

The rectangle with these coordinates is centered at the origin, with side length 2. Note that any equation y = mx will bisect the area of the square, so the y-intercept is 0. (Try drawing it on graph paper too) 

f(x) = a{trig}(bx - c) + d

a is the amplitude

2pi/|b| = the period

c/b = the horizontal phase shift

d = midline

Following this generalization, the midline is y = 1.8

0.03 + 0.0014 repeating = 0.0314 repeating

Let 0.0014repeating = S, 100S = 0.141414repeating = S + 0.14

Solving, you have 99S = 0.14, S = 0.14/99 = 14/(99)(100) = 14/9900

0.03 + 7/4950 = 297/9900 + 14/9900 = 311/9900

A cubic polynomial, by the fundamental theorem of algebra, will have EXACTLY three roots (which generalizes to an nth degree polynomial to have n complex roots). Clearly, in this problme, there are four roots, so a cubic polynomial f does not exist, and all coefficients would be 0. Thus, the sum of the coefficients is 0.

Let the center of the small circle be O, and the point where circle O is tangent to the circle with diameter AC be D, and let the radius of circle O be r. Lastly, let the midpoints of AB and BC be points M and N, respectively.

Since B is the center of a semicircle, BD must also be the same length as AB (because they are the radius) = 2.

MO = 1 + r, and MB = 1. OMB forms a right triangle with angel OBM = 90 degrees, since O lies on the perpendicular bisector of AC.

This means that OB, by the pythagorean theorem, has length \(\sqrt{(1+r)^2-1}=\sqrt{r^2+2r}\).

Additionally, DO = r, and lines on the same line as OB. Recalling that DB = 2, we can set an equation: DO + OB = 2:

\(r + \sqrt{r^2+2r}=2\)

\(\sqrt{r^2+2r}=2-r\)

\(r^2+2r=(2-r)^2=r^2-4r+4\)

\(2r = -4r+4\)

\(6r=4\)

\(r={2\over3}\)

Because triangle ABC is a right triangle with B = 90 degrees, then we know that the pythagorean theorem is a possible strategy.

AB = 3, and BC = 4, and they are both the legs of the right triangle because they include the right angle, and so by pythagorean theorem:

\(AC^2=AB^2+BC^2=3^2+4^2=25\)

So, AC = 5 units.

Through construction, AB = AF + FB = 3.

AF = EF = x. Since ADEF is a square, DE is parallel to AB, and thus angle CED is congruent to angle CBA by corresponding angles, and ADE = angle EFB because they are both 90 degrees. Thus, triangle CDE is similar to EFB is similar to CAB, and CA : AB = 1 : 3, likewise, FE : FB = 1 : 3. Substituting, we have x : FB = 1 : 3, and therefore FB = 3x. Substituting this into the original equation, AB = x + 3x = 4x = 3. And thus, x = 3/4. The area of the square would be x^2 = 9/16.