Please help for my geometry test
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OMG THANKS FOR THE POINTS

Step-by-step explanation:

You can use the sin() and cos() and tan() and SOHCAHTOA to solve for the angle.

If you think that problem is hard, see this:

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10 of the Toughest Math Problems Ever Solved

They're guaranteed to make your head spin.

BY DAVE LINKLETTER

SEP 11, 2019

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Earlier this week, a math puzzle that had stumped mathematicians for decades was finally solved. It’s called a Diophantine Equation, and it’s sometimes known as the “summing of three cubes”: Find x, y, and z such that x³+y³+z³=k, for each k from 1 to 100.

On the surface, it seems easy. Can you think of the integers for x, y, and z so that x³+y³+z³=8? Sure. One answer is x = 1, y = -1, and z = 2. But what about the integers for x, y, and z so that x³+y³+z³=42?

That turned out to be much harder—as in, no one was able to solve for those integers for 65 years until a supercomputer finally came up with the solution to 42. (For the record: x = -80538738812075974, y = 80435758145817515, and z = 12602123297335631. Obviously.)

That’s the beauty of math: There’s always an answer for everything, even if takes years, decades, or even centuries to find it. So here are nine more brutally difficult math problems that once seemed impossible until mathematicians found a breakthrough.

The Poincaré Conjecture

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In 2000, the Clay Mathematics Institute, a non-profit dedicated to “increasing and disseminating mathematical knowledge,” asked the world to solve seven math problems and offered $1,000,000 to anybody who could crack even one. Today, they’re all still unsolved, except for the Poincaré conjecture.

Henri Poincaré was a French mathematician who, around the turn of the 20th century, did foundational work in what we now call topology. Here’s the idea: Topologists want mathematical tools for distinguishing abstract shapes. For shapes in 3-D space, like a ball or a donut, it wasn’t very hard to classify them all. In some significant sense, a ball is the simplest of these shapes.

Poincaré then went up to 4-dimensional stuff, and asked an equivalent question. After some revisions and developments, the conjecture took the form of “Every simply-connected, closed 3-manifold is homeomorphic to S^3,” which essentially says “the simplest 4-D shape is the 4-D equivalent of a sphere.”

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A century later, in 2003, a Russian mathematician named Grigori Perelman posted a proof of Poincaré’s conjecture on the modern open math forum arXiv. Perelman’s proof had some small gaps, and drew directly from research by American mathematician Richard Hamilton. It was groundbreaking, yet modest.

After the math world spent a few years verifying the details of Perelman’s work, the awards began. Perelman was offered the million-dollar Millennium Prize, as well as the Fields Medal, often called the Nobel Prize of Math. Perelman rejected both. He said his work was for the benefit of mathematics, not personal gain, and also that Hamilton, who laid the foundations for his proof, was at least as deserving of the prizes.

Fermat's Last Theorem

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Pierre de Fermat was a 17th-century French lawyer and mathematician. Math was apparently more of a hobby for Fermat, and so one of history’s greatest math minds communicated many of his theorems through casual correspondence. He made claims without proving them, leaving them to be proven by other mathematicians decades, or even centuries, later. The most challenging of these has become known as Fermat’s Last Theorem.

It’s a simple one to write. There are many trios of integers (x,y,z) that satisfy x²+y²=z². These are known as the Pythagorean Triples, like (3,4,5) and (5,12,13). Now, do any trios (x,y,z) satisfy x³+y³=z³? The answer is no, and that’s Fermat’s Last Theorem.

may i see the picture? it would a lot, xox

u = -(6/7)w

step-by-step explanation:

place all values with 'u' on one side of the equation.

4u + 8w = - 3u + 2w

4u + 3u = 2w - 8w

further simplify the equation by solving the addition and subtraction.

7u = - 6w

make 'u' the independent variable by placing its co-efficient on the right-hand side.

u = - (6/7)w