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Thepotemich [5.8K]
4 years ago
13

Find the length of the third side to the nearest tenth.3 6​

Mathematics
1 answer:
timurjin [86]4 years ago
6 0

i may not know how the triangle looks like but i assume that its a right angle triangle.

if we want to find the missing length example hypotenuese.. 3 is adjacent and 6 is opposite. Then use the formula:

c²= a²+b²

c²= 3²+6²

c²= 45

c= √45

c= 7 (nearest tenth)

hope this helps

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Answer:B. 1

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Answer:

  Practice -- lots of it

Step-by-step explanation:

I usually choose a solution method that I believe will find the desired answer for me in the least number of simple steps. What constitutes a "simple" step depends on a number of things:

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I often choose a graphing calculator as my first approach to solving a quadratic. It easily shows the roots, and can show the vertex. (Sometimes, scaling the graph is required, so I actually prefer to use a touch-screen for the purpose. It is easier to "pinch" than to type in possible graph limits by trial and error.) The answer is generally provided to 3 or 4 significant figures, which is usually enough to tell if the solution is rational or irrational.

The easiest way to describe an integer (or rational) solution to someone else (on Brainly, for example) is usually to show the factoring of the quadratic. Finding the roots with a graphing calculator significantly aids the factoring process. (It is much easier to "find" the solution if you know the solution before you start looking.)

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As we said at the beginning, the answer to your question is "practice." Many curricula try to offer opportunities to practice. I find most of these pretty annoying. What I suggest is that you find your own internal motivation to practice solving different kinds of quadratics in several different ways. That is, <em>solve the same quadratic</em> by ...

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  • factoring
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  • scaling and/or translation and/or transformation*

It might take you a couple of hours of work to learn the best way to solve an equation in under 5 minutes.

In this process, you will discover what you need to know to choose a method that works nicely for you. I believe you will find that total familiarity with arithmetic facts (sums, products, squares) and operations on all the different ways to represent numbers (integer, decimal, fraction, percent, scientific notation) will help you significantly.

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<em>* Additional comment</em>

Some "area" problems give you the difference of dimensions (length and width), and ask you to find the dimensions that give a specific area. This can be solved by (x)(x -d) = A. If you don't immediately recognize factors of A that differ by d, then you might have to resort to the quadratic formula to solve this.

Another approach is to <em>let the variable represent the average dimension</em>. Then the equation becomes (x -d/2)(x +d/2) = A, which is a "difference of squares" form that is much more easily solved. The solution for x is √(A +(d/2)²), and all you have to do is add and subtract d/2 from that solution to find the dimensions. This is an example of what I mean by <em>transforming the problem.</em>

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<h2><u>Answer:</u></h2>

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