Tutor profile: Corey N.
In a right triangle, where x° is one of the angles and sin(x°) = 3/6, what is the cos(90-x°)?
Sin = Opposite / Hypotenuse Cos = Adjacent / Hypotenuse Tan = Opposite / Adjacent *the above equations can be remembered using the word: SOH-CAH-TOA *A right triangle has one 90° right angle and two complimentary acute angles that add up to the other 90°, making a total of 180° in a right triangle. The cos(90-x°) is referring to the other acute angle in the triangle. If you subtract the right angle (90°) from our angle (x°), we are left with the last angle of the triangle. *We know that the cosine of one complimentary angle and the sine of the other complimentary angle are equal. Given that our two angles are complimentary, we can conclude, without needing to do any calculations, that cos(90-x°) = 3/6, because cos(90-x°) = sin(x°).
Subject: Basic Math
Write 4 equations that fit the solution using basic arithmetic operators (+,-,*,/). The equations can be as complicated as you'd like, just as long as they use one of each of the 4 basic operators. 4 = __ + __ 4 = __ - __ 4 = __ * __ 4 = __ / __
4 = 2 + 2 4 = 8 - 4 4 = 4 * 1 4 = 8 / 2 An equal sign means the "same as" so we know that 2+2 = 4 just like we know that 2+2 = 8-4 = 4*1 = 8/2 = 4. They are all the same thing.
1.) Simplify: y = (x + 1)(x + 1) 2.) Expand: y = x² + 2x + 1
(x + 1)(x + 1) = x² + 2x + 1 These two are the same thing and let's look at why: 1.) y = (x + 1)(x + 1) to multiply the two parenthesis, we would solve like this: (a + b)(a + b) = (a*a)+(a*b)+(b*a)+(b*b) so in our case: (x*x)+(x*1)+(x*1)+(1*1) first, we know that x*x = x², so start simplifying x² + x + x + 1 you'll see that there are 2 x's by themselves, so x + x = 2x x² + 2x + 1 and that's it! we now have what's called a quadratic equation! we use this knowledge to now think of the problem in reverse. 2). y = x² + 2x + 1 We can recognize a quadratic equation using this simple patter of (x²) + (x) + (constant) The way we would the set up our problem: ( ± )( ± ) With our symbols in place, we then look for what numbers to fill. The first part is easy. x^2 can be created by multiplying 2 x's together: x * x = x^2 Place your 2 x's in the first part of each parenthesis: ( x ± )( x ± ) The second problem is tricky, but fun. Find 2 numbers that can be added to equal the "2" from our "2x" AND can also be multiplied to equal the constant, or the "1" in our case. Sounds strange right? Let's try a couple: 3 + -1 = 2 AND 3 * -1 = -3 nope... thats not it. The 2 works, but the -3 is not the same as +1 from our equation above. This one should work: 1 + 1 = 2 AND 1 * 1 = 1 the 1 and 1 can be added to equal 2 and multiplied to equal 1 so we will write the 1's in: ( x ± 1 )( x ± 1 ) Now we also know that the 1's are both positive so our "±" will be positive in both cases. y = ( x + 1 )( x + 1 ) and that's it! You can now simplify and expand a quadratic equation!
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