<p>Oalrite, fellows, I have my midterm tommorrow, and I am reviewing, but there is a few questions that I can not solve, any help would be greatly thanked…</p>
<li><p>sq(3x-2)=4-x</p></li>
<li><p>sq(2x+3)+sq(x-2)=2</p></li>
<li><p>tan of 390…how do I find the exact value?</p></li>
<li><p>And what is even and odd functions…</p></li>
<li><p>tan(sin^-1 -sq(3)/2)</p></li>
</ol>
<p>thanks a lot, people, I cannt get a hold of my math teacher and I really need those question answered…</p>
<p>sq(3x-2)=4-x square both sides
3x-2 = 16-8x+x^2
x^2 - 11x + 18=0
(x-9)(x-2) so x = 2,9
they both work because the squareroot needs to be atleast 0 and both cases its greater</p>
<p>number 2 put one sqrt on one side then square and then put all non sqrt terms on one side and square again. Then you should have no more squareroots. Solve for x by using quadratic eq or w/e and then make sure if u plug x into the original its sqrt of a pos number.</p>
<p>tan of 390 = tan of 30 which is sin30/cos 30
draw a 30, 60, 90 triangle and just look at the side relations, x, xroot3 2x.</p>
<p>Even is y axis symmetry, plug in negative x for every x, equation should equal. IE no x to an odd exponent
Even i forget</p>
<p>On a ferris wheel, you travel through a central angel of 3,000. If the radius is 57 feet, how far have you traveled?</p>
<p>--- can someone enlight me on that problem, I thought its similar to an arc length problem, but the answer is 2984.5 feet and I just cannt get anywhere close....</p>
<p>For the ferris wheel, 3000degrees /360 degrees = 8.333 rotations. Each rotation is 2pi<em>R (circumference of the ferris wheel)= 2(3.14)</em>57 = 2983 ft</p>
<p>Say if cosine is root 3/ 2 then secant is just 1/cos so you just switch the cosine value. So root3/2 is to cosine as 2/root3 is to secant. Similiar, you can do the same to sin, because 1/sin = cosecant. However, cosecant is undefined whenever sin=0, values of 0 and 180 degrees or in radians, 0 and pi.
Cot is just 1/tan</p>
<ol>
<li><p>tan(sin^-1 -sq(3)/2)
it is asking you to take the tan of an angle whose sin is equal to root3/2
The sin of 60 = sq3/2, so you have to make that negative. You move it into the third or fourth quadrant. Then you have 60+180 and 360-60. The answer is the tan of 240 and the tan of 300.</p></li>
<li><p>sq(2x+3)+sq(x-2)=2
sq(2x+3)=2-sq(x-2) square both sides
2x+3 = 4-4sq(x-2) + x-2
4sq(x-2) = -x - 1 divide by 4
sq(x-2) =-(x+1)/4 square both sides
x-2 = -(x^2 + 2x +1) /16 multiply by 16
16x-36 = -x^2 + 2x +1
X^2 + 14x - 35 =0</p></li>
</ol>
<p>use quadratic -3.25 and -10.75 Now you have to check
sq(2x+3) for x= both of these, since they both fail, there are no solutions. They would both have to pass this sq, and also the sq(x-2). You cant take the sqrt of a negative number (no imaginary). Also be sure to check your solutions for values of x is you have say (x^2-4) /(x-3) = a number. If x =3 seems to be a solution, you must check it and see it is not, (/ by 0 error).</p>