<ol>
<li>a^10=4900 find a^5
2 change log10 10,000+4 to exponential</li>
<li>give the domain of f(x) (3x+1)/(x^2-2x)</li>
<li>give the domain of: f(x)=radical over 3-x
5.simplify: (csc^2u-1)tan^2u=</li>
<li>T or F? cos^2x-sin^2x=1</li>
<li>simplify (sinx)(tanx)(csc^2x)=</li>
</ol>
<p>Could you kind of explain your answers.</p>
<ol>
<li> Factor out x from the denominator and set it equal to zero since the domain is everything we can put besides what is undefined. x^2-2x becomes x(x-2) so x=0,2 So domain is all real numbers except 0 and 2.</li>
</ol>
<ol>
<li>10 radical sign 4900. finish by putting the answer (2.3389…) to the 5th power… which should equal 70.
i believe… … does my explanation make sense? i can’t speak in mathy language for my lifeeeee.
if im not. ima dumbass. haha</li>
</ol>
<p>
</p>
<p>This question is a bit hard for me to read. What does the +4 have to do with the question?</p>
<p>
</p>
<p>False. The identity is cos²x**+**sin²x=1</p>
<p>
</p>
<p>tan x=(sin x/cos x), but that fact is sort of irrelevant now. Multiplying sin x with tan x leaves you with sin²x/cos x. csc²x just so happens to be equal to 1/sin²x. Thus, cancellation is possible and you’re left with 1/cos x, which is sec x.</p>
<ol>
<li><p>4900^(1/2) = 70. [the root of x^2 is x, so the root of a^10 is a^5]
Alternatively, 4900^(1/10) = a, a=2.339. 2.339^5 = 70.</p></li>
<li><p>By + I’m assuming you meant “=”. log10(10,000) = 4.
This can be rewritten as 10^4 = 10,000.</p></li>
<li><p>(3-x)^.5. You cannot square root a negative number, so set (3-x) >= 0.
x >= 3, so domain is XE[3, infinity)</p></li>
</ol>
<p>Thanks but can you explain to me why it’s 2pi over 3.</p>
<p>I advise that you go to yahoo answers for your pre-calculus hw or go to the SAT II Math level 2 forum.</p>
<p>Is there trig on the SAT? What?</p>
<p>I would advise never going to Yahoo answers.</p>