<h1>3</h1>
<p>(x-y)/r can be written as x/r - y/r then x/r is cos theta and y/r is sin theta</p>
<p>1) A range is a set that contains all the values of the function. We have an open interval (-5;5) which means that the function doesn’t include 5 or -5. So only function (B) has this domain. </p>
<p>2) Let h<em>1 be the height of the pyramid and S</em>1 be the area of the base of the pyramid.
h<em>2 is the height of the box and S</em>2 is the area of the base of the box.
From the problem we obtain h<em>2=1/3 h</em>1 and S<em>2=2S</em>1.
As we know, the volume of the pyramid can be calculated using the following formula
V=1/3* h<em>1* S</em>1<br>
The volume of the box is
V<em>box=S</em>2* h<em>2=1/3* h</em>1<em>2S_1=2</em>1/3* h<em>1* S</em>1=2V.
So the correct answer is (C).</p>
<p>3) We have the right triangle with catheti x, y and hypotenuse r. From the definition of the sine and cosine functions we get
sinθ=y/r, cosθ=x/r
So we get
(x-y)/r=x/r-y/r=cosθ-sinθ.
Therefore, the correct answer is (B).</p>
<p>4) We have the equation A=A<em>0<em>e^(-t/300). We need to find the value of t at which A=0.5</em>A</em>0. Substituting this into the given equation, we get
0.5*A<em>0=A</em>0 e^(-t/300)
0.5=e^(-t/300)
ln0.5=-t/300
t=-300 ln0.5=300 ln2≈208.
Therefore, the correct answer is (D).</p>
<p>5) From the definition of inverse function we get
y=F(x) ⇔ x=F^(-1) (y)
From this we can find
y=2x ⇔ x=y/2 ⇒ f^(-1) (x)=x/2.
To find f^(-1) (3x+6) we should substitute (3x+6) instead of x into the inverse function.
f^(-1) (3x+6)=(3x+6)/2=3x/2+3.
Therefore, the correct answer is (B).</p>
<p>Yes, #1 is definitely B. The range of the function in choice A is [-5, 5).</p>
<p>Also #12 is B, II only. x^2 + 2 only covers the positive numbers at least 2, and 1 - cos x only hits the numbers in the set [0,2].</p>
<ol>
<li><p>As 3/2>1, we have that (3/2)^n tends to infinity, as n—>\infinity. The answer is (D).</p></li>
<li><p>Take a look at (C): 0=\sqrt(1/\cos(0)-1) (1) .
As \cos(0)=1, we obtain in (1): \sqrt(1/1-1)=\sqrt(0)=0. We cant take heta=\pi/2 or -\pi/2, because \sec( heta) doesnt exist. If we take heta=-\pi/4, we get: \sqrt(1/cos(-\pi/4)-1)=\sqrt(1/\sqrt(2)-1) , which doesnt equal -\pi/4. So, the answer is (C).</p></li>
<li><p>You have 5 possible outcomes: HHH, HTH, HTH, HHHH, HHHH. So, the probability that the coin landed heads up all four times (its the last outcome) is 1/5.</p></li>
<li><p>An arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant:
a<em>(n)=a</em>(1)+(n-1)d.
We have that a<em>(3)=-2. On the other hand, a</em>(3)=a<em>(1)+2x. And a</em>(5)=a<em>(1)+4x, a</em>(10)=a<em>(1)+9x. So, we have a system of equations:
a</em>(1)+2x=-2 and a<em>(1)+4x=a</em>(1)+9x-3. → 5x=3 → x=3/5. And a<em>(1)=-6/5-2=-16/5. Find a</em>(6) and a<em>(7):
a</em>(6)=a<em>(1)+5x=-16/5+3=-1/5<0;
a</em>(7)=a<em>(1)+6x=-16/5+18/5=2/5>0
So, zero lies between a</em>(6) and a_(7).The right answer is (C).</p></li>
</ol>