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<p>I need help with the first one (Skill 1 - Example). I have no idea how to find the answer.</p>
<p>x/y=r
ry=x</p>
<p>Now substitute.</p>
<p>ry+y/x =</p>
<p>Then distribute y</p>
<p>y(r+1)/ry =</p>
<p>(r+1)/r </p>
<p>D</p>
<p>Hope that helps !</p>
<p>It does, thanks a lot!</p>
<p>How do you do the first one [url=<a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools]here[/url</a>]?</p>
<p>And, last question, how do you do the first one [url=<a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools]here[/url</a>]?</p>
<p>2nd question
We know that the sum of sq root of x and the sq root of y is 10.
So list out the numbers the numbers that will add up to 10.
Lets just list all the single digits
1,2,3,4,5,6,7,8,9</p>
<p>The question states that x and y are odd, thus their roots must also be odd.
the odds are 1,3,5,7,9
Which one of these makes 10?
Lets choose 9 and 1.
so 9^2=81=x
1^2=1=y
1+81=82. </p>
<p>Sorry i don’t know how to solve that one algebraically so i just reasoned it out.</p>
<p>3rd question.</p>
<p>we are given that the circles are tangent, so rsp is 90 degrees.
The radius of the small one is 1, and the radius of the alrher one is 6.
We can just get segment PQ by adding the 2 radius because they create the triangles hypotenuse.</p>
<p>The radius of the bigger circle, pr, is 6, and is also the leg of the triangle.</p>
<p>Use pythagorean theorem</p>
<p>7^2-6^2 = 13</p>
<p>We want RS, which is on RQ.</p>
<p>This part is easy, we already know QS, qhich is the radius. so to find rs</p>
<p>the length of RQ = sq root of 13 - 1. </p>
<p>B.</p>
<p>**If x and y are positive odd integers and squareroot of x plus squareroot of y equal 10, what is one possible value of x + y ? **</p>
<p>Knowing that x and y are odd integers, possibilities for sq(x), sq(y) would be:</p>
<p>1, 9
3, 7
5, 5</p>
<p>(Obviously you can reverse them for y and x, but since you’re looking for the sum, those would be duplicates.)</p>
<p>Therefore, x+y:
1, 9 => 1 + 81 = 82
3, 7 => 9 + 49 = 58
5, 5 => 25 + 25 = 50</p>
<p>**In the figure above, the circle with center P has radius 6 and the circle with center Q has radius 1. The circles are tangent to each other. Segment QR is tangent to the larger circle and intersects the smaller circle at S. What is the length of segment RS ? **</p>
<p>First of all, since QR is tangent to the larger circle, therefore PRQ is a right triangle. We know:</p>
<p>PR = radius of larger circle = 6
PQ = radius of larger circle + radius of smaller circle = 7</p>
<p>Therefore, via Pythagorean theorem, RQ = sq(49-36) = sq(13)</p>
<p>Since RQ = RS + SQ
Therefore RS = RQ - SQ</p>
<p>Notice that SQ = radius of smaller circle
Therefore RS = sq(13) - 1</p>
<p>Thanks a lot!</p>
<p>I assumed the triangle question was more difficult than it was because it said that the figure wasn’t drawn to scale. I realize now that PR and RS have to form a right angle because RS is tangent to the larger circle. Thanks!</p>