More Math Questions NEED HELP PLEASE

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1

<p>For all positive integers x, let xΔ be defined to be (x-1)(x+1). Which of the following is equal to 6Δ-5Δ?</p>

<p>2Δ+1Δ
3Δ+2Δ
4Δ+3Δ
5Δ+4Δ
6Δ+5Δ</p>

<p>The square of x is equal to 4 times the square of y. If x is 1 more than twice y, what is the value of x?</p>

<p>-4
-1/2
-1/4
1/4
1/2</p>

<pre><code> ▄(1 square)
▄▄(2 squares)
▄▄▄(3 squares)
▄▄▄▄(4 squares)
</code></pre>

<p>^^ The figure above shows an arrangement of 10 squares, each with side length k *inches. The perimeter of the figure is *p inches. The area of the figure is a square inches. If p=a, what is the value of k?</p>

<p>The answer for ^^ = 1.6 (not sure how)</p>

<p><a href="http://i32.tinypic.com/4ha4ja.jpg%5B/url%5D"&gt;http://i32.tinypic.com/4ha4ja.jpg&lt;/a&gt;&lt;/p>

<p>In the figure above, rectangles PQRS and WXYZ each have perimeter 12 and are inscribed in the circle. How many other rectangles with perimeter 12 can be inscribed. </p>

<p>Answer: More than 4 (but why?)</p>

<p>Next</p>

<p>Image</a> - TinyPic - Free Image Hosting, Photo Sharing & Video Hosting</p>

<p>The triangle above is isosceles and AB > AC. Which of the following must be FALSE?</p>

<p>AB=BC
BC=AC
x=y
x=z
y=z</p>

<p>Answer : E</p>

2

<p>First problem:</p>

<p>6Δ-5Δ= (6-1)(6+1)-(5-1)(5+1)=11
3Δ+2Δ= (3-1)(3+1)+(2-1)(2+1)=11
Answer is B.</p>

3

<p>bump 10char</p>

4

<p>2) The square of x is equal to 4 times the square of y. If x is 1 more than twice y, what is the value of x?
First equation:</p>

<p>x^2 = (4)*(y^2)</p>

<p>Second equation:</p>

<p>x=2y+1
square both sides
x^2=(2y+1)(2y+1)
x^2=4y^2 + 4y + 1</p>

<p>set the bold equations equal to each other
solve and you get 4y+1 = 0
solve this equation
4y = -1
y = -1/4</p>

<p>3) there are 10 squares of side length k
the area of 1 square is k^2
the area of 10 squares is 10*k^2</p>

<p>since a = p, and a = the area of the figure,</p>

<p>the area of the figure is also a
the area of the figure is also p, the perimeter of the figure</p>

<p>so</p>

<p>10*k^2 = a = p</p>

<p>p = perimeter, which is the sum of all the side’s lengths
if you count the number of uniform sides (the squares’ sides) you get 16 sides of k length each, or 16k</p>

<p>so perimeter = 16k = p</p>

<p>10<em>k^2 = 16k
solve for k:
k^2=1.6k
k</em>k=1.6*k
k=1.6</p>

<p>the key to this problem is just knowing how to get the perimeter and the area. you add up all the side’s lengths to get the perimeter. you find the area of each square and multiply that by the number of squares to get the area.</p>

<p>4) if you rotate the circle, or any circle about its center, then the circle will stay the same size, but the rectangles inscribed inside it will rotate amid the circle. this visualization shows that you can inscribe more rectangles all around the circle, not just up down or left right across the center</p>

<p>so the answer is infinitely many, or more than 4</p>

<p>5) if an angle is relatively big, then the side opposite that angle is also relatively big.
if a side is relatively big, then the angle opposite that side is also relatively big.</p>

<p>so if 1 side is bigger than another, its corresponding ANGLE is also bigger than the other’s</p>

<p>so since AB > AC, the angle opposite AB > the angle opposite AC, or y > z
so y=z is false</p>

5

<p>Sorry for nearly 2 year bump, but I need help on problem 5 as well. Basically, it looks sort of like this:
(ignore the periods, and imagine if it is a triangle, with uppercase letter (i.e. A, B, C) as the angle name, and lower case letter in parenthesis (i.e. (x), (y), (z)) as the variable representing their respective angles in degrees)</p>

<p>…A (y)
…^
…/…
…/…
…/…
(x)B ______ C (z)</p>

<p>The triangle above is isosceles and AB > AC. Which of the following must be FALSE?</p>

<p>(A) AB = AC
(B) BC = AC
(C) x = y
(D) x = z
(E) y = z</p>

<p>The answer is supposedly (E) y = z, but I don’t see why (C) x = y and (B) BC = AC are false as well, due to the property of isosceles triangles. Since AB > AC, that also means that AB and AC are not the congruent sides of the isosceles triangle, leaving AB = BC to be the only other option. And since AB = BC, it means that BC > AC, too. So doesn’t that also make (B) and (C) wrong as well? =&lt;/p>

6

<p>Three year bump now. I also need help with number 5.</p>

<p>You see, since AB is bigger than AC and it’s an isosceles tranlge, that means AC and BC are the two equal sides right?</p>

<p>I don’t get it, can someone explain it further</p>

7

<p>I believe the diagrams provided in the above posts are mis-labeled.</p>

<p>For an earler discussion of this question, look through this thread:</p>

<p><a href=“http://talk.collegeconfidential.com/sat-preparation/919126-tricky-math-questions.html[/url]”>http://talk.collegeconfidential.com/sat-preparation/919126-tricky-math-questions.html&lt;/a&gt;&lt;/p&gt;

<p>The diagrams posted in that thread show that x is the angle at vertex A, y is the angle at vertex B and z iat C.</p>

<p>Also, though AB>AC, that does NOT tell you which of the two sides are the equal ones. But, as you will see as you read through the thread, you don’t need to know that to know that one of the answers MUST be false…</p>