<p>can someone explain to me On the SAT blue book practice test 5 in section 3 numbers 7, 8, and 17,18 were done.</p>
<p>7 is the only one i feel like doing</p>
<ol>
<li>Call the two numbers A and B, where A > B. </li>
</ol>
<p>A - B = 1
A + B = T</p>
<p>Add the two, so 2A = 1 + T, A = (1+T)/2</p>
<h1>7</h1>
<p>let x be the smaller number and x + 1 be the larger number
x + (x +1) = t
2x +1 = t
2x = t - 1
x = (t - 1)/2
x + 1 = (t - 1)/2 +1
x + 1 = (t/2) - (1/2) +1
x + 1 = (t/2) + (1/2) = (t +1)/2</p>
<h1>8</h1>
<p>First find the average number of siblings for the class before the student came.</p>
<p>[(0<em>3) + (1</em>6) + (2<em>2) + (3</em>1)]/12 = 13/12
This is about 1, and 1 is a choice for the number of siblings. Since we're only adding one student, and median is always going to be one. So to make 13/12 into 13/13, the student must have no siblings so that he only adds to the bottom part of the average, and does not have any siblings.</p>
<p>Somebody ripped out the other pages so could you post the other questions?</p>
<p>sorry by the other question has diagrams on them,
but thanks for the explanations anyway</p>
<h1>17: "If p, r, and s are three different prime numbers greater than 2, and n = p x r x s, how many positive factors, including 1 and n does n have?</h1>
<p>p, r, and s are all odd and prime.
Factors of n are:
1, n, p, r, s, p x r, p x s, r x s.<br>
That's 8 factors altogether.</p>
<h1>18: h(t) = c - (d - 4t )^2</h1>
<p>At time t = 0, a ball was thrown upward from an initial height of 6 feet. Until the ball hit the ground, its height, in feet after t seconds was given by the function h above, in which c and d are positive constants. If the ball reached its maximum height of 106 feet at time t = 2.5, what was the maximum height, in feet, of the ball at time t = 1?</p>
<p>I don't know if this is the easiest way to solve the problem. It takes a bit of time:</p>
<p>Step 1: Substitute t = 0 and h = 6 into the formula:
6 = c - (d -4(0))^2
Therefore c - d^2 = 6</p>
<p>Step 2:
Now substitute t = 2.5 and h = 106 into the formula:
106 = c - (d - 4(2.5))^2
106 = c - d^2 + 8(2.5)d - (4(2.5))^2
Into the above, substitute c - d^2 = 6 and you obtain:
106 = 6 + 8(2.5)d - (4(2.5))^2
d = 10</p>
<p>Step 3: Substitute d = 100 into Step 1:
c - d^2 = 6
Therefore: c = 106</p>
<p>Step 4: Substitute t = 1, c = 106 and d = 10 into the original formula:
h = 106 - (10 - 4(1))^2
h = 70 feet</p>
<p>Find different ways of solving #17 and similar questions in this thread:
<a href="http://talk.collegeconfidential.com/showthread.php?t=76332&page=1&pp=40%5B/url%5D">http://talk.collegeconfidential.com/showthread.php?t=76332&page=1&pp=40</a></p>
<h1>18.</h1>
<p>First, correction:
The question in the book reads
"If the ball reached its maximum height of 106 feet at time t = 2.5, what was the height, in feet, of the ball at time t = 1?" </p>
<p>h(t) = c - (d - 4t )^2
h(t) = -16*(t - d/4)^2 + c</p>
<p>For any parabola y(x) = m*(x-p)^2 + q
coordinates of the vertex are (p,q) (sorry, this HAS to be memorized).
For our parabola it's
(d/4, c).</p>
<p>Maximum height of the ball's trajectory h=106 is reached at t=2.5,
so the vertex is at (2,5; 106)
(d/4, c) = (2,5; 106)
d=10.
c=106.
Straight to step 4 in EllenF solution now.</p>
<p>gfc101, where did u get 16 from in h(t)= -16*(t-d/4)^2+c? If u could clear this up that would be great because of out the 5 methods shown about this question, this is the easiest and quickest.</p>
<p>Interesting.
First I arrived at EllenF's first step
Step 1: Substitute t = 0 and h = 6 into the formula:
6 = c - (d -4(0))^2
Therefore c - d^2 = 6
then i figure it was too complicated.
so i brought out my calculus knowledge.
i derived the equation, which gives you 2(d-4t)(-4)=0, at time=2.5, there is a turning point. so plugging in to find out what that d is ,to make the function equal zero. 2(d-10)(-4)=0 , clearly, d is 10.
then you subtitute the numbers, you get C=106.
sorry i know this is unnecessary, especially when it comes to SAT.
but in terms of effectiveness and understanding, i think this is easier a bit, at least,for me ^^</p>
<p>please disregard this if u're not taking or haven't taken calculus yet.
thanks.</p>
<p>Ellen F i don't get why you placed 106 as the maximum height when it STARTS at a height of 6. Shouldn't it have traveled 100 feet in 2.5 seconds?</p>
<p>No. 18 does not need any calculus at all...it is actually a pretty fast one:</p>
<p>h(t) = c - (d - 4t)^2 = - 16(t - .25 d )^2 + c
when t = .25d, the function has its maximum c
Because this occurs when t = 2.5, plug it in ,you get 2.5 = .25d, so d = 10
The maximum c is given as 106
the function becomes h(t) = 106 - (10-4t)^2
plug in t = 1, you get h = 70</p>
<p>Wavvy and gcf101: I am also wondering how you both got the equation </p>
<p>h(t) = -16(t - .25d)^2 + c</p>
<p>from the original equation. Could either one of you (or someone else) please clarify?</p>
<p>The key part is noticing that if you have [number] - [other number]^2, the thing you are subtracting is always positive or zero, because of the squaring. So to get a maximum value, you must have subtracted ZERO. That's how you know that d -4t = 0 at the peak height, when t = 2.5...</p>
<p>Just elaborating on what pckeller already said.</p>
<p>Let us take the first condition of when the ball was at maximum height
h(t) = c - (d - 4t )^2
or, 106=c - (d - 4*2.5)^2
or, 106=c - (d-10)^2
Now even if you don't remember the thing about parabola what you could do is understand that since the ball is reaching a MAX. HEIGHT at 2.5 seconds and the (d-10) expression is squared, the d must be 10 so that (d-10) becomes zero.
Substituting this value you'll get
106=c</p>
<p>Now you know c and d then you can easily put t = 1 in the equation and then get
h(t)=106 - (10 - 4*1)^2
h(t)=106-36=70</p>
<p>The reason we assumed that d-10= 0 is because there could be only one value of max. height and this can only happen on 1 particular time.For instance let's say d was 11 then in that case the value of (d-4t)^2 at t=2.5 sec. would had gone on to become (11-10)^2=1
and the height would had been (c-1). But the same value could had been obtained when t=3sec. Now this would be an impossibility,since there cannot be a max. height at two particular instances of time. So to make sure that there is only one time when the ball could achieve maximum height we need to assume d-4t=0 at t =2.5 sec.</p>
<p>@quix:the equation is not supposed to give the height TRAVELLED by the ball in the time but instead it is actually supposed to give the height of the ball at any instant of time.So if you plug t=0 you'll get height of the ball at that instant is 6m.</p>
<p>@bvg1100,drgonbkwrm:
h(t) = c - (d - 4t )^2
Taking 4 common from d-4t
or,h(t) = c - [ 4 (0.25d-1)]^2
or,h(t) = c - (4)^2*(0.25d-1)^2
or,h(t) = c - 16 (0.25d-1)^2</p>