Bundle of Math Questions

Test Preparation SAT Preparation
1

<p>1) [photo</a> 4 | Flickr - Photo Sharing!](<a href=“http://www.■■■■■■■■■■/photos/94894113@N07/9364696547/in/photostream/]photo”>photo 4 | Simha Gulkarov | Flickr)</p>

<p>2) [photo</a> 3 | Flickr - Photo Sharing!](<a href=“http://www.■■■■■■■■■■/photos/94894113@N07/9367475054/in/photostream/]photo”>photo 3 | Simha Gulkarov | Flickr)</p>

<p>3) [photo</a> 2 | Flickr - Photo Sharing!](<a href=“http://www.■■■■■■■■■■/photos/94894113@N07/9367476790/in/photostream/]photo”>photo 2 | Simha Gulkarov | Flickr)</p>

<p>4) [photo</a> 1 | Flickr - Photo Sharing!](<a href=“http://www.■■■■■■■■■■/photos/94894113@N07/9367471230/in/photostream/]photo”>photo 1 | Simha Gulkarov | Flickr)</p>

<p>5) [photo-7</a> | Flickr - Photo Sharing!](<a href=“http://www.■■■■■■■■■■/photos/94894113@N07/9364636429/in/photostream/]photo-7”>photo-7 | Simha Gulkarov | Flickr)</p>

<p>Please help ! Thanks in advance!</p>

2

<p>1) Since p=ks and s=nt, substitute the second equation into the first equation to get p=knt. k and n are constants, so you can say that p is directly proportional to t. A is correct.</p>

<p>2) In A, the dog could have been in the yard. In B, you know that the deer ate the flowers, so the dog could not have been in the yard because the dog would have scared them away. In C, the Jacksons might not be growing daisies or roses. In D, the two events are unrelated. In E, the dog might be in the yard. Thus, the answer is B.</p>

<p>3) Find the slope of the line between P and Q. You find the slope to be 1/2, which just means to go across 2 and up 1 from the previous point, in this case (4,2). Adding two to the x value and 1 to the y value gives you (6,3), which is answer choice A.</p>

<p>4) You can navigate this problem based on the position of A. A cannot be at the beginning or the end, because it is between C and D. However, A cannot be in the box 3 because in that situation, B would not be next to C. Therefore, A can only exist in boxes 2 and 4. C cannot be first or last, so that further narrows down the choices. The only switch that is allowed for each position of A is to swap B and E. Thus, the possible arrangements are DACBE, DACEB, EBCAD, and BECAD; answer choice C is correct.</p>

<p>5) This problem is just a matter of trying to flip the way your mind looks at this. Look closely at the concave sides of the figure, and try to imagine moving the horizontal side up to meet the other, and moving the vertical side left to meet its match. You can now see that this figure has the same perimeter as a normal 12*10 rectangle. So, the perimeter is 44.</p>

3

<p>Every explanation was perfect except the last one, can you like draw it out or something :(</p>

4

<p>Nevermind I got what you said, you’re the best!</p>

5

<ol>
<li>You could also answer the question in 2 seconds just by looking for patterns. In each coordinate, the Y value is half of the X value. Tada, thus it has to be (6,3)</li>
</ol>

<p>And shalooky explained the others very well so there you are.</p>