answer: e.<\/strong><\/p>\nfirst of all, we need to remember what rational and irrational numbers are. rational numbers can be expressed as a fraction; irrational numbers cannot.<\/p>\n
now let\u2019s think through the question. do you really have to go through every number from 1 to 50? the act will never expect you to do that, so there must be a better way.<\/p>\n
let\u2019s take a look at the pattern: \u221a1, \u221a2, \u221a3, \u221a4. as soon as we get to \u221a4, we have something that can be simplified to an integer. \u221a4 = 2. \u221a4 is a perfect square, and guess what? the perfect squares are the only square roots that are rational numbers. something like \u221a5 gets really messy because there are not two equal fractions that can be multiplied together to equal 5.<\/p>\n
so we just need to count the perfect squares before 50. there are seven: 1, 4, 9, 16, 25, 36, 49. so that means out of our 50 numbers, 7 can be reduced to integers (or fractions) and 43 are irrational. so our answer is e.<\/p>\n
bonus hint:<\/strong> if you come across a problem like this on the act, and don\u2019t know how to solve it, make sure you eliminate some answer choices! the moment you find an irrational number as you count through the series, you can eliminate answer choice a, for example. and this happens really quickly: \u221a2 is an irrational number. you can also take a logical guess if you just work through the first 11 numbers. you should find that out of the first 11, 8 numbers are irrational. this definitely eliminates b, and thinking logically, you could probably take a guess that the overall probability is going to be high.<\/p>\n4. answer: c<\/strong><\/p>\nthis question is difficult primarily because you need to have some higher-level knowledge of the equations of circles and ellipses. if you just studied these in school, then it wouldn\u2019t be nearly as tough. but because equations of ellipses and circles are fair game for higher-level questions on the act, we thought we\u2019d feature one here in our challenge series!<\/p>\n
so let\u2019s get started with the equation for an ellipse:<\/p>\n
![\"(x-h)^2](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_977_c0cec8d800d4fcfb1c0af35c60333b1c.png\" "\\"(x-h)^2")
+ ![\"(y-k)^2](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_977_297649818603f1c6ee04de88ebe1995a.png\" "\\"(y-k)^2")
= 1<\/p>\n
![\"(y-k)^2](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_977_d66aeae7d6a0b3f9e110d9959ee321b2.png\" "\\"(y-k)^2")
+ ![\"(x-h)^2](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_977_37c02bcb81241e4d1489339740eecb54.png\" "\\"(x-h)^2")
= 1<\/p>\n
if an ellipse is wider than it is tall, our equation looks like the first one. if it is taller than it is wide, it looks like the second one. the a always goes with the variable whose axis parallels the wider direction of the ellipse, and the b always goes with the variable whose axis parallels the narrower direction, hence the reason for the difference in the equations.<\/p>\n
in our problem, we have an ellipse that is taller than it is wide. what\u2019s important is that you visualize what an ellipse like this looks like:<\/p>\n
![\"ellipse\"](\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/act\/files\/2015\/06\/05160024\/ellipse1.png\")
<\/p>\n
if we put a circle inside this shape, it can only have a diameter that is as wide as the ellipse is wide, otherwise it wouldn\u2019t fit.<\/p>\n
![\"circle](\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/act\/files\/2015\/06\/05160024\/circle-inscribed-in-ellipse.png\")
<\/p>\n
okay! back to the ellipse equation:<\/p>\n
h<\/em> and k<\/em> tell us where the center of the ellipse is, so our center is (2,-4). a and b tell us how many units away from the center each vertex is. since a is 3 and b<\/em> is 4, we know that our ellipse is 6 units wide and 8 units tall.<\/p>\nif we put a circle in this ellipse, this means it cannot have a radius greater than 3.<\/p>\n
the equation of a circle is:<\/p>\n
(x-h)2<\/sup> + (y-k)2<\/sup> = r2<\/sup><\/p>\nso we need to see 32<\/sup>, or 9, in our answer choice for the radius of the circle, making our answer c.<\/p>\n5. answer: a<\/strong><\/p>\nprobably the most elegant way to solve this problem is to remember the factor theorem. this is a useful trick for problems like this on the act. the factor theorem states that a polynomial p(x) is divisible by binomial (x \u2013 c) if and only if p(c) = 0. in order for a polynomial to be divisible by a linear binomial, the polynomial and the binomial must have the same root.<\/p>\n
if we want polynomial p(x) to be divisible by (x + 2), it must be true that p(\u2013 2) = 0. in other words, x = -2 must be a root of the equation.<\/p>\n
p(x) = x3<\/sup> + 5x2<\/sup> + sx + 6
\np(-2) = -23<\/sup> + 5(-2)2<\/sup> + s(-2) + 6 = 0
\n-8 + 5*4 – 2s + 6 = 0
\n-8 + 20 + 6 = 2s
\n18 = 2s
\n9 = s<\/p>\nso our answer is a.<\/p>\n
6. answer: d<\/strong><\/p>\nin a geometric series, each term is the product of previous term times some common ratio. we multiply by the common ratio to change any one term into its successor.<\/p>\n
here, we don’t know the common ratio, so let it be r.<\/p>\n
a2<\/sub>=2
\na3<\/sub>=2r
\na4<\/sub>=2r2<\/sup><\/p>\nbut we know the numerical value of the fourth term.<\/p>\n
2r2<\/sup>=10
\nr2<\/sup>=5
\nr=\u221a5<\/p>\nbecause every term is positive, we don’t have to consider the negative square root. now that we have the common ratio, we can move from term to term. each time, we multiply by the square root of 5.<\/p>\n
a4<\/sub>=10
\na5<\/sub>=10\u221a5
\na6<\/sub>=(10\u221a5)=10*5=50
\na7<\/sub>=50\u221a5<\/p>\nso our answer is d.<\/p>\n
7. answer: d<\/strong>
\naverage speed = total distance \/ total time.<\/p>\n40 miles + 30 miles so the total distance was 70 miles. suzanne drove for 2 hours + 3 hours so the total time was 5 hours. 70\/5 = 14.<\/p>\n
the average speed for the whole trip was 14 mph. the correct answer is (d).<\/p>\n
notice how the test-maker has made this problem tricky! the average speed in this problem is 14 mph, which is different<\/i> from simply taking the mean of the two speeds. if we had just averaged the two speeds (10mph and 20mph) we would have gotten 15mph. average speed is a weighted average. since suzanne spent more time in the problem going 10mph than 20mph, it makes sense that the average speed would be closer to 10mph.<\/p>\n
8. answer: d<\/strong>
\nto find the average speed of the bus, we know we will need to find the total distance and the total time, so we can start by using another formula (distance = rate x time<\/b>) to help us find the pieces we\u2019re missing for each part of the trip.<\/p>\nfor the first part of the field trip: 30 miles = 15mph x t, so we know that t = 2 hours. for the middle part of the trip, we know that d = 10mph x 3 hours, so we know that d = 30 miles. for the last part of the trip, we know that 40 miles = r x 2 hours, so we know that r = 20mph. now we can find the total distance and the total time:<\/p>\n
total distance = 30 miles + 30 miles + 40miles = 100 miles.<\/p>\n
total time = 2 hours + 3 hours + 2 hours = 7 hours.<\/p>\n
the average speed = 100 miles\/ 7 hours = 14.28mph. since the question used the word \u201capproximately,\u201d we can round to the nearest integer: 14. the correct answer is (d).<\/p>\n
9. answer: e<\/strong>
\nthis is a hard problem. first of all, notice that the angle must be in qiii or qiv, since the sine is negative. it has to be a value on the line ![\"y\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_f15df4ceab77f403bc3fed686a8c8edf.png\" "\\"y\\"\/")
= \u2013 ![\"(3\/5)\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_982_b7e492a02df70ce1835c42d30747a7f2.png\" "\\"(3\/5)\\"\/")
. the first three choices are all in qi and qii, so the angle can’t possibly be in any of those. we can easily eliminate (a) & (b) & (c).<\/p>\nlet’s think about the last two. first, option (d). this option is in qii and qiii, so the part in qiii might go down low enough to contain the angle. the lowest ray in that region is at ![\"(7pi)\/6\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_db981eddfed12a9cc75c8166deb795ae.png\" "\\"(7pi)\/6\\"\/")
.<\/p>\n
technically, this ray isn’t even included, because it’s the endpoint of an inequality, but we will just use this value. sine is positive in qii, zero at the negative x-axis, and starts to get more and more negative as we go around through qiii. how low does this go by this last value?<\/p>\n
![\"sin](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_d5d12a27d7a3fe75746d538491af509f.png\" "\\"sin")
= –![\"1\/2\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_7827c38498fd06c395fac2a30b119c50.png\" "\\"1\/2\\"\/")
= ![\"-0.5\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_4bbfbc4b23202d67946032b4f377a9a5.png\" "\\"-0.5\\"\/")
<\/p>\n
this does not go down as far as<\/p>\n
![\"sin(theta)](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_ab7495c2d646ce241ee927909c2f66e1.png\" "\\"sin(theta)")
<\/p>\n
here’s a diagram with region (d) and the line = \u2013 ![\"3\/5\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_4138a6373721eab2032917ca4af6cfe2.png\" "\\"3\/5\\"\/")
.<\/p>\n
![\"act](\"\/\/www.catharsisit.com\/hs\/files\/2015\/08\/act-1-600x503.png\")
<\/p>\n
so, we can eliminate (d).<\/p>\n
this leaves (e) by the process of elimination, but let’s verify that this works. region (e) is in qiv. the sine is negative one at the bottom, where the unit circle intersects the negative y-axis. as we rise from this bottom, we get negative values of smaller and smaller absolute value, until it equals zero at the positive x-axis. we have to know how low the lowest values in this region are. well, the limiting value is ![\"(5pi)\/3\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_62c1ce825a79e9233b5b77371fd58873.png\" "\\"(5pi)\/3\\"\/")
.<\/p>\n
we know that ![\"sin\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_084686dc4dd22f3fd30f663330e3550c.png\" "\\"sin\\"\/")
= – ![\"sqrt{3}\/2\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_ccd1b0ec1e83edfa03849e0a8f6f9380.png\" "\\"sqrt{3}\/2\\"\/")
<\/p>\n
now, if you happen to have the decimal approximations memorized, you will see that:
\n![\"sin(5pi)\/3\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_ec1edd068c6c7281ad09cd2c42ad4d93.png\" "\\"sin(5pi)\/3\\"\/")
= – ![\"approx\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_651c61db359612b0d056ddb248009d9f.png\" "\\"approx\\"\/")
![\"-0.866](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_ba3972003836781fdc9c1cc6529012a9.png\" "\\"-0.866")
<\/p>\n
if you don’t have this decimal equivalent memorized, you could think about the triangles in qiv:<\/p>\n
![\"act](\"\/\/www.catharsisit.com\/hs\/files\/2015\/08\/act-2-600x598.png\")
<\/p>\n
even if we can’t directly compare the sizes of the two vertical legs, we definitely can compare the two horizontal legs, and 0.8 is definitely bigger than 0.5. that means the purple {0.6, .8, 1} triangle is further around in the counterclockwise direction from the green 30-60-90 triangle, and the point with a sine of \u2013 0.6 would clearly be included in the arc from the 30-60-90 triangle up to the positive x-axis.
\nanswer = (e)<\/p>\n
10. answer: d<\/strong>
\nthe graph clearly has a hole, a point discontinuity, at x = 3. this means that the function cannot be defined at x = 3. it must have (x \u2013 3) in the denominator. on this basis, we can eliminate (a)<\/strong>, (c)<\/strong>, and (e)<\/strong>.<\/p>\nnotice that the graph is identical to the line y = 2x \u2013 3 at every point except for the point discontinuity at x = 3. also, to get a point discontinuity instead of an asymptote, we need to have a factor of (x \u2013 3) in both the numerator and the denominator. choice (b)<\/strong> at x = 3 has a zero denominator and a non-zero numerator, which is the condition for a vertical asymptote. we have eliminated the other four answers, so it would seem that (d)<\/strong> is the answer, but let’s verify this.<\/p>\nin order to get a graph that is identical to y = 2x \u2013 3 at every point except for x = 3 and has a point discontinuity at x = 3, we would have to multiply (2x \u2013 3) by (x \u2013 3) over itself.![\"acq17_img3.png\"](\"\/\/www.catharsisit.com\/hs\/files\/2015\/08\/acq17_img3.png.jpg\")
<\/p>\n
so everything checks out and our answer is (d)!<\/p>\n
11. answer: e<\/strong>
\nwhen we have functions inside functions, we call these \u201cnested\u201d functions \u2013 it\u2019s important to always start from the innermost parentheses and work your way out. the expression m(m(n)) starts with an inntermost \u201cm(n)\u201d that we know equals -n3<\/sup>, so let\u2019s start by substituting that in. now the question is asking, what is the value of m(-n3<\/sup>). just like we normally do for any f(x) function, we will plug whatever in inside the parentheses in for the variable in the function. here, the testmaker is trying to confuse you by using the same variable \u201cn,\u201d but we\u2019re too smart for the act\u2019s tricks! \ud83d\ude42<\/p>\nm(-n3<\/sup>) = -(-n3<\/sup>)3<\/sup><\/p>\naccording to the order of operations, or pemdas, we\u2019d start with what is inside the parentheses. since we don\u2019t know the value of \u201cn\u201d we cannot simplify -n3<\/sup> any further. next, we simplify the exponents. here\u2019s where knowing your exponent rules really come in handy! when two exponents are separated by a parenthesis, we must always multiply<\/i> them.<\/p>\nm(-n3<\/sup>) = -(-n9<\/sup>)<\/p>\nremember that in number properties, any negative number raised to an odd exponent will stay negative, since it takes a pair of negatives to cancel each other out and become positive. -n9<\/sup> will stay negative until it is multiplied by the -1 in front of the parentheses. finally, we arrive at our answer:<\/p>\nm(-n3<\/sup>) = n9<\/sup><\/p>\nthe correct answer is (e).<\/p>\n
12. answer: f<\/strong>
\nthe function values for p(x) vary directly as x for all real numbers. the graph of y=p(x) in a standard (x,y) coordinate plane is a line with a y-intercept at 0. since y = p(x) doesn’t have any coefficients, and p(x) varies directly with x (like p(x) = x, as opposed to something like p(x) = 2x), this line will cross the origin (0,0). the correct answer is (f).<\/p>\nwant more sample problems? check out this act math practice<\/a>!<\/p>\n","protected":false},"excerpt":{"rendered":"working on improving your act math score? try these hard act math problems to test and hone your skills!<\/p>\n","protected":false},"author":228,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[90],"tags":[],"ppma_author":[24872],"acf":[],"yoast_head":"\n
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kapelke-dale","image":{"@type":"imageobject","inlanguage":"en-us","@id":"\/\/www.catharsisit.com\/hs\/#\/schema\/person\/image\/1fa8237f9d15e21c17ec0cc6264dd6e6","url":"https:\/\/secure.gravatar.com\/avatar\/a68bd32a77dfbc38df6b7158ca0d4fcb?s=96&d=mm&r=g","contenturl":"https:\/\/secure.gravatar.com\/avatar\/a68bd32a77dfbc38df6b7158ca0d4fcb?s=96&d=mm&r=g","caption":"rachel kapelke-dale"},"description":"rachel has helped students around the world prepare for various standardized tests, including the sat, act, toefl, gre, and gmat, and she is one of the authors of our magoosh act prep book. rachel has a bachelor of arts in comparative literature from brown university, an ma in cinematography from the universit\u00e9 de paris vii, and a ph.d. in film studies from university college london.","honorificsuffix":"phd","knowsabout":["sat","act","toefl","gre","gmat"],"knowslanguage":["english","french"],"jobtitle":"content creator","worksfor":"magoosh","url":"\/\/www.catharsisit.com\/hs\/author\/rachelkd\/"}]}},"authors":[{"term_id":24872,"user_id":228,"is_guest":0,"slug":"rachelkd","display_name":"rachel kapelke-dale","avatar_url":"https:\/\/secure.gravatar.com\/avatar\/a68bd32a77dfbc38df6b7158ca0d4fcb?s=96&d=mm&r=g","user_url":"http:\/\/rachelkapelkedale.com\/","last_name":"kapelke-dale","first_name":"rachel","description":"rachel is a magoosh content creator. she writes and updates content on our <a href=\"\/\/www.catharsisit.com\/hs\/\">high school <\/a>and <a href=\"\/\/www.catharsisit.com\/gre\/\">gre blogs<\/a> to ensure students are equipped with the best information during their test prep journey. as a test-prep instructor for more than five years in there different countries, rachel has helped students around the world prepare for various standardized tests, including the sat, act, toefl, gre, and gmat, and she is one of the authors of our <a href=\"https:\/\/www.amazon.com\/act-prep-magoosh-schedules-strategies\/dp\/1610660692\" rel=\"noopener noreferrer\">magoosh act prep book<\/a>. rachel has a bachelor of arts in comparative literature from brown university, an ma in cinematography from the universit\u00e9 de paris vii, and a ph.d. in film studies from university college london. for over a decade, rachel has honed her craft as a fiction and memoir writer and public speaker. her novel,<a href=\"https:\/\/read.macmillan.com\/lp\/the-ballerinas\/\" rel=\"noopener noreferrer\"> the ballerinas<\/a>, is forthcoming in december 2021 from<a href=\"https:\/\/us.macmillan.com\/author\/rachelkapelkedale\/\" rel=\"noopener noreferrer\"> st. martin's press<\/a>, while her memoir,<a href=\"https:\/\/www.penguinrandomhouse.com\/books\/315080\/graduates-in-wonderland-by-jessica-pan\/\" rel=\"noopener noreferrer\"> graduates in wonderland<\/a>, co-written with jessica pan, was published in 2014 by penguin random house. her <a href=\"http:\/\/rachelkapelkedale.com\/\" rel=\"noopener noreferrer\">work<\/a> has appeared in over a dozen online and print publications, including vanity fair hollywood. when she isn't strategically stringing words together at magoosh, you can find rachel riding horses or with her nose in a book. join her on<a href=\"https:\/\/twitter.com\/rkapelkedale\" rel=\"noopener noreferrer\"> twitter<\/a>,<a href=\"https:\/\/www.instagram.com\/rachelkapelkedale\/\" rel=\"noopener noreferrer\"> instagram<\/a>, or<a href=\"https:\/\/www.facebook.com\/rkapelkedale\" rel=\"noopener noreferrer\"> facebook<\/a>!"}],"_links":{"self":[{"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/posts\/9930"}],"collection":[{"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/users\/228"}],"replies":[{"embeddable":true,"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/comments?post=9930"}],"version-history":[{"count":0,"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/posts\/9930\/revisions"}],"wp:attachment":[{"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/media?parent=9930"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/categories?post=9930"},{"taxonomy":"post_tag","embeddable":true,"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/tags?post=9930"},{"taxonomy":"author","embeddable":true,"href":"\/\/www.catharsisit.com\/hs\/wp-json\/wp\/v2\/ppma_author?post=9930"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
![\"magoosh](\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/act\/files\/2015\/04\/magoosh-act-challenge-question-600x600.png\")
<\/p>\n![\"110pi\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_50b5e89e0d5a1b704200455ac28a5464.png\" "\\"110pi\\"\/")
square centimeters. what is the measure of angle b?<\/p>\n![\"act](\"\/\/www.catharsisit.com\/act\/files\/2015\/04\/act-challenge-question-3.png\")
<\/p>\n![\"sqrt{1}\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_74b05759a9fa4e4753d2e0d3bcc833d9.png\" "\\"sqrt{1}\\"\/")
is written on the first card, ![\"sqrt{2}\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_52ef8013c05ff62d72ce781af35d4c9e.png\" "\\"sqrt{2}\\"\/")
on the second, ![\"sqrt{3}\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_49dcc68f31fb15fe1f2f90e2ba5399d6.png\" "\\"sqrt{3}\\"\/")
on the third and so on through ![\"sqrt{50}\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_3a1e2d48864280f8d4f3b809242fe848.png\" "\\"sqrt{50}\\"\/")
, with no numbers repeated. a card is drawn at random from the box. what is the probability that the number on the card is an irrational number?<\/p>\n![\"\"](\"\/\/www.catharsisit.com\/hs\/files\/2017\/04\/screen-shot-2017-04-28-at-6.59.51-pm.png\")
<\/p>\n![\"{x^3}](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_cff25a9a3fc889dac9a64331e4c4ab18.png\" "\\"{x^3}")
is divisible by (x + 2)?![\"theta\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_dfcdced295ae300b5c8b4b2f956174b0.png\" "\\"theta\\"\/")
= – ![\"2pi\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_0a59d05e806caf6476c7437cec00e605.png\" "\\"2pi\\"\/")
\/3![\"pi\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_8edb2cf68079344a2edd739531259f6c.png\" "\\"pi\\"\/")
\/4 < ![\"5pi\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_ccc803e6e1de980203fbc10bcebcab5b.png\" "\\"5pi\\"\/")
\/6 < ![\"7pi\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_3bf5660fab7dc2b5b45c9e5ba0db852d.png\" "\\"7pi\\"\/")
\/6![\"3pi\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_90fa909c371b6b1d9727b8803ed81b2a.png\" "\\"3pi\\"\/")
<\/p>\n![\"x\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_137b8dad9acade63b507e67878d3b94b.png\" "\\"x\\"\/")
, ![\"2x-3\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_c5072e79ae7610b61ce67f4e887aa82c.png\" "\\"2x-3\\"\/")
![\"(2x-3)\/(x-3)\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_979_13d5b2088f51b5f404fee12a7b8b2a24.png\" "\\"(2x-3)\/(x-3)\\"\/")
![\"(2x-3)\/(x+3)\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_979_02ad4b7247ffb9ceabdb43bdb5929c83.png\" "\\"(2x-3)\/(x+3)\\"\/")
![\"(2x^2-9x+9)\/(x-3)\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_979_64709e2a96ef165bbe14d53cf8b980a2.png\" "\\"(2x^2-9x+9)\/(x-3)\\"\/")
![\"2x^2-9x+9\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_c631d520e7e5d00e5529c8f200349ef3.png\" "\\"2x^2-9x+9\\"\/")
\/![\"(x+3)\"](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_990.5_acab6c6c70673eeef3b3cf49794ae7e7.png\" "\\"(x+3)\\"\/")
<\/p>\n![\"\"](\"\/\/www.catharsisit.com\/hs\/files\/2017\/05\/q10-graph-600x372.png\")
<\/p>\n![\"m(n)](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_991.5_df4cfd159589f1b2b53968db1620deb9.png\" "\\"m(n)")
. which of the following is the value of m(m(n))?<\/p>\n![\"a=pi](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993.5_dca7e992caf2ea36d892be2607949443.png\" "\\"a=pi")
.<\/p>\n![\"a](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_0bdff722df8a9fe1c875ff6ec9b88ba1.png\" "\\"a")
![\"a](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_c68eaad9cb449161ce91ed91c546f8c0.png\" "\\"a")
<\/p>\n![\"110](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_fd84e327a175a731dac1bdfb696a589a.png\" "\\"110")
. this means that the area of the unshaded sector of the circle inside the triangle must be ![\"11](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_6fc03f72a4fe54e6db999e7b70c3dc75.png\" "\\"11")
, since these two regions must add up to the total area of the circle: ![\"121](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_993_aba826af1e42582b8c0c6d6ee36e718b.png\" "\\"121")
.<\/p>\n![\"(11](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_979_58f93739bdff18b18635eb799644e565.png\" "\\"(11")
= ![\"(x](\"\/\/www.catharsisit.com\/hs\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_979_8c55f6d5b0069344d860799ab6aa0bdd.png\" "\\"(x")
<\/p>\n