\u00dcSL\u00dc SAYILAR<\/p>\n
x . an + y . an \u2013 z . an = (x + y \u2013 z) . an
\n am . an = am + n
\n am . bm = (a . b)m
\n am : an = am – n KARE’N\u0130N ALANI: <\/p>\n
A=a.a
\n (a karenin bir kenar\u0131)<\/p>\n
D\u0130KD\u00d6RTGEN’\u0130N ALANI:<\/p>\n
A = a.b
\n (a k\u0131sa kenar\u0131, b uzun kenar\u0131)<\/p>\n
YAMUK’UN ALANI:<\/p>\n
A = (a+c).h \/ 2 (a alt taban uzunlu\u011fu, c \u00fcst taban uzunlu\u011fu, h y\u00fckseklik)
\nPARALELKENAR’IN ALANI:<\/p>\n
A = a.h (a taban kenar\u0131, h tabana inen y\u00fckseklik) S\u0130L\u0130ND\u0130R’\u0130N HACM\u0130:
\n H = taban alan.y\u00fckseklik
\n H = \u03c0.r.r.h
\n (\u03c0=3,14 al\u0131r\u0131z, r taban yar\u0131\u00e7ap\u0131, h y\u00fckseklik)
\n (konserve tenekesi)
\nK\u00dcP’\u00dcN HACM\u0130:
\n H = a.a.a
\n (a k\u00fcp\u00fcn bir kenar\u0131n\u0131n uzunlu\u011fu)
\n (k\u00fcp \u015feker)<\/p>\n
D\u0130KD\u00d6RTGENLER PR\u0130ZMASI’NIN HACM\u0130:
\n H = a.b.c
\n (a en, b boy, c y\u00fcksekli\u011fi)
\n (kibrit kutusu)<\/p>\n
KARE PR\u0130ZMA’NIN HACM\u0130:
\n H = taban alan.y\u00fcksekli\u011fi H = a.a.b
\n (a kare olan taban\u0131n bir kenar\u0131, b y\u00fckseklik)<\/p>\n
D\u0130K PR\u0130ZMALARIN HACM\u0130: V= (taban alan\u0131) X (y\u00fckseklik)
\n\u00c7EMBER’\u0130N VE DA\u0130RE’N\u0130N \u00c7EVRES\u0130:
\n \u00c7 = 2.\u03c0.r
\n (\u03c0=3,14 al\u0131r\u0131z r daire veya \u00e7emberin yar\u0131\u00e7ap\u0131)<\/p>\n
DA\u0130RE’N\u0130N ALANI:
\n A = \u03c0.r.r
\n (\u03c0=3,14 al\u0131r\u0131z r dairenin yar\u0131\u00e7ap\u0131)
\nDA\u0130RE D\u0130L\u0130M\u0130N\u0130N ALANI:
\n A = \u03c0.r.r.x \/ 360\u00ba
\n (\u03c0=3,14 al\u0131r\u0131z r dairenin yar\u0131\u00e7ap\u0131, x a\u00e7\u0131s\u0131 daire diliminin aras\u0131nda kalan merkez a\u00e7\u0131)<\/p>\n
\u00c7EMBER YAYININ UZUNLU\u011eU:
\n \u00c7 = 2.\u03c0.r.x \/ 360\u00ba
\n (\u03c0=3,14 al\u0131r\u0131z r \u00e7emberin yar\u0131\u00e7ap\u0131, x a\u00e7\u0131s\u0131 \u00e7ember par\u00e7as\u0131n\u0131n aras\u0131nda kalan merkez a\u00e7\u0131)
\n\u00dc\u00c7GEN\u0130N ALANI VE \u00c7EVRES\u0130 <\/p>\n
\u00dc\u00e7genin \u00e7evresini bulabilmek i\u00e7in
\n kenarlar toplan\u0131r.
\n \u00c7 = a + b + c
\n \u00dc\u00e7genin alan\u0131n\u0131 bulmak i\u00e7in y\u00fckseklikle
\n kenar \u00e7arp\u0131l\u0131r ve ikiye b\u00f6l\u00fcn\u00fcr.
\n Alan=(a x Ha)\/2<\/p>\n
\u00c7OKGENDE i\u00e7 a\u00e7\u0131lar toplam\u0131: D\u0131\u015f b\u00fckey bir \u00e7okgenin n tane kenar\u0131 var ise i\u00e7 a\u00e7\u0131lar\u0131n\u0131n toplam\u0131
\n(n – 2) . 180\u00b0<\/p>\n
D\u0131\u015f a\u00e7\u0131lar toplam\u0131: B\u00fct\u00fcn d\u0131\u015fb\u00fckey \u00e7okgenlerde<\/p>\n
D\u0131\u015f a\u00e7\u0131lar toplam\u0131 =360\u00b0<\/p>\n
K\u00f6\u015fegenlerin say\u0131s\u0131: n kenarl\u0131 d\u0131\u015fb\u00fckey bir \u00e7okgenin<\/p>\n
n.(n-3) \/ 2
\n Bir k\u00f6\u015feden (n \u2013 3) tane k\u00f6\u015fegen \u00e7izilebilir.
\n n kenarl\u0131 d\u0131\u015fb\u00fckey bir \u00e7okgenin i\u00e7erisinde, bir k\u00f6\u015feden k\u00f6\u015fegenler \u00e7izilerek
\n (n \u2013 2) adet \u00fc\u00e7gen elde edilebilir.
\n n kenarl\u0131 d\u00fczg\u00fcn bir \u00e7okgende bir i\u00e7 a\u00e7\u0131n\u0131n \u00f6l\u00e7\u00fcs\u00fc
\n(n – 2) . 180\u00b0\/ n
\n Konveks \u00e7okgenlerin d\u0131\u015f a\u00e7\u0131lar\u0131 toplam\u0131 360\u00b0 oldu\u011fundan d\u00fczg\u00fcn \u00e7okgenin bir d\u0131\u015f a\u00e7\u0131s\u0131n\u0131n \u00f6l\u00e7\u00fcs\u00fc<\/p>\n
360\u00b0 \/ n
\nDO\u011eRUNUN E\u011e\u0130M\u0130<\/p>\n
E\u011fim kar\u015f\u0131n\u0131n kom\u015fuya b\u00f6l\u00fcm\u00fcd\u00fcr.
\n E\u011fim=tanx<\/p>\n
E\u011fim=b\/c Kar-Zarar Problemleri <\/p>\n
Maliyet:100 %20 kar Sat\u0131\u015f:100+20=120
\n Maliyet:100 %20 \u0130ndirimli Sat\u0131\u015f:
\n 100-20=80
\n \u0130ndirimli sat\u0131\u015f\u0131n \u00fczerinden %20 karl\u0131 sat\u0131\u015f:<\/p>\n
80.%120=(80.120):100=96 Y\u00dcZDE PROBLEMLER\u0130
\nY\u00fczde, paydas\u0131 100 olan kesirlere denir. <\/p>\n
\u00d6rne\u011fin, y\u00fczde 50 (%50)= 50\/100 = 1\/2<\/p>\n
Y\u00fczde 20 (%20) = 20\/100 = 1\/5<\/p>\n
FA\u0130Z PROBLEMLER\u0130
\n f = a.n.t \/ 100 (y\u0131ll\u0131k faiz)
\n f = a.n.t \/ 1200 (ayl\u0131k faiz)
\n f = a.n.t \/ 36000 (g\u00fcnl\u00fck faiz)
\n (a anapara, n faiz y\u00fczdesi, t zaman, f faiz) <\/p>\n
SAAT PROBLEMLER\u0130<\/p>\n
|30.saat(akrep)-5,5.dakika(yelkovan|
\n =kollar aras\u0131ndaki a\u00e7\u0131
\nHAREKET PROBLEMLER\u0130
\n Yol: x
\n H\u0131z: v
\n Zaman: t
\n Yol= H\u0131z . Zaman x=v.t<\/p>\n
H\u0131z = Yol \/ Zaman v=x\/t
\n Zaman= Yol \/ H\u0131z t=x\/v
\n Hareketliler ayn\u0131 anda ve z\u0131t y\u00f6nde ise x = (v1 + v2). t
\n Hareketliler ayn\u0131 anda ve ayn\u0131 y\u00f6nde
\n ise x = (v1 – v2). t
\n Nehir problemlerinde ise her zaman kay\u0131\u011f\u0131n h\u0131z\u0131ndan ak\u0131nt\u0131n\u0131n h\u0131z\u0131 \u00e7\u0131kart\u0131l\u0131r. YA\u015e PROBLEMLER\u0130
\n Bir ki\u015finin ya\u015f\u0131 a olsun,
\n T y\u0131l \u00f6nceki ya\u015f\u0131 : x-T
\n T y\u0131l sonraki ya\u015f\u0131 : x + T olur. <\/p>\n
\u0130ki ki\u015finin ya\u015flar\u0131 oran\u0131 y\u0131llara
\n g\u00f6re orant\u0131l\u0131 de\u011fildir.
\n n ki\u015finin ya\u015flar\u0131 toplam\u0131 b ise
\n T y\u0131l sonra b + n.T
\n T y\u0131l \u00f6nce b – n.T
\n Ki\u015filer aras\u0131ndaki ya\u015f fark\u0131
\n her zaman ayn\u0131d\u0131r.
\n x y\u0131l \u00f6ncede ya\u015f fark\u0131 a-b
\n x y\u0131l sonrada ya\u015f fark\u0131 a-b
\n Katlar ve oranlar hangi y\u0131lda verildiyse<\/p>\n
denklem o y\u0131lda kurulur. \u0130\u015e\u00c7\u0130 – HAVUZ PROBLEMLER\u0130
\n Bir i\u015fi; <\/p>\n
A i\u015f\u00e7isi tek ba\u015f\u0131na a saatte,
\n B i\u015f\u00e7isi tek ba\u015f\u0131na b saatte,
\n C i\u015f\u00e7isi tek ba\u015f\u0131na c saatte
\n yapabiliyorsa;
\n \u0130\u015f t saatte bitiyorsa
\n 1\/a + 1\/b + 1\/c = 1\/t olur.
\n A i\u015f\u00e7isi 1 saatte i\u015fin 1\/a s\u0131n\u0131 bitirir.
\n A ile B birlikte t saatte i\u015fin
\n (1\/a + 1\/b).t sini bitirir.
\n A i\u015f\u00e7isi x saatte, B i\u015f\u00e7isi y saatte
\n C i\u015f\u00e7isi z saatte
\n \u00e7al\u0131\u015farak i\u015fin tamam\u0131n\u0131 bitirdiklerine g\u00f6re \u00fc\u00e7\u00fc birlikte i\u015fi k saatte bitiriyorsa,
\n k\/x + k\/y + k\/z = 1 olur.
\n Havuz problemleri i\u015f\u00e7i problemleri
\n gibi \u00e7\u00f6z\u00fcl\u00fcr.
\n A muslu\u011fu havuzun tamam\u0131n\u0131 a saatte
\n doldurabiliyor.
\n Tabanda bulunan B muslu\u011fu dolu havuzun
\n tamam\u0131n\u0131 tek ba\u015f\u0131na b saatte bo\u015faltabiliyor
\n olsun.
\n Bu iki musluk birlikte bu havuzun t saatte
\n (1\/a – 1\/b).t sini doldurur.<\/p>\n
Bu havuzun dolmas\u0131 i\u00e7in b > a olmal\u0131d\u0131r.
\n E\u011fer havuz t saatte doluyorsa
\n 1\/a – 1\/b = 1\/t
\n Havuz dolduruluyorsa dolduran musluk (+), bo\u015faltan musluk (-) al\u0131n\u0131r.
\n Havuz bo\u015falt\u0131l\u0131yorsa dolduran musluk (-), bo\u015faltan musluk (+) al\u0131n\u0131r. <\/p>\n
TR\u0130GONOMETR\u0130
\n SinC = kar\u015f\u0131 \/ hipoten\u00fcs
\n SinC = c \/ a
\n CosC = kom\u015fu \/ hipoten\u00fcs
\n CosC = b \/ a
\n TanC = kar\u015f\u0131 \/ kom\u015fu
\n TanC = c \/ b
\n CotC = kom\u015fu \/ kar\u015f\u0131
\n CotC = b \/ c<\/p>\n
tanx = sinx \/ cosx
\n cotx = cosx \/ sinx
\n tanx . cotx = 1
\n sinx.sinx + cosx.cosx = 1
\n\u00d6ZDE\u015eL\u0130KLER
\n\u0130ki Kare Fark\u0131 – Toplam\u0131
\n I) a2 \u2013 b2 = (a \u2013 b) (a + b)
\n II) a2 + b2 = (a + b)2 \u2013 2ab ya da
\n a2 + b2 = (a \u2013 b)2 + 2ab dir.<\/p>\n
\u0130ki K\u00fcp Fark\u0131 – Toplam\u0131
\n I) a3 \u2013 b3 = (a \u2013 b) (a2 + ab + b2 )
\n II) a3 + b3 = (a + b) (a2 \u2013 ab + b2 )
\n III) a3 \u2013 b3 = (a \u2013 b)3 + 3ab (a \u2013 b)
\n IV) a3 + b3 = (a + b)3 \u2013 3ab (a + b)
\nTam Kare \u0130fadeler
\n I) (a + b)2 = a2 + 2ab + b2
\n (a + b)2 = (a \u2013 b)2 + 4ab
\n II) (a \u2013 b)2 = a2 \u2013 2ab + b2
\n (a \u2013 b)2 = (a + b)2 \u2013 4ab
\n III) (a + b + c)2 = a2 + b2 + c2 + 2(ab + ac + bc)
\n IV) (a + b \u2013 c)2 = a2 + b2 + c2 + 2(ab \u2013 ac \u2013 bc)<\/p>\n
(a + b)3 = a3 + 3a2b + 3ab2 + b3
\n (a \u2013 b)3 = a3 \u2013 3a2b + 3ab2 \u2013 b3
\n (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 +b4
\n (a \u2013 b)4 = a4 \u2013 4a3b + 6a2b2 \u2013 4ab3 + b4<\/p>\n
P\u0130SAGOR BA\u011eINTISI<\/p>\n
a2=b2+c2
\n a.a=b.b+c.c <\/p>\n
OLASILIK
\n P(A)=S(A) \/ S(E)
\n Bir olay\u0131n olas\u0131l\u0131\u011f\u0131=istenilen durumlar\u0131n say\u0131s\u0131 \/ t\u00fcm durumlar\u0131n say\u0131s\u0131
\n p(A)=0 ise imkans\u0131z olay=ger\u00e7ekle\u015fmesi m\u00fcmk\u00fcn de\u011fil
\n P(A)=1 ise kesin olay=ger\u00e7ekle\u015fmesi kesin
\n Herhangi bir olay\u0131n olmama olas\u0131l\u0131\u011f\u0131:
\n P'(A) = 1 – P(A)<\/p>\n
Ba\u011f\u0131ms\u0131z olay:
\nBirbirlerini etkilemiyorlarsa(para-zar)
\n P(A \u00c7 B)= P(A) . P(B)<\/p>\n
Ayr\u0131k iki olay\u0131n birle\u015fiminin olas\u0131l\u0131\u011f\u0131:
\n P(AUB)= P(A) + P(B)<\/p>\n
Ayr\u0131k olmayan iki olay\u0131n birle\u015fiminin olas\u0131l\u0131\u011f\u0131:
\n P(AUB)= P(A) + P(B) – P(A \u00c7B)<\/p>\n
n elemanl\u0131 bir k\u00fcmenin r elemanl\u0131 perm\u00fctasyonu:
\n P(n,r)=n! \/ (n-r)!
\n P(n,n)= n! p(0,0)= 1
\n P(n,0)= 1 P(n,1)= n
\n Dairesel Perm\u00fctasyon: (n-2)!
\nKOMB\u0130NASYON
\n n elemanl\u0131 k\u00fcmenin r ‘ li kombinasyonlar\u0131 say\u0131s\u0131n\u0131n form\u00fcl\u00fc,<\/p>\n
FAKT\u00d6R\u0130YEL
\nn!=1.2.3.4.5………n
\n 6!=1.2.3.4.5.6=720 <\/p>\n
ORANTI
\n1) a\/b=c\/d ise a.d= b.c
\n2) a : b : c = x : y : z ise,
\n Burada, a = x . k
\n b = y . k
\n c = z . k d\u0131r.<\/p>\n","protected":false},"excerpt":{"rendered":"
\u00dcSL\u00dc SAYILAR x . an + y . an \u2013 z . an = (x + y \u2013 z) . an am . an = am + n am . bm = (a . b)m am : an = am – n KARE’N\u0130N ALANI: A=a.a (a karenin bir kenar\u0131) D\u0130KD\u00d6RTGEN’\u0130N ALANI: A = a.b (a …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1404,1403],"tags":[7313,10021,10022,3565,10020],"class_list":["post-4618","post","type-post","status-publish","format-standard","hentry","category-matematik-odevleri","category-odevler","tag-cember","tag-daire","tag-kar-zarar-problemleri","tag-kare","tag-matematik-formulleri"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/4618","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/comments?post=4618"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/4618\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=4618"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=4618"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=4618"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}