Jika 2 log ⁡ 3 = a d a n 3 log ⁡ 5 = b , 2_{\log3\ }=a\ \ dan\ 3_{\log_{ }5\ }=b,\ \ 2 lo g 3 = a d a n 3 lo g 5 = b , maka nilai 1 5 log ⁡ 20 = 15_{\log20\ }= 1 5 lo g 2 0 = ....

2 + a b a ( 1 + b ) \frac{2+ab}{a\left(1+b\right)} a ( 1 + b ) 2 + a b

b + 1 2 a b + 1 \frac{b+1}{2ab+1} 2 a b + 1 b + 1

a ( 1 + b ) 2 + a b \frac{a\left(1+b\right)}{2+ab} 2 + a b a ( 1 + b )

Diberikan fungsi f ( x ) = 2 + sin ⁡ 3 x , 0 ° ≤ x ≤ 360 ° f\left(x\right)=2+\sin3x,\ 0\degree\le x\le360\degree f ( x ) = 2 + sin 3 x , 0 ° ≤ x ≤ 3 6 0 ° . Fungsi tersebut naik pada interval ...

0 ° ≤ x < 30 ° 0\degree\le x<30\degree\ 0 ° ≤ x < 3 0 °

30 ° < x < 90 ° 30\degree<x<90\degree\ 3 0 ° < x < 9 0 °

150 ° < x < 210 ° 150\degree<x<210\degree 1 5 0 ° < x < 2 1 0 °

270 ° < x < 330 ° 270\degree<x<330\degree 2 7 0 ° < x < 3 3 0 °

0 o < x < 9 0 o 0^o<x<90^o 0 o < x < 9 0 o

f ( x ) = sin ⁡ ( 4 x − π 2 ) f\left(x\right)=\sin\left(4x-\frac{\pi}{2\ }\right)\ f ( x ) = sin ( 4 x − 2 π ) Hubungan terjadinya stasioner (x) dan titik stasioner (x, y) pada kurva dalam interval 0 ≤ x ≤ π 0\le x\le\pi 0 ≤ x ≤ π adalah ...

x = π 4 d a n ( π 4 , 1 ) x=\frac{\pi}{4}\ dan\ \left(\frac{\pi}{4\ },\ 1\right)\ x = 4 π d a n ( 4 π , 1 )

x = π 6 d a n ( π 6 , 1 2 ) x=\frac{\pi}{6}\ dan\ \left(\frac{\pi}{6},\frac{1}{2}\right) x = 6 π d a n ( 6 π , 2 1 )

x = π 3 d a n ( π 3 , 1 2 ) x=\frac{\pi}{3}\ dan\ \left(\frac{\pi}{3},\ \frac{1}{2}\right)\ x = 3 π d a n ( 3 π , 2 1 )

x = π 4 d a n ( π 4 , 0 ) x=\frac{\pi}{4}\ dan\ \left(\frac{\pi}{4},\ 0\right)\ x = 4 π d a n ( 4 π , 0 )

x = π d a n ( π , 0 ) x=\pi\ dan\ \left(\pi,\ 0\right)\ x = π d a n ( π , 0 )

Jika kubus diatas memiliki panjang rusuk 6 cm, dan titik x merupakan titik ditengah-tengah AB, Jarak titik H ke x pada bagun ruang diatas adalah...

6 2 c m 6\sqrt{2}\ cm 6 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> c m

3 5 c m 3\sqrt{5}\ cm 3 5 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> c m

3 2 c m 3\sqrt{2}cm 3 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> c m

Segitiga ABC dibentuk dari titik A (3, 2), titik B (6, 2), titik C (5, 4) yang dicerminkan terhadap sumbu Y dilanjutkan dengan perputaran sejauh [O, 90 o ]. Koordinat bayangan segitiga ABC adalah ...

A'' (-2, -3), B'' (-2, -6), C'' (-4, -5)

A'' (2, -3), B'' (-2, 6), C'' (-4, 5)

A'' (-2, -3), B'' (2, 6), C'' (-4, -5)

A'' (-2, -3), B'' (2, -6), C'' (-4, -5)

A'' (-2, 3), B'' (-2, -6), C'' (-4, -5)

α + β = π 6 \alpha\ +\ \beta\ =\frac{\pi}{6} α + β = 6 π dan cos ⁡ α cos ⁡ β = 3 4 \cos\alpha\cos\ \beta=\frac{3}{4} cos α cos β = 4 3 . Nilai dari cos ⁡ ( α − β ) = . . . \cos\left(\alpha-\beta\right)=... cos ( α − β ) = . . .

3 2 − 3 2 \frac{3}{2}-\frac{\sqrt{3}}{2} 2 3 − 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

1 9 + 3 2 \frac{1}{9}+\frac{\sqrt{3}}{2} 9 1 + 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 4 − 3 2 \frac{3}{4}-\frac{\sqrt{3}}{2} 4 3 − 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 3 + 3 2 \frac{2}{3}+\frac{\sqrt{3}}{2} 3 2 + 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

4 3 + 3 2 \frac{4}{3}+\frac{\sqrt{3}}{2} 3 4 + 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Himpunan penyelesaian dari 2 sin 2x > 1 dalam interval 0 o ≤ x ≤ 360 o adalah ...

{x |15 o ≤ x ≤ 75 o } atau {x |195 o ≤ x ≤ 225 o }

{x |25 o ≤ x ≤ 75 o } atau {x |195 o ≤ x ≤ 225 o }

{x |15 o ≤ x ≤ 75 o } atau {x |175 o ≤ x ≤ 225 o }

{x |25 o ≤ x ≤ 75 o } atau {x |175 o ≤ x ≤ 225 o }

{x |15 o ≤ x ≤ 75 o } atau {x |215 o ≤ x ≤ 225 o }

Kiper kesebelasan Persib dapat menggagalkan tendangan penalti pemain lawan sebesar 0,8. Jika dalam suatu pertandingan terjadi 5 kali tendangan penalti, peluang terjadi 2 gol pada kiper Persib adalah ...

128 625 \frac{128}{625} 6 2 5 1 2 8

32 625 \frac{32}{625} 6 2 5 3 2

64 625 \frac{64}{625} 6 2 5 6 4

256 625 \frac{256}{625} 6 2 5 2 5 6

512 625 \frac{512}{625} 6 2 5 5 1 2

Diketahui a → = 2 i + 1 j \overrightarrow{a\ }=2\ i\ +\ 1\ j a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = 2 i + 1 j dan b → = − 3 i + 2 j \overrightarrow{b\ }=-3i\ +\ 2j b <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = − 3 i + 2 j . Vektor proyeksi a → \overrightarrow{a\ } a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> pada b → \overrightarrow{b\ } b <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> adalah ...

12 13 i + − 8 13 j \frac{12}{13}i\ +\frac{-8}{13}j 1 3 1 2 i + 1 3 − 8 j

12 13 i + − 8 − 13 j \frac{12}{13}i+\frac{-8}{-13}j 1 3 1 2 i + − 1 3 − 8 j

− 12 13 i + − 8 13 j \frac{-12}{13}i+\frac{-8}{13}j 1 3 − 1 2 i + 1 3 − 8 j

− 12 13 i + − 8 − 13 j \frac{-12}{13}i+\frac{-8}{-13}j 1 3 − 1 2 i + − 1 3 − 8 j

− 12 − 13 i + − 8 − 13 j \frac{-12}{-13}i+\frac{-8}{-13}j − 1 3 − 1 2 i + − 1 3 − 8 j

a → = 3 i − 4 j + 2 k \overrightarrow{a\ }=3i\ -4j+2k\ a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> = 3 i − 4 j + 2 k dan titik P ( 2 , − 1 , 3 ) P\left(2,\ -1,\ 3\right)\ P ( 2 , − 1 , 3 ) . Jika P Q → \overrightarrow{PQ} P Q <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> berlawanan dengan a → \overrightarrow{a\ } a <path d="M0 241v40h399891c-47.3 35.3-84 78-110 128 -16.7 32-27.7 63.7-33 95 0 1.3-.2 2.7-.5 4-.3 1.3-.5 2.3-.5 3 0 7.3 6.7 11 20 11 8 0 13.2-.8 15.5-2.5 2.3-1.7 4.2-5.5 5.5-11.5 2-13.3 5.7-27 11-41 14.7-44.7 39-84.5 73-119.5s73.7-60.2 119-75.5c6-2 9-5.7 9-11s-3-9-9-11c-45.3-15.3-85 -40.5-119-75.5s-58.3-74.8-73-119.5c-4.7-14-8.3-27.3-11-40-1.3-6.7-3.2-10.8-5.5 -12.5-2.3-1.7-7.5-2.5-15.5-2.5-14 0-21 3.7-21 11 0 2 2 10.3 6 25 20.7 83.3 67 151.7 139 205zm0 0v40h399900v-40z"> , tentukan koordinat titik Q Q Q ! (Essai)

− i + 3 j + k -i+3j+k − i + 3 j + k

i − 3 j + k i-3j+k i − 3 j + k

i − 3 j − k i-3j-k i − 3 j − k

− i − 3 j − k -i-3j-k − i − 3 j − k

Sebuah tangga memiliki panjang 6 m. Tangga tersebut disandarkan pada tembok rumah dengan membentuk sudut 6 0 o 60^o 6 0 o terhadap tanah. Ketinggian tembok yang dapat dicapai oleh ujung tangga dari permukaan tanah adalah… m.

3 3 3\sqrt{3} 3 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 2 2\sqrt{2} 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 2 3\sqrt{2} 3 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 3 2\sqrt{3} 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

6 3 6\sqrt{3} 6 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Diketahui Panjang rusuk kubus ABCD.EFGH adalah 6 cm. Jika S adalah titik potong EG dan FH, maka jarak DH ke AS adalah ... cm.

3 2 3\sqrt{2} 3 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 3 2\sqrt{3} 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 6 2\sqrt{6} 2 6 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Aisyah membantu orang tua untuk berjualan kue di sekolahnya, ia tidak merasa malu karena ada hadist yang berbunyi “Pedagang yang jujur dan terpercaya akan dibangkitkan bersama para Nabi, orang-orang shiddiq dan para syuhada” (HR. Tirmidzi). Tempat kue yang digunakan Aisyah berjualan hanya mampu menampung 100 jenis kue. Aisyah akan menjual 3 jenis kue yakni donat, roti isi, dan brownies, donat dibeli dengan harga Rp. 2.000,00 per biji, roti isi dibeli dengan harga Rp. 3.500,00 per biji, dan brownies dibeli dengan harga Rp. 3.000,00 per potong, sedangkan Aisyah hanya memiliki modal sebesar Rp. 250.000,00. Jika x, y, z berturut-turut menyatakan banyak donat, roti isi, dan brownies maka model matematika yang tepat untuk permasalahan tersebut adalah … .

x + y + z = 100 dan 4x + 7y + 6z = 500

x + y + z = 100 dan 4x + 7y + 6z = 100

x + y + z = 100 dan 7x + 4y + 6z = 500

x + y + z = 500 dan 7x + 4y + 6z = 100

x + y + z = 500 dan 4x + 7y + 6z = 100

US MATEMATIKA WAJIB 2020

Authored by Ruqy Faishal

Mathematics

12th Grade

CCSS covered

Used 8+ times

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MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Jika
$2_{\log3\ }=a\ \ dan\ 3_{\log_{ }5\ }=b,\ \$ maka nilai
$15_{\log20\ }=$ ....

$\frac{2}{a}$

$\frac{a}{2}$

$\frac{2+ab}{a\left(1+b\right)}$

$\frac{b+1}{2ab+1}$

$\frac{a\left(1+b\right)}{2+ab}$

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Persamaan grafik fungsi logaritma berikut adalah

$y\ =\ ^2\ \log\ x$

$y\ =\ ^{3\ }\log\ x$

$y=^{\frac{1}{2}}\ \log\ x$

$y\ =\ 2\ \log\ x$

$y\ =\ ^a\ \log\ x$

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

penyelesaian dari $4^{x-3}=\sqrt{8^{x-5}}$ adalah ...

-3

$\frac{3}{7}$

$\frac{9}{7}$

Hasil bagi polinom 2x³-3x²+ x + 13 oleh 2x + 3 adalah ....

2x²- 6x + 8

2x²- 6x + 10

x²- 3x + 5

x²- 2x + 5

x²- 2x + 4

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Diketahui dua buah bilangan. Hasil selisihnya adalah 5, sedangkan jumlah kuadratnya adalah 73. jika simbol bilangan pertama x dan bilangan kedua dinotasikan y, maka x + y adalah …

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Perhatikan sistem pertidaksamaan berikut: $y<x+2$ dan $y>x^2-2x-8$ . Kedua titik potong yang merupakan batas daerah penyelesaian kedua pertidaksamaan diatas adalah ...

( - 2, 0 ) dan ( 4, 0 )

( - 2, 0 ) dan ( 5, 7 )

( 0, 2 ) dan ( 0, - 8 )

( 0, 2 ) dan ( 4, 0 )

( 0, 2 ) dan ( 5, 7 )

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Himpunan penyelesaian dari persamaan diatas adalah ...

$\left\{x\left|-3<x<2\right|\right\}$

$\left\{x\left|-2<x<3\right|\right\}$

$\left\{x\left|2<x<3\right|\right\}$

$\left\{x\left|3<x<5\right|\right\}$

$\left\{x\left|3<x<6\right|\right\}$

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Pada umumnya sungai akan meluap pada musim hujan dan menyusut pada musim kemarau. Tinggi air sungai pada cuaca normal adalah 400 cm dan perubahan tinggi air sungai akibat perubahan cuaca menyimpang paling banyak 50 cm. Misalkan tinggi air akibat perubahan adalah T, model matematika yang sesaui untuk situasi tersebut adalah …

$\left|T-50\right|\ \le400$

$\left|T+50\right|\le400$

$\left|T-400\right|\le50$

$\left|T+400\right|\le50$

$\left|T+350\right|\le350$

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US MATEMATIKA WAJIB 2020

Jika 2log⁡3 =a dan 3log⁡5 =b, 2_{\log3\ }=a\ \ dan\ 3_{\log_{ }5\ }=b,\ \ 2log3 ​=a dan 3log​5 ​=b, maka nilai 15log⁡20 =15_{\log20\ }=15log20 ​= ....

Persamaan grafik fungsi logaritma berikut adalah

penyelesaian dari 4x−3=8x−54^{x-3}=\sqrt{8^{x-5}}4x−3=8x−5​ adalah ...

Hasil bagi polinom 2x3 -3x2 + x + 13 oleh 2x + 3 adalah ....

Diketahui dua buah bilangan. Hasil selisihnya adalah 5, sedangkan jumlah kuadratnya adalah 73. jika simbol bilangan pertama x dan bilangan kedua dinotasikan y, maka x + y adalah …

Perhatikan sistem pertidaksamaan berikut: y<x+2y<x+2y<x+2 dan y>x2−2x−8y>x^2-2x-8y>x2−2x−8 . Kedua titik potong yang merupakan batas daerah penyelesaian kedua pertidaksamaan diatas adalah ...

Himpunan penyelesaian dari persamaan diatas adalah ...

Suatu persegi panjang luasnya adalah (2x - 3) satuan persegi dan lebarnya (x + 1) satuan. Agar panjangnya selalu lebih besar dari 1, maka batas x yang memenuhi adalah ...

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Jika
$2_{\log3\ }=a\ \ dan\ 3_{\log_{ }5\ }=b,\ \$ maka nilai
$15_{\log20\ }=$ ....

penyelesaian dari $4^{x-3}=\sqrt{8^{x-5}}$ adalah ...

Hasil bagi polinom 2x³-3x²+ x + 13 oleh 2x + 3 adalah ....

Perhatikan sistem pertidaksamaan berikut: $y<x+2$ dan $y>x^2-2x-8$ . Kedua titik potong yang merupakan batas daerah penyelesaian kedua pertidaksamaan diatas adalah ...