Soal

1. Calculate:

a.(14.82+6.52)

3.82 + 55√2+14

b. (-3.5 )3+ e6

ln524+206

13

2. Calculate:

a.16.52(8.4−√70)

4.32−17.3

b.5.23−6.42+3

1.68−2+(13.3

5 )1.5

3. Calculate:

a. 15( √10+3.72

log10 (1365 )+1.9 )b.

2.53(16−21622 )

1.74+14+ 4√2050

4. Calculate:

a.2.32 .1.7

√(1−0.82 )2+(2 – √0.87 )2

b. 2.34+ 12

2.27 (5.92−2.42 )+9.8 ln 51

5. Calculate:

a.

sin7 π9

cos2( 57

π )+

17

tan ( 512

π )

b.tan 640

cos2 140 −3 sin 800

3√0.9+cos 550

sin 110

6. Define the variabel x as x = 2.34, then evaluate:a. 2 x4−6 x3+14.8 x2+9.1

b.e2 x

√14+x2−x

7. Define the variable t as t = 6.8, then evaluate:

a. ln (|t2− t3|)

b.752 t

cos ( 0.8t−3 )

8. Define the variables x and y as x = 8.3 and y = 2.4, then evaluate:

a. x2+ y2− x2

y2

b. √ xy−√ x+ y+( x− yx−2 y )

2

−√ xy

9. Define the variables a, b, c, and d as:

a=13 , b=4.2 , c=(4 b )

a,∧d= abc

a+b+c, t h en :|

a. ab

c+d+ d

cab−(a−b2 ) (c+d )

b. √a2+b2

(d−c )+ ln (|b−a+c−d|)

10. A cube has a side of 18 cm.a. Determine the radius of a sphere that has the same surface area

as the cube.b. Determine the radius of a sphere that has the same volume as

the cube.

11. The perimeter P of an ellipse with semi-minor axes a and b is given

approximately by: P=2 π √ 12

( a2+b2 )

a. Determine the perimeter of an ellipse with a = 9 in. and b = 3 in.

b. An ellipse with b = 2a has a perimeter of P = 20cm. Determine a and b.

12. Two trigonometric identities are given by:a. sin 4 x=4 sin x cos x−8sin3 x cos x

b. cos2 x=1 – tan2 x1+tan 2 x

For each part, verify that the identity is correct by calculating the

values of the left and right sides of the question, subtituting x=π9

.

a b

Jawab:

1. (a) (14.82+6.52)

3.82 + 55√2+14

>>( (14.8^2+6.5^2)/3.8^2)+(55/(sqrt(2)+14))

ans =

21.6630

(b) (-3.5 )3+ e6

ln524+206

13

>>( -3.5)^3+(exp(6)/log(524))+206^(1/3)

ans =

27.4611

2. (a)16.52(8.4−√70)

4.32−17.3

>> (16.5^2*(8.4-sqrt(70)))/(4.3^2-17.3)

ans =

7.6412

(b)5.23−6.42+3

1.68−2+(13.3

5 )1.5

>> ((5.2^3-6.4^2+3)/(1.6^8-2))+(13.3/5)^1.5

ans =

6.8450

3. (a) 15( √10+3.72

log10 (1365 )+1.9 )>> 15*((sqrt(10)+3.7^2)/(log10(1365)+1.9))

ans =

50.2041

(b)2.53(16−216

22 )1.74+14

+ 4√2050

>> (2.5^3*(16-(216/22)))/(1.7^4+14)+nthroot(2050,4)

ans =

11.0501

4. (a)2.32 .1.7

√(1−0.82 )2+(2 – √0.87 )2

>> (2.3^2*1.7)/sqrt((1-(0.8^2))^2+(2-sqrt(0.87))^2)

ans =

7.9842

(b) 2.34+ 12

2.27 (5.92−2.42 )+9.8 ln 51

>> 2.34+(1/2)*2.7*(5.9^2-2.4^2)+9.8*log(51)

ans =

80.0894

5. (a)sin

7 π9

cos2( 57

π )+

17

tan ( 512

π )

>> (sin(7*pi/9)/cos((5/7)*pi)^2)+((1/7)*tan((5/12)*pi))

ans =

2.1867

(b)tan 640

cos2 140 −3 sin 800

3√0.9+cos 550

sin 110

>>(tand(64)/cosd(14)^2)-((3*sind(80))/nthroot(0.9,3))+cosd(55)/sind(11)

ans =

2.1238

6. x = 2.34

(a) 2 x4−6 x3+14.8 x2+9.1

>> 2*(x^4)+6*(x^3)+14.8*(x^2)+9.1

ans =

226.9807

(b)e2 x

√14+x2−x

>> exp(2*x)/sqrt(14+x^2-x)

ans =

26.0345

7. t = 6.8

(a) ln (|t2−t3|)

>> log(abs(t^2-t^3))

ans =

5.5917

(b)752 t

cos ( 0.8t−3 )

>> (75/(2*t))*cos((0.8*t)-3)

ans =

-4.2122

8. x = 8.3 and y = 2.4

(a) x2+ y2− x2

y2

>> x^2+y^2-(x^2/y^2)

ans =

62.6899

(b) √ xy−√ x+ y+( x− yx−2 y )

2

−√ xy

>> sqrt(x*y)-sqrt(x+y)+((x-y)/(x-(2*y)))^2-sqrt(x/y)

ans =

2.1741

9. a = 13 , b = 4.2 , c = 4 ba , d =

abca+b+c

>> a=13;

>> b=4.2;

>> c=(4*b)/a;

>> c

c =

1.2923

>> d=(a*b*c)/(a+b+c);

>> d

d =

3.8156

(a) ab

c+d+ d

cab−(a−b2 ) (c+d )

>> a*(b/(c+d))+((d/a)*(a/b))-((a-b^2)*(c+d))

ans =

35.2986

(b) √a2+b2

(d−c )+ ln (|b−a+c−d|)

>> (sqrt(a^2+b^2)/(d-c))+log(abs(b-a+c-d))

ans =

7.8410

10. S = 18cm

(a) luas permukaan bola = luas permukaan kubus

4 π r2=6 s2

r2=6 s2

4 π

r=2√ 3 s2

2 π

>> s=18;

>>r=sqrt((3*(s^2))/2*pi)

r =

12.4377

(b) volume kubus = volume bola

s3= 43

π r3

Sehingga, r=3√ 3 s3

4 π

>> s=18;

>> r=((3*s^3)/4*pi)^1/3;

r =

11.1663

11. P=2 π √ 12

( a2+b2 )

(a) a = 9 , b = 3

>> a=9;

>> b=3;

>> p=2*pi*sqrt((1/2)*(a^2+b^2));

>> p

p =

42.1489

(b) b = 2a , P = 20

P=2 π √ 12

( a2+b2 )

P2 π

=√ 12

(a2+b2 )

( P2π )

2

=(√ 12

( a2+b2 ))2

P2

(2π )2=1

2(a2+b2 )

2 P2

(2π )2=(a2+b2 )

2 P2

(2π )2=(a2+(2a )2 )

2 P2

(2π )2=(a2+4 a2 )

2 P2

(2π )2=5 a2

2P2

5 (2 π )2=a2

√ 2 P2

5 (2 π )2=a

>> p=20;

>> a=sqrt((2*p^2)/(5*(2*pi)^2));

>> a

a =

2.0132

>> b=2*a;

>> b

b =

4.0263

Jadi, P = 20 a = 2.0132=2 dan b = 4.0263=4

12. x=π9

>> x=pi/9;

>> x

x =

0.3491

(a) sin 4 x=4 sin x cos x−8sin3 x cos x

>> sin(4*x)

ans =

0.9848

>> (4*sin(x)*cos(x))-(8*sin(x)^3*cos(x))

ans =

0.9848

Jadi, sin 4 x=4 sin x cos x−8sin3 x cos x adalah benar

(b) cos2 x=1 – tan2 x1+tan 2 x

>> cos(2*x)

ans =

0.7660

>> (1-tan(x)^2)/(1+tan(x)^2)

ans =

0.7660

Jadi, cos2 x=1 – tan 2 x1+tan 2 x

adalah benar