Assignment 1 – BD309 – Otniel Feliks Putra Wahyudi – 2481417024

Pertanyaan:

1. A law firm seeks to recruit top-quality experienced lawyers. The total package offered is

the sum of three separate components: a basic salary which is 1.2 times the candidate’s

current salary together with an additional $3000 for each year worked as a qualified

lawyer and an extra $1000 for every year that they are over the age of 21.

Work out a formula that could be used to calculate the total salary, S, offered to

someone who is A years of age, has E years of relevant experience and who currently

earns $N. Hence work out the salary offered to someone who is 30 years old with five

years’ experience and who currently earns $150 000.

2. Write down a formula for each situation:

  1. A plumber has a fixed call-out charge of $80 and has an hourly rate of $60. Work

out the total charge, C, for a job that takes L hours in which the cost of materials

and parts is $K.

  1. An airport currency exchange booth charges a fixed fee of $10 on all transactions

and offers an exchange rate of 1 dollar to 0.8 euros. Work out the total charge, C,

(in $) for buying x euros.

  1. A firm provides 5 hours of in-house training for each of its semi-skilled workers and

10 hours of training for each of its skilled workers. Work out the total number of

hours, H, if the firm employs a semi-skilled and b skilled workers.

  1. A car hire company charges $C a day together with an additional $c per mile. Work

out the total charge, $X, for hiring a car for d days and travelling m miles during

that time.

Without using a calculator, evaluate

(a) 10 × (−2) (b) (−1) × (−3) (c) (−8) ÷ 2 (d) (−5) ÷ (−5)

(e) 24 ÷ (−2) (f) (−10) × (−5) (f) (−6) × 5 × (−1)

 

3. Without using a calculator, evaluate

(a) 5 − 6 (b) −1 − 2 (c) 6 − 17 (d) −7 + 23

(e) −7 − (−6) (f) −4 − 9 (g) 7 − (−4) (h) −9 − (−9)

(i) 12 − 43 ( j) 2 + 6 − 10

 

4. Without using a calculator, evaluate

(a) 5 × 2 − 13 (b) 5 × (1 − 4)

(c) 1 − 6 × 7 (d) −5 + 6 ÷ 3 (e) 2 × (−3)2 (f) −10 + 22

(g) (−2)2 − 5 × 6 + 1 ( j)

 

5. Simplify each of the following algebraic expressions:

(a) 2 × P × Q (b) I × 8 (c) 3 × x × y

(d) 4 × q × w × z (e) b × b (f) k × 3 × k

Keterangan: Saya telah menyelesaikan tugas ini dengan baik dan benar
Status: 100% Tercapai
Bukti:
jawab

1. Total paket gaji S terdiri dari: gaji pokok 1.2N ditambah $3000 untuk setiap tahun pengalaman E dan $1000  untuk setiap tahun di atas umur           21 (yaitu A−21)

Jadi Rumus:
S=1.2N+3000E+1000(A21)

Catetan: jika ingin menangani kasus A≤21 secara ketat, bagian 1000(A−21) harus diganti dengan   1000max⁡(0,A−21)

Untuk A= 30,  E= 5,  N= 150000

S= 1.2(150000)+3000(5)+1000(30−21)
= 180000+15000+9000
   = 204 000

Jadi gaji yang ditawarkan adalah $204 000

2) Tuliskan rumus untuk tiap situasi

a) Tukang ledeng: biaya tetap $80 + tarif per jam $60 + biaya material $K untuk pekerjaan L jam.

C = 80 + 60 L + K

b) Booth penukaran mata uang: biaya tetap $10 per transaksi, kurs 1 dollar=0.8 euro. Untuk membeli xx euro, jumlah dollar yang perlu dibayar (tanpa fee) adalah x/(0.8)=1.25x. Jadi

C = 10 + x/0.8  =  10 + 1.25x

c) Pelatihan: 5 jam untuk setiap pekerja semi-skilled (jumlah a) dan 10 jam untuk setiap pekerja skilled (jumlah b).

H = 5 a + 10 b

d) Sewa mobil: biaya harian $C per hari dan tambahan $c per mil. Untuk sewa d hari dan m mil:

X= C ⋅ d + c ⋅ m

Hitungan tanpa kalkulator

Berikut evaluasi:

(a) 10×(−2)=−20
(b) (−1)×(−3)=3
(c) (−8)÷2=−4
(d) (−5)÷(−5)=1
(e) 24÷(−2)=−12
(f) (−10)×(−5)=50
(f lagi) (−6)×5×(−1)=(−30)×(−1)=30

3) Hitungan tanpa kalkulator (operasi pengurangan/penjumlahan dengan negatif)

(a) 5−6=−1
(b) −1−2=−3
(c) 6−17=−11
(d) −7+23=16
(e) −7−(−6)=−7+6=−1
(f) −4−9=−13
(g) 7−(−4)=7+4=11
(h) −9−(−9)=0
(i) 12−43=−31
(j) 2+6−10=−2

4) Hitungan tanpa kalkulator (dengan perkalian/pembagian)

(a) 5×2−13=10−13=−3
(b) 5×(1−4)=5×(−3)=−15
(c) 1−6×7=1−42=−41
(d) −5+6÷3=−5+2=−3
(e) 2×(−3)2=2×9=18 (karena (−3)2=9(-3)^2=9)
(f) −10+22=12
(g) (−2)2−5×6+1=4−30+1=−25

5) Sederhanakan ekspresi aljabar

(a) 2×P×Q=2PQ
(b) I×8=8I
(c) 3×x×y=3xy
(d) 4×q×w×z=4qwz
(e) b×b=b2
(f) k×3×k=3k2

 

 

 

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