Round off each of the following to the number of significant figures given in brackets.
(a) 289.0456 (4 s.1)
(b) 1.236 (3 s. f.)
(c) *0.06573 (2 s.)
(d) *2408 (2 s. f.)

a) 289

b)1.24

c)0.066

d)2400

Step-by-step explanation:

(a) 289.0456 (4 s.1)

We want to round, 289.0456 to four significant figures.

We start counting from 2 because it is a significant digit and end on zero.

Therefore 289.0456=289.0 to 4 s.f

Or 289

(b) 1.236 (3 s.f.)

We want to round 1.236 to three significant figures.

We start counting from 1 and end on 3, but we need to run up because 6≥5

Therefore 1.236=1.24 to three significant figures.

(c) 0.06573 (2 s.)

To round 0.06573 to two significant figures, we need to start counting from the first non-zero digit, which is 6 and end on 5 and then round up because 7≥5

0.06573=0.066 to 2 s.f

(d) *2408 (2 s.f.)

To round 2408 to two significant figures, we start counting from 2 and end on 4.

We round off because 0<5

In this case we need to replace all digits after 4 with zeros.

2408=2400 to 2 s.f

it should be a and d. lmk if i'm wrong xd

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.