5. Four towns R, T, K and G are such that T is 84 km directly to the north of R, and K is on bearing of 295° from R at a distance of 60 km. G is on a bearing of 340° from K and a
distance of 30 km. Using a scale of 1 cm to represent 10 km, make an accurate scale
drawing to show the relative positions of the towns.
Find: (a) the distance and the bearing of T from K.
a)the distance and the bearing of T and K
(b) the distance and the bearing of G from T.

(c) the bearing of R from G.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.

answer:   expression= p+p*d%

price reduction + sale price = $625

sale price: $p+$p*d%

where d% is the percentage of discount

step-by-step explanation:

original price = p

price went down by $150 price reduction $p*d%

sale price= $ 475

therefore the price before sale is: $625

and the expression that shows the price before sale is: p+p*d%