The ratio of the price of a pen to the price of a ruler is 4:3. If the price of a ruler increases by 70p, then the ratio becomes 4:5.
Complete these sentences.
(a) The price of a pen is £
(b) The price of a ruler before the increase is £

(a) The price of the pen is $1.4

(b) The price of the ruler before the increase is $1.05

Step-by-step explanation:

∵ The ratio of the price of a pen to the price of a ruler is 4: 3

→ Multiply each term of the ratio by m, where m is a constant

∴ The cost of the pen = 4m

∴ The cost of the ruler = 3m

∵ The price of a ruler increases by 70p

→ Change 70 p to dollars by divide it by 100

∴ The price of a ruler increases by $0.7

∵ The old price of the ruler = 3m

∴ The new price of the ruler = 3m + 0.7

∵ The ratio between the pen and the ruler becomes 4: 5

∴ The price of the pen/the new price of the ruler =  

∴ The new ratio of the prices =

→ Equate the fraction of the ratio by the fraction of the prices

∴ =

→ By using cross multiplication

∵ 5 × (4m) = 4 × (3m + 0.7)

∴ 20m = 4(3m) + 4(0.7)

∴ 20m = 12m + 2.8

→ Subtract 12 m from both sides

∴ 8m = 2.8

→ Divide both sides by 8 to find m

∴ m = 0.35

(a)

∵ The price of the pen = 4m

∵ m = 0.35

∴ The price of the pen = 4(0.35)

∴ The price of the pen = $1.4

(b)

∵ The price of the ruler before the increase = 3m

∵ m = 0.35

∴ The price of the ruler before the increase = 3(0.35)

∴ The price of the ruler before the increase = $1.05

the answer is a

step-by-step explanation:

because if you put 1 in to the square root of x it equals 300.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}