All categories

2 years ago

A shop sells compost in 20 litre bags and in 40 litre bags.

Mathematics

1 answer:

vlada-n [284]2 years ago

5 0

Answer:

Step-by-step explanation:

3.50/2 = 1.75 and 9/3= 3

so 2 bags for £3.50 is better than 3 bags for £9

You might be interested in

Guys i got an easy question for u...How is 3.098 written in words?(Easy)

IRISSAK [1]

3 and 98 thousandths

4 0

3 years ago

Read 2 more answers

21.

jasenka [17]

Answer:

b = y intercept

Step-by-step explanation:

y = mx +b

This is the slope intercept form of a line

The m is the slope and the b is the y intercept

8 0

3 years ago

What is the slope-intercept equation of this line?

PIT_PIT [208]

Answer:

Y=-4/2 + 6

Step-by-step explanation:

It is +6 because the line intercepts the Y axis at the point 6. It is a negative slope. It is a slope of -4/2 because from point (0,6) it goes down 4 and 2 right.

8 0

3 years ago

Read 2 more answers

mr.browns salary is 32,000 and imcreases by $300 each year, write a sequence showing the salary for the first five years when wi

chubhunter [2.5K]

Hello!

We have the following data:

a1 (first term or first year salary) = 32000

r (ratio or annual increase) = 300

n (number of terms or each year worked)

We apply the data in the Formula of the General Term of an Arithmetic Progression, to find in sequence the salary increases until it exceeds 34700, let us see:

formula:

$a_n = a_1 + (n-1)*r$

* second year salary

$a_2 = a_1 + (2-1)*300$

$a_2 = 32000 + 1*300$

$a_2 = 32000 + 300$

$\boxed{a_2 = 32300}$

* third year salary

$a_3 = a_1 + (3-1)*300$

$a_3 = 32000 + 2*300$

$a_3 = 32000 + 600$

$\boxed{a_3 = 32600}$

* fourth year salary

$a_4 = a_1 + (4-1)*300$

$a_4 = 32000 + 3*300$

$a_4 = 32000 + 900$

$\boxed{a_4 = 32900}$

* fifth year salary

$a_5 = a_1 + (5-1)*300$

$a_5 = 32000 + 4*300$

$a_5 = 32000 + 1200$

$\boxed{a_5 = 33200}$

We note that after the first five years, Mr. Browns' salary has not yet surpassed 34700, let's see when he will exceed the value:

* sixth year salary

$a_6 = a_1 + (6-1)*300$

$a_6 = 32000 + 5*300$

$a_6 = 32000 + 1500$

$\boxed{a_6 = 33500}$

* seventh year salary

$a_7 = a_1 + (7-1)*300$

$a_7 = 32000 + 6*300$

$a_7 = 32000 + 1800$

$\boxed{a_7 = 33800}$

* eighth year salary

$a_8 = a_1 + (8-1)*300$

$a_8 = 32000 + 7*300$

$a_8 = 32000 + 2100$

$\boxed{a_8 = 34100}$

* ninth year salary

$a_9 = a_1 + (9-1)*300$

$a_9 = 32000 + 8*300$

$a_9 = 32000 + 2400$

$\boxed{a_9 = 34400}$

* tenth year salary

$a_{10} = a_1 + (10-1)*300$

$a_{10} = 32000 + 9*300$

$a_{10} = 32000 + 2700$

$\boxed{a_{10} = 34700}$

we note that in the tenth year of salary the value equals but has not yet exceeded the stipulated value, only in the eleventh year will such value be surpassed, let us see:

* eleventh year salary

$a_{11} = a_1 + (11-1)*300$