All categories

3 years ago

If 8 cents are needed to buy 10 pebbles, how many dollars are needed to buy 205 pebbles?

Mathematics

1 answer:

stira [4]3 years ago

7 0

Answer:

Hello There!!

Step-by-step explanation:

If 8 cents=10 pebbles then do 205 divided by 10 which is 20.5 . So, 8×20.5=164 dollars.

(8×20.5=164) 8 cents=10 (10×20.5=205)

164 dollars=205

Therefore,the answer is 164 dollars.

hope this helps,have a great day!!

~Pinky~

You might be interested in

Please help, thank you!

andrew11 [14]

Answer:

Step-by-step explanation:

11 13 18 19 22 25 29 32 37 37 38

find the median (25)

with 11,13,18,19,22 find the middle for that (18) Q1

with 29,32,37,37,38 find the middle for that (37)Q3

Q3-Q1

37-18=19

7 0

3 years ago

One vertex of a polygon is located at (3, –2). After a rotation, the vertex is located at (2, 3).

motikmotik

yes it is right when the vertex of the first polygon is located on 3 - 2 after rotation the work vertex will be there on (2 , 3) it is right

4 0

3 years ago

Read 2 more answers

The area of a rectangular garden is 7722 m“.

Slav-nsk [51]

The width of the garden is 78
You just divide 7722 by 99 (which is 78)

7 0

3 years ago

The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th

inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4 -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

$y(y-2)=100$

$y^2-2y=100$

subtracting 100 from both sides

$y^2-2y-100=0$

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

$ax^2+bx+c=0$

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

$y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

Solving first

$y=\frac{2+\sqrt{4+400}{2}$

$y=\frac{2+\sqrt{404}{2}$

$y=\frac{2+20.099}{2}$

$y=\frac{22.099}{2}$

$y=11.049$

Solving second one

$y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}$

$y=\frac{2-\sqrt{4+400}{2}$

$y=\frac{2-\sqrt{404}{2}$

$y=\frac{2-20.099}{2}$

$y=\frac{-18.99}{2}$

$y=-9.045$

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

4 0

3 years ago

Find the average rage of change of each of the following functions over the interval 1 < × < 5.

tamaranim1 [39]

not sure maybe 45 or 10

5 0

2 years ago