All categories

2 years ago

PLZ HELP by what percent does the price change if the price was $100 and now it is $1250

Mathematics

2 answers:

Lana71 [14]2 years ago

8 0

1150% i think it’s the right answer.

Cerrena [4.2K]2 years ago

4 0

Answer:

1150% increase

how to solve

(V2 - V1)

------------ x 100

| V1 |

You might be interested in

Do #6 for me pls and thank you!

emmainna [20.7K]

Answer: function a

Hope this helped!

4 0

3 years ago

Think About the Process

mina [271]

Answer:

Step-by-step explanation:

i passed grade 6 lol

4 0

3 years ago

Read 2 more answers

Write the equation of the graph shown below in factored form. a graph that starts at the top left and continues down through the

Brut [27]

Answer:

(d) f(x) = (x − 3)^2(x − 2)(x − 1)

Step-by-step explanation:

In this context, a crossing of the axis at x=p means there is a factor of (x-p). A "touch" of the axis at x=q means there is a factor of (x -q)^2.

A crossing at x=1 and x=2, and a touch at x=3 means the factors are ...

f(x) = (x -1)(x -2)(x -3)^2 . . . . . matches the last choice

8 0

2 years ago

The bar graph shown here provides the numbers of scoops of different ice cream flavors that were sold during school lunch today.

Mila [183]

C is the answer because 10 - 6 = 4 and 6 + 4 = 10

3 0

3 years ago

Read 2 more answers

Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +

GenaCL600 [577]

<u>Answer: </u>

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

<u> Solution: </u>

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3) ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2) ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)

On solving above equation for x

(x+6) (x+3) =2(x+3) (2x+2)

x+6 = 4x + 4

x-4x = 4 – 6

x = $\frac{2}{3}$

Dimension of pool A

Length = x+6 = ( $\frac{2}{3}$ ) +6 = 6.667m

Width = x +3 = ( $\frac{2}{3}$ ) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2( $\frac{2}{3}$ ) + 6 = $\frac{22}{3}$ = 7.333m

Width = 2x + 2 = 2( $\frac{2}{3}$ ) + 2 = $\frac{10}{3}$ = 3.333m

Verifying the area:

Area of pool A = ( $\frac{20}{3}$ ) x ( $\frac{11}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Area of pool B = ( $\frac{22}{3}$ ) x ( $\frac{10}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

5 0

3 years ago