The dimensions of a rectangular garden were 3m by 10 m. When both dimensions were increased by equal amounts, the area of the garden doubled. Find the dimensions of the new garden

5m by 12m

Step-by-step explanation:

Let A represent the area, L, the Length and W, the width of the garden with dimension 3m by 10m

Area of a rectangle, A = LW

Area of the garden with the old dimension = 3(10) = 30 sq meters

Let A1 represent the area, L1 represent the length and W1, the width of the new dimension of the garden

Since the dimensions were increased by equal amounts, lets call this amount x, the area of the garden doubled

This means,

A1 = 2A

Put in the value of A

A1= 2 (30)

A1 = 60 sq units

and

L1 = L + x

W1 =W + x

Therefore, to find the new area, we have;

A1 = ( L + x ) ( W+x )

60 = ( 3 + x ) ( 10 + x )

expand the bracket

60 = 30 + 3x + 10x + x^2

60 = 30 + 13x + x^2

Subtract 30 from both sides

60 - 30 = 30 -30 + 13x + x^2

30 = 13x +x^2

Equate to zero to form a quadratic equation

x^2 + 13x - 30 = 0

Now, lets think of two numbers that when multiplied they give -30 and when summed give +13

Yes! it's +15 and -2

Noe lets solve the equation

x^2 + 15x - 2x - 30 = 0

x( x + 15 ) -2 ( x + 15 ) = 0

( x - 2 ) ( x + 15) = 0

x - 2= 0

x = 2

x + 15 = 0

x = -15

Since we can only use the positive value of x, therefore x = 2m

To get the new dimension

Remember that;

L1 = L + x

slot in the value of L

L1 = 3 + 2

L1 = 5m

slot in the value of W

W1 = W + x

W1 = 10 + 2

W1 = 12m

Hence, the new dimension is 5m by 12m

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier

your exact question is kind of difficult to understand, so maybe an example will explain what you're supposed to do? i'll work through the first problem on the worksheet.

suppose you have a tray of 652 brownies, and you have 5 friends that you want to feed without picking favorites. how many brownies can you dole out to each of your friends in equal amounts? how many will be left over for you to eat?

what you're doing here is finding a way to arrange the number 652 as the sum of integer multiple of 5 and some remainder:

here ("q" for quotient) is the number of brownies each of your friends get, and is the remainder. that's the basic idea behind division.

we break up the number 652 into parts so that the leading digit stands on its own:

652 = 600 + 52

first we check if it's possible to hand out exactly 100 brownies to each friend. with 5 friends, this means we want to know if we can take out 500 brownies from the tray. we can do this 1 time because we have 600 brownies to take from, and when we do that we're left with 100 + 52 brownies to divvy up. so we know that

652 = 1*500 + 152

next we want to know if it's possible to hand out exactly 10 brownies per friend. if we can, we have to remove 50 brownies from the tray each time until we can't. we have 152 brownies left, and we know that 3(50) = 150, so

152 = 3*50 + 2

now we have 2 brownies left in the tray. is it possible to hand 1 out to each of your 5 friends? no, you'd need a minimum of 5 brownies to do that. so the last 2 are yours.

putting everything together, we found that

652 = 1*500 + 3*50 + 2

and we can rearrange this to write

652 = 5*100 + 5*30 + 2

652 = 5(100 + 30) + 2

652 = 5*130 + 2

so and .

you might be more accustomed to the table depicting the algorithm. that's just a compact way of organizing what we've done above. you can consult the image i've attached below.