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3 years ago

Pls help pls pls than you

Mathematics

1 answer:

azamat3 years ago

7 0

Answer:

Step-by-step explanation:

8*5=40

40*1/3=40/3

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John enlarged one side of his square flower bed by 8 ft to make it into a rectangle. The area of the new flower bed is 3 times t

maks197457 [2]

Answer:

4 ft.

Step-by-step explanation:

We can consider the orignial square's sides to be x, so area of the original figure: $x^{2}$

The new rectangle's sides will be x and x+8, so the area of the new figure will be x^2+8x.

We know that the new area is 3 times the old one. So our equation will be:

x^2+8x=3x^2

our answer will be x=4.

Therfore, the length of the side of the old flower bed is 4 ft.

4 0

3 years ago

Read 2 more answers

What is the answer to these question 1:1/3y-1/2=5 2:5n+1/2=1/6

Natasha2012 [34]

$\frac {1}{3}y-\frac{1}{2}=5\\2y-3 =30\\2y=33\\y=16.5\\\\\\5n+\frac{1}{2}=\frac {1}{6}\\30n+3=1\\30n=-2\\n=-\frac {2}{30} \\n= -\frac{1}{15}$

3 0

3 years ago

Joan wants to reduce the area of her posters by one-third. Draw lines to match the original dimensions in the left column with t

3241004551 [841]

Answer:

1) $30\ in\ by\ 12\ in$ --------> $120\ in^{2}$

2) $30\ in\ by\ 18\ in$ --------> $180\ in^{2}$

3) $12\ in\ by\ 15\ in$ --------> $60\ in^{2}$

4) $18\ in\ by\ 15\ in$ --------> $90\ in^{2}$

Step-by-step explanation:

Verify each case

case 1) $30\ in\ by\ 12\ in$

Find the area with the original dimensions

$A=30(12)=360\ in^{2}$

Reduce the area of the poster by one-third

$360/3=120\ in^{2}$

therefore

$30\ in\ by\ 12\ in$ --------> $120\ in^{2}$

case 2) $30\ in\ by\ 18\ in$

Find the area with the original dimensions

$A=30(18)=540\ in^{2}$

Reduce the area of the poster by one-third

$540/3=180\ in^{2}$

therefore

$30\ in\ by\ 18\ in$ --------> $180\ in^{2}$

case 3) $12\ in\ by\ 15\ in$

Find the area with the original dimensions

$A=12(15)=180\ in^{2}$

Reduce the area of the poster by one-third

$180/3=60\ in^{2}$

therefore

$12\ in\ by\ 15\ in$ --------> $60\ in^{2}$

case 4) $18\ in\ by\ 15\ in$

Find the area with the original dimensions

$A=18(15)=270\ in^{2}$

Reduce the area of the poster by one-third

$270/3=90\ in^{2}$

therefore

$18\ in\ by\ 15\ in$ --------> $90\ in^{2}$

6 0

4 years ago

Find the common difference of the arithmetic sequence 2, 8, 14, ...

BlackZzzverrR [31]

It skips by 6 if that's what you mean

3 0

3 years ago

Which is larger, 2/3 x 6 or 4/5 x 7?

Aneli [31]

2/3 × 6 is 6/3, which would simplify to 2. 4/5 × 7 is 28/5, which would simplify to 5 3/5. so 4/5×7 is larger

6 0

3 years ago