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3 years ago

If a rectangle has a length of 4x and a width of x+5 what is the perimeter?

Mathematics

1 answer:

Paul [167]3 years ago

7 0

Answer:

10x + 10 units

Step-by-step explanation:

To find the perimeter of a rectangle, you can simply add up the length and width, and then multiply that sum by 2. The sum of the dimensions is 4x + (x + 5) = 5x + 5 and two times that would then be 2(5x + 5) = 10x + 10.

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3 years ago

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One cubic meter of water weighs approximately 2200 pounds. There are 100 cm in one meter. If you have a box-shaped fish tank tha

Diano4ka-milaya [45]

Answer:

158.4 pounds

Step-by-step explanation:

first find the volume of the tank

$V = 30 \times 30 \times 80$

$V = 72000 cm^3$

to convert this to m^3 we know that 1m = 100cm

$V = 72000 cm^3 \times \left(\dfrac{1\,\text{m}}{100\,\text{cm}}\right)^3$

$V = 72000 cm^3 \times \left(\dfrac{1\,\text{m}^3}{100^3\,\text{cm}^3}\right)$

the cm^3 cancel out.

$V = \dfrac{72000}{100^3} m^3$

$V = 0.072\,m^3$

by simply converting the side-lengths to meters beforehand can give you this answer directly:

$V = 0.3 \times 0.3 \times 0.8 = 0.072 m^3$

it is given that 1 meter cube of water weighs about 2200 pounds.

$1\,m^3 = 2200 \,\text{lb}$

multiply both sides with 0.072 to find our answer:

$1\,m^3 \times 0.072 = 2200 \times 0.072 \,\text{lb}$

$0.072\,m^3 = 158.4 \,\text{lb}$

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3 0

3 years ago

PLS ANSWER WHAT IS 3 and 14 hundredths minus 1 and 9 hundredths?

k0ka [10]

Answer:

2 and 5 hundredths

Step-by-step explanation:

3 - 1 = 2

14 hundredths - 9 hundredths = 5 hundredths

hence you get 2 and 5 hundredths

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3 years ago

Find the distance between A and B.

Papessa [141]

Answer:

The result is $d = 2\sqrt{5}$ units.

Step-by-step explanation:

The coordinates for A and B:

A = (-2, 0)

B = (2, 2)

-To find the distance between A and B, you need the distance formula:

$d = \sqrt{(x_{2} - x_{1})^2 + (y_{2}-y_{1})^2}$ Where the first coordinate is $(x_{1},y_{1})$ and the second coordinate is $(x_{2}, y_{2})$ .

-Use the coordinates A and B for the equation:

$d = \sqrt{(2+2)^2 + (2-0)^2}$

-Then, you solve the equation:

$d = \sqrt{(2+2)^2 + (2-0)^2}$

$d = \sqrt{(4)^2 + (2)^2}$

$d = \sqrt{16 + 4}$

$d = \sqrt{20}$

$d = 2\sqrt{5}$

So, therefore the distance is $2\sqrt{5}$ units.

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3 years ago

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Sonbull [250]

The answer would be 65

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