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2 years ago

10

The width of a swimming pool is 12 feet less than the pool's length. The area of the pool is 540 square feet. What are the dimen

sions of the pool?

1 answer:

fredd [130]2 years ago

8 0

100ft long and 42 wide

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What is the area of this figure?

ludmilkaskok [199]

Answer:

$\large\boxed{\large\boxed{42 unit^2}}$

Explanation:

The figure is not a regular hexagon. It is an irregular hexagon.

Please, find attached the picture with the original question and the figure.

You can split the figure into two triangles and one rectangle.

The rectangle has dimensions: 7units × 4units, thus its area is 28 units².

Both the upper triangle and lower triangle have base 7 units and height 2 units.

Hence the area of each triangle is:

$area=(1/2)base\times height$

$area=(1/2)7units\times 2units = 7units^2$

Hence, the area of the hexagon is:

$28units^2+7units^2+7units^2=42units^2$

3 0

3 years ago

scoray [572]

The answer to the problem is 5.

6 0

1 year ago

HELPPPP PLASSSSSSSS

meriva

Answer:

$0.30 per ounce for the small popcorn

Step-by-step explanation:

Divide 4.5 (price) by 15 (oz) to get the unit price.

Have a nice day!!!

7 0

2 years ago

Two planes start from los angeles international airport and fly in opposite directions. the second plane starts 1/2 hour after t

mixer [17]

Let x = first plane's speed.
We have the equation:
2 hours * x km/h + 1.5 hours * (x+60) km/h = 3135 km.
3.5 hours*x km/h + 90 km = 3135km
3.5hours*x km/h = 3045km
x = 870 km/h.
The first plane is traveling at 870 km/h, therefore since it is 60 km/h faster, the second plane is traveling at 930 km/h.

6 0

3 years ago

Simplify completely the quantity 14 times x to the 5th power times y to the 4th power plus 21 times x to the third power times y

timofeeve [1]

We have to simplify the given expression:

$\frac{14x^5y^4+21x^3y^2}{7x^3y}$

Dividing the terms of the numerator by the given term of denominator individually, we get

= $\frac{14x^5y^4}{7x^3y}+\frac{21x^3y^2}{7x^3y}$

By using the laws of exponent, $a^m \div a^n = a^{m-n}$ , we get

= $\frac{7 \times 2 x^5y^4} {7x^3y} + \frac{7 \times 3 x^3y^2}{7x^3y}$

= $\frac{2 x^5y^4} {x^3y} + \frac{3 x^3y^2}{x^3y}$

= ${2 x^{5-3}y^{4-1}} + 3 y^{2-1}$

= ${2 x^{2}y^{3}} + 3 y$

Therefore, the simplification of the given expression is ${2 x^{2}y^{3}} + 3 y$ .

So, Option 3 is the correct answer.

6 0

2 years ago

Read 2 more answers