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3 years ago

What is the lowest common denominator of 1/3 and 2/5

Mathematics

1 answer:

Citrus2011 [14]3 years ago

8 0

Answer:

Step-by-step explanation:

3 and 5 are not common denominators so you have to find the lowest common denominator which is 15 because 3 times 5 is 15

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You are building a shelf that fits in a corner. In the figure, the entire shelf is △XYZ. Each unit in the coordinate plane repre

natima [27]

Complete Question: You are building a shelf that fits in a corner in the figure the entire shelf is XYZ each unit in the coordinate plane represents one inch find the area of the shelf.

Answer:

112.5 units²

Step-by-step explanation:

Area of the shelf = ½*base*height.

The base is the distance between Z(0, 5) and Y(15, 5)

The height is the distance between X(15, 20) and Y(15, 5).

Therefore area of the shelf = ½*ZY*XY

Use the distance formula to find ZY and XY.

Distance between Z(0, 5) and Y(15, 5):

$ZY = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$Z(0, 5) = (x_1, y_1)$

$Y(15, 5) = (x_2, y_2)$

$ZY = \sqrt{(15 - 0)^2 + (5 - 5)^2}$

$AB = \sqrt{(15)^2 + (0)^2}$

$AB = \sqrt{225 + 0} = \sqrt{225}$

$AB = 15$

Distance between X(15, 20) and Y(15, 5):

$XY = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$X(15, 20) = (x_1, y_1)$

$Y(15, 5) = (x_2, y_2)$

$XY = \sqrt{(15 - 15)^2 + (5 - 20)^2}$

$XY = \sqrt{(0)^2 + (-15)^2}$

$XY = \sqrt{0 + 225} = \sqrt{225}$

$XY = 15$

Area of shelf = ½*15*15 = ½*225 = 112.5 units²

4 0

3 years ago

REAL LIFE A cylindrical hazardous waste container has a diameter of 1.5 feet and a height of 1.6 feet. Abo

aivan3 [116]

Answer:

It can hold up to $1.256ft^3$ of waste

Step-by-step explanation:

Given

$D = 1.5ft$ -- Diameter

$h = 1.6ft$ -- height

Required

Determine the amount of waste it can hold

This simply means that we calculate the volume of the cylindrical container.

And this is calculated using:

$Volume = \pi r^2h$

Where

$\pi = 3.14$

and

$r = \frac{D}{2}$

$r = \frac{1.5}{2}$

$r = 0.75$

So, we have:

$Volume = 3.14 * 0.5^2 * 1.6$

$Volume = 1.256ft^3$

Hence, it can hold up to $1.256ft^3$ of waste

3 0

3 years ago

A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Four hu

castortr0y [4]

Answer: h = 120 ft; w = 80 ft

A = 9600 ft^2

Step-by-step explanation: Let h and w be the dimensions of the playground. The area is given by:

A = h*w (eq1)

The total amount of fence used is:

L = 2*h + 2*w + w (eq2) (an extra distance w beacuse of the division)

Solving for w:

w = L - 2/3*h = 480 - 2/3*h (eq3) Replacing this into the area eq:

We derive this and equal zero to find its maximum:

Solving for h:

h = 120 ft. Replacing this into eq3:

w = 80ft

Therefore the maximum area is:

A = 9600 ft^2

5 0

3 years ago

The system of equations has no solution: y=2/3x-5 4x - 6y = 30 true or false?

Serjik [45]

1. y = 2/3x - 5

2. 4x - 6y = 30

Divide 2. by 2

3. 2x - 3y = 15

Substitute 1. into 3.

4. 2x - 3(2/3x - 5) = 15

5. 2x - 2x + 15 = 15

6. 15 = 15

False. There are an infinite number of solutions.

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3 years ago

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jek_recluse [69]

Answer:

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3 years ago

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