The floor of a rectangular room will be tiled with square tiles that measure 3 inches on each side. The room is 109 inches long and 103 inches wide. How many tiles, including parts of tiles, are needed to complete the job?

You have 22+3x/3x+7=2. That 3x/3x raises eyebrows; is this what you meant, with the result that 3x/3x = 1? or did you mean

3x

22 + ------------- + 7 = 2?

3x+7

If you meant the latter, then simplify the equation by subtracting 2 from both sides:

3x

27 + ---------- = 0

3x+7

Multiply all three terms by (3x+7), obtaining

27(3x+7) + 3x = 0

Then 21x + 3x + 189 = 0, so that 24x = -189, or x = -189/24 = -7 7/8

The scale of the architectural drawing is such that 1/4" represents 1'. This means, if the room measures 1.5" by 2.25" on the drawing, it measures 6 feet by 9 feet in real life.

In this scenario, the unit scale indicates that 1/4" on the drawing represents 1' in real life. So, to find the real-life dimensions of the room, you would need to convert the drawing dimensions to real-life dimensions.

For the length, 1.5" on the drawing would equate to 1.5 * 4 feet because every 1/4" equals 1 foot. Therefore, the length of the room is 6 feet.

The width of the room is similarly calculated. 2.25" on the drawing would equate to 2.25 * 4 feet which gives us a width of 9 feet.

So, the bedroom in real life is 6 feet by 9 feet.

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The bedroom is 6 feet by 9 feet in real life.

To determine the actual size of the bedroom, we can use the scale given in the architectural drawing. The scale 1/4" = 1' means that every 1/4 inch on the drawing represents 1 foot in real life. In this case, the bedroom measures 1.5 inches by 2.25 inches on the drawing. To find the actual size, we can multiply these dimensions by the scale factor: 1.5 inches * (1 foot / 1/4 inch) = 6 feet, and 2.25 inches * (1 foot / 1/4 inch) = 9 feet. Therefore, the bedroom is 6 feet by 9 feet in real life.

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The precise measurement of distance between Junction A at -6 and Junction B at 0 is indispensable in urban planning, optimizing traffic flow, and ultimately enhancing the quality of life for city residents.

One real-world problem where finding the number of units between -6 and 0 on a number line is crucial is in urban planning and transportation engineering.

Imagine a city with two major traffic junctions, Junction A at position -6 and Junction B at position 0 on the number line. City officials need to optimize traffic flow between these junctions to reduce congestion and commute times.

To achieve this, engineers must calculate the exact distance between these junctions. This information is vital for designing efficient road networks, determining signal timings, and planning for public transportation routes. It also helps in estimating travel times for commuters and enables the implementation of traffic management strategies.

Accurate measurements between -6 and 0 on the number line ensure that resources are allocated efficiently, minimizing environmental impact, fuel consumption, and overall transportation costs.

This problem highlights how fundamental mathematical concepts like distance on a number line play a critical role in shaping the urban landscape and improving quality of life for city residents.

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Marcus's wall has an area of 140 square feet. If he paints half of it blue, 70 square feet will be blue.

To find the area of the wall, we multiply its height by its length:

[tex]\[ \text{Area of the wall} = \text{Height} \times \text{Length} \][/tex]

First, convert the mixed numbers to improper fractions:

- Height: [tex]\(8 \frac{2}{5} = \frac{42}{5}\) feet[/tex]

- Length: [tex]\(16 \frac{2}{3} = \frac{50}{3}\) feet[/tex]

Now, multiply:

[tex]\[ \text{Area} = \left( \frac{42}{5} \right) \times \left( \frac{50}{3} \right) \][/tex]

[tex]\[ \text{Area} = \frac{42 \times 50}{5 \times 3} = \frac{2100}{15} = 140 \text{ square feet} \][/tex]

If Marcus paints half of the wall blue, the area painted blue is:

[tex]\[ \text{Blue area} = \frac{1}{2} \times 140 = 70 \text{ square feet} \][/tex]

Therefore, 70 square feet of the wall will be blue.

Answer:

At this rate, which expression can be used to determine how fast a mule deer runs in miles per hour. A 25 seconds/ 1/4 mile = 36 miles per hour

Answer:

B) one-fourth mile over twenty-five seconds = 36 miles per hour

The question involves finding the dimensions of a rectangular prism packed with 84 smaller cubes of a certain size. The answer requires understanding of volume and consideration of the physical characteristics of a rectangular prism.

The problem presents a rectangular prism completely packed with 84 cubes of edge length 1 over 3 inch. To find the dimensions of the prism, we must first understand the volume. Given the volumes of the small cubes (edge length cubed), we can multiply this by the number of cubes (84) to find the total volume of the prism.

Next, we use this total volume to derive the possible dimensions of the rectangular prism. The dimensions should be factors of the total volume, keeping in mind the shape of a physical rectangular prism (which usually has length longer than width and height).

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Applying the cosine and sine ratios, the value of cos C × sin A = 3/4.

Cosine ratio: cos ∅ = adj/hyp

Sine ratio: sin ∅ = opp/hyp.

Find cos C using the cosine ratio:

cos C = √3/2

Find sin A using the sine ratio:

sin A = √3/2

cos C × sin A = √3/2 × √3/2

= 3/4

cos C × sin A = 3/4

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Answer: d. s ≥ 23

Step-by-step explanation:

Let s be the side of the square.

We know that the perimeter of a square =4s

Let f(s)=4s

If the cardboard should have a perimeter equal to at least 92 inches, the inequality would look like

[tex]4s\geq 92[/tex]

Divide both sides by 4, we get

[tex]s\geq\frac{92}{4}\\\Rightarrow\ s\geq23[/tex]

Thus, d. is the right answer. The reasonable domain of f(s) is [tex]s\geq23[/tex]

Answer:

The difference between the volumes is 728 ft³.

Step-by-step explanation:

Our first step will be to find the volume of the smaller pyramid. Notice that we have all the necessary dimensions. The formula for the volume of a pyramid is

[tex]V = \frac{A_bh}{3},[/tex]

where [tex]h[/tex] stands for the height and [tex]A_b[/tex] for the area of the basis. In this case [tex]h=4 ft[/tex], and the area of the basis, which is a rectangle, is [tex]A_b = 3 ft * 7 ft = 21 ft².[/tex] Then,

[tex]V = \frac{(21 ft²)(4 ft)}{3} = 28 ft³.[/tex]

Now, two calculate the volume of the second pyramid, recall that it has dimensions three times larger. This means, [tex]h=3*4 ft=12 ft[/tex] and [tex]A_b = (3*3 ft) *(3* 7 ft) = 9*21 ft² = 189 ft².[/tex] Then,

[tex]V = \frac{(189 ft²)(12 ft)}{3} = 756 ft³.[/tex]

Finally, we only need to substract the values of the volumes:

756 ft³-28 ft³= 728 ft³.

5/9 / 5/7 =

5/9 * 7/5 = 35/45 reduce to 7/9

In this exercise we have to use the knowledge of cosine quadrants, like this:

Letter B

We have that the quadrant of the cosine is given by:

So we know that:

[tex]cos (2\pi/3)[/tex] If it's in quadrant three, that's negative.

[tex]cos(5\pi/3)[/tex] If you're in quadrant four, that's positive.

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The correct answer is option B. Cosine is negative in the second quadrant and positive in the fourth quadrant.

To determine why [tex]cos(\frac{2\pi }{3})[/tex] is not equal to [tex]cos(\frac{5\pi }{3})[/tex], we can analyze the properties of cosine in different quadrants.

Thus, because cosine is negative in the second quadrant and positive in the fourth quadrant, [tex]cos(\frac{2\pi }{3})[/tex] is not equal to [tex]cos(\frac{5\pi }{3})[/tex]. Therefore, the best explanation is: Cosine is negative in the second quadrant and positive in the fourth quadrant.