suppose the university bursar said that 600 student entered for business mathematics 300 entered for statistics while 173 students entered for both business mathematics and statistics how many student enrolled in the two courses altogether

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:

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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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:

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.

Final answer:

To factor quadratic expressions in general, we can use the AC Method, which involves breaking down the middle term and finding two numbers that multiply to give a certain product and add up to the coefficient of the middle term.

Explanation:

To factor quadratic expressions in general, we can use the method known as the AC Method. This method involves breaking down the middle term of the quadratic expression and finding two numbers that multiply to give a certain product (the product of the coefficient of the leading term and the constant term) and add up to the coefficient of the middle term. Once we find these two numbers, we can rewrite the quadratic expression as a product of two binomials.

For example, let's say we have the quadratic expression [tex]2x^2[/tex] + 5x + 3. The product of the leading coefficient (2) and the constant term (3) is 6. We need to find two numbers that multiply to 6 and add up to the coefficient of the middle term (5). The numbers 3 and 2 fit this criteria. Therefore, we can rewrite the quadratic expression as (2x + 3)(x + 1).

Note that if the quadratic expression cannot be factored using this method, it is considered a prime polynomial and cannot be factored further.

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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5/9 / 5/7 =

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

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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