A formula that approximates​ real-world phenomena is call​ a/an

A square pyramid's height is three times the side of the base of a cube if they have the same base and volume, due to their differing volume formulas.

If a square pyramid and a cube have the same base and volume, the relationship between their heights can be determined using the formulas for their volumes. The volume of a cube is given by [tex]V = s^3,[/tex] where s is the length of a side of the cube's base. The volume of a square pyramid is [tex]V = (1/3) * b^2 * h,[/tex] where b is the length of the base side and h is the pyramid's height. Since both have the same volume, we can set these equations equal to each other and solve for h.

Setting the volumes equal: [tex]s^3 = (1/3) * s^2 * h,[/tex] simplifying to h = 3s. Thus, the height of the pyramid is three times the side of the base of the cube. This is because a cube's volume grows as a cube of its side length, while a pyramid's volume is one third of the base area times the height. Therefore, since they have the same volume, the height of the pyramid must be three times larger to compensate for the 1/3 factor in its volume formula.

Answer:

5

Step-by-step explanation:

A little trick: the imaginary part ([tex]i[/tex]) in the coordinates of complex numbers act same as the y-coordinate in a rectangular plane.

So we want the distance between the coordinate points (4,2) and (7,-2).

How do we get the distance between 2 points? We use the distance formula.

Distance Formula = [tex]\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]

Where [tex](x_1,y_1)=(4,2)[/tex]  and  [tex](x_2,y_2)=(7,-2)[/tex]

Let's solve for distance:

[tex]\sqrt{(-2-2)^2+(7-4)^2}\\ =\sqrt{(-4)^2+(3)^2} \\=\sqrt{16+9} \\=\sqrt{25} \\=5[/tex]

Hence, the length of the segment in the complex plane with endpoints at 4 + 2i and 7 – 2i is 5.

a. The lowest total number of cars is ( x = 3 ) Fiats and ( y = 7 ) Rolls Royces, totaling 10 cars.

b. the cheapest solution is to use 3 Fiats and 7 Rolls Royces, costing £240.

To solve this problem, we need to use a system of linear equations and optimization techniques.

Let's define ( x ) as the number of Fiats and ( y ) as the number of Rolls Royces.

The constraints are:

1. People constraint: [tex]\( 2x + 2y \geq 20 \)[/tex] (at least 20 people)

2. Sheep constraint: [tex]\( x + 4y \geq 32 \)[/tex] (at least 32 sheep)

3. Car constraint: [tex]\( x + y \leq 15 \)[/tex] (no more than 15 cars)

We also want to minimize the total number of cars, so the objective function is simply ( x + y ).

(a) To find the lowest total number of cars, we'll solve the system of equations:

[tex]\[\begin{cases}2x + 2y \geq 20 \\x + 4y \geq 32 \\x + y \leq 15\end{cases}\][/tex]

We solve these inequalities by  algebraically:

1. Solve the first two equations simultaneously:

 [tex]\[ \begin{cases} 2x + 2y = 20 \\ x + 4y = 32 \end{cases} \][/tex]

  Multiply the second equation by 2 to match coefficients with the first equation:

 [tex]\[ \begin{cases} 2x + 2y = 20 \\ 2x + 8y = 64 \end{cases} \][/tex]

Subtract the first equation from the second:

[tex]\[ \begin{aligned} (2x + 8y) - (2x + 2y) &= 64 - 20 \\ 6y &= 44 \\ y &= \frac{44}{6} \\ y &= 7.33 \end{aligned} \][/tex]

However, ( y ) must be a whole number, so we take ( y = 7).

Substitute ( y = 7 ) into the first equation:

  [tex]\[ \begin{aligned} 2x + 2(7) &= 20 \\ 2x + 14 &= 20 \\ 2x &= 6 \\ x &= 3 \end{aligned} \][/tex]

Therefore, the lowest total number of cars is ( x = 3 ) Fiats and ( y = 7 ) Rolls Royces, totaling 10 cars.

(b) To find the cheapest solution, we calculate the cost for each option:

Cost of Fiats: £10 x 3 = £30

Cost of Rolls Royces:  £30 x 7 = £210

Total cost:  £30 + £210 = £240

So, the cheapest solution is to use 3 Fiats and 7 Rolls Royces, costing £240.

Complete Question:

A farmer has to transport 20 people and 32 sheep to a market. He can use either Fiats (x) which take 2 people and 1 sheep, or Rolls Royces ( y) which take 2 people and 4 sheep. He must not use more than 15 cars altogether.

(a) What is the lowest total numbers of cars he could use?

(b) If it costs £10 to hire each Fiat and £30 for each Rolls Royce, what is the cheapest solution?

Answer:

6

Step-by-step explanation:

If he sold half of 6 thats 3, then he bought 8 more that equals 11.

Hope that helps!

Answer:

Multiply the Rate by the Time

Step-by-step explanation:

Hiya! So you would multiply R(rate) by T(time) Then you would divide by I(interest) and you should get your answer for P(principle)

Ex 24= P x 0.04 x 3   0.04 x 3= 0.12   24 divided by 0.12 = 200 So 200= P

                  (4%)

Hope this helped you!

Answer:

-9814

Step-by-step explanation:

P.E.M=D.A=S- The order of operations

P.erenthisis

E.xponent

M.ultiplication

D.ivision

A.ddition

S.ubtraction

first multiplication, -99•99, = -9801

now we work left to right, as subtraction is the same step as subtraction.

89-9801=9712

-9712-102=-9814

Answer:

[tex]5.24-8.85=-3.61[/tex]

Step-by-step explanation:

We have been given an equation. We are asked to fill in the blanks for the given equation.

5._4 - (_.8_) = -3.61

We will find each part separately as:

For hundredths digit:

[tex]4-?=-1[/tex]  

[tex]4-4-?=-1-4[/tex]

[tex]-?=-5[/tex]

[tex]?=5[/tex]  

So hundredths digit would be 5.

Tenth digit:

[tex]?-8=-6[/tex]

[tex]?-8+8=-6+8[/tex]

[tex]?=2[/tex]

Ones digit:

[tex]5-?=-3[/tex]

[tex]5-5-?=-3-5[/tex]

[tex]-?=-8[/tex]

[tex]?=8[/tex]

Substitute the values in given equation.

5.24 - (8.85) = -3.61

We can verify our answer by subtracting 8.85 from 5.24.

[tex]5.24-8.85=-3.61[/tex]

Therefore, our equation would be [tex]5.24-8.85=-3.61[/tex].

To describe the dimensions of the frame, we can represent the width as x inches and the length as (x+3) inches. The perimeter is less than 52 inches. We can solve an inequality using this information to find the possible values for the width.

To describe the dimensions of the picture frame, let's represent the width of the frame as x inches. According to the information given, the length of the frame is 3 inches greater than the width, so it can be represented as (x+3) inches. The perimeter of a rectangle is given by the formula: Perimeter = 2(width + length). Since the perimeter is less than 52 inches, we can form the following inequality: 2(x + (x+3)) < 52. Solving this inequality will give us the range of possible values for the width, and we can then determine the corresponding length of the frame.

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To increase the probability from 1/5 to 1/4, we need to remove 24 blue marbles from the bag.

To increase the probability from 1/5 to 1/4, we need to find the number of blue marbles that need to be removed. Let's assume there are x blue marbles in the bag. Initially, the total number of marbles in the bag is 30, and the probability of picking a red marble is 1/5. This means that (30 - x)/30 = 1/5. Solving this equation, we get x = 24. Therefore, 24 blue marbles need to be removed from the bag to increase the probability to 1/4.

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Because the initial probability of picking a red marble from 30 marbles is 1/5, there are 6 red marbles in the bag. If we want to increase the probability of drawing a red marble to be 1/4, we need to reduce the total number of marbles in the bag to 24. So, we must remove 6 non-red marbles, which could be blue.

The subject of your question is probability in Mathematics. Initially, we have 30 marbles, and the probability of picking a red one is 1/5. So, there are 1/5 * 30 = 6 red marbles in the bag. Thus, the bag also contains 24 non-red marbles, which can be blue or other colors.

Next, we want the probability of picking a red marble to be 1/4. Since the number of red marbles will remain the same, we need to change the total number of marbles in the bag. If 'x' is the number of marbles in the bag, then 6/x = 1/4, which means x = 24. Since initially we had 30, to get to 24 we need to remove 6 marbles.

Therefore, if we remove 6 blue marbles (or any non-red marbles), the probability of picking a red marble will increase to 1/4.

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The scale factor is 3.5 and the length of the banner is 84 inches. Then the correct option is A.

Dilation means the changing of the size of the object without changing the shape. The size of the object may be increased or decreased based on the scale factor.

A picture of a school’s mascot is 18 in. wide and 24 in. long.

It is enlarged proportionally to banner size.

If the width is enlarged to 63 in.

Then the scale factor will be

[tex]\rm S.F. = \dfrac{63}{18}\\\\S.F. = 3.5[/tex]

Then the length of the banner will be

[tex]\rm Banner \ length = S.F. \times 24\\\\Banner \ length = 3.5 \times 24\\\\Banner \ length = 84[/tex]

The length of the banner is 84 inches.

More about the dilation link is given below.

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