The coordinates of △ABC△ABC are A(12,8), B(10,18), C(4,16)A(12,8), B(10,18), C(4,16). After a dilation, the coordinates are A'(6,4), B'(5,9), C'(2,8)A′(6,4), B′(5,9), C′(2,8). Find the scale factor.

The area f the helicoid ramp  is:

∫∫A) | r(u) *r( v) | dudv

The solution is:

A = 10.35×π  square units

r ( u , v ) = u×cosv i +  u×sinv j +v k

To get

r (u ) = δ(r ( u , v ) ) / δu   = [ cosv , sinv , 0 ]

r ( v ) = δ(r ( u , v ) ) / δv  = [ -u×sinv , u×cosv , 1 ]

The vectorial product is:

                                 i                    j              k

r (u ) * r ( v )            cosv             sinv           0

                            -u×sinv           u×cosv       1

r (u ) * r ( v )  =  i × ( sinv - 0 )  - j × ( cosv - 0 ) + k ( u×cos²v + u× sin²v )

r (u ) * r ( v )  = sinv i - cosv j + u k

Now

| r (u ) * r ( v ) |  = √sin²v + cos²v + u²  = √ 1 + u²

Then  

A = ∫₀ (9π) dv    ∫₀¹  √ 1 + u²  du    

 ∫₀¹  √ 1 + u²  du    = 1.15  

A = 1.15 × v |( 0  , 9π )

A = 10.35×π  square units

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Answer:

48%

Step-by-step explanation:

In order to find the percentage we need to divide the sold cans by total cans and multiply the result by 100.

Total cans = 150

Sold cans = 72

→ 72/150 = 0.48

→ 0.48 * 100 = 48

The percentage of the cans of popcorn stocked were sold that weekend is 48%

The given parameters are:

Total can of popcorn = 150

Sold can of popcorn = 72

The percentage of can sold is then calculated as:

[tex]\%Sold = \frac{72}{150} *100\%[/tex]

Multiply 72 and 100

[tex]\%Sold = \frac{7200}{150}\%[/tex]

Divide 7200 by 150

[tex]\%Sold = \%48[/tex]

Hence, the percentage of the cans of popcorn stocked were sold that weekend is 48%

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It will takes 3  mexican pesos to buy one brazilian real .

According to the given condition

It takes 0.20 dollars to buy a mexican peso and 0.60 dollars to buy a brazilian real,

We have to determine that  it takes how many mexican pesos to buy one brazilian real.

This question can be solved by  applying the principles of unitary method

One peso will be bought in  $ 0.20

One real will be bought in $ 0.60

1 dollar is equivalent to

[tex]\rm 1 \; dollar = \dfrac{1}{0.2} peso \\\\\rm 1 \; dollar = \dfrac{1}{0.6 } \; real[/tex]

[tex]\rm 1/0.2\; peso = 1/0.6 \; real \\1 \; real = 0.6/0.2 = 3 \; peso[/tex]

So it will take 3 pesos to buy one real

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By setting the given height (6 feet) in the height equation and using the quadratic formula to solve for time 't', we get two solutions. Since the ball reaches 6 feet twice in its ascension and decension, the latter value of t = 3.79 seconds would be the time it is caught.

The question is regarding the time at which a baseball, hit with an initial upward velocity and caught at 6 feet above the ground, is caught. Firstly, input the given height of 6 feet into the height equation h= -16t^2 + 70t + 4 and solve for

t

. Based on the quadratic formula, we receive two solutions: t = 3.79 s and t = 0.54 s. Since the ball has two points at which it reaches the height of 6 feet during its trajectory - once while going up and once while coming down - the time when it is caught would be the larger value,

t = 3.79 s

. Therefore, approximately 3.79 seconds after being hit, the ball is caught.

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We can find the lengths of the legs of the right triangle using the Pythagorean theorem. One leg is x and the other leg is x+20. A quadratic equation can be solved to find x.

The problem involves a right triangle, and we are given the length of the hypotenuse and a relationship between the lengths of the legs. We can solve it using the Pythagorean theorem, which for a right triangle with legs of lengths 'a' and 'b' and hypotenuse 'c' is stated as a² + b² = c².

Let's assign 'x' to the shorter leg. Given that the other leg is 20ft longer, it would be 'x + 20'. The hypotenuse is given as 24, hence the equation becomes: x² + (x + 20)² = 24².

By solving this equation, we find two potential values for 'x', but since a length can't be negative, we exclude the negative value. Hence, the length of the shorter leg is 'x' and of the longer leg is 'x + 20'.

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rugby 7 has 3 out of 5 sold out = 3/5 = 0.6

Junior Athletics has 3 out of 5 sold out = 3/5 = 0.6

Volley ball has 2 out of 5 sold out = 2/5 = 0.4

Volleyball is the answer for the first part

 Part 2 =  3/5 x 3/5  =9/25

The volume [tex]\( V \)[/tex] of the solid [tex]\( S \)[/tex] is 32 cubic units.

To find the volume [tex]\( V \)[/tex] of the solid [tex]\( S \),[/tex] where the base is an elliptical region defined by the equation [tex]\( 4x^2 + 9y^2 = 36 \),[/tex] and the cross-sections perpendicular to the [tex]\( x \)[/tex]-axis are isosceles right triangles with their hypotenuse lying on the base, we proceed as follows:

  The equation [tex]\( 4x^2 + 9y^2 = 36 \)[/tex] represents an ellipse centered at the origin with semi-major axis [tex]\( \sqrt{9} = 3 \)[/tex] along the [tex]\( y \)[/tex]-axis and semi-minor axis [tex]\( \sqrt{4} = 2 \)[/tex] along the [tex]\( x \)[/tex]-axis.

Each cross-section perpendicular to the [tex]\( x \)[/tex]-axis is an isosceles right triangle with its hypotenuse on the elliptical base. The height [tex]\( h(x) \)[/tex] of each triangle at a given [tex]\( x \)[/tex] is determined by the elliptical equation.

  For a fixed [tex]\( x \),[/tex] the corresponding [tex]\( y \)[/tex] values on the ellipse satisfy [tex]\( 4x^2 + 9y^2 = 36 \).[/tex] Solving for [tex]\( y \)[/tex], we get:

[tex]\[ y = \frac{2}{3} \sqrt{36 - 4x^2} \][/tex]

The height  of the triangle is [tex]\( \frac{2}{3} \sqrt{36 - 4x^2} \).[/tex]

  To find the volume [tex]\( V \)[/tex] of the solid [tex]\( S \),[/tex] integrate the area of each triangular cross-section along the [tex]\( x \)[/tex]-axis from [tex]\( x = -3 \) to \( x = 3 \):[/tex]

[tex]\[ V = \int_{-3}^{3} \text{Area of triangle at } x \, dx \][/tex]

The area of each triangle is [tex]\( \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2h(x) \cdot h(x) = h(x)^2 \).[/tex]

Thus,

[tex]\[ V = \int_{-3}^{3} h(x)^2 \, dx = \int_{-3}^{3} \left( \frac{2}{3} \sqrt{36 - 4x^2} \right)^2 \, dx \][/tex]

[tex]\[ V = \int_{-3}^{3} \frac{4}{9} (36 - 4x^2) \, dx \][/tex]

[tex]\[ V = \frac{4}{9} \int_{-3}^{3} (36 - 4x^2) \, dx \][/tex]

[tex]\[ V = \frac{4}{9} \left[ 36x - \frac{4x^3}{3} \right]_{-3}^{3} \][/tex]

Solving further,

[tex]\[ V = \frac{4}{9} \left[ \left( 36 \cdot 3 - \frac{4 \cdot 27}{3} \right) - \left( 36 \cdot (-3) - \frac{4 \cdot (-27)}{3} \right) \right] \][/tex]

[tex]\[ V = \frac{4}{9} \left[ (108 - 36) - (-108 + 36) \right] \][/tex]

[tex]\[ V = \frac{4}{9} \left[ 72 \right] \][/tex]

[tex]\[ V = \frac{4 \cdot 72}{9} \][/tex]

[tex]\[ V = 32 \][/tex]

Therefore, the volume [tex]\( V \)[/tex] of the solid [tex]\( S \)[/tex] is 32 cubic units.

Answer:

Area of an ellipse(a), ,having x and y being the  lengths of the largest and smallest diameters of the ellipse = π xy

  The  lengths of the largest and smallest diameters of the ellipse is called Major Axis and Minor axis of the ellipse.

  [tex]\rightarrow a=\pi x y\\\\\rightarrow y=\frac{a}{\pi \times x}[/tex]

Answer:

6 miles.

Step-by-step explanation:

We have been given that on a city map 2.5 inches represents 5 miles the library and the bank are 3 inches apart on the map. We are asked to find the actual distance between the library and the bank.

[tex]\frac{\text{Actual distance}}{\text{Map distance}}=\frac{\text{5 miles}}{\text{2.5 inches}}[/tex]

[tex]\frac{\text{Actual distance}}{\text{3 inches}}=\frac{\text{5 miles}}{\text{2.5 inches}}[/tex]

[tex]\frac{\text{Actual distance}}{\text{3 inches}}*\text{3 inches}=\frac{\text{5 miles}}{\text{2.5 inches}}*\text{3 inches}[/tex]

[tex]\text{Actual distance}=\frac{\text{5 miles}}{2.5}*3[/tex]

[tex]\text{Actual distance}=\text{2 miles}*3[/tex]

[tex]\text{Actual distance}=\text{6 miles}[/tex]

Therefore, the actual distance between the library and the bank is 6 miles.

Crystal would read 37.5 pages in [tex]\( \frac{3}{4} \)[/tex] hour.

To represent the relationship between the number of pages Crystal reads and the time she spends reading, we can use the formula:

[tex]\[ \text{Pages Read} = \text{Reading Rate} \times \text{Time Spent Reading} \][/tex]

In this case, Crystal reads 25 pages in 1/2 hour, so her reading rate can be calculated as follows:

[tex]\[ \text{Reading Rate} = \frac{\text{Pages Read}}{\text{Time Spent Reading}} \][/tex]

[tex]\[ \text{Reading Rate} = \frac{25 \text{ pages}}{\frac{1}{2} \text{ hour}} \][/tex]

[tex]\[ \text{Reading Rate} = 25 \times 2 \][/tex]

[tex]\[ \text{Reading Rate} = 50 \text{ pages per hour} \][/tex]

Now, we can substitute this reading rate into the equation to represent the relationship:

[tex]\[ \text{Pages Read} = 50 \times \text{Time Spent Reading} \][/tex]

This equation describes the relationship between the number of pages Crystal reads and the time she spends reading.

To illustrate how to use this equation, let's say Crystal reads for [tex]\( \frac{3}{4} \)[/tex] hour. We can plug this value into the equation to find out how many pages she reads:

[tex]\[ \text{Pages Read} = 50 \times \frac{3}{4} \][/tex]

[tex]\[ \text{Pages Read} = 37.5 \][/tex]

So, Crystal would read 37.5 pages in [tex]\( \frac{3}{4} \)[/tex] hour.

The equation of the line that describes the baby's weight is y = (1/2)x + 10.

To find the equation of the line that describes the baby's weight, we need to determine the slope and y-intercept of the line. The slope represents the rate at which the baby's weight increases, and the y-intercept represents the initial weight of the baby.  

Since the baby gains 2 pounds every 4 months, the slope of the line is 2/4 = 1/2. This means that for every month that passes, the baby's weight increases by 1/2 pound.  

To find the y-intercept, we can use the initial weight of the baby, which is 10 pounds. So the equation of the line is y = (1/2)x + 10.

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The difference is a multiple of 9, so it is always divisible by 9.

An expression contains one or more terms with addition, subtraction, multiplication, and division.

We always combine the like terms in an expression when we simplify.

We also keep all the like terms on one side of the expression if we are dealing with two sides of an expression.

Example: so

1 + 3x + 4y = 7 is an expression.com

3 + 4 is an expression.

2 x 4 + 6 x 7 – 9 is an expression.

33 + 77 – 88 is an expression.

We have,

The difference between the larger of the two numbers and the smaller of the two numbers.

A - B or B - A (whichever is greater)

If we transpose the digits of a two-digit number A to form B, then:

A = 10a + b, where a is the tens digit and b is the one's digit

B = 10b + a, where b is the tens digit and a is the one's digit

The difference between the two numbers.

= A - B

= (10a + b) - (10b + a)

= 9a - 9b

= 9(a - b)

or

B - A

= (10b + a) - (10a + b)

= 9b - 9a

= 9(b - a)

Either way, the difference is a multiple of 9, so it is always divisible by 9.

Therefore,

The difference is a multiple of 9, so it is always divisible by 9.

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x+x-1+x+1 =120

3x=120

x=40

40-1=39

40+1 =41

39 +40 +41 = 120

 ages are 39 40 & 41

The arrangement of the 864 cubes would be 18 6 by 8 layers.

All 48 would be touching the bottom of the box on the bottom layer and all 48 would be touching the top of the box on the top layer.

The cubes along both lengths would be touching the sides of the box for the remaining 16 layers. That would be16 cubes per layer or 256 cubes after counting both sides.

The cubes along the width would be touching the ends of the box for those 16 layers. Those need to be eliminated from the count since the corner cubes were already counted as part of the ones touching the sides. 4 cubes have not been previously counted for each width of 6 cubes. Both ends of the 16 layers has 8 cubes per layer or 128 cubes.

Therefore, that is 256 (sides) + 128 (ends) + 48 (top) + 48 (bottom) and that totals to 480 cubes touching the box.

The probability of rolling a sum greater than 7 with a pair of dice is [tex]\( \frac{5}{12} \).[/tex]

To find the probability of getting a sum greater than 7 when rolling a pair of dice, let's first list all the possible outcomes when rolling two dice:

1. (1,1)

2. (1,2)

3. (1,3)

4. (1,4)

5. (1,5)

6. (1,6)

7. (2,1)

8. (2,2)

9. (2,3)

10. (2,4)

11. (2,5)

12. (2,6)

13. (3,1)

14. (3,2)

15. (3,3)

16. (3,4)

17. (3,5)

18. (3,6)

19. (4,1)

20. (4,2)

21. (4,3)

22. (4,4)

23. (4,5)

24. (4,6)

25. (5,1)

26. (5,2)

27. (5,3)

28. (5,4)

29. (5,5)

30. (5,6)

31. (6,1)

32. (6,2)

33. (6,3)

34. (6,4)

35. (6,5)

36. (6,6)

Out of these 36 possible outcomes, the sums greater than 7 are:

12, 17, 18, 22, 23, 24, 27, 28, 29, 30, 32, 33, 34, 35 and 36.

There are 15 favorable outcomes. So, the probability of getting a sum greater than 7 is:

[tex]\[ \frac{15}{36} = \frac{5}{12} \][/tex]

Adding a constant to an exponential function results in shifting the graph vertically or horizontally, depending on where the constant is added. The shift will be positive for a positive constant and negative for a negative constant.

In Mathematics, when dealing with an exponential function, adding a constant can affect the function in two different ways, depending on where the constant is being added. If the constant is added to the exponent, this results in shifting the graph horizontally. However, if the constant is added outside the exponent (as in f(x) = 2x + k), this will result in the entire graph being shifted upward or downward vertically, based on whether the constant is positive or negative.
For example, consider the simple exponential function f(x) = 2x. If a constant 'c' is added - resulting in f(x) = 2x + c, the resulting graph will be the same as the original, but shifted 'c' units upward if 'c' is positive and downward if 'c' is negative. This is a fundamental principle of exponential functions.

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