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2 years ago

1.

2 answers:

8 0

1. The answer is 76.

If the group all together is only being charged once per ride, and the whole group is going twice, then you will double the cost of each ride:

4.50 x 2 = 9
4 x 2 = 8
3.50 x 2 = 7

In total, the rides will cost $24. So, if you're paying with a $100 bill, you will receive $76 in change.

2.2 The answer is 58.

You will do the same procedure as the first question, but this time multiply the costs by 3.

5.50 x 3 = 16.5
2 x 3 = 6
6.50 x 3 = 19.5

The total will be 42. Subtracted from 100, you will get 58.

3.3 The answer is 2.

If you have 15 dollars and the coaster costs 5.50, the first time you ride it you will have 9.50 left. The second time, you will have 4 dollars left, and then you will not be able to pay for a third time.

4.4 The answer is 23.25

If you add up all the correlating costs of the tickets (6.75 + 4.50 + 4.50 + 10.50 + 10.50), you will get 36.75. Then you must subtract that from 60, which will give you 23.25 dollars in change.

5.5 The answer is 54.75

You will do the same procedure here as you did above. Add the corresponding costs together (you're done after this part, because the question isn't asking for how much change you'll receive).

6.6 The answer is 7.

If the tickets cost $10.50 and you have $80, then you want to divide 80 by 10.50. You will get 7.6. Since you can't buy a sixth of a ticket, your answer will be 7.

7.7 The answer is to pay separately.

In this question, you will add all the costs together (except the day pass) and compare it to the cost of the day pass. Once you add all the costs together, you will get 17.50. This is cheaper than the day pass, which is 19.95.

zheka24 [161]2 years ago

3 0

I took the test

1. 4

2. 168

3. 2

4. 23.25

5. 54.75

6. 7

7. Pay separately

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Answer: Check the explanation for the solution.

Step-by-step explanation:

Given that the garden is a right triangle with base 25m start and height 30 m

The length of the hypothenus can be achieved by using pythagorean theorem.

Please find the attached file for the diagram.

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2 years ago

The amphitheater has two types of tickets available, reserved seats and lawn seats. The maximum capacity of the venue is 20,000

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Let the number of reserved tickets = x

Let the number of lawn seats = y

Constraint functions:

Maximum capacity means $x+y\leq 20000$

For concert to be held $x+y\geq 5000$

$lawn seats\leq reserved$ means $y\leq x$

Objective functions :

Maximum profit equation p = 65x +40y

Intersection points :

(10000,10000) (20000,0)(2500,2500)(5000,0)

p at (10000,10000) = 65(10000) + 40(10000) = $1050000

p at (20000,0) = 65(20000) + 40(0) = $1300000

p at (2500,2500) = 65(2500) + 40(2500) = $262500

p at (5000,0) = 65(5000) + 40(0) = $325000

Hence maximum profit occurs when all 20000 reserved seats are sold and the profit is $1300000

Please find attached the graph of it.

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2 years ago

Suppose that five ones and four zeros are arranged around a circle. Between any two equal bits you insert a 0 and between any tw

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Answer:

Using backward reasoning we want to show that "We can never get nine 0's".

Step-by-step explanation:

Basically in order to create nine 0's, the previous step had to have all 0's or all 1's. There is no other way possible, because between any two equal bits you insert a 0.

If we consider two cases for the second-to-last step:

There were 9 0's:

We obtain nine 0's if all bits in the previous step were the same, thus all bit were 0's or all bits were 1's. If the previous step contained all 0's, then we have the same case as the current iteration step. Since initially the circle did not contain only 0's, the circle had to contain something else than only 0's at some point and thus there exists a point where the circle contained only 1's.

There were 9 1's:

A circle contains only 1's, if every pair of the consecutive nine digits is different. However this is impossible, because there are five 1's and four 0's (we have an odd number of bits!), thus if the 1's and 0's alternate, then we obtain that 1's that will be next to each other (which would result in a 1 in the next step). Thus, we obtained a contradiction and thus assumption that the circle contains nine 0's after iteratins the procedure is false. This then means that you can never get nine 0's.

To summarize, in order to create nine 0's, the previous step had to have all 0's or al 1's. As we didn't start the arrange with all 0's, the only way is having all 1's, but having all 1's will not be possible in our case since we have an odd number of bits.

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Which statements about the graph of the function f(x) = –x2 – 4x + 2 are true? Select three options.

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Answer:

The true statements about the graph of the function are:

The range of the function is {y : y ≤ 6} ⇒ 2nd

The function is increasing over the interval (–∞ , –2) ⇒ 3rd

The function has a positive y-intercept ⇒ 5th

Step-by-step explanation:

* Lets explain how to solve the problem

- The function f(x) = -x² - 4x + 2 is a quadratic function represented

graphically by a downward parabola with maximum vertex

∵ The form of the quadratic function is f(x) = ax² + bx + c

∵ The coordinates of the vertex of the parabola are (h , k) where

h = -b/2a and k = f(h)

∵ a = -1 , b = -4

∴ h = -(-4)/2(-1) = 4/-2 = -2

∵ k = f(h)

∴ k = -(-2)² - 4(-2) + 2 = -(4) - (-8) + 2

∴ k = -4 + 8 + 2 = 6

∴ The vertex of the parabola is (-2 , 6)

* Look to the attached figure to find the true statements

∵ The greatest value of the function is 6 ⇒ y-coordinate of the vertex

∵ The range of the function is the values y-coordinates of the points

on the parabola

∴ The range of the function is {y : y ≤ 6}

- The domain of the function is {x : x ∈ R} or (-∞ , ∞)

∵ The value of y increasing after -∞ to the x-coordinate of the vertex

∵ x-coordinate of the vertex is -2

∴ The function is increasing over the interval (–∞ , –2)

- The parabola intersect the y-axis at point (0 , 2)

∵ The y-intercept is the intersection between the parabola and the

y-axis

∵ The parabola intersect the y-axis at point (0 , 2)

∴ The y-intercept is 2

∴ The function has a positive y-intercept

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