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cestrela7 [59]
3 years ago
8

A comedy show sells lower level tickets for $77 and upper level tickets for $99. On the opening night 2615 tickets were sold for

a total $247071.
Mathematics
2 answers:
babunello [35]3 years ago
7 0

Let x represent lower level tickets that cost $77

Let y represent upper level tickets that cost $99

Cost equation: 77x + 99y = 247,071

Tickets equation: x + y = 2615

Using the elimination method, multiply the second equation by -77:

77x + 99y = 247,071

-77x - 77y = -201,355

--> 22y = 45,716

--> y = 2,078

Now plug "y" into either equation and solve for "x". I chose the Tickets equation. x + y = 2615 → x = 2615 - y → x = 2615 - 2078 → x = 537

Answer: lower level = 537 tickets, upper level = 2078 tickets.


rusak2 [61]3 years ago
5 0

Let U = upper

Let L = lower

U + L = 2615

99U + 77L = 247071 You can make your life slightly easier by dividing this equation through by 11

9U + 7L =24461 Multiply the first equation by 7

7U + 7L = 2615 * 7

7U +7L = 18305

====================

Combine the 2 new equations

9U + 7L = 24461

<u>7U + 7L = 18305</u> Subtract

2U = 6156 Divide by 2

U = 6156 / 2

U = 3078

The Number of Upper Level Tickets sold was 3078

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Which description does NOT guarantee that a quadrilateral is a square?
Ivahew [28]
Let's go through the choices one by one

------------------------------------------
Choice A

If all sides are congruent, then this figure is a rhombus (by definition). If all angles are congruent, then we have a rectangle. Combine the properties of a rhombus with the properties of a rectangle and we have a square.

In terms of "algebra", you can think
rhombus+rectangle = square

Or you can draw out a venn diagram. One circle represents the set of all rhombuses; another circle represents the set of all rectangles. The overlapping region is the set of all squares. The overlapping region is inside both circles at the same time.

So we can rule out choice A. This guarantees we have a square when we want something that isn't a guarantee.

------------------------------------------
Choice B

If we had a parallelogram with perpendicular diagonals, then we can prove that we have a rhombus (all four sides congruent). However, we don't know anything about the four angles of this parallelogram. Are they congruent? We don't know. So we can't prove this figure is a rectangle. The best we can say is that it's a rhombus. It may or may not be a rectangle. There isn't enough info about the rectangle & square part.

This is why choice B is the answer. We have some info, but not enough to be guaranteed everytime.

------------------------------------------
Choice C

This is a repeat of choice A. Having "all right angles" is the same as saying "all angles congruent". This is because "right angle" is the same as saying "90 degrees". So we can rule out choice C for identical reasons as we did with choice A.

------------------------------------------
Choice D

As mentioned before in choice A, if we know that a quadrilateral is a rectangle and a rhombus at the same time, then the figure is also a square. This is always true, so we are guaranteed to have a square. We can cross choice D off the list.

------------------------------------------

Once again, the final answer is choice B


3 0
3 years ago
Can you please explain this?​
Leni [432]

Answer:

The question is giving you pairs of points in space which can be used to define lines. It is then asking you to determine if the lines defined by those points are parallel, perpendicular, or neither.

Step-by-step explanation:

Two key things you need to know to solve this is that the lines will be parallel if their slopes are the same, and perpendicular if one slope is the negative reciprocal of the other (i.e. s_{1} = -s_{2}^{-1})

Let's start with question 11. You are given two pairs of points, each of which describes a distinct line:

(3,5)-(1,1) and (0, 2)-(5, 12)

To find the slope of each pair, take the vertex with the lesser x co-ordinate, and subtract it from the vertex with the greater x co-ordinate.  That will give us a valid Δx and Δy to get the slope.

In the first pair, 3 > 1, so we'll subtract the second point from the first:

s = \Delta y / \Delta x\\s = \frac{5 - 1}{3 - 1}\\s = 4/2\\s = 2

So the first pair of vectors describe a line with a slope of 2.  Let's look at the other pair:

s = \Delta y / \Delta x\\s = (12 - 2) / (5 - 0)\\s = 10 / 5\\s = 2

That also gives us a slope of 2, meaning that the two lines are parallel.

This same process will need to be done for the other three questions.  We can't answer questions 12 or 14 here, as the last point is cut off on the edge of the image.  For question 3 though, one line has a slope of 7/3, and the other 3/7. That puts them in the "neither" category, as one is not the negative reciprocal of the other, but instead the positive reciprocal.

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The answer to this question is 5.
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You had 250 minutes left on your cellphone and you talk an hour a week
hjlf
Let w = the number of weeks you talked on your cell phone

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Answer:

$50 per hour

Step-by-step explanation:

$10+$40=$50

8 0
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