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crimeas [40]
3 years ago
14

The municipal swimming pool in Charlotte, North Carolina has three different ways of paying for individual open swimming. Nick i

s trying to decide which way to pay. • Early Pay: Pay $38.00 before Memorial Day; swim any number of days • Deposit Plus: $13.00 deposit plus $3.00 per day • Daily Pay: $5.00 per day
Mathematics
2 answers:
alekssr [168]3 years ago
8 0

Answer:

Table of swimming costs using the three different payment methods

Payment Method        Number of days

                           0 days  5 days  10 days  15 days  20 days

Early Pay           $38.00  $38.00  $38.00  $38.00    $38.00

Deposit Plus      $13.00  $28.00  $43.00  $58.00     $73.00

Daily Pay           $00.00  $25.00  $50.00  $75.00   $100.00

Mumz [18]3 years ago
4 0

Answer:

$13.00 deposit plus $3 per day.

Step-by-step explanation:

Nick is deciding to acquire membership in municipal swimming pool in Charlotte, North Carolina. There are three different ways for the payment. The lump sum initial is $38 which is difficult for Nick to finance. Also if Nick is unable to go for swim he will lose $38. The per day deposit is suitable option for Nick as he will only have to pay when he goes for swim.

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A triangle has vertices at A(–2, 4), B(–2, 8), and C(6, 4). If A prime has coordinates of (–0. 25, 0. 5) after the triangle has
babunello [35]

The true statements about the dilation are as follows;

The scale factor is One-eighth

The coordinates of the C prime are (0.75, 0.5).

The coordinates of B prime are (–0.25, 1).

Given

A triangle has vertices at A(–2, 4), B(–2, 8), and C(6, 4).

If A prime has coordinates of (–0. 25, 0. 5) after the triangle has been dilated with a center of dilation about the origin.

<h3>What is transformation?</h3>

Transformation is the movement of a point from its initial location to a new location. If an object is transformed then all its points are also transformed. Types of transformation are reflection, dilation, rotation, and translation.

Dilation is the enlargement or reduction in the size of an object. If a point X(x, y)  is dilated about the origin by a factor k, its new location is X'(kx, ky).

Triangle ABC with vertices at A(–2, 4), B(–2, 8), and C(6, 4) is dilated to give A'(-0.25, 0.5)

kA = A'

(-2k, 4k) = (-0.25, 0.5)

-2k = -0.25, hence k = 1/8

Therefore the scale factor is 1/8 (one- eight)

The new location of the vertices is B'(-0.25, 1), C'(0.75, 0.5)

To know more about Transformation click the link given below.

brainly.com/question/11709244

3 0
2 years ago
Justin is constructing a line through point Q that is perpendicular to line n. He has already constructed the arcs shown. He pla
krek1111 [17]
<span>B. It must be the same as when he constructed the arc centered at point A. This problem would be a lot easier if you had actually supplied the diagram with the "arcs shown". But thankfully, with a few assumptions, the solution can be determined. Usually when constructing a perpendicular to a line through a specified point, you first use a compass centered on the point to strike a couple of arcs on the line on both sides of the point, so that you define two points that are equal distance from the desired intersection point for the perpendicular. Then you increase the radius of the compass and using that setting, construct an arc above the line passing through the area that the perpendicular will go. And you repeat that using the same compass settings on the second arc constructed. This will define a point such that you'll create two right triangles that are reflections of each other. With that in mind, let's look closely at your problem to deduce the information that's missing. "... places his compass on point B ..." Since he's not placing the compass on point Q, that would imply that the two points on the line have already been constructed and that point B is one of those 2 points. So let's look at the available choices and see what makes sense. A .It must be wider than when he constructed the arc centered at point A. Not good. Since this implies that the arc centered on point A has been constructed, then it's a safe assumption that points A and B are the two points defined by the initial pair of arcs constructed that intersect the line and are centered around point Q. If that's the case, then the arc centered around point B must match exactly the setting used for the arc centered on point A. So this is the wrong answer. B It must be the same as when he constructed the arc centered at point A. Perfect! Look at the description of creating a perpendicular at the top of this answer. This is the correct answer. C. It must be equal to BQ. Nope. If this were the case, the newly created arc would simply pass through point Q and never intersect the arc centered on point A. So it's wrong. D.It must be equal to AB. Sorta. The setting here would work IF that's also the setting used for the arc centered on A. But that's not guaranteed in the description above and as such, this is wrong.</span>
8 0
3 years ago
Read 2 more answers
Help pleaseeeeeeeeeeeeeeeeeeeeeee
bixtya [17]

Answer:  \bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}

<u>Step-by-step explanation:</u>

(1)           (12, 18, 27, ...)

The common ratio is:

r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\  r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}

(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad  r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\  r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}

3 0
3 years ago
Graph the system of equations on graph paper to answer the question. Y=2/5x+4 and y=2x+12 what is the solution for this system o
riadik2000 [5.3K]

Answer:

Solution of the given equations is (-5,2)

Step-by-step explanation:

We have been given the following system of equations:

y = (2/5)x + 4

y = 2x + 12

The solution of the system of equation can be found by finding the point of intersection of both lines on the graph. The graph of both equations is attached below.

As we can see that both the lines intersect each other at one point, and that point is (-5,2). So the solution of the given equations is (-5,2)

6 0
3 years ago
A carpenter is constructing staircase in a house. The distance from the floor to the basement is 10.8 feet. The staircase will b
konstantin123 [22]

Answer:

\theta = 40.59^\circ

Step-by-step explanation:

Given

The question is illustrated with the attached image

Required

Find \theta

To do this, we use sine formula

i.e.

\sin(\theta) = \frac{Opposite}{Hypotenuse}

So, we have:

\sin(\theta) = \frac{10.8}{16.6}

\sin(\theta) = 0.6506

Take arc sin

\theta = \sin^{-1}(0.6506)

\theta = 40.59^\circ

4 0
2 years ago
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