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3 years ago

The graph shows the total monthly sales as a function of the number of cake tiers. Which statement is true?

Mathematics

2 answers:

liraira [26]3 years ago

6 0

as tiers increases, sales increases

since there seems to be no assemtote (horizontal or vertical), As the number of cake tiers increases without bound, the total monthly sales increase without bound.

answer is 3rd option

kobusy [5.1K]3 years ago

5 0

Answer:

As the number of cake tiers increases without bound, the total monthly sales increase without bound.

Step-by-step explanation:

We can see from the graph, that as the independent variable (number of cake tiers) increases, the dependent variable (the monthly sales) increases.

There is no asymptote shown on the graph; this means that both variables increase without bound.

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I really don't understand how roots work with these. Please help

zheka24 [161]

Answer: This is geometry. Put point C at (-4,-3). Distance from A to C = AC = 5. Distance from C to B = CB = 9. Then ACB is a right triangle with legs AC and CB, and hypotenuse AB, so

AC^2 + CB^2 = AB^2, so

AB = √(AC^2 + CB^2) = √(5^2+9^2) = √(25+81) = √106.

However, in vector algebra, A=(-4,3) and B=(5,-2), and (A-B)=(-9,5).

distance(A,B) = √((A-B) dot (A-B)),

where (x,y) dot (x,y) = x×x + y×y.

So the answer is √((-9,5)dot(-9,5)) = √((-9)^2+5^2) = √(81+25) = √106.

This works in 3D, (x,y,z) dot (x,y,z) = x×x + y×y + z×z. distance(A,B) gives √(x^2+y^2+z^2) for distances in 3 dimensions.

In 4D, distance is √(x^2+y^2+z^2+w^2)

And in infinite dimensional Hilbert space, (x,y,z,a,b,c,...) dot (x,y,z,a,b,c,...) = x×x + ... c×c + .... distance(A,B) gives

√(x^2+...+c^2+...).

And for real valued functions, f(x) dot g(x) is roughly the sum of f(x)×g(x) over uncountably many points x from -infinity to +infinity. It's the area under h(x)=f(x)g(x).

8 0

4 years ago

The data show the average monthly temperatures for two cities over a 6-month period. City 1: {20, 24, 40, 63, 76, 89} City 2: {4

Mrac [35]

Answer:

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

Step-by-step explanation:

For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,

So for city 1:

$Mean = x1 = \frac{20+24+40+63+76+89}{6}$

$x1 = \frac{312}{6}$

$x1 = 52$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)

$20-52 = -32=32\\24-52=-28=28\\40-52=-12=12\\63-52=11\\76-52=24\\89-52=37$

The deviations will be added then. $Mean Absolute Deviation = \frac{32+28+12+11+24+37}{6} \\=\frac{144}{6}\\=24$

So the mean absolute deviation for city 1 is 24 ..

For city 2:

$Mean = x2 = \frac{41+50+58+62+72+83}{6}$

$x2 = \frac{366}{6}$

$x2 = 61$

Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)

$41-61=-20=20\\50-61=-11=11\\58-61=-3=3\\62-61=1\\72-61=11\\83-61=22$

The deviations will be added then. $Mean Absolute Deviation = \frac{20+11+3+1+11+22}{6} \\=\frac{68}{6}\\=11.33$

So the MAD for city 2 is 11.33 ..

So,

The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.

3 0

3 years ago

Read 2 more answers

expeople1 [14]

B 6x - 3

help theres a character limit

7 0

3 years ago

Find the slope of the line that passes through the two given points.

777dan777 [17]

Answer:

12/4 = 3/1

Step-by-step explanation:

x = 7-3=4

y = 14-2=12

3 0

2 years ago

Prove that \P(A) \cup \P(B) \subseteq \P(A \cup B) and find a counter-example to show that we don't always have equality

pentagon [3]

Answer:

P(A) ∪ P(B) ⊆ P(A ∪ B) can be proved when $X$ ∈ P ( A U B )

Step-by-step explanation:

To Prove that P(A) ∪ P(B) ⊆ P(A ∪ B) is attached below and also a counter example to prove that we do not always get an equality is attached below as well

3 0

3 years ago