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liubo4ka [24]
3 years ago
7

The graph of the sales of a new product at a candy store, where x is time in months and y is number sold in hundreds, goes throu

gh the points (4, 2) and (6, 8). What is the rate of change?
Mathematics
1 answer:
tamaranim1 [39]3 years ago
3 0

Answer:

Step-by-step explanation:

the rate of change = difference in y / difference in x

for points (4, 2) and (6, 8)

the rate of change =( 2-8) /(4-6) = -4 / -2 = 2

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Verify that the conclusion of Clairaut’s Theorem holds, that is, uxy = uyx, u=tan(2x+3y)
choli [55]

Answer: Hello mate!

Clairaut’s Theorem says that if you have a function f(x,y) that have defined and continuous second partial derivates in (ai, bj) ∈ A

for all the elements in A, the, for all the elements on A you get:

\frac{d^{2}f }{dxdy}(ai,bj) = \frac{d^{2}f }{dydx}(ai,bj)

This says that is the same taking first a partial derivate with respect to x and then a partial derivate with respect to y, that taking first the partial derivate with respect to y and after that the one with respect to x.

Now our function is u(x,y) = tan (2x + 3y), and want to verify the theorem for this, so lets see the partial derivates of u. For the derivates you could use tables, for example, using that:

\frac{d(tan(x))}{dx} = 1/cos(x)^{2} = sec(x)^{2}

\frac{du}{dx}  =  \frac{2}{cos^{2}(2x + 3y)} = 2sec(2x + 3y)^{2}

and now lets derivate this with respect to y.

using that \frac{d(sec(x))}{dx}= sec(x)*tan(x)

\frac{du}{dxdy} = \frac{d(2*sec(2x + 3y)^{2} )}{dy}  = 2*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*3 = 12sec(2x + 3y)^{2}tan(2x + 3y)

Now if we first derivate by y, we get:

\frac{du}{dy}  =  \frac{3}{cos^{2}(2x + 3y)} = 3sec(2x + 3y)^{2}

and now we derivate by x:

\frac{du}{dydx} = \frac{d(3*sec(2x + 3y)^{2} )}{dy}  = 3*2sec(2x + 3y)*sec(2x + 3y)*tan(2x + 3y)*2 = 12sec(2x + 3y)^{2}tan(2x + 3y)

the mixed partial derivates are equal :)

7 0
3 years ago
The price of the book is $4. which expression can be used to show how much money the library will make if it sells 289 books
kogti [31]
If 1 book costs $4 then 289 books will make 289 times as much
so...
(price of book) x (number of books sold) = (amount of money made)
in this case...
4 x 289 = 1156
4 0
3 years ago
368.54 = ( _ x 100) + ( _ x 10) + ( _ x 1/10) + ( _ x 1/100)
SCORPION-xisa [38]

Answer:

368.54 = 3*100 + 6.8*10 +5*1/10 + 4*100

Step-by-step explanation:

Given

368.54 = ( [\ ] * 100) + ( [\ ] * 10) + ( [\ ] * 1/10) + ( [\ ] * 1/100)

Required

Fill in the gap

The number can be expressed as:

368.54 = 300 + 60 + 8 +0.5 + 0.04

Factorize:

368.54 = 3*100 + 6*10 + 0.8*10 +5*1/10 + 4*100

Factor out 10 in 6 * 10 + 0.8 * 10

368.54 = 3*100 + [6 + 0.8]*10 +5*1/10 + 4*100

368.54 = 3*100 + 6.8*10 +5*1/10 + 4*100

<em><u>Solved</u></em>

4 0
3 years ago
Given the sequence −2,−8/3,−103,−4,−14/3,... find the recursive formula.
Alinara [238K]

I suppose the third term should say -10/3, not -103.

Notice that

-2 = -6/3

-4 = -12/3

so that, starting with the first term <em>a</em>(1) = -6/3, the every following term is obtained by subtracting 2/3.

-2 - 2/3 = -6/3 - 2/3 = -8/3

-8/3 - 2/3 = -10/3

-10/3 - 2/3 = -12/3 = -4

and so on.

So the recursive rule is

<em>a</em>(1) = -2,

<em>a</em>(<em>n</em> + 1) = <em>a</em>(<em>n</em>) - 2/3, for <em>n</em> ≥ 1

or C.

4 0
3 years ago
Select the correct answer from each drop-down menu. The equation (y-2)^2/3^2 - (x-2)^2/4^2=1 represents a hyperbola whose foci a
aev [14]

Answer:

The foci are (2 , 7) and (2 , -3)

Step-by-step explanation:

* lets revise the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the y-axis is

  (y - k)²/a² - (x - h)²/b² = 1

- The coordinates of the vertices are ( h ± a , k )  

- The coordinates of the co-vertices are ( h , k ± b )

- The coordinates of the foci are (h , k ± c), where c² = a² + b²

* Now lets solve the problem

∵ The equation of the hyperbola of vertex (h , k) is

    (y - k)²/a² - (x - h)²/b² = 1

∵ The equation is (y - 2)²/3² - (x - 2)²/4² = 1

∴ k = 2 , h = 2 , a = 3 , b = 4

∵ The foci of it are (h , k + c) and (h , k - c)

- Lets find c from the equation c² = a² + b²

∵ a = 3

∴ a² = 3² = 9

∵ b = 4

∴ b² = 4² = 16

∴ c² = 9 + 16 = 25

∴ c = √25 =  5

- Lets find the foci

∵ The foci are (h , k + c) and (h , k - c)

∵ h = 2 , k = 2 , c = 5

∴ The foci are (2 , 2 + 5) and (2 , 2 - 5)

∴ The foci are (2 , 7) and (2 , -3)

5 0
4 years ago
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