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3 years ago

Consider the differential equation: dy/dx = (y-2)(x^2+1)

Mathematics

1 answer:

Klio2033 [76]3 years ago

3 0

Answer:

Step-by-step explanation:

Given,

$\dfrac{dy}{dx}=(y-2)(x^2+1)$ ...(i)

Differentiating w.r. to x.

$\dfrac{d^2y}{dx^2}= (y-2)(2x)+ (x^2+1)\dfrac{dy}{dx}$

From equation (1)

$\dfrac{d^2y}{dx^2}= (y-2)(2x)+ (x^2+1)^2(y-2)$

Now, at the point (1,3)

$\dfrac{d^2y}{dx^2}= (3-2)(2\times 1)+ (1^2+1)^2(3-2)$

$\dfrac{d^2y}{dx^2}=6$

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Jenny and Natalie are selling cheesecakes for a school fundraiser. Customers can buy chocolate cakes and vanilla cakes. Jenny so

IrinaVladis [17]

The cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

Solution:

Let "c" be the cost of 1 chocolate cake

Let "v" be the cost of 1 vanilla cake

Jenny sold 14 chocolate cakes and 5 vanilla cakes for 119 dollars

Therefore, we can frame a equation as:

14 x cost of 1 chocolate cake + 5 x cost of 1 vanilla cake = 119

$14 \times c + 5 \times v=119$

14c + 5v = 119 ------- eqn 1

Natalie sold 10 chocolate cakes and 10 vanilla cakes for 130 dollars

Therefore, we can frame a equation as:

10 x cost of 1 chocolate cake + 10 x cost of 1 vanilla cake = 130

$10 \times c + 10 \times v = 130$

10c + 10v = 130 -------- eqn 2

Let us solve eqn 1 and eqn 2

Multiply eqn 1 by 2

28c + 10v = 238 ------ eqn 3

Subtract eqn 2 from eqn 3

28c + 10v = 238

10c + 10v = 130

( - ) --------------------------

18c = 108

c = 6

Substitute c = 6 in eqn 1

14(6) + 5v = 119

84 + 5v = 119

5v = 119 - 84

5v = 35

v = 7

Thus cost of 1 chocolate cake is $ 6 and cost of 1 vanilla cake is $ 7

8 0

3 years ago

a regular rectangular pyramid has a base and lateral faces that are congruent equilateral triangles. it has a lateral surface ar

Margarita [4]

A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)

Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.

The lateral faces are equilateral triangles of side length x.

The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.

now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.

by the Pythagorean theorem:

$h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x$

thus,

$Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2$

thus:

$x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\ \sqrt{3} }{3}=24 \sqrt{3}$ (cm^2)

but $x^{2}$ is exactly the base area, since the base is a square of sidelength = x cm.

So, the total surface area = base area + lateral area = $24 \sqrt{3}+72$ cm^2

Answer: $24 \sqrt{3}+72$ cm^2

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3 years ago

Solve the system using substitution. Show your work. x = -4y + 9 x - y = 4

FromTheMoon [43]

Answer:

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3 years ago

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The sum of three and four tenths and a number is six and six tenths.

Solnce55 [7]

Hopefully I drew what you described.

The 6.6 represents the total of N + 3.4
We subtract the part we know from the whole /sum/total we know to get the other part.
So 6.6 - 3.4 = N
3.2 = N

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3 years ago

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Let g(x)=5x-1 and h(x)=X^2-1 Solve: g(h(x))=74

AURORKA [14]

Answer:

Let's replace the h(x) function in g(x) and then use 74 as a result on the axis y. The correct answer is 4 .

4 0

3 years ago