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Hunter-Best [27]
3 years ago
5

The graph of a linear equation has a slope of -2 of passes through (-4,3)

Mathematics
1 answer:
vovikov84 [41]3 years ago
7 0

Answer:

I think D. y= -2x+5....

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Solve: (3x^2-y)dx + (4y^3-x)dy =0 and find the solution passing through (1,1).
nordsb [41]

Step-by-step explanation:

The given equation is

(3x^{2}-y)dx+(4y^{3}-x)dy=0\\M(x,y)dx+N(x,y)dy=0

As a check for exactness we have

\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial (4y^{3}-x)}{\partial x} =-1\\\\\frac{\partial M}{\partial y}=\frac{\partial (3x^{3}-y)}{\partial y} =-1\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}=-1

Hence the given equation is an exact differential equation and thus the solution is given by

thus the solution is given by

u(x,y)=\int M(x,y)\partial x+\phi (y)\\\\u(x,y)=\int (3x^{2}-y)\partial x+\phi (y,c)\\\\u(x,y)=x^{3}-xy+\phi (y,c)\\\\

Similarly we have

u(x,y)=\int N(x,y)\partial y+\phi (x,c)\\\\u(x,y)=\int (4y^{3}-x)\partial y+\phi (x,c)\\\\u(x,y)=y^{4}-xy+\phi (x,c)\\\\

Comparing both the solutions we infer

\phi (x,c)=x^{3}+c

Hence the solution becomes

u(x,y)=x^{3}+y^{4}-xy=c

given boundary condition is that it passes through (1,1) hence

1^{3}+1^{4}-1=c\\\\\therefore c=1

thus solution is

u(x,y)=x^{3}+y^{4}-xy=1

4 0
3 years ago
What is x squared divided by x?
GaryK [48]
\frac{x^{2}}{x} =x^{2-1}=x


7 0
3 years ago
Read 2 more answers
Raul deposited $3000 into a bank account that earned simple interest each year. After 3.5 years, he had earned $262.50 in intere
lana [24]

Answer:

75

Step-by-step explanation:

divide the interest by the amounts of years

7 0
3 years ago
Read 2 more answers
All zeros of polynomial function 3x^4 +14x^2 -5
Marizza181 [45]

Answer:

Correct answer:  x₁ = 1 / √3 = √3 / 3  or  x₂ = - 1 / √3 = - √3 / 3

Step-by-step explanation:

Given:

3 x⁴ + 14 x² - 5 = 0   biquadratic equation

this equation is solved by a shift  x² = t and get:

3 t² + 14 t - 5 = 0

t₁₂ = (-14 ± √14² - 4 · 3 · 5) / 2 · 3 = (-14 ± √196 + 60) / 6

t₁₂ = (-14 ± √256) / 6 = (-14 ± 16) / 6

t₁ = -5   or  t₂ = 1 / 3

the solution t₁ = -5 is not accepted because it cannot be x² = -5

we accepted  t₂ = 1 / 3

x² =  1 / 3  ⇒  

x₁ = 1 / √3 = √3 / 3  or  x₂ = - 1 / √3 = - √3 / 3

God is with you!!!

5 0
3 years ago
Angle θ is in standard position. if sin(θ) = − 1/3, and π < θ < 3π/2 , find cos(θ).
scoray [572]
Use pythagorean theorem:
sin^2 + cos^2 = 1

(-1/3)^2 + cos^2 = 1
1/9 + cos^2 = 1
cos^2 = 8/9
cos = +- sqrt(8)/sqrt(9) = +- 2sqrt(2)/3

Determine whether cos is positive or negative by looking at which quadrant the angle is in.
pi < theta < 3pi/2 ---> this is 3rd quadrant where x or cos is negative

Therefore cos(theta) = -2sqrt(2)/3
7 0
3 years ago
Read 2 more answers
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