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3 years ago

12

If you were to use the substitution method to solve the following system, choose the new equation after the expression equivalen

t to y from the first equation is substituted into the second equation.
6x − y = 1
4x − 3y = −11

1 answer:

Korolek [52]3 years ago

3 0

Answer:x=1, y=5

Step-by-step explanation:

You might be interested in

If dB=4 and dc=6 find ad

Neporo4naja [7]

See the attached figure
DB = 4 and DC = 6 , We need to find AD

Using <span>Euclid's theorem for the right triangle
</span><span>
</span><span>∴ DB² = AD * DC
</span><span>
</span><span>∴ 4² = AD * 6
</span><span>
</span><span>∴ 6 AD = 16
</span><span>
</span><span>
</span><span>∴ AD = 16/6 = 8/3 ≈ 2.67
</span>

7 0

2 years ago

Read 2 more answers

Consider the following initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1 Let ∂f ∂x = (x + y)2 = x2 + 2xy + y2

IRISSAK [1]

$(x+y)^2\,\mathrm dx+(2xy+x^2-2)\,\mathrm dy=0$

Suppose the ODE has a solution of the form $F(x,y)=C$ , with total differential

$\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

This ODE is exact if the mixed partial derivatives are equal, i.e.

$\dfrac{\partial^2F}{\partial y\partial x}=\dfrac{\partial^2F}{\partial x\partial y}$

We have

$\dfrac{\partial F}{\partial x}=(x+y)^2\implies\dfrac{\partial^2F}{\partial y\partial x}=2(x+y)$

$\dfrac{\partial F}{\partial y}=2xy+x^2-2\implies\dfrac{\partial^2F}{\partial x\partial y}=2y+2x=2(x+y)$

so the ODE is indeed exact.

Integrating both sides of

$\dfrac{\partial F}{\partial x}=(x+y)^2$

with respect to $x$ gives

$F(x,y)=\dfrac{(x+y)^3}3+g(y)$

Differentiating both sides with respect to $y$ gives

$\dfrac{\partial F}{\partial y}=2xy+x^2-2=(x+y)^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies x^2+2xy-2=x^2+2xy+y^2+\dfrac{\mathrm dg}{\mathrm dy}$

$\implies\dfrac{\mathrm dg}{\mathrm dy}=-y^2-2$

$\implies g(y)=-\dfrac{y^3}3-2y+C$

$\implies F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y+C$

so the general solution to the ODE is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=C$

Given that $y(1)=1$ , we find

$\dfrac{(1+1)^3}3-\dfrac{1^3}3-2=C\implies C=\dfrac13$

so that the solution to the IVP is

$F(x,y)=\dfrac{(x+y)^3}3-\dfrac{y^3}3-2y=\dfrac13$

$\implies\boxed{(x+y)^3-y^3-6y=1}$

5 0

2 years ago

How do I solve!!!!!!!!!! plsss

Solnce55 [7]

I think it might be -3

5 0

2 years ago

Read 2 more answers

Taylor and Natasha are both saving money.

patriot [66]

Answer:

The number of completed weeks when Natasha and Taylor will have an equal amount of saved is 8 weeks

Step-by-step explanation:

The starting amount of Taylor's savings = $10

The amount Taylor plans to save a week = $20

Therefore, the function that models Natasha's saving plan is y = 10 + 20·x

The function that models Natasha's saving plan is y = 50 + 15·x

Where;

y is the amount in savings

x is the number of completed weeks

When the amount of savings, y, of both Natasha and Taylor are equal, we have;

y for Natasha = y for Taylor

Therefore

10 + 20·x = 50 + 15·x

20·x - 15·x = 50 - 10

5·x = 40

x = 40/5 = 8

The number of completed weeks when Natasha and Taylor will have an equal amount of saved, x = 8 weeks

6 0

2 years ago

Y = -6, W = 7, j = -2<br><br> - 2y +w-4j

Reika [66]

Answer:

answer is 27

Step-by-step explanation:

-2(-6) + 7 -4(-2)

12+7+8

19+8

27

Hope this helped

7 0

2 years ago