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photoshop1234 [79]

3 years ago

6

5x^2-26x+5 factor the trinomial

1 answer:

ss7ja [257]3 years ago

4 0

The factored form for this question is (5x-1)(x-5)

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Find the solution to the system of equations: x + 3y = 7 and 2x + 4y = 8 1. Isolate x in the first equation: 2. Substitute the v

klemol [59]

X + 3y = 7
x = -3y + 7

2x + 4y = 8
2(-3y + 7) + 4y = 8
-6y + 14 + 4y = 8
-2y = 8 - 14
-2y = - 6
y = -6/-2
y = 3

x + 3y = 7
x + 3(3) = 7
x + 9 = 7
x = 7 - 9
x = -2

solution is (-2,3)

6 0

3 years ago

Read 2 more answers

3(x-7) = 2 (x - 12) <br> i need help fast

gulaghasi [49]

Are you solving for X? If so, your answer will be x=-3. Hope this helps!

7 0

2 years ago

How many turning points can a polynomial with a degree of 7 have?

aleksklad [387]

That would be 7 - 1 = 6 turning points

5 0

3 years ago

A deer population is estimated to be 4,350. The growth rate of the population is given by the equation y = 4350 What is the perc

nirvana33 [79]

I’m not sure actullay

5 0

3 years ago

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