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BlackZzzverrR [31]
3 years ago
11

Solve this system algebraically.

Mathematics
1 answer:
seraphim [82]3 years ago
4 0
From y-2x+3=0 we have y=2x-3
So we can replace y by 2x-3 in the equation: 9x+2y=5
And we have 9x+2(2x-3)=5
Or 9x+4x-6=5
and 13x=5+6
Or x=11/13, then y=2x-3= 2. 11/13-3=22/13-3= -17/13

So we have the answer <span>{(11/13, -17/13)}</span>
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Solve the following differential equations or initial value problems. In part (a), leave your answer in implicit form. For parts
shepuryov [24]

Answer:

(a) (y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) y = arctan(t(lnt - 1) + C)

(c) y = -1/ln|0.09(t + 1)²/t|

Step-by-step explanation:

(a) dy/dt = (t^2 + 7)/(y^4 - 4y^3)

Separate the variables

(y^4 - 4y^3)dy = (t^2 + 7)dt

Integrate both sides

(y^5)/5 + y^4 = (t^3)/3 + 7t + C

(b) dy/dt = (cos²y)lnt

Separate the variables

dy/cos²y = lnt dt

Integrate both sides

tany = t(lnt - 1) + C

y = arctan(t(lnt - 1) + C)

(c) (t² + t) dy/dt + y² = ty², y(1) = -1

(t² + t) dy/dt = ty² - y²

(t² + t) dy/dt = y²(t - 1)

(t² + t)/(t - 1)dy/dt = y²

Separating the variables

(t - 1)dt/(t² + t) = dy/y²

tdt/(t² + t) - dt/(t² + t) = dy/y²

dt/(t + 1) - dt/(t(t + 1)) = dy/y²

dt/(t + 1) - dt/t + dt/(t + 1) = dy/y²

Integrate both sides

ln(t + 1) - lnt + ln(t + 1) + lnC = -1/y

2ln(t + 1) - lnt + lnC = -1/y

ln|C(t + 1)²/t| = -1/y

y = -1/ln|C(t + 1)²/t|

Apply y(1) = -1

-1 = ln|C(1 + 1)²/1|

-1 = ln(4C)

4C = e^(-1)

C = (1/4)e^(-1) ≈ 0.09

y = -1/ln|0.09(t + 1)²/t|

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3 years ago
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Masteriza [31]

Answer:

17

Step-by-step explanation:

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6 0
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Benny Brought $5 dozen Eggs from the Grocery store to bake some cakes how many did benny buy
user100 [1]

Answer:

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Step-by-step explanation:

a dozen is 12. So 5 times 12 is 60.

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Answer:

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