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Keith_Richards [23]
3 years ago
14

Is there any option to to 1 not be equal to 1?

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
4 0

No.  1 is is always equal to to 1 .

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This question on math and needs to be written a proportion
guapka [62]

Answer:

g = 3

Step-by-step explanation:

miles / gallon = miles / gallon

104 / 4 = 78 / g

cross-multiply to solve for 'g':

78(4) = 104g

312 = 104g

312/104 = g

3 = g

4 0
2 years ago
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6 2/3 divided by 2 6/7
Arturiano [62]

Answer:

7/3 in decimal form it's 2.3 repeated

Step-by-step explanation:

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3 years ago
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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|

8 0
3 years ago
Solve 33 1/2 of 48<br><br> Plz thanks &lt;3
SIZIF [17.4K]

Answer:

1608

Step-by-step explanation:

5 0
2 years ago
Can anyone help? Please??!!
goldfiish [28.3K]

Hi! The answer to that question is B and D cause if you read the promblem it is it!

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