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Shtirlitz [24]
3 years ago
5

Plz help on these questions u have to simply to find the answer plz i need help

Mathematics
1 answer:
masya89 [10]3 years ago
6 0

Answer:

1,3,6,8,9

Step-by-step explanation:

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What is the answer to this question and how do I get the answer
Svetlanka [38]
I think it is something

7 0
3 years ago
Can you please help
Verizon [17]

Answer:

used formula to find Area of parallelogram

area of parallelogram =b× h

Step-by-step explanation:

solution :

given ,

h = 7cm

b = 9 cm

we know ,

Area of parallelogram = b× h

=63cm²

6 0
2 years ago
A sequence is defined by the function F(n) = F(n -1) +5, where n represents the number of the term for n >1, and f(1) = -4. W
Anit [1.1K]

Answer:

4th option

Step-by-step explanation:

Using the recursive rule and f(1) = - 4 , then

f(2) = f(1) + 5 = - 4 + 5 = 1

f(3) = f(2) + 5 = 1 + 5 = 6

f(4) = f(3) + 5 = 6 + 5 = 11

The first 4 terms are - 4, 1, 6, 11

5 0
3 years ago
Help plz !! I will give you points or anything !!
pashok25 [27]

Answer:

Answer C is correct

Step-by-step explanation:

-200(3/4) = -600/4 = -150 ft

3 0
2 years ago
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
4 years ago
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