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mihalych1998 [28]
2 years ago
11

HELP! ASAP!

Mathematics
1 answer:
masha68 [24]2 years ago
6 0

Step-by-step explanation:

2.

the volume of the box is

length × width × height = 13×13×13 = 13³ = 2197 cm³

the ratio of the density indignation tells us that every cm³ of soil weighs 1.33 g (or every unit of 1.33g soil fits into 1 cm³).

we have 2197 cm³.

their weight is

2197 × 1.33 = 2,922.01 g

so, the filled box clearly exceeds the max. weight of the window ledge, and it is NOT safe to put it there.

b

yes, it would be a little bit less dense.

because every bit of weight you put on top of something increases the pressure on and therefore the density of that something.

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URGENT!!!Can someone give me the answer and explanation plz
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Answer: D

Step-by-step explanation: Divide 110 and 1700 to get 15 then multiply that 15 with 110 to get 1600. I hope this helps ;)

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3 years ago
What is the value of x?<br> Enter your answer in the box.<br><br> x =
LUCKY_DIMON [66]

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A scale on a map shows that 2 inches = 25 miles
ycow [4]
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3 years ago
Simplify using distributive property 4(z+2)+2(3-z)
riadik2000 [5.3K]

Answer:

2z+14

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5 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
3 years ago
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