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olasank [31]
3 years ago
11

A girl is swimming 1 meter below sea level and dives down another .5 meters. How far below sea level is she now?

Mathematics
1 answer:
GaryK [48]3 years ago
4 0

Answer:

1.5 meters

Step-by-step explanation:

If she is already below sea level by 1 meter and she goes down .5 meters more, she would be 1.5 meters below the sea level.

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Step-by-step explanation:

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Can someoe help? <br> 7x-24=6x-2
just olya [345]

Answer:

7x - 24 = 6x - 2 \\ collecting \: like \: terms \:  \\ 7x - 6x =  - 2 + 24 \\ 1x = 22 \\ x =  \frac{22}{1}  \:  \: or \:  \: x = 22

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2 years ago
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(1 point) Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference betw
slavikrds [6]

Answer:

a. \frac{dT}{dt}=k(T-Tm); T(0)=190

b. C_{0}=122

c. k=-0.00259

d. t=153.39838\\ minutos

Step-by-step explanation:

a. Newton's law of cooling states that the speed with which a body is cooled is proportional to the difference between its temperature and that of the medium in which it is found. Then, the initial value problem is given by:

Tm=68

\frac{dT}{dt}=k(T-Tm); T(0)=190

b. The differential equation obtained is a differential equation of separable variables:

\frac{dT}{T-Tm}=kdt\\\\\int {\frac{dT}{T-Tm}}=\int{kdt}\\\\Ln|T-Tm|=kt+C\\\\T(t)=C_{0}e^{kt}+Tm=C_{0}e^{kt}+68\\\\T(0)=C_{0}e^{k(0)}+68=190\\\\C_{0}=122

c. After 33 minutes of serving the coffee has cooled to 180°:

T(33)=122e^{33k}+68=180\\\\e^{33k}=\frac{112}{122}\\\\33k=Ln(\frac{112}{122})\\\\k=-0.00259

d.

150=122e^{-0.00259t}+68\\\\Ln(\frac{150-68}{122})=-0.00259t\\\\t=153.39838\\\\

8 0
3 years ago
What combination of transformations is shown below
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7 0
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Andreas93 [3]

Answer:

Either one of these.

Step-by-step explanation:

volumepile=1/3 (PI r^2)h

but r=h/2, so

volume=1/12 PI h^3

dv/dt=10 ft^3/min

but dv/dt=1/12 PI 3h^2 dh/dt

solve for dh/dt

This assumes you mean by "altitude" the height. If you mean altitude as slant height, you have to adjust the fromula

_______________________________________________

given: r = h/2

V = (1/3)π r^2 h

= (1/3)π (h/2)^2 (h)

= (1/12) π h^3

dV/dt = (1/4)π h^2 dh/dt

for the given data ...

10 = (1/4)π(25)dh/dt

dh/dt = 10(4)/((25π) = 1.6/π feet/min

3 0
3 years ago
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