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skad [1K]
3 years ago
6

How can caregivers give children self confidence safely?

Mathematics
1 answer:
Ann [662]3 years ago
8 0
Telling them that no mater what they should love them for who they are love your self no matter what
You might be interested in
Using the properties of exponents and logarithms, find the value of x in 19 + 2 ln x = 25.
REY [17]
19+2\ln x=25\ \ \ |-19\\\\2\ln x=6\ \ \ |:2\\\\\ln x=3\iff x=e^3

\text{Used de.finition of the logarithm:}\ \ \ \log_ab=c\iff a^c=b



8 0
3 years ago
Read 2 more answers
I need some help please
Veronika [31]
The definition of the tangent function tells you
   tan(angle) = (300 ft) / (distance to mountain)
This equation can be rearranged to
   (distance to mountain) = (300 ft) / tan(angle)

For the far end of the river,
   distance to far end = (300 ft) / tan(24°) ≈ 673.8 ft

For the near end of the river
   distance to near end = (300 ft) / tan(40°) ≈ 357.5 ft

Then the width of the river can be calculated by finding the difference of these distances:
   width of river = distance to far end - distance to near end
   width of river = 673.8 ft - 357.5 ft
   width of river = 316.3 ft

The appropriate answer choice is
   316 ft.
8 0
3 years ago
Evaluation: Flight of the Albatross
aev [14]

Answer:

Step-by-step explanation:

400/8=50

8 0
3 years ago
Read 2 more answers
Find the reciprocal of 3​
butalik [34]
1/3 because all you have to do is flip it
4 0
3 years ago
The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t),
pishuonlain [190]

Answer:

S(3)=22

Step-by-step explanation:

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t).

\dfrac{dS}{dt}=k(30-S(t))\\ \dfrac{dS}{dt}+kS(t)=30k\\$The integrating factor: e^{\int k dt}=e^{kt}\\$Multiply all through by the integrating factor\\ \dfrac{dS}{dt}e^{kt}+kS(t)e^{kt}=30ke^{kt}

(Se^{kt})'=30ke^{kt} dt\\$Integrate both sides\\ Se^{kt}=\dfrac{30ke^{kt}}{k}+C$ (C a constant of integration)\\Se^{kt}=30e^{kt}+C\\$Divide both sides by e^{kt}\\S(t)=30+Ce^{-kt}

When t=0, S(t)=15

15=30+Ce^{-k*0}\\C=15-30\\C=-15

When t = 2, S(t)=20

20=30-15e^{-2k}\\20-30=-15e^{-2k}\\-10=-15e^{-2k}\\e^{-2k}=\dfrac23\\$Take the natural log of both sides$\\-2k=\ln \dfrac23\\k=-\dfrac{\ln(2/3)}{2}

Therefore:

S(t)=30-15e^{\frac{\ln(2/3)}{2}t}\\$When t=3$\\S(t)=30-15e^{\frac{\ln(2/3)}{2} \times 3}\\S(3)=21.8 \approx 22

8 0
3 years ago
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