The answer option which is considered to be true of analogies in supporting problem solving is: Relevant analogies usually help people solve problems, and people do spontaneously think of relevant analogies.
An analogy can be defined as a process through which a situation is compared with another for the purpose of clarity or explanation. Thus, an analogy can be used to explain and illustrate the relationship that exist between two or more things.
In the problem-solving process, an analogy uses relevant or similar situations to proffer a solution to a problem since there exist a relationship between the two circumstances.
In conclusion, relevant analogies usually help people in proffering solution to problems.
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The fact that Saanvl deposits the $4,000, means that initially, the money supply remains the same.
<h3>What happened to the money supply after Saanvl's deposit?</h3>
After Saanvl deposited, the money was simply converted from cash to cash deposits at the bank.
Both of these things are considered to be a part of money supply so there is no change in money supply initially.
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Answer:
not enough information but based off other ppl's questions..
Explanation:
4,10,11 = not a right traingle
4,4, square root of 32 = right traingle.
8,7, square root of 113 = right triangle
4,4, square root of 24 = not right triangle
7,8,15 = not a right triangle
Answer: See below
Explanation:
For this question, we are given the volume, and we want to find the diameter. We have to work backwards to get our radius, then to find the diameter.
(Volume of cylinder)
(The π cancel out)
(Plug in h=8)
(Divide 8)
(Square root of 9 is 3)
Now that we have our radius, we can find our diameter.
(Equation for diameter)
(Plug in r=3)
(Multiply)
Answer: Twice the previous time would be taken to reach the same speed v with the puck of mass 2m.
Explanation:
Let a Force pushes the hockey puck of mass m.
Then acceleration, a= \frac{F}{m}a=mF
From the equation of motion,
\begin{gathered}\➪ v=u+at\\ v=0+\frac{F}{m}\Delta t\end{gathered}⇒v=u+atv=0+mFΔt ......(1)
In the second case, when mass is 2m, then acceleration,
a'=\frac{F}{2m}a′=2mF
and t' is the time taken.
The final speed is v,
\begin{gathered}\➪ v=0+ a't'\\ \➪ \frac {F}{m}\Delta t=\frac{F}{2m}t'\\ \➪ t'= 2\Delta t\end{gathered}⇒v=0+a′t′⇒mFΔt=2mFt′⇒t′=2Δt using equation (1)
Hence, it would take two times the previous amount of time to push the pluck of double mass.
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