Answer:
14
Step-by-step explanation:
Let n represent the number of nickels. Then the number of quarters is 36-n. The difference is ...
(36-n) -n = 8 . . . . . 8 more quarters than nickels
18 -n = 4 . . . . . . . divide by 2
14 = n . . . . . . . . add n-4 to both sides
There are 14 nickels in the jar.
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<em>Additional comment</em>
The problem supplies more information than is needed for a solution. You can work this as a "sum and difference" problem, as we have above, or you can work it as a "mixture" problem where the total coin value comes into play. Any two of the three given relations will give a solution.
n + q = 36 . . . . . . . number of coins
5n +25q = 620 . . . value of coins (in cents)
q -n = 8 . . . . . . . . difference in number of coins
Answer:
What is the Picture?
Step-by-step explanation:
Answer:
B. 1.2 x 
Step-by-step explanation:
Scientific notation is a way of writing very large or very small numbers. A number is written in scientific notation when a number between 1 and 10 is multiplied by a power of 10. For example, 650,000,000 can be written in scientific notation as 6.5 ✕ 10^8. Or, in this case it's the opposite since we have a negative number. Since it's a decimal it's multiplied by 10 still but the exponent is negative. All we need to do to find what the exponent would be in this case is to count the number of zeros before the "12." There are 4 zeros before the 12. So your exponent would be -4. Now all we need to do is write out "equation" per say.
1.2 x 10^-4
<u>Hope this helps and have a nice day!</u>
Check the picture below.
so the object hits the ground when h(x) = 0, hmmm how long did it take to hit the ground the first time anyway?
now, we know the 2nd time around it hit the ground, h(x) = 0, but it took less time, it took 0.5 or 1/2 second less, well, the first time it took 3/2, if we subtract 1/2 from it, we get 3/2 - 1/2 = 2/2 = 1, so it took only 1 second this time then, meaning x = 1.
![\bf ~~~~~~\textit{initial velocity in feet} \\\\ h(x) = -16x^2+v_ox+h_o \quad \begin{cases} v_o=\textit{initial velocity}&0\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&\\ \qquad \textit{of the object}\\ h=\textit{object's height}&0\\ \qquad \textit{at "t" seconds}\\ x=\textit{seconds}&1 \end{cases} \\\\\\ 0=-16(1)^2+0x+h_o\implies 0=-16+h_o\implies 16=h_o \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x) = -16x^2+16~\hfill](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%20in%20feet%7D%20%5C%5C%5C%5C%20h%28x%29%20%3D%20-16x%5E2%2Bv_ox%2Bh_o%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Ctextit%7Binitial%20velocity%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h_o%3D%5Ctextit%7Binitial%20height%7D%26%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h%3D%5Ctextit%7Bobject%27s%20height%7D%260%5C%5C%20%5Cqquad%20%5Ctextit%7Bat%20%22t%22%20seconds%7D%5C%5C%20x%3D%5Ctextit%7Bseconds%7D%261%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%200%3D-16%281%29%5E2%2B0x%2Bh_o%5Cimplies%200%3D-16%2Bh_o%5Cimplies%2016%3Dh_o%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20h%28x%29%20%3D%20-16x%5E2%2B16~%5Chfill)
quick info:
in case you're wondering what's that pesky -16x² doing there, is gravity's pull in ft/s².
Answer:
The answer would be B
Step-by-step explanation:
C=2(pi)r
C=25.12
