The probability that he will guess correctly is:
1/5
Well, from the sequence, notice <span>70,63,56,49,…
is simply dropping on the next term by 7 units, so,
70 - 7, 63
63 - 7, 56
and so on
thus, the "common difference", or the number you "add" to get the next term is -7..... as from the sequence itself, you can see the first term's value is 70.
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![\bf n^{th}\textit{ term of an arithmetic sequence}\\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference}\\ ----------\\ a_1=70\\ d=-7\\ n=49 \end{cases} \\\\\\ a_{49}=70+(49-1)(-7)\implies a_{49}=70-336\implies a_{49}=-266](https://tex.z-dn.net/?f=%5Cbf%20n%5E%7Bth%7D%5Ctextit%7B%20term%20of%20an%20arithmetic%20sequence%7D%5C%5C%5C%5C%0Aa_n%3Da_1%2B%28n-1%29d%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0An%3Dn%5E%7Bth%7D%5C%20term%5C%5C%0Aa_1%3D%5Ctextit%7Bfirst%20term%27s%20value%7D%5C%5C%0Ad%3D%5Ctextit%7Bcommon%20difference%7D%5C%5C%0A----------%5C%5C%0Aa_1%3D70%5C%5C%0Ad%3D-7%5C%5C%0An%3D49%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0Aa_%7B49%7D%3D70%2B%2849-1%29%28-7%29%5Cimplies%20a_%7B49%7D%3D70-336%5Cimplies%20a_%7B49%7D%3D-266)
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The triangle that has side lengths of 7, 10, and 12 is not a right triangle. The correct answer is B. No.
The reason for this can be found in the Pythagorean theorem, which is a2 + b2 = c2 (this number 2 means square)
So, let's test this.
7*7 + 10*10 = 12*12
49+100 = 144
149 = 144
Since this is not equal, this isn't a right triangle.
Answer:
1. x=2 2. x=23
Step-by-step explanation:
1: 7(x+3) = 39-2x
7x+21 = 39-2x
9x=18; x=2
2. 2x-5 = 1/2(3x-13)
2x-5 = 1.5x-13/2
0.5x=11.5; x=23
Answer:
Step-by-step explanation:
Given that:
mass m = 100 g = 0.1 kg
Length of the spring = 5 cm = 0.05 m
The set in motion from the equilibrium position u(0) = 0
The set in motion from its equilibrium position with a downward velocity u'(0) = 50 cm/s = 0.5 m/s
The spring constant (k) = ![\dfrac{0.1 \times 9.8}{0.05}](https://tex.z-dn.net/?f=%5Cdfrac%7B0.1%20%5Ctimes%209.8%7D%7B0.05%7D)
The equation of the system is expressed as:
![\dfrac{1}{10} u'' + \dfrac{98}{5} u =0](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B10%7D%20u%27%27%20%2B%20%5Cdfrac%7B98%7D%7B5%7D%20u%20%3D0)
By estimating the characteristics equation, we have r = ± 14![i](https://tex.z-dn.net/?f=i)
Thus; the general solution is:
![u(t) = c_1cos \ 14t + c_2 sin 14 \ t](https://tex.z-dn.net/?f=u%28t%29%20%3D%20c_1cos%20%20%5C%2014t%20%2B%20c_2%20sin%2014%20%5C%20t)
By applying the initial condition:
u(0) = 0
⇒ 0 =
∴
![\dfrac{du}{dt} = ( - c_1 \ sin 14 t )\times 14 +14c_2 \ cos 14 t](https://tex.z-dn.net/?f=%5Cdfrac%7Bdu%7D%7Bdt%7D%20%3D%20%28%20-%20c_1%20%5C%20sin%2014%20t%20%29%5Ctimes%2014%20%2B14c_2%20%5C%20cos%2014%20t)
u'(0) = 0.5
0.5 = 14 × c₂
c₂ = 0.5/14
c₂ = 1/28
∴
![\mathbf{u(t) = \dfrac{1}{28} sin 14 t}](https://tex.z-dn.net/?f=%5Cmathbf%7Bu%28t%29%20%3D%20%5Cdfrac%7B1%7D%7B28%7D%20sin%2014%20t%7D)
Equating u(t) = 0, we have t = π/14sec as the time when the mass first returns to its equilibrium position.