Answer:
r is -7
Step-by-step explanation:
(I believe that you mistyped and meant that the fourth term in the sequence should be -686)
The easiest way to find the common ratio is to divide the numbers that are adjacent to each other
![\frac{-14}{2} =-7\\\\\frac{98}{-14} =-7\\\\\frac{-686}{98} =-7](https://tex.z-dn.net/?f=%5Cfrac%7B-14%7D%7B2%7D%20%3D-7%5C%5C%5C%5C%5Cfrac%7B98%7D%7B-14%7D%20%3D-7%5C%5C%5C%5C%5Cfrac%7B-686%7D%7B98%7D%20%3D-7)
As you can see, when we divide a term by its previous term, we get -7 every time. This is our common ratio.
there are 12 months in a year, so then 4 months is really 4/12 of a year.
![~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & 1200\\ r=rate\to 5\%\to \frac{5}{100}\dotfill &0.05\\ t=\to \frac{4}{12}years\dotfill &\frac{1}{3} \end{cases} \\\\\\ I = (1200)(0.05)(\frac{1}{3})\implies I=20~GHC](https://tex.z-dn.net/?f=~~~~~~%20%5Ctextit%7BSimple%20Interest%20Earned%7D%20%5C%5C%5C%5C%20I%20%3D%20Prt%5Cqquad%20%5Cbegin%7Bcases%7D%20I%3D%5Ctextit%7Binterest%20earned%7D%5C%5C%20P%3D%5Ctextit%7Boriginal%20amount%20deposited%7D%5Cdotfill%20%26%201200%5C%5C%20r%3Drate%5Cto%205%5C%25%5Cto%20%5Cfrac%7B5%7D%7B100%7D%5Cdotfill%20%260.05%5C%5C%20t%3D%5Cto%20%5Cfrac%7B4%7D%7B12%7Dyears%5Cdotfill%20%26%5Cfrac%7B1%7D%7B3%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20I%20%3D%20%281200%29%280.05%29%28%5Cfrac%7B1%7D%7B3%7D%29%5Cimplies%20I%3D20~GHC)
If the sides of the triangle have lengths of: 2, 1, and 2√3
Then, the triangle is a
c. neither a 45-45-90 triangle nor a 30-60-90 triangle.
Looking at the lengths, if the triangle is a right triangle
2 and 1 must be the measure of the legs
2√3, the measure of the hypotenuse
If we use the Pythagorean theorem,
2^2 + 1^2 = (2√3)^2
5 = 12 (not correct)
Therefore, the triangle is not a right trangle
<span>You must get p by itself, so first subtract B from both sides: SA-B=1/2lp. Then multiply both sides by 2, which cancels out the 1/2: 2SA=lp. Then divide both sides by l: (2SA)/l=p.</span>
Answer:
![\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7Dy%3D-%5Cdfrac%7B1%7D%7B9%7Dx%5E2%20%2B%5Cdfrac%7B2%7D%7B9%7Dx%2B%5Cdfrac%7B89%7D%7B9%7D%5C%5C%20%5C%5Cy%3D%5Cdfrac%7B1%7D%7B8%7Dx%5E2%20-7%5Cend%7Barray%7D%5Cright.)
Step-by-step explanation:
1st boat:
Parabola equation:
![y=ax^2 +bx+c](https://tex.z-dn.net/?f=y%3Dax%5E2%20%2Bbx%2Bc)
The x-coordinate of the vertex:
![x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=1\\ \\b=-2a](https://tex.z-dn.net/?f=x_v%3D-%5Cdfrac%7Bb%7D%7B2a%7D%5CRightarrow%20-%5Cdfrac%7Bb%7D%7B2a%7D%3D1%5C%5C%20%5C%5Cb%3D-2a)
Equation:
![y=ax^2 -2ax+c](https://tex.z-dn.net/?f=y%3Dax%5E2%20-2ax%2Bc)
The y-coordinate of the vertex:
![y_v=a\cdot 1^2-2a\cdot 1+c\Rightarrow a-2a+c=10\\ \\c-a=10](https://tex.z-dn.net/?f=y_v%3Da%5Ccdot%201%5E2-2a%5Ccdot%201%2Bc%5CRightarrow%20a-2a%2Bc%3D10%5C%5C%20%5C%5Cc-a%3D10)
Parabola passes through the point (-8,1), so
![1=a\cdot (-8)^2-2a\cdot (-8)+c\\ \\80a+c=1](https://tex.z-dn.net/?f=1%3Da%5Ccdot%20%28-8%29%5E2-2a%5Ccdot%20%28-8%29%2Bc%5C%5C%20%5C%5C80a%2Bc%3D1)
Solve:
![c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}](https://tex.z-dn.net/?f=c%3D10%2Ba%5C%5C%20%5C%5C80a%2B10%2Ba%3D1%5C%5C%20%5C%5C81a%3D-9%5C%5C%20%5C%5Ca%3D-%5Cdfrac%7B1%7D%7B9%7D%5C%5C%20%5C%5Cb%3D-2a%3D%5Cdfrac%7B2%7D%7B9%7D%5C%5C%20%5C%5Cc%3D10-%5Cdfrac%7B1%7D%7B9%7D%3D%5Cdfrac%7B89%7D%7B9%7D)
Parabola equation:
![y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}](https://tex.z-dn.net/?f=y%3D-%5Cdfrac%7B1%7D%7B9%7Dx%5E2%20%2B%5Cdfrac%7B2%7D%7B9%7Dx%2B%5Cdfrac%7B89%7D%7B9%7D)
2nd boat:
Parabola equation:
![y=ax^2 +bx+c](https://tex.z-dn.net/?f=y%3Dax%5E2%20%2Bbx%2Bc)
The x-coordinate of the vertex:
![x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0](https://tex.z-dn.net/?f=x_v%3D-%5Cdfrac%7Bb%7D%7B2a%7D%5CRightarrow%20-%5Cdfrac%7Bb%7D%7B2a%7D%3D0%5C%5C%20%5C%5Cb%3D0)
Equation:
![y=ax^2+c](https://tex.z-dn.net/?f=y%3Dax%5E2%2Bc)
The y-coordinate of the vertex:
![y_v=a\cdot 0^2+c\Rightarrow c=-7](https://tex.z-dn.net/?f=y_v%3Da%5Ccdot%200%5E2%2Bc%5CRightarrow%20c%3D-7)
Parabola passes through the point (-8,1), so
![1=a\cdot (-8)^2-7\\ \\64a-7=1](https://tex.z-dn.net/?f=1%3Da%5Ccdot%20%28-8%29%5E2-7%5C%5C%20%5C%5C64a-7%3D1)
Solve:
![a=-\dfrac{1}{8}\\ \\b=0\\ \\c=-7](https://tex.z-dn.net/?f=a%3D-%5Cdfrac%7B1%7D%7B8%7D%5C%5C%20%5C%5Cb%3D0%5C%5C%20%5C%5Cc%3D-7)
Parabola equation:
![y=\dfrac{1}{8}x^2 -7](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B1%7D%7B8%7Dx%5E2%20-7)
System of two equations:
![\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5Cbegin%7Barray%7D%7Bl%7Dy%3D-%5Cdfrac%7B1%7D%7B9%7Dx%5E2%20%2B%5Cdfrac%7B2%7D%7B9%7Dx%2B%5Cdfrac%7B89%7D%7B9%7D%5C%5C%20%5C%5Cy%3D%5Cdfrac%7B1%7D%7B8%7Dx%5E2%20-7%5Cend%7Barray%7D%5Cright.)