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stiks02 [169]
2 years ago
14

The 12th term in a sequence with a common difference of-9 is -106. which of the following formulas can be used to represent this

sequence
Mathematics
1 answer:
Effectus [21]2 years ago
6 0

Answer:

T12=a-99=106

Step-by-step explanation:

that's the answer

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-x + 4y when x = -4/5 and y = 1/3
Usimov [2.4K]

we have the expression

-x+4y

Evaluate for x=-4/5 and y=1/3

substitute the values of x and y in the given expression

\begin{gathered} -(-\frac{4}{5})+4(\frac{1}{3}) \\ \frac{4}{5}+\frac{4}{3} \\ \frac{3\cdot4+5\cdot4}{15} \\ \frac{32}{15} \end{gathered}

the answer is 32/15

8 0
11 months ago
The general solution of 2 y ln(x)y' = (y^2 + 4)/x is
Sav [38]

Replace y' with \dfrac{\mathrm dy}{\mathrm dx} to see that this ODE is separable:

2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}

Integrate both sides; on the left, set u=y^2+4 so that \mathrm du=2y\,\mathrm dy; on the right, set v=\ln x so that \mathrm dv=\dfrac{\mathrm dx}x. Then

\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v

\implies\ln|u|=\ln|v|+C

\implies\ln(y^2+4)=\ln|\ln x|+C

\implies y^2+4=e^{\ln|\ln x|+C}

\implies y^2=C|\ln x|-4

\implies y=\pm\sqrt{C|\ln x|-4}

4 0
2 years ago
Point One = (-1,-1)<br> Point Two = (0,4)<br> Point-Slope Form:
-Dominant- [34]

Answer:

y+2=m(x+2)

Step-by-step explanation:

y - y1 = m(x - x1)

y- (-1) = m( x-(-1) )

y+2 = mx +2m

4 0
3 years ago
given the points a (-1,2) and (7,14), find the coordinates of the point p on directed line segment ab that partitions ab in the
Vinvika [58]
Check the picture below.

\bf \left. \qquad  \right.\textit{internal division of a line segment}&#10;\\\\\\&#10;a(-1,2)\qquad b(7,14)\qquad&#10;\qquad 1:3&#10;\\\\\\&#10;\cfrac{aP}{Pb} = \cfrac{1}{3}\implies \cfrac{a}{b} = \cfrac{1}{3}\implies 3a=1b\implies 3(-1,2)=1(7,14)&#10;\\\\&#10;-------------------------------\\\\&#10;{ P=\left(\cfrac{\textit{sum of "x" values}}{r1+r2}\quad ,\quad \cfrac{\textit{sum of "y" values}}{r1+r2}\right)}

\bf -------------------------------\\\\&#10;P=\left(\cfrac{(3\cdot -1)+(1\cdot 7)}{1+3}\quad ,\quad \cfrac{(3\cdot 2)+(1\cdot 14)}{1+3}\right)&#10;\\\\\\&#10;P=\left( \cfrac{-3+7}{4}~,~\cfrac{6+14}{4} \right)

3 0
3 years ago
create a sketch of the graph of a function whose gradient is linear and that gradient has an x-intercept at x=4
valentina_108 [34]
Here is one that meets your requirements.

7 0
3 years ago
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