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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

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.

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

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

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