Which line appears to have a y-intercept of 5 and an x-intercept of 3

Mathematics

1 answer:

s344n2d4d5 [400]3 years ago

8 0

Answer:

the vertical line appears to represent the y intercept and the horizontal appears to represent the x intercept

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ANTONII [103]

Answer:a)9

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Step-by-step explanation:

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Delicious77 [7]

$\bf \displaystyle\int~sec(4x-1)tan(4x-1)dx \\\\[-0.35em] ~\dotfill\\\\ sec(4x-1)tan(4x-1)\implies \cfrac{1}{cos(4x-1)}\cdot \cfrac{sin(4x-1)}{cos(4x-1)}\implies \cfrac{sin(4x-1)}{cos^2(4x-1)} \\\\[-0.35em] ~\dotfill\\\\ \displaystyle\int~\cfrac{sin(4x-1)}{cos^2(4x-1)}dx \\\\[-0.35em] ~\dotfill\\\\ u=cos(4x-1)\implies \cfrac{du}{dx}=-sin(4x-1)\cdot 4\implies \cfrac{du}{-4sin(4x-1)}=dx \\\\[-0.35em] ~\dotfill$

$\bf \displaystyle\int~\cfrac{~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }{u^2}\cdot \cfrac{du}{-4~~\begin{matrix} sin(4x-1) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies -\cfrac{1}{4}\int\cfrac{1}{u^2}du\implies -\cfrac{1}{4}\int u^{-2}du \\\\\\ -\cfrac{1}{4}\cdot \cfrac{u^{-2+1}}{-1}\implies \cfrac{1}{4}\cdot u^{-1}\implies \cfrac{1}{4u}\implies \cfrac{1}{4cos(4x+1)}+C$