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Flauer [41]
2 years ago
12

Please help. Links=Report

Mathematics
1 answer:
NARA [144]2 years ago
4 0
Download SLADER and take a picture of the barcode at the bottom, on the app and you will have all the answers on the textbook.
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They are so amazing!

Step-by-step explanation:

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3 years ago
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Which is an equation in point-slope form of the line that passes through the points (4,5) and (-3,-1)?
Annette [7]
(4,5),(-3,-1)
slope = (-1 - 5) / (-3 - 4) = -6/-7 = 6/7

y - y1 = m(x - x1)
slope(m) = 6/7
using points (-3,-1)...x1 = -3 and y1 = -1
now we sub...pay attention to ur signs
y - (-1) = 6/7(x - (-3) =
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6 0
3 years ago
I need help with 42,43,44 they’re all based off the chart right there!
ELEN [110]

Answer:

43 4445 itmean near divide by 5

8 0
3 years ago
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5 0
2 years ago
Lim x approaches 0 (1+2x)3/sinx
jok3333 [9.3K]

Interpreting your expression as

\dfrac{3(1+2x)}{\sin(x)}

when x approaches zero, the numerator approaches 3:

3(1+2x) \to 3(1+2\cdot 0) = 3(1+0) = 3\cdot 1 = 3

The denominator approaches 0, because \sin(0)=0

Moreover, we have

\displaystyle \lim_{x\to 0^-} \sin(x) = 0^-,\quad \displaystyle \lim_{x\to 0^+} \sin(x) = 0^+

So, the limit does not exist, because left and right limits are different:

\displaystyle \lim_{x\to 0^-} \dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^-} = -\infty,\quad \displaystyle \lim_{x\to 0^+}\dfrac{3(1+2x)}{\sin(x)}= \dfrac{3}{0^+} = +\infty

8 0
3 years ago
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