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masha68 [24]
3 years ago
7

Write any equation in slope intercept form of the line that passes through the given point and it's parallel to the graph of the

given equation
(2,-2) ; y= -x-2

Y=-2x
Y=2x
Y=1/2x
Y=-x
Mathematics
1 answer:
Orlov [11]3 years ago
4 0
Paralalell means that is has the same slope
y=mx+b
m=slope
y=-1x-2
slope=-1

the equation of a line that passes through the point (x1,y1) and has a slope of m is y-y1=m(x-x1)
given
(2,-2) and slope is -1
y-(-2)=-1(x-2)
y+2=-x+2
minus 2
y=-x
answer is y=-x
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Step-by-step explanation:

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Which expression is equivalent to 17s - 10 + 3(2s +1)?
klemol [59]

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Bob constructed an equilateral triangle PQR as shown. He then drew a straight line ST THROUGH POINT P. Determine the sum of the
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tresset_1 [31]

Answer:

See Below.

Step-by-step explanation:

We are given the graph of <em>y</em> = f'(x) and we want to determine the characteristics of f(x).

Part A)

<em>f</em> is increasing whenever <em>f'</em> is positive and decreasing whenever <em>f'</em> is negative.

Hence, <em>f</em> is increasing for the interval:

(-\infty, -2) \cup (-1, 1)\cup (3, \infty)

And <em>f</em> is decreasing for the interval:

\displaystyle (-2, -1) \cup (1, 3)

Part B)

<em>f</em> has a relative maximum at <em>x</em> = <em>c</em> if <em>f'</em> turns from positive to negative at <em>c</em> and a relative minimum if <em>f'</em> turns from negative to positive to negative at <em>c</em>.

<em>f'</em> turns from positive to negative at <em>x</em> = -2 and <em>x</em> = 1.

And <em>f'</em> turns from negative to positive at <em>x</em> = -1 and <em>x</em> = 3.

Hence, <em>f</em> has relative maximums at <em>x</em> = -2 and <em>x</em> = 1, and relative minimums at <em>x</em> = -1 and <em>x</em> = 3.

Part C)

<em>f</em> is concave up whenever <em>f''</em> is positive and concave down whenever <em>f''</em> is negative.

In other words, <em>f</em> is concave up whenever <em>f'</em> is increasing and concave down whenever <em>f'</em> is decreasing.

Hence, <em>f</em> is concave up for the interval (rounded to the nearest tenths):

\displaystyle (-1.5 , 0) \cup (2.2, \infty)

And concave down for the interval:

\displaystyle (-\infty, -1.5) \cup (0, 2.2)

Part D)

Points of inflections are where the concavity changes: that is, <em>f''</em> changes from either positive to negative or negative to positive.

In other words, <em>f </em>has an inflection point wherever <em>f'</em> has an extremum point.

<em>f'</em> has extrema at (approximately) <em>x</em> = -1.5, 0, and 2.2.

Hence, <em>f</em> has inflection points at <em>x</em> = -1.5, 0, and 2.2.

7 0
2 years ago
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