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3 years ago

Can a local maximum also be a point of inflection? Explain.

1 answer:

7 0

<span>Inflection points are where the function changes concavity. Since concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative, then when the function changes from concave up to concave down (or vise versa) the second derivative must equal zero at that point. So the second derivative must equal zero to be an inflection point. But don't get excited yet. You have to make sure that the concavity actually changes at that point.</span>

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A radar gun was used to record the speed of a swimmer (in meters per second) during selected times in the first 2 seconds of a r

Natalka [10]

Trapezoidal is involving averageing the heights
the 4 intervals are
[0,4] and [4,7.2] and [7.2,8.6] and [8.6,9]

the area of each trapezoid is (v(t1)+v(t2))/2 times width

for the first interval
the average between 0 and 0.4 is 0.2
the width is 4
4(0.2)=0.8

2nd
average between 0.4 and 1 is 0.7
width is 3.2
3.2 times 0.7=2.24

3rd
average betwen 1.0 and 1.5 is 1.25
width is 1.4
1.4 times 1.25=1.75

4th
average betwen 1.5 and 2 is 1.75
width is 0.4
0.4 times 1.74=0.7

add them all up
0.8+2.24+1.75+0.7=5.49

5.49
t=time
v(t)=speed
so the area under the curve is distance
covered 5.49 meters

8 0

3 years ago

Find all complex solutions of 5x^2+5x+3=0.

oksian1 [2.3K]

Answer:

see explanation

Step-by-step explanation:

5x² + 5x + 3 = 0 ← in standard form

with a = 5 , b = 5 , c = 3

Solve using the quadratic formula

x = $\frac{-b+/-\sqrt{b^2-4ac} }{2a}$

= $\frac{-5+/-\sqrt{5^2-(4(5)(3)} }{2(5)}$

= $\frac{-5+/-\sqrt{25-60} }{10}$

= $\frac{-5+/-\sqrt{-35} }{10}$

= $\frac{-5+/-\sqrt{-1(35)} }{10}$

= $\frac{-5+/-i\sqrt{35} }{10}$

Then

x = $\frac{-5}{10}$ ± $\frac{1}{10}$ i $\sqrt{35}$ , that is

x = - $\frac{1}{2}$ - $\sqrt{x} \frac{1}{10}$ i $\sqrt{35}$ , - $\frac{1}{2}$ + $\frac{1}{10}$ i $\sqrt{35}$

5 0

3 years ago

#4 home do I find which one it is?

lina2011 [118]

$\bf \begin{array}{lllll}
solutions&graphs&slopes\\
----&----&----\\
\textit{exactly one}&
\begin{array}{llll}
\textit{the two lines intersect}\\
\textit{at one point}
\end{array}&\textit{different slopes}\\\\
infinitely\quad many&
\begin{array}{llll}
\textit{the two lines coincide}\\
\textit{one is right on top}\\
\textit{ of the other}
\end{array}&
\begin{array}{llll}
\textit{equal slopes}\\
\textit{equal y-intercepts}
\end{array}
\end{array}$

$\bf \textit{no solution}\qquad\quad &\textit{lines are parallel} \qquad &
\begin{array}{llll}
\textit{equal slopes}\\
\textit{different y-intercepts}
\end{array}
\end{array}$

for example, let's look at the first set

y+3x =5 or y = -3x+ 5
and y = -3x + 2
y = m + b

the slopes are equal, the y-intercepts differ
that means, they're just parallel lines, no solution

8 0

4 years ago

Please help me and explain please!

Ann [662]

It would be 15 greater than or equal to 14

5 0

3 years ago

Which category does each card go into?

WINSTONCH [101]

Step-by-step explanation:

E3, E6, E11, E7: no solution

E1, E12 : 1 solution

E2, E8, E10 : infinite solutions

7 0

3 years ago

Read 2 more answers