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klio [65]
3 years ago
11

Calculate the average rate of change for the function f(x) = x4 + 3x3 - 5x2 + 2x - 2, from x = -1 to x = 1

Mathematics
2 answers:
kolezko [41]3 years ago
5 0
In interval <a, b>  the average rate of change the function you count from this formula:
x=\frac{f(b)-f(a)}{b-a} \qquad b>a
Here you've got a=-1 and b=1:
f(a)=f(-1)=(-1)^4+3 \cdot (-1)^3 - 5 \cdot (-1)^2 +2 \cdot (-1)-2= \\ = 1-3-5-2-2=-11 \\ \\ f(b)=f(1)=1^4+3 \cdot 1^4 - 5 \cdot 1^2 + 2 \cdot 1 -2=1+3-5+2-2=-1 \\ \\ \hbox{So average rate of change this function is:} \\ \\ x= \frac{-1+11}{1+1}=\frac{10}{2}=5
Georgia [21]3 years ago
3 0
Workings and answer in the attachments below.

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3 years ago
GEOMETRY PROOFS!
yarga [219]

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<em>Additional</em><em> comment</em><em>:</em><em>-</em>

SAS property:-

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4 0
2 years ago
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A ship traveling east at 45 mph is 15 mi from a harbor when another ship leaves the harbor traveling east at 60 mph. How long do
Harman [31]

Answer:

  1 hour

Step-by-step explanation:

I find these easiest to work by considering the initial difference in distance and the speed at with that gap is closing.

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The rate of closure is the difference in the speeds of the two ships:

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Then the closure time is ...

  time = distance/speed

  time = (15 mi)/(15 mi/h) = 1 h

It will take the second ship 1 hour to catch up to the first ship.

7 0
2 years ago
F/4 + 16 &lt; 49<br> Solve the inequality
Mnenie [13.5K]

Answer:

f < 132

Step-by-step explanation:

Greetings!

I'm~Isabelle~Williams~and~I~will~be~answering~your~question!

Subtract~16~from~both~sides:

\frac{f}{4} +16-16 < 49-16

Simplify:

\frac{f}{4} < 33

Now,~multiply~both~sides~by~4:

\frac{4f}{4} < 33\times4

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5 0
3 years ago
The center of a hyperbola is located at the origin. One focus is located at (−50, 0) and its associated directrix is represented
leva [86]

The equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

The center of a hyperbola is located at the origin that means at (0, 0) and one of the focus is at (-50, 0)

As both center and the focus are lying on the x-axis, so the hyperbola is a horizontal hyperbola and the standard equation of horizontal hyperbola when center is at origin: \frac{x^{2}}{a^{2}}  - \frac{y^{2}}{b^{2}}    = 1

The distance from center to focus is 'c' and here focus is at (-50,0)

So, c= 50

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d= \frac{a^{2}}{c}

Here the directrix line is given as : x= 2304/50

Thus, \frac{a^{2}}{c}  = \frac{2304}{50}

⇒ \frac{a^{2}}{50}  = \frac{2304}{50}

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