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djyliett [7]
3 years ago
12

Which of the functions listed below will produce the graph shown here? HELP!!!!

Mathematics
1 answer:
aivan3 [116]3 years ago
8 0
A) -3x^2/2x-1 is the right answer
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Given the polynomial 6x3 + 4x2 − 6x − 4, what is the value of the constant 'k' in the factored form? 6x3 + 4x2 − 6x − 4 = 2(x +
CaHeK987 [17]
First you graph it using a graphing calculator, you look at the table of values to find out one point in which y= 0. The first one that comes up is when x=1.
If you don't have a graphing calculator you can use trial and error by inputing some numbers into x until you get y= 0. 

Once you have an x value which makes y=0, you can start factorizing it. 
you divide 6x3 +4x2 -6x - 4 into (x-1) which is when y =0 
to get 6x2+10x+4 

This can be used to write the polynomial as (x-1)(6x2 +10x+4) 
you then factorize the second bracket, 6x2 +10x+4.
you can take the 2 outside to give you 2(3x2 +5x+2) 
you can factorize this to become 2(3x+2)(x+1) 

Now you just substitute your factorized second bracket into your unfactorized second bracket to give you 2(3x+2)(x+1)(x-1).

From this you can deduce that k= 1 
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3 years ago
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This is a question on mean and mediator thats due in 15min
luda_lava [24]

Answer:

mean = 15. 14 or 15

median= 14

range = 25

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Erica was given $300 for her birthday and decided to put it in a savings account that earns 3.75% interest. If she makes no othe
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In circle C, what is mArc F H?<br><br><br> 31°<br><br> 48°<br><br> 112°<br><br> 121°
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We are standing on the top of a 320 foot tall building and launch a small object upward. The object's vertical altitude, measure
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Answer:

The highest altitude that the object reaches is 576 feet.

Step-by-step explanation:

The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be h(t) = -16\cdot t^{2} + 128\cdot t + 320, the first and second derivatives are, respectively:

First Derivative

h'(t) = -32\cdot t +128

Second Derivative

h''(t) = -32

Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:

-32\cdot t +128 = 0

t = \frac{128}{32}\,s

t = 4\,s (Critical value)

The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:

h(4\,s) = -16\cdot (4\,s)^{2}+128\cdot (4\,s) +320

h(4\,s) = 576\,ft

The highest altitude that the object reaches is 576 feet.

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3 years ago
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