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mina [271]
2 years ago
13

"Sketch a possible graph for a negative Quintic polynomial with one real zero." Can anyone explain how this is done? And if poss

ible provide a graph?

Mathematics
2 answers:
Sidana [21]2 years ago
6 0

quintic means degree of 5.

a negative quintic would be: f(x) = -x⁵ <u>+              </u>

an example with one real zero could be: -x⁵ + x³ + x + 3   (see attachment)

mart [117]2 years ago
4 0

Well let's start with y = -x^5 and y = - x^5 +1

The graph looks like the one on the right for these two.  The red one is y = - x^5 and the blue one is y = - x^5 + 1

See below on the right

You can tell that these two are correct because there is only 1 x axis intercept for each of them.  That means there is only 1 real solution which is what you are calling for.

Another one could be something like

y = -(x - 5)(x^2 + 2)(x^2 + 3)

The graph of that is on the left.


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Find p and q for which the linear eqn has infinite solutions
jekas [21]

Answer:

p= 2.5

q= 7

Step-by-step explanation:

The lines should overlap to have infinite solutions, slopes should be same and y-intercepts should be same.

Equations in slope- intercept form:

6x-(2p-3)y-2q-3=0 ⇒ (2p-3)y= 6x -2q-3 ⇒ y= 6/(2p-3)x -(2q+3)/(2p-3)

12x-( 2p-1)y-5q+1=0 ⇒ (2p-1)y= 12x - 5q+1 ⇒  y=12/(2p-1)x - (5q-1)/(2p-1)

Slopes equal:

6/(2p-3)= 12/(2p-1)

6(2p-1)= 12(2p-3)

12p- 6= 24p - 36

12p= 30

p= 30/12

p= 2.5

y-intercepts equal:

(2q+3)/(2p-3)= (5q-1)/(2p-1)

(2q+3)/(2*2.5-3)= (5q-1)/(2*2.5-1)

(2q+3)/2= (5q-1)/4

4(2q+3)= 2(5q-1)

8q+12= 10q- 2

2q= 14

q= 7

7 0
3 years ago
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DaniilM [7]

Answer:

A) x^{3} y^{4}

B) -\frac{5x^{2} }{y}

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D)\frac{1}{x^{3} }

Step-by-step explanation:

Hope this helps.

6 0
2 years ago
Which number is irrational? A. 0.3 B. [5 C. 0.777 D. 00.454445
Yanka [14]
C.0.777 decuase it is smart

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2 years ago
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motikmotik

Answer:

D

Step-by-step explanation:

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Answer:

8x^6y^3

Step-by-step explanation:

(2x2y)3

=8x6y3

3 0
3 years ago
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