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Ganezh [65]
3 years ago
6

PLEASE HELP!! 99 POINTS!! I HAVE NOOOO IDEA!

Mathematics
2 answers:
Otrada [13]3 years ago
7 0
This is x^3+x^2+1=-x in math terms, where x is the number. So x^3+x^2+x+1=0. We can write this: x^2(x+1)+(x+1)=(x^2+1)(x+1)=0. The only real solution is x+1=0, so x=-1.
Hopefully that makes sense
olchik [2.2K]3 years ago
6 0
X+1=0, so x=-1.



:)
................
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Can someone please answer. There is only one problem. There is a picture. Thank you!
KATRIN_1 [288]
5Red + 6Blue + 4 Black

Total = 5 + 6 = 4 = 15

Probality of Blue, Blue.

P(Blue) = 6/15 = 2/5

P(Blue, Blue) = (2/5)*(2/5) = 4/25
6 0
3 years ago
P(x) =x and q(x) = x-1Given:minimum x and Maximum x: -9.4 and 9.4minimum y and maximum y: -6.2 and 6.2Using the rational functio
Arte-miy333 [17]

we have the following function

\frac{p(x)}{g(x)}=\frac{x}{x\text{ -1}}

where x is between -9.4 and 9.4 and y is between -6.2 and 6.2.

We will first draw the function

from the graph, we can see that the zeroes are all values of x for which the graph crosses the x -axis

In this case, we see that that the only zero is at x=0.

Now, we have that the asymptotes are lines to which the graph of the function get really close to. On one side, we can see that as x goes to infinity or minus infinity, the values of the function get really close to 1. So the graph has a horizontal asymptote at y=1. Also, we can see that as x gets really close to 1, the graph gets really close to the vertical line x=1. So the graph has a vertical asymptote at x=1.

Recall that the domain of a function is the set of values of x for which the function is defined. From our graph, we can see that graph is not defined when x=1. So the domain of the function is the set of real numbers except x=1. Now, recall that the range of the function is the set of y values of the graph. From the picture we can see that the graph has a y coordinate for every value of y except for y=1. So, this means that the range of the function is the set of real numbers except y=1.

From the graph, we can see that we cannot draw the graph having a continous drawing. That is, imagine we take a pencil and start on one point on the graph on the left side. We can draw the whole graph on the left side, but we cannot draw the graph on the right side without lifting the pencil up. As we have to "lift the pencil up" this means that the graph is not continous

Finally note that as we have a vertical asymptote at x=1 and horizontal asymptote at y=1 we have that when y is 1 or x is 1, the function y=f(x)/g(x) is undefined

5 0
1 year ago
I don't understand need help?
QveST [7]

9514 1404 393

Answer:

  2. (only)

Step-by-step explanation:

The Pythagorean theorem tells you the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. To determine if these are right triangles, determine if that condition is met.

  1. 3^2 +5^2 = 9 + 25 = 34 ≠ (√35)^2 . . . . not a right triangle

  2. 5^2 +4^2 = 25 +16 = 41 = (√41)^2 . . . a right triangle

  3. 6^2 +8^2 = 36 +64 = 100 ≠ (√10)^2 . . . . not a right triangle

  4. 3^2 +3^2 = 9 +9 = 18 ≠ (3√3)^2 = 27 . . . . not a right triangle

3 0
2 years ago
Which equation best represents the graph above
Svet_ta [14]

Answer:

x = -1

Step-by-step explanation:

Since the line is completely vertical there is no slope.

When a line is completely vertical, the setup is x = ___ (fill in the blank with the number that the line intersects on the x axis, in this case -1.)

When a line is completely horizontal (not applicable in this problem) the setup is y = ___ (fill in the blank with the number the line intersects on the y axis.)

7 0
3 years ago
 10  points for this                                                                                                            
kow [346]
<span>The range would be all real numbers such that 0 ≤ y ≤ 40.

hope this helps!</span>
3 0
3 years ago
Read 2 more answers
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