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loris [4]
2 years ago
9

(SOS )WHO HELP ME I'LL GIVE ''16'' PIONT!!!

Mathematics
1 answer:
SashulF [63]2 years ago
7 0

y1 = 0.5 or 1/2

y2 = 2

y3 = 3.5

y3 = 5

y5 = 6.5

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Find all solutions to the following system of equations -y²+6y+x-9=0 ; 6y=x+27.Illustrate with a graph.
Dimas [21]

Answer:

The solution to the system is the pair (9, 6)

Step-by-step explanation:

Hi!

First, let´s write the system of equations:

-y² + 6y + x -9 = 0

6y = x +27

The solutions of the system are the pairs (x, y) that satisfy both equations.

Let´s take the second equation and solve it for x:

6y = x +27

Subtract 27 from both sides of the equation

6y - 27 = x

Now, we can replace x in the first equation and solve it for y:

-y² + 6y + x -9 = 0

-y² + 6y + 6y - 27 -9 = 0

-y² + 12y - 36 = 0

Notice that -y² + 12y - 36 = -(y - 6)², then:

-(y - 6)² = 0

y - 6 = 0

y = 6

(alternatively you can solve the quadratic equation using the quadratic formula).

Now let´s find the value of x:

x = 6y -27

x = 6·6 -27

x = 9

The solution to the system is the pair (9, 6)

Please see the attached figure. The point where the curves intersect is the solution to the system.

6 0
3 years ago
Prove that the equation <br> x^3 + x + 3 = 0 <br> has exactly one real root on the interval [-2,-1].
maw [93]

Answer:

  • the value of the function changes sign in the interval
  • the function is monotonic in the interval

Step-by-step explanation:

All polynomial functions are continuous, so we know from the intermediate value theorem that if the expression on the left changes sign in the interval [-2, 1] then there will be a zero in that interval. If the function is monotonic in the interval, there can only be one zero.

a) For f(x) = x^3 +x +3 = (x^2 +1)x +3, the values at the ends of the interval are ...

  f(-2) = (4+1)(-2) +3 = -7

  f(-1) = (1 +1)(-1) +3 = 1

The function value goes from -7 to +1 in the interval, so there exists at least one root in that interval.

__

b) The derivative of the function is ...

  f'(x) = 3x^2 +1

This is positive for any x, so is positive in the interval [-2, -1]. That is, the function is continuously increasing in that interval, so cannot have more than one crossing of the x-axis. There is exactly one root in the interval [-2, -1].

8 0
3 years ago
2(a-3)÷(a-4)(a-5)+a-1÷(3-a)(a-4)+a-2÷(5-a)(a-3) simplify.​
Arlecino [84]

<u><em>I'm almost confident that I did this wrong, but i tried...</em></u>

Answer:

4a-9 ÷ -19

Step-by-step explanation:

(It was easier for me to put it in fractions)

\frac{2(a-3)}{(a-4)(a-5)} + \frac{a-1}{(3-a)(a-4)} + \frac{a-2}{(5-a)(a-3)}        (3-a) (a-4) + (5-a) (a-3)

\frac{2a-6}{(a-4)(a-5)} + \frac{a-1}{(3-a)(a-4)} +  \frac{a-2}{(5-a)(a-3)}      3 - a · a - 4 + 5 - a · a - 3

    \frac{2a-6}{(2a-20)} + \frac{a-1}{(3-a)(a-4)} + \frac{a-2}{(5-a)(a-3)}               3 - a + 1 - a - 3

   \frac{2a-6+a-1+a-2}{(2a-20)+(3-a)(a-4)+(5-a)(a-3)}                        1 - a - a

                 \frac{4a-9}{2a-20+1-2a}                                       1 - 2a

                   \frac{4a-9}{2a-19-2a}

                      \frac{4a-9}{-19}

8 0
3 years ago
Nick has 189 trading cards. Kyle has 198 trading cards. Who has fewer cards ?
Darya [45]
Nick has fewer trading cards. 189 < 198; there is a difference of 9.
7 0
4 years ago
Read 2 more answers
Eduardo thinks of a number between 1 and 20 that has exactly 5 factors what number is the thinking of?
babunello [35]

Eduardo is thinking of the number ' 16 '.

The factors of  16 are  1,  2,  4,  8, and  16.

Here's a cool little factoid:

Almost every number has an even number of factors. The
only ones that have an odd number are perfect squares.
(They still have an even number of factors, but two of the factors
are the same number, so the list looks like an odd number long.)

3 0
3 years ago
Read 2 more answers
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