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salantis [7]
2 years ago
8

НА

Mathematics
1 answer:
Oksanka [162]2 years ago
4 0

????????????????????

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What is the quadratic formula
LUCKY_DIMON [66]

Answer:

x = -b +or- the square root of (b^2-4ac) ALL / 2a

sing to "pop goes the weasel" lol

7 0
3 years ago
Read 2 more answers
Simplify. Write your answer using whole numbers and variables.
Andre45 [30]

Answer:

q + 7

Step-by-step explanation:

(q² + 4q − 21) / (q − 3)

Factor the numerator:

(q + 7) (q − 3) / (q − 3)

Divide:

q + 7

5 0
3 years ago
How do I solve both questions?
diamong [38]
1. divide 8.75 by 3.5 and divide 9.75 by 3.75
2. do 2.5 ×2 plus 1.25
5 0
2 years ago
Quadratics need help making parabola
timama [110]

Answer:

The equation of the parabola is y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4, whose real vertex is (x,y) = (2, -5.333), not (x,y) = (2, -5).

Step-by-step explanation:

A parabola is a second order polynomial. By Fundamental Theorem of Algebra we know that a second order polynomial can be formed when three distinct points are known. From statement we have the following information:

(x_{1}, y_{1}) = (-2, 0), (x_{2}, y_{2}) = (6, 0), (x_{3}, y_{3}) = (0, -4)

From definition of second order polynomial and the three points described above, we have the following system of linear equations:

4\cdot a -2\cdot b + c = 0 (1)

36\cdot a + 6\cdot b + c = 0 (2)

c = -4 (3)

The solution of this system is: a = \frac{1}{3}, b = - \frac{4}{3}, c = -4. Hence, the equation of the parabola is y = \frac{1}{3}\cdot x^{2}-\frac{4}{3}\cdot x -4. Lastly, we must check if (x,y) = (2, -5) belongs to the function. If we know that x = 2, then the value of y is:

y = \frac{1}{3}\cdot (2)^{2}-\frac{4}{3}\cdot (2) - 4

y = -5.333

(x,y) = (2, -5) does not belong to the function, the real point is (x,y) = (2, -5.333).

5 0
2 years ago
Which equation has the solutions x = 1+
BaLLatris [955]

Answer:

The last equation x2 - 2x -4 = 0

has solution  (x - 1)^2 - 5 = 0,   x  =  1 + root(5)  or  x = 1 - root(5)

Step-by-step explanation:

If a quadratic function has roots 1 and 5

f(x) = (x -1)(x- 5)

f(x) = x^2 - 6x + 5

Unless you meant.  -4 and 6  ?

g(x) = (x + 4)(x - 6)

g(x) = x^2 -2x -24

-------------------------

Or did you mean  x = 1 and x =4 ?...

x^2 + 2x + 4 = 0  :   complete square  x^2 + 2x + 1 + 3 = 0,   (x+1)^2 + 3 = 0

x^2 - 2x + 4 = 0 :  complete square:  (x -1)^2 + 3 = 0

0x^2 + 2x - 4 = 0,    2x - 4 = 0,  x = 2

x^2 - 2x - 4 = 0  becomes:   x^2 - 2x + 1 - 1 -4 = 0 ;  (x - 1)^2 - 5 = 0

5 0
3 years ago
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