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Tasya [4]
3 years ago
13

What is the minimum value

Mathematics
1 answer:
IgorC [24]3 years ago
3 0
It's the smallest value in this equation which should at x=-10, the equation is x=-b/(2a)
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Solve this absolute value /x+2/-/x/=6
stira [4]

Answer:

x € ø

Step-by-step explanation:

-(x + 2) -(-x) =6,x +2 <0,x<0

x € ø, x<-2, x<0

x € ø, x €❮-∞-,2❯

x € ø

6 0
3 years ago
Divide the number by 3 and adds 2
anzhelika [568]

aiejrhejekwhvqiekwnww

8 0
2 years ago
Let f be a functions of degree 4 whose coefficients are real numbers: two of its zeros are - 3 and 4 - i. Explain why one of the
kvasek [131]

Answer:

Step-by-step explanation:

We have the following theorem, if f is a polynomial with real coefficients, we can factor it completely in factors of the degree at most 2.

Consider first a polynomial of degree two, hence it is a polynomial of the form ax^2+bx+c. The cuadratic formula tells us that the solutions are of the form

x = \frac{-b\pm \sqrt[]{b^2-4ac}}{2a}.

Note that square root, over the reals, tells us that the are only real solutions if b^2-4ac \geq 0. If that is not the case, say it's negative, the solution are complex. Then, the solutions are of the form

x = \frac{-b \pm i \sqrt[]{4ac-b^2}}{2a}. NOte that this means that if we have a complex number of the form a+bi that is a solution, then the number a-bi (who is called the complex conjugate) is also a solution.

Recall that when we have a polynomial f(x) whose a zero is the number c, then we can factor f as follows f(x) = (x-c) * p(x) where p(x) is another polynomial of lesser degree .

So far, we know that -3 and 4-i are zeros of the function f. Note that we are missing two zeros. But, since complex numbers are zeros of polynomial only by pairs (that is the number and its conjugate are zeros), then, we must have that one of the missing 2 zeros is a real number. We have 4-i as a zero, then, its complex conjugate must be also a zero, i.e 4+i is a zero.

8 0
3 years ago
GCSE maths question about the turning point of the graph.<br> please help!
Neporo4naja [7]

Answer:

The coordinates of the turning point are (2, -9)

Step-by-step explanation:

The coordinates of the turning points of the quadratic equation

y = ax² + bx + c, are (h, k), where

  • h = \frac{-b}{2a}
  • k is the value of y at x = h

∵ The equation of the curve is y = x² + bx + c

→ By comparing it with the form above

∴ a = 1

∵ The point (0, -5) lies on the curve

→ Substitute x by 0 and y by -5 in the equation to find the value f c

∵ -5 = (0)² + b(0) + c

∴ -5 = 0 + 0 + c

∴ -5 = c

→ Substitute it in the equatin

∴ y = x² + bx - 5

∵ The point (5, 0) lies on the curve

→ Substitute x by 5 and y by 0 in the equation to find the value f c

∵ 0 = (5)² + b(5) - 5

∴ 0 = 25 + 5b - 5

→ Add the like terms in the right side

∴ 0 = 20 + 5b

→ Subtract 5b from bth sides

∵ 0 - 5b = 20 + 5b - 5b

∴ -5b = 20

→ Divide both sides by -5 to find b

∴ b = -4

→ Substitute it in the equatin

∴ y = x² - 4x - 5

∵ a = 1 and b = -4

→ Substitute them in the rule of h above t find it

∵ h = \frac{-(-4)}{2(1)} = \frac{4}{2}

∴ h = 2

→ To find k, substitute x by 2 and y by k

∵ k = (2)² - 4(2) - 5

∴ k = 4 - 8 - 5

∴ k = -9

∴ The cordinates of the turning point are (2, -9)

6 0
2 years ago
What whole number is equivalent to 8/2
svetoff [14.1K]
4 is the whole number equivalent.  When dealing with an improper fraction or one that is "top heavy" you take the denominater (the bottom number) unto the numerator.
 Two(2)goes into 8, 4 times, making it whole!
7 0
3 years ago
Read 2 more answers
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