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elixir [45]
4 years ago
11

1. for what constant k must f(k) always equal the constant term of f(x) for any polynomial f(x) 2. If we multiply a polynomial b

y a constant, is the result a polynomial? 3. if deg(f+g) is less than both deg f and deg g, then must f and g have the same degree?
Mathematics
1 answer:
DIA [1.3K]4 years ago
5 0

Answer:

1. k=0

2. yes, result is still a polynomial.

3. yes, f and g must have the same degree to have deg(f+g) < deg(f) or deg(g)

Step-by-step explanation:

1. for what constant k must f(k) always equal the constant term of f(x) for any polynomial f(x)

for k=0 any polynomial f(x) will reduce f(k) to the constant term.

2. If we multiply a polynomial by a constant, is the result a polynomial?

Yes, If we multiply a polynomial by a constant, the result is always a polynomial.

3. if deg(f+g) is less than both deg f and deg g, then must f and g have the same degree?

Yes.  

If

deg(f+g) < deg(f) and

deg(f+g) < deg(g)

then it means that the two leading terms cancel out, which can happen only if f and g have the same degree.

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pashok25 [27]

Answer:

320√5 it simpleflied

Step-by-step explanation:

4 0
3 years ago
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PLEASE HELP URGENT WILL MARK AND 5 STARS
Anna007 [38]

Answer: a

Step-by-step explanation: go back to 1 grade (sorry)

5 0
3 years ago
What’s 90% as a fraction in simplest form
larisa86 [58]
90% = 90/100 which simplifies to 9/10
4 0
3 years ago
(Q6) Decide if the function is an exponential function. If it is, state the initial value and the base. y= -4.8·4^x
topjm [15]

Answer:

D

Step-by-step explanation:

Exponential equation takes the form  y=a*b^x  where

  1. a is the initial value ( a ≠ 0), and
  2. b is the base ( b ≠ 1)

The equation given in the problem can be written as  y=-4.8*4^x, so it is <em>an exponential equation, </em>  where a = -4.8 and b = 4.

Thus we can say that the initial value = -4.8 and the base is 4

The correct answer is  D

3 0
3 years ago
Someone help me out please
uysha [10]

1 step (B): raise both sides of the equation to the power of 2.

(\sqrt{x+3}-\sqrt{2x-1})^2=(-2)^2,\\   (x+3)-2\sqrt{x+3}\cdot \sqrt{2x-1}+(2x-1)=4,\\ 3x+2-2\sqrt{x+3}\cdot \sqrt{2x-1}=4.

2 step (A): simplify to obtain the final radical term on one side of the equation.

-2\sqrt{x+3}\cdot \sqrt{2x-1}=4-3x-2,\\ -2\sqrt{x+3}\cdot \sqrt{2x-1}=2-3x,\\ 2\sqrt{x+3}\cdot \sqrt{2x-1}=3x-2.

3 step (F): raise both sides of the equation to the power of 2 again.

(2\sqrt{x+3}\cdot \sqrt{2x-1})^2=(3x-2)^2,\\ 4(x+3)(2x-1)=(3x-2)^2.

4 step (E): simplify to get a quadratic equation.

4(2x^2-x+6x-3)=(3x)^2-2\cdot 3x\cdot 2+2^2,\\ 8x^2+20x-12=9x^2-12x+4,\\ x^2-32x+16=0.

5 step (D): use the quadratic formula to find the values of x.

D=(-32)^2-4\cdot 16=1024-64=960, \\ \sqrt{D} =8\sqrt{5} ,\\ x_{1,2}=\dfrac{32\pm 8\sqrt{5}}{2} =16\pm 4\sqrt{5}.

6 step (C): apply the zero product rule.

x^2-32x+16=(x-16-4\sqrt{5}) (x-16+4\sqrt{5}) ,\\ (x-16-4\sqrt{5}) (x-16+4\sqrt{5}) =0,\\ x_1=16+4\sqrt{5} ,x_2=16-4\sqrt{5}.

Additional 7 step: check these solutions, substituting into the initial equation.

3 0
3 years ago
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