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maw [93]
4 years ago
12

On what interval is the function f(x) = -2|x + 1| − 3 decreasing?

Mathematics
1 answer:
Murrr4er [49]4 years ago
6 0

Answer:

answer is b iam sure this is correct

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Find the standard form of the equation of the parabola with a vertex at the origin and a focus at (0, -7).
AnnyKZ [126]

Answer: D

Step-by-step explanation:

I got (-7,0) for the focus & Origin: (0,0)

3 0
3 years ago
Crystal wants to add the fractions 1/16 and 3/18 what is the lowest common denominator of the two fractions?
zhenek [66]

Answer:

288

Step-by-step explanation:

First, simplify the fraction 3/18 to 1/6, this will help a lot. Then, you just, find the LCM, and you can list out the mutliples to help you

6, 12, 18, 24, 39, 36, 42, 48, 54, 60

16, 32, 48, 64, 80

It would be easiest to do 48

But that's not exactly what the question is asking

It's asking the LCM of 1/16 and 3/18

I wanted to show you the other way so you would be able to think of it in a different way which would be easier

You would just list out the multiples again.  

18, 36, 54, 72, 90, 108, 126, 135, 153, 171, 189

16, 32, 48, 64, 80, 96, 112, 130, 146, 162, 178, 194

Ok, I give up on that

Let's just do 16*18 which is 288.

288 is your LCM

Sometimes multiplying is easier and other times listing it out is easier.

I hope this helped

Let me know if you have any questions

4 0
3 years ago
What's the solution x+1/4x=15
lianna [129]

Answer:

x = 12

Step-by-step explanation:

1.25x = 15

x = 12

3 0
3 years ago
What’s 1+2842-40+7263-1000
SpyIntel [72]

Answer:

9,066 is your answer

3 0
2 years ago
Read 2 more answers
For each, list three elements and then show it is a vector space.
Butoxors [25]

Answer:

(a) Three polynomials of degree 1 with real coefficients belong to the set P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}, then:

2+3x \in P_1

4.5+\sqrt2 x \in P_1

\log5+78x \in P_1

(b) Three polynomials of degree 1 with real coefficients that hold the relation a_0 - 2a_1 = 0 belong to the set P_2=\{a_0+a_1x\ | a_0-2 a_1 =0 \}. The relation between the coefficients is equivalent to a_1 = \frac{a_0}{2}, then:

4+2x \in P_2

13+6.5x \in P_2

10.5+5.25x \in P_2

Step-by-step explanation:

(a) Three polynomials of degree 1 with real coefficients belong to the set P_1=\{a_0+a_1x\ | a_0, a_1 \in \mathbb{R} \}, then:

  • 2+3x \in P_1
  • 4.5+\sqrt2 x \in P_1
  • \log5+78x \in P_1

A vector space is any set whose elements hold the following axioms for any \vec{u}, \vec{v} and \vec{w} and for any scalar a and b:

  1. (\vec{u} + \vec{v} )+\vec{w} = \vec{u} +( \vec{v} +\vec{w})
  2. There is the <em>zero element </em>such that: \vec{0} + \vec{u} = \vec{u} + \vec{0}
  3. For all element \vec{u}of the set, there is an element -\vec{u} such that: -\vec{u} + \vec{u} = \vec{u} + (-\vec{u}) = \vec{0}
  4. \vec{u} + \vec{v} = \vec{v} + \vec{u}
  5. a(b\vec{v}) = (ab)\vec{v}
  6. 1\vec{u} = \vec{u}
  7. a(\vec{u} + \vec{v} ) = a\vec{u} + a\vec{v}
  8. (a+b)\vec{v} = a\vec{v}+b\vec{v}

Let's proof each of them for the first set. For the proof, I will define the polynomials a_0+a_1x, b_0+b_1x and c_0+c_1x and the scalar h and g.

  1. (a_0+a_1x + b_0+b_1x)+c_0+c_1x = a_0+a_1x +( b_0+b_1x+c_0+c_1x)\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x and defining a_0+b_0+c_0 = \alpha_0 and a_1+b_1+c_1 = \alpha_1, we obtain \boxed{\alpha_0+\alpha_1x= \alpha_0+\alpha_1x} which is another polynomial that belongs to P_1
  2. A null polynomial is define as the one with all it coefficient being 0, therefore: \boxed{0 + a_0+a_1x = a_0+a_1x + 0 = a_0+a_1x}
  3. Defining the inverse element in the addition as -a_0-a_1x, then -a_0-a_1x + a_0 + a_1x = a_0+a_1x + (-a_0-a_1x)\\\boxed{(-a_0+a_0)+(-a_1+a_1)x = (a_0-a_0)+(a_1-a_1)x = 0}
  4. (a_0+a_1x) +( b_0+b_1x) =( b_0+b_1x) +( a_0+a_1x)\\(a_0+b_0)+(a_1+b_1)x = (b_0+a_0)+(b_1+a_1)x\\\boxed{(a_0+b_0)+(a_1+b_1)x = (a_0+b_0)+(a_1+b_1)x}
  5. a[b(a_0+a_1x)] = ab (a_0+a_1x)\\a[ba_0+ba_1x] = aba_0+aba_1x\\\boxed{aba_0+aba_1x = aba_0+aba_1x}
  6. \boxed{1 \cdot (a_0+a_1x) = a_0+a_1x}
  7. \boxed{a[(a_0+a_1x)+(b_0+b_1x)] = a(a_0+a_1x) + a(b_0+b_1x)}
  8. (a+b)(a_0+a_1x)=aa_0+aa_1x+ba_0+ab_1x\\\boxed{(a+b)(a_0+a_1x)= a(a_0+a_1x) + b (a_0+a_1x)}

With this, we proof the set P_1 is a vector space with the usual polynomial addition and scalar multiplication operations.

(b) Three polynomials of degree 1 with real coefficients that hold the relation a_0 - 2a_1 = 0 belong to the set P_2=\{a_0+a_1x\ | a_0-2 a_1 =0 \}. The relation between the coefficients is equivalent to a_1 = \frac{a_0}{2}, then:

  • 4+2x \in P_2
  • 13+6.5x \in P_2
  • 10.5+5.25x \in P_2

Let's proof each of axioms for this set. For the proof, I will define again the polynomials a_0+a_1x, b_0+b_1x and c_0+c_1x and the scalar h and g. Again the relation a_1 = \frac{a_0}{2} between the coefficients holds

  1. [(a_0+a_1x) +( b_0+b_1x)]+(c_0+c_1x) = (a_0+a_1x) +[( b_0+b_1x)+(c_0+c_1x)]\\(a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x and considering the coefficient relation and defining a_0+b_0+c_0 = \alpha_0 and a_1+b_1+c_1 = \alpha_1, we have (a_0+b_0+c_0) + (a_1+b_1+c_1)x = (a_0+b_0+c_0) + (a_1+b_1+c_1)x\\(a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x = (a_0+b_0+c_0) + \frac{1}{2} (a_0+b_0+c_0)x\\\boxed{\alpha_0 + \alpha1x = \alpha_0 + \alpha1x} which is another element of the set since it is a degree one polynomial whose coefficient follow the given relation.

The proof of the other axioms can be done using the same logic as in (a) and checking that the relation between the coefficients is always the same.

6 0
4 years ago
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