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Aneli [31]
4 years ago
6

400000 to the nearest ten thousand

Mathematics
1 answer:
Anton [14]4 years ago
4 0
The answer is 4, in my opinion
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PLZ HELP 100PTS!!!!!!!!!!!
Sonja [21]

Answer:

GJK

Step-by-step explanation:

An inscribed angle is an angle formed by two line segments in a circle  that intersect on the circle. The line segments must touch the circle at two points The angle must be on the circle itself.

GIH  is not inscribed because I is in the center of the circle  not on the circle

GJK  is an inscribed angle  GJ and JK   touch the circle at two point and are inside the circle.

IGJ  IG does not touch the circle at two points

JKL   KL  is outside the circle

8 0
3 years ago
Read 2 more answers
Mariah swam her leg of the race in 24.98 seconds and Janet swam her leg of the relay race in 25.874 seconds. What was their comb
yawa3891 [41]

Answer:

50.854 seconds

Step-by-step explanation:

24.98 + 25.874 = 50.854 seconds

5 0
3 years ago
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
3.) Use inductive reasoning to find the next two terms in the sequence. 32,
tatuchka [14]
The answer would be -10
5 0
4 years ago
If yo travel56mph how far will you go in2hr15min how long does it take to cover 168 miles
Alex73 [517]
2 hr 15 min = 2 \frac{1}{4} hours = 2.25 hours
"how far" means distance ... so distance = rate x time
distance = 56 x 2.25 = 126 miles

"how long" means time ... so using the same formula 
168 = 56 x time
divide both sides of the equation by 56 to solve
3 hours = time

5 0
3 years ago
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