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Ipatiy [6.2K]
3 years ago
15

499,553 round to nearest thousand

Mathematics
2 answers:
velikii [3]3 years ago
5 0
The answer would be 500,000 because...

499,553 has a nine just before the four which tells the four to round up

~Hope I Helped




Lisa [10]3 years ago
3 0
The answer is 500,000. 
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Solve for x(need help)
bija089 [108]

Answer:

x = 18

Step-by-step explanation:

I the given triangle, it appears that M and N are the midpoints of the segments BG and BD respectively. If it so, then let us solve it.

By mid segment theorem:

2MN = GD

2(6x - 51) = 114

12x - 102 = 114

12x = 114 + 102

12x = 216

x = 216/12

x = 18

4 0
3 years ago
Qual é o coeficiente linear da funçao f(x) =2x - 1?
raketka [301]

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7 0
3 years ago
A line has a slope of 10 and includes the points (3, n) and (4,0). What is the value of n?<br> n =
ExtremeBDS [4]

Answer:

The value of n = -10  

Step-by-step explanation:

Given the points

  • (3, n)
  • (4,0)

Finding the slope between (3, n) and (4,0)

\left(x_1,\:y_1\right)=\left(3,\:n\right),\:\left(x_2,\:y_2\right)=\left(4,\:0\right)

As slope = 10

using the formula

Slope  =  [y₂ - y₁] /  [x₂ - x₁]

so substituting the values

      10   =  [0 - n] / [4 - 3]

      10   = -n / 1

     10 = -n

     n = -10    

Therefore, the value of n = -10  

8 0
2 years ago
What is the equation of the axis of symmetry for the parabola y equals start fraction one over three end fraction left parenthes
wariber [46]
The correct axis of symmetry is x = -1.

Explanation:

Our equation is
y=\frac{1}{3}(x+1)^2-3.

This is in vertex form, which is 
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In our equation, h corresponds with -1 and k corresponds with -3, making the vertex (-1, -3).

The axis of symmetry is the x-coordinate of the vertex; this makes the axis of symmetry for this equation x = -1.
6 0
2 years ago
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Brut [27]

Answer:

Step-by-step explanation:                

\displaystyle\Large\boldsymbol{} x^4+2x^3-2x^2+2x-3= \\\\\\x^4+2x^2(x-1)+2x-\underbrace{2-1}_{-3}=  \\\\\\x^4-1+2x^2(x-1)+2x-2 = \\\\\\(x^2-1)(x^2+1)+2x^2(x-1)+2(x-1) = \\\\\\\underline{(x-1)}(x+1)(x^2+1)+2x^2\underline{(x-1)} +2\underline{(x-1)} =\\\\\\(x-1)((x+1)(x^2+1)+2x^2+2)=\\\\\\(x-1)(x^3+x^2+2x^2+x+2+1)=\\\\\\(x-1)(x^3+3x^2+x+3) =\\\\\\(x-1)( \underline{(x+3)}x^2+\underline{(x+3)})=(x^2+1)\underbrace{(x-1)(x+3)}_{x^2+2x-3}= \\\\\\(x^2+1)(x^2+2x-3) : (x^2+2x-3)=x^2+1

7 0
3 years ago
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