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Nataly [62]
3 years ago
10

Enter the value of 5.2+(-5)-(-0.4).

Mathematics
2 answers:
Sunny_sXe [5.5K]3 years ago
6 0

Answer:

0.6

Step-by-step explanation:

MrRissso [65]3 years ago
3 0

Answer:

0.6

Step-by-step explanation:

5.2+(-5)-(-0.4) \\ 5.2 - 5 + 0.4 = 0.6 \\

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MA_775_DIABLO [31]
Answer:226
Step-by-step explanation:
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8 0
2 years ago
True or false? A histogram and a relative frequency histogram, constructed from the same data, always have the same basic shape.
alexandr402 [8]

Answer:

A. False

C. False

B. False

D. True

Step-by-step explanation:

Frequency histogram is denote the actual frequency of each observation or different class- intervals whereas Relative frequency distribution shows the fraction of that frequency to the total frequencies or the percent of that frequency to the total.

Thus the shape of both histogram are same.

A. False.

C. False

Since, Scale of the y-axis is different. But they has same shape.

B. False

Since, vertical scale is different for both histogram and horizontal scale is same.

D. True

4 0
3 years ago
_________ 5) What is the average rate of change of the function g(x) = -2x +1 over the interval [5, 6]?
yulyashka [42]

Answer:

-2

Step-by-step explanation:

They are asking for the slope. y2-y1/x2-x1

Make a table of results for the equation

x      y

5      -2(5)+1=-10+1=-9 Y1

6      -2(6)+1=-12+1=-11  Y2

-11--9=-11+9=-2

6-5=1

Average rate pf change os -2/1=-2

6 0
3 years ago
Identify the standard form of the equation by completing the square.
OLEGan [10]

Answer:

\dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1

Step-by-step explanation:

<u>Given equation</u>:

4x^2-9y^2-8x+36y-68=0

This is an equation for a horizontal hyperbola.

<u>To complete the square for a hyperbola</u>

Arrange the equation so all the terms with variables are on the left side and the constant is on the right side.

\implies 4x^2-8x-9y^2+36y=68

Factor out the coefficient of the x² term and the y² term.

\implies 4(x^2-2x)-9(y^2-4y)=68

Add the square of half the coefficient of x and y inside the parentheses of the left side, and add the distributed values to the right side:

\implies 4\left(x^2-2x+\left(\dfrac{-2}{2}\right)^2\right)-9\left(y^2-4y+\left(\dfrac{-4}{2}\right)^2\right)=68+4\left(\dfrac{-2}{2}\right)^2-9\left(\dfrac{-4}{2}\right)^2

\implies 4\left(x^2-2x+1\right)-9\left(y^2-4y+4\right)=36

Factor the two perfect trinomials on the left side:

\implies 4(x-1)^2-9(y-2)^2=36

Divide both sides by the number of the right side so the right side equals 1:

\implies \dfrac{4(x-1)^2}{36}-\dfrac{9(y-2)^2}{36}=\dfrac{36}{36}

Simplify:

\implies \dfrac{(x-1)^2}{9}-\dfrac{(y-2)^2}{4}=1

Therefore, this is the standard equation for a horizontal hyperbola with:

  • center = (1, 2)
  • vertices = (-2, 2) and (4, 2)
  • co-vertices = (1, 0) and (1, 4)
  • \textsf{Asymptotes}: \quad y = -\dfrac{2}{3}x+\dfrac{8}{3} \textsf{ and }y=\dfrac{2}{3}x+\dfrac{4}{3}
  • \textsf{Foci}: \quad  (1-\sqrt{13}, 2) \textsf{ and }(1+\sqrt{13}, 2)

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