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laiz [17]
3 years ago
12

Oliver wants to run at least 30 miles this week she has already run 14 miles by Tuesday how many miles does she need to run each

day if she wants to reach her goal in 4 more days
Mathematics
1 answer:
Olin [163]3 years ago
3 0

Answer:

If she wants to run 16 more miles in four days (30 miles minus already done 14 miles) she can run  4 miles a day for the next four days. 4 miles a day.

Step-by-step explanation:

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Which of the following expressions is equivalent to x+14−3x−27−11y
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The answer is D)-2x-11y-13

Step-by-step explanation:

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The place value of 4 in 149 is???​
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Ten

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For what value of x is the rational expression below equal to zero 20+2x/5-x
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\frac{20+2x}{5-x}=0;\ \ \ \ D:x\neq5\\\\\frac{20+2x}{5-x}=0\iff20+2x=0\\\\2x=-20\ \ \ \ /:2\\\\x=-10\in D
4 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)

4 0
2 years ago
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