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Goryan [66]
3 years ago
7

1000 more than the smallest 4 digit number is​

Mathematics
1 answer:
melamori03 [73]3 years ago
3 0

Answer:

1112

Step-by-step explanation:

please make me brainlis

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Which of the following equations is of a parabola with a vertex at (1, -1)?
Helen [10]
\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
\boxed{y=a(x-{{ h}})^2+{{ k}}}\\\\
x=a(y-{{ k}})^2+{{ h}}
\end{array} \qquad\qquad  vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
(1,-1)\implies \begin{cases}
h=1\\
k=-1
\end{cases}\implies y=a(x-1)^2-1
\\\\\\
\textit{and if we assume \boxed{a=1} then}\implies y=(x-1)^2-1
5 0
4 years ago
A satellite is at a distance of 1.3776 x 108 feet from the center of Earth. There are approximately
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Answer:4.2 x 10^4 your welcome

Step-by-step explanation:

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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
You are going to draw 4 cards from a standard deck of 52 cards. How many different combinations of cards could you draw?
zloy xaker [14]

Answer:

Step-by-step explanation:

To start with we can consider how many possibilities there are for the first card which is obviously 52, as any one of the cards in the deck can be found on the top of the pile. The next card is a bit trickier as there are only 51 possible cards that it could be as the  first card takes up a spot. The third card now only has 50 possible cards it could and this continues all the way down to the last card.

is that good

5 0
3 years ago
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Brut [27]

Answer:

6,7,8,9,10

Step-by-step explanation:

3 0
3 years ago
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