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MrMuchimi
3 years ago
13

The coordinates of the vertices of AABC are A(2,5), B(6, -1), and C(-4,-2). Find the perimeter of AABC, to the nearest tenth.​

Mathematics
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer:

S...tep-by-step explanation:

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You are studying for your final exam of the semester. up to this point, you received 3 exam scores of 67%, 68%, and 78℅. to rece
scoray [572]
(67 + 68 + 78 + x) / 4 = 70
(213 + x) / 4 = 70
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213 + x = 280
x = 280 - 213
x = 67

(213 + x) / 4 = 79
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lowest u can make is 67, highest u can make is 100
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2 years ago
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4x/3 + 2(3- x) = 5 Solve for x as an FRACTION
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Answer:

<h2>The answer is .... <u><em> x = 3/2</em></u><em> </em>and if you want to write it as a mixed fraction i will be x= <u><em>1 and 1/2</em></u></h2>

Step-by-step explanation:

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Is it A? If not, why?
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I just need help with #16 … can someone please help me
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7 0
3 years ago
A small plane took 3 hours to fly 960 km from Ottawa to Halifax with a tail wind. On the return trip, flying into the wind, the
Rina8888 [55]

Answer:

  • Wind speed: \rm 40\; km \cdot h^{-1}.
  • Speed of the plane in still air: \rm 320\; km \cdot h^{-1}.

Step-by-step explanation:

This problem involves two unknowns:

  • wind speed, and
  • speed of the plane in still air.

Let the speed of the wind be x \rm \; km \cdot h^{-1}, and the speed of the plane in still air be y\rm \; km \cdot h^{-1}. It takes at least two equations to find the exact solutions to a system of two variables.

Information in this question gives two equations:

  • It takes the plane three hours to travel \rm 960\; km from Ottawa to with a tail wind (that is: at a ground speed of x + y.)
  • It takes the plane four hours to travel \rm 960\; km from Halifax back to Ottawa while flying into the wind (that is: at a ground speed of -x + y.)

Create a two-by-two system out of these two equations:

\left\{ \begin{aligned}&3(x + y) = 960 && (1) \\ &4(-x + y) = 960 && (2) \end{aligned}\right..

There can be many ways to solve this system. The approach below avoids multiplying large numbers as much as possible.

Note that this system is equivalent to

\left\{ \begin{aligned}&4 \times 3 (x + y) = 4\times960 && 4 \times (1) \\ &3\times 4(-x + y) = 3\times 960 && 3 \times (2) \end{aligned}\right..

\left\{ \begin{aligned}&12 x + 12y = 4\times960 && 4 \times (1) \\ &- 12x + 12y = 3\times 960 && 3 \times (2) \end{aligned}\right..

Either adding or subtracting the two equations will eliminate one of the variables. However, subtracting them gives only 1 \times 960 on the right-hand side. In comparison, adding them will give 7 \times 960, which is much more complex to evaluate. Subtracting the second equation (3 \times (2)) from the first (4 \times (1)) will give the equation

(12 - (-12) x = 1 \times 960.

24 x = 960.

x = 40.

Substitute x back into either equation (1) or (2) of the original system. Solve for y to obtain y = 320.

In other words,

  • Wind speed: \rm 40\; km \cdot h^{-1}.
  • Speed of the plane in still air: \rm 320\; km \cdot h^{-1}.
3 0
3 years ago
Read 2 more answers
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