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Lyrx [107]
3 years ago
13

PLZ HELP IN THIS QUESTION !!! Please

Mathematics
2 answers:
ANTONII [103]3 years ago
4 0
The answer is 5/12 a mile :) Hope this helps!!
mestny [16]3 years ago
3 0
1/12 over her goal is the answer
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I'm In need Of help guys on math this assignment is due at 1:45 i'm stuck on these couples of questions I'm willing to send $3 p
kati45 [8]
What is it if i know i will hell
4 0
3 years ago
Which of the following statements would be correct to use when proving that limx→4(3x−4)=8?
Fudgin [204]

Answer:

Option c

Step-by-step explanation:

given that limit x tending to 4 of the function (3x-4) is 8

This implies for all values of x such that for epsilon >0 arbitrary small ,

||x-4||, we get

|f(x)-8|<3epsilon

this is equivalent to the option c.

Proof:

Consider

||x-4||

Hence it follows that option C is right.

3 0
3 years ago
Solve the following equation simultaneously 1/x-5/y=7, 2/x+1/y=3​
maria [59]

Answer:

  (x, y) = (1/2, -1)

Step-by-step explanation:

Subtracting twice the first equation from the second gives ...

  (2/x +1/y) -2(1/x -5/y) = (3) -2(7)

  11/y = -11 . . . . simplify

  y = -1 . . . . . . . multiply by y/-11

Using the second equation, we can find x:

  2/x +1/-1 = 3

  2/x = 4 . . . . . . . add 1

  x = 1/2 . . . . . . . multiply by x/4

The solution is (x, y) = (1/2, -1).

_____

<em>Additional comment</em>

If you clear fractions by multiplying each equation by xy, the problem becomes one of solving simultaneous 2nd-degree equations. It is much easier to consider this a system of linear equations, where the variable is 1/x or 1/y. Solving for the values of those gives you the values of x and y.

A graph of the original equations gives you an extraneous solution of (x, y) = (0, 0) along with the real solution (x, y) = (0.5, -1).

6 0
2 years ago
The picture below...
Flura [38]

Answer:

27

Step-by-step explanation:

9+8+10=27

7 0
3 years ago
In the above figure, AE = 2, CE = 3, and DE = 4. What is the length of BE?
Gre4nikov [31]
BE= 6. I had this question not too long ago. Hope this helps. :)
3 0
3 years ago
Read 2 more answers
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