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butalik [34]
3 years ago
13

Please help with this limit :(((

Mathematics
2 answers:
salantis [7]3 years ago
7 0

Answer:

doesn't exist is the answer.

V125BC [204]3 years ago
6 0
Does not exist is the correct answer
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Solve the system of linear equations below. <br><br> X+y=4<br> 2x+3y=0
deff fn [24]
X + y = 4...x = 4 - y

2x + 3y = 0
2(4-y) + 3y = 0
8 - 2y + 3y = 0
-2y + 3y = 0 - 8
y = -8

x + y = 4
x - 8 = 4
x = 4 + 8
x = 12

solution is : (12,-8)
7 0
3 years ago
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If tu=18 and u=11 what is tv
yuradex [85]

If the point U is between points T and V, then the numerical length of TV is 29 units

<h3>How to determine the numerical length of segment TV?</h3>

From the question, we have the following lengths that can be used in our computation:

  • Length TU = 18 units
  • Length UV = 11 units

The above parameters and representations implies that the point U is between endpoints T and V

This also means that the length TV is longer than the other lengths TU and TV

So, we have the following length equation

TV = TU + UV

Substitute the known values in the above equation

So, we have the following equation

TV = 18 + 11

Evaluate the sum of the like terms in the above equation

So, we have the following equation

TV = 29

Hence, the numerical length of segment TV is 29 units

Read more about lengths at

brainly.com/question/19131183

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<u>Possible question</u>

If tu = 18 and uv = 11 what is tv, if point u is between points t and v

8 0
1 year ago
Are the triangles similar? Justify your answer​
cluponka [151]

Answer:

Yes

Step-by-step explanation:

If you divide all the sides of the larger triangle by 3, it's equivalent to the sides of the smaller triangle. Therefore, they are similar.

4 0
2 years ago
Which is a correct first step for solving this equation? X+7=2x+5-4x
shepuryov [24]

Step-by-step explanation:

you collect like terms from both sides that is, the variables on one side and the constants on the other side

8 0
3 years ago
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What equation would u use for the one that says find the distance
sladkih [1.3K]
You would use the distance formula

Hope this helped
7 0
3 years ago
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