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Nina [5.8K]
3 years ago
6

AD, BE, and CF intersect at point G.

Mathematics
1 answer:
EastWind [94]3 years ago
5 0
Wait none of those are the answers the answer had to be 55
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Please help i will mark brainliest if your answer is correct
alukav5142 [94]

Answer:

C

Step-by-step explanation:

180-123=57

3 0
3 years ago
Use the definition of continuity to determine whether f is continuous at a.
dmitriy555 [2]
f(x) will be continuous at x=a=7 if
(i) \displaystyle\lim_{x\to7}f(x) exists,
(ii) f(7) exists, and
(iii) \displaystyle\lim_{x\to7}f(x)=f(7).

The second condition is immediate, since f(7)=8918 has a finite value. The other two conditions can be established by proving that the limit of the function as x\to7 is indeed the value of f(7). That is, we must prove that for any \varepsilon>0, we can find \delta>0 such that

|x-7|

Now,


|f(x)-f(7)|=|5x^4-9x^3+x-8925|

Notice that when x=7, we have 5x^4-9x^3+x-8925=0. By the polynomial remainder theorem, we know that x-7 is then a factor of this polynomial. Indeed, we can write

|5x^4-9x^3+x-8925|=|(x-7)(5x^3+26x^2+182x+1275)|=|x-7||5x^3+26x^2+182x+1275|

This is the quantity that we do not want exceeding \varepsilon. Suppose we focus our attention on small values \delta. For instance, say we restrict \delta to be no larger than 1, i.e. \delta\le1. Under this condition, we have

|x-7|

Now, by the triangle inequality,


|5x^3+26x^2+182x+1275|\le|5x^3|+|26x^2|+|182x|+|1275|=5|x|^3+26|x|^2+182|x|+1275

If |x|, then this quantity is moreover bounded such that

|5x^3+26x^2+182x+1275|\le5\cdot8^3+26\cdot8^2+182\cdot8+1275=6955

To recap, fixing \delta\le1 would force |x|, which makes


|x-7||5x^3+26x^2+182x+1275|

and we want this quantity to be smaller than \varepsilon, so


6955|x-7|

which suggests that we could set \delta=\dfrac{\varepsilon}{6955}. But if \varepsilon is given such that the above inequality fails for \delta=\dfrac{\varepsilon}{6955}, then we can always fall back on \delta=1, for which we know the inequality will hold. Therefore, we should ultimately choose the smaller of the two, i.e. set \delta=\min\left\{1,\dfrac{\varepsilon}{6955}\right\}.

You would just need to formalize this proof to complete it, but you have all the groundwork laid out above. At any rate, you would end up proving the limit above, and ultimately establish that f(x) is indeed continuous at x=7.
5 0
3 years ago
Which of the following functions represent exponential decay?
Alexandra [31]
Y=(3/4)^x represents exponential decay since 3/4 is less than one, so y will become increasing smaller as the exponent increases
7 0
3 years ago
2,000 mL= L<br><br> Please show work
mel-nik [20]
2000 mL= 2 L
    Hope this helps
              :D

3 0
3 years ago
Read 2 more answers
Choose an expression that is equivalent to (-2)^4/(-2)^2. out of these ​
Marianna [84]

The expression that is equivalent to (-2)^4/(-2)^2 is (-2)^6/(-2)^4

<h3>How to determine the expression that is equivalent to (-2)^4/(-2)^2?</h3>

The expression is given as:

(-2)^4/(-2)^2

Apply the law of indices

(-2)^4/(-2)^2 = (-2)^(4 -2)

4 - 2 has the same value as 6 - 4

So, we have:

(-2)^4/(-2)^2 = (-2)^(6-4)

Apply the law of indices

(-2)^4/(-2)^2 = (-2)^6/(-2)^4

Hence, the expression that is equivalent to (-2)^4/(-2)^2 is (-2)^6/(-2)^4

Read more about equivalent expressions at:

brainly.com/question/2972832

#SPJ1

3 0
2 years ago
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