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rusak2 [61]
3 years ago
10

Help me please see me!!!

Mathematics
1 answer:
Mkey [24]3 years ago
5 0
Its not really hard you just have to explain when like its not really complicated
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marcos had 1 1/3 gallons of punch left over. He poured all of it into several containers for family members to take home. use fr
Ivenika [448]
4/3 = 16/12
One container could be 5/16 one could be 4/16 and one could be 7/16

Hope I helped! Best of luck!
3 0
4 years ago
A shelf holds 2 cans of tomato soup, 8 cans of vegetable soup, 1 can of chicken noodle soup, and 8 cans of potato soup. Without
yanalaym [24]

You have an 8 out of 19 chance to grab a potato soup.


4 0
4 years ago
What is the value of 3/4 + 1/16 in lowest terms
lidiya [134]
To get operations easier lets get their denominators equal, that would be converting both to 16, lets start with that:
(3/4) = (4/4)<span>(3/4) = 4*3/*4*4 = 12/16
so, 3/4 = 12/16 they are equivalent, we can now perform the sum easily:
</span><span>3/4 + 1/16
= </span>12/16 <span>+ 1/16
= 13/16
that is the result in the lowest terms because we cannot reduce that fraction any more.</span>
7 0
3 years ago
Read 2 more answers
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
Write an expression to represent:Nine less than the quotient of two and a number xxx.
Illusion [34]

Answer:

2/x - 9

Step-by-step explanation:

Hi there!

"9 less than" means we'll be subtracting 9 from our expression

"the quotient of 2 and a number x" means we'll be dividing 2 and x:

2/x - 9

I hope this helps!

3 0
3 years ago
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