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KatRina [158]
3 years ago
14

Solve: a.) 5(x - 2) > -15 AND b.) x/1 - 9 < -10

Mathematics
1 answer:
uysha [10]3 years ago
6 0

Answer:

a.) x>-1

b.) x/1 - 9 < -10

Step-by-step explanation:

a.)

5(x - 2) > -15

(divide both sides)

x-2>-3

(move the constant to the right)

x>-3+2

(calculate)

x>-1

(answer)

b.)

x/1 - 9 < -10

(divide)

x-9<-10

(move the constant to the right)

x<-10+9

(calculate)

x/1 - 9 < -10

(answer)

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The answer is three because the only thing u have to do is substitute your values in and multiply then subtract. have a nice day 
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3 years ago
Please help I need help on this please please please!!!!!!
olganol [36]
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Explanation: the first thee are the fractions times the number in the parentheses. The last two are doing 10 divided by the fraction to get x. It's just reverse of the first three, because you have to find x in a different order, so you use division. Hope this helps. 

3 0
4 years ago
D) Twelve percent of the price of a bike is $64.8. What is the price of a bike?
wel

Answer:

12% of the price of a bike is $64.8.

100% of the price of a bike is a.

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5 0
4 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.
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3 years ago
Write an equation of the line that passes through the points, (0, 4), (2, 1)
Vlad1618 [11]

Answer:

y= --3/2x + 4

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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