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iren2701 [21]
4 years ago
12

Answer the question in the image

Mathematics
1 answer:
Vlad1618 [11]4 years ago
4 0

X=9

Y=3

Look at the picture to see what I did

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What are the solutions to f(x) = 2x^2 – 7x – 15
professor190 [17]

Answer:

x=-\frac{3}{2}\textrm{ or }x=5

Step-by-step explanation:

Assuming you want to solve f(x)=0,

2x^2-7x-15=0\\(2x+3)(x-5)=0\\x=-\frac{3}{2}\textrm{ or }x=5

3 0
3 years ago
Which of the following relations is a function?
Mila [183]

Answer:

Option A

Step-by-step explanation:

A does not have an x that has 2 or more outputs

3 0
3 years ago
Read 2 more answers
Calculate the area of parallelogram. <br><br>​
Nesterboy [21]

Answer:

84. What you did is right.

Step-by-step explanation:

A parallellogram's area is b x h. So the equation is 7 x 12=84.

8 0
4 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
Solve 2(3 + m)<br> What us the answer
spayn [35]

Answer:  2m+6

im sure its right :)

6 0
3 years ago
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