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nikitadnepr [17]
3 years ago
10

Solve the following systems of linear equations using any algebraic method. If possible, check your solution.

Mathematics
1 answer:
julia-pushkina [17]3 years ago
5 0
-8x+18y=-12
24x-18y=18
16x=6
X=3/8
Y=3-12)/6
y=-3/2
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Write an equation of the line.
jarptica [38.1K]

Answer:

y + 5 = 0(x + 1)

y + 5 = 0

y = -5

standard form

0x + y = -5

7 0
3 years ago
Help please!
Stella [2.4K]

Answer:

1.What is the y intercept of the function?

y-coordinate when x = 0, take from the table.

  • (0, 5)

2.What is the first difference?

Difference of y-values. This is the slope

  • 7-5 = 2

3.Write an equation to represent the function given in the table

Use the first difference and the y-intercept: b = 5, m = 2

  • y = 2x + 5
4 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
Please help me if you can
nikdorinn [45]

answer

b3 is d long so 40 -d shape gives 36

8 0
3 years ago
Alana runs a game at the fair and needs to count the people who play. At the beginning of her shift 56 people had played. This t
Bingel [31]

Answer:

y=14x+56

Step-by-step explanation:

Question says to find the equation in slope intercept form that means we have to use y=mx+b form.

Where b is the initial value or you can say y-value when x=0

Given that at shift x=0, number of people y=56

so that means b=56

now find the difference between consecuitve value of y

70-56=14

84-70=14

98-84=14

We are getting constant difference of 14 so that mean slope m=14

now plug values of m and b into y=mx+b

Then final answer will be y=14x+56

8 0
4 years ago
Read 2 more answers
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