All categories

3 years ago

What is the answer to this been struggling need help ASAP -4 (y+1) > 16

Mathematics

1 answer:

bekas [8.4K]3 years ago

4 0

Answer:

y < -5

Step-by-step explanation:

-4(y+1) > 16

y+1 < -4

y < -5

You might be interested in

Someone please help i can’t find the answer.

AveGali [126]

Answer:

whats the question

Step-by-step explanation:

3 0

3 years ago

GUYS I NEED HELP ASAP I"LL MARK BRAINLYIST TO first answer. i need to know wht is the value of y in the last unlabelled workshee

Nuetrik [128]

Answer:

don't know this answer closely relates to

6 0

3 years ago

Read 2 more answers

Please help I’ll mark you as brainliest if correct

Dominik [7]

Answer:

25%

Step-by-step explanation:

5 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

Find the 12th term of the arithmetic sequence 2x + 8,6x + 5, 10x + 2, ...

lawyer [7]

Answer: The 12th term is 46x -25

Step-by-step explanation:

It is increasing by 4x and decreasing by 3 through each term, this represents the d. a1 has already been identified as 2x +8, so you just have to write down the formula and the solve form there by plugging in the 12 in (n-1) to find the answer.

8 0

3 years ago