Is y=5x-25 proportional

abruzzese [7]

140,000. the hundred is under 5 so you round down

8 0

3 years ago

Tom and Becky can paint Jim's fence together in 12 hours. If all people paint at the same rate, how many hours would it take to

Delvig [45]

Answer: 8 hours

Step-by-step explanation:

Let x be the time taken by each person to paint the fence.

Given: Tom and Becky can paint Jim's fence together in 12 hours.

Since all people paint at the same rate, then we have the following equation ;-

$\dfrac{1}{x}+\dfrac{1}{x}=\dfrac{1}{12}\\\\\Rightarrow\ \dfrac{2}{x}=\dfrac{1}{12}\\\\\Rightarrow\ x=24$

Thus, the time taken by each person to paint the fence alone = 24 hours

Now, the time taken by all three paint together:-

$\dfrac{1}{t}=\dfrac{1}{24}+\dfrac{1}{24}+\dfrac{1}{24}\\\\\Rightarrow\ \dfrac{1}{t}=\dfrac{3}{24}\\\\\Rightarrow\ t=\dfrac{24}{3}=8\text{ hours}$

Hence, it would take 8 hours if all three paint together.

3 0

3 years ago

The number 123,456 is divisible by...

Montano1993 [528]

Answer:

2

Step-by-step explanation:

Because the # ends in a even number, it can be divisible by 2

7 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 $x=7$ .

5 0

3 years ago

1

Digiron [165]

Answer:

105°

Step-by-step explanation:

105°

7 0

3 years ago