Whats 5 - x = 12 please respond fast someone

Mathematics

1 answer:

Elan Coil [88]3 years ago

3 0

5-x=12
Subtract 5 on both sides -1x=7 than divide 7 by -1 and you get x=-7

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Please help me Please

icang [17]

1. Y=1
2.Y=6
3.Y=3
4.Y=5

Explanation:
In The first one our x values are

-2 -1 0 1 2 3

And our y values are

-2 -1 0 1 2 3
3 6

This one is easy cause you can see that it’s counting by 1. For example it goes 3, 4, 5, 6, 7, 8.

THE SECOND ONE

the table is

-1 0 1 2 3 4
4 16

Our y values are 4 and 16

Subtract 16 - 4

divided by 2
12 divided by 2 is 6.

4 + 6 = 10

10 + 6 = 16

So it checks out.

To find the numbers after 4, just subtract 4 from 6 and then subtract 6 from the answer you get.

THE THRID ONE

The table is

0 1 2 3 4 5
-1 8 14

14 - 8 = 6

6 divided by 2 is 3

-1 + 3 = 2

2 + 3 = 5

5 + 3 = 8

8 + 3 = 11

11 + 3 = 14 ✔️

So this works

THE FOURTH ONE

the table is

6 7 8 9 10 11
32 47 57

47 - 57 = 10

10 divided by 2 is 5

So let’s try it

32 + 5 = 37

37 + 5 = 42

42 + 5 = 47

47 + 5 = 52

52 + 5 = 57 ✔️

So this works

3 0

3 years ago

Evaluate the limits

julsineya [31]

$x > \ln(x)$ for all $x$ , so

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}$

Similarly, $\displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}$

We can of course see the limits are identical by replacing $x\mapsto e^x$ , so that

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)$

You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,

$\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)$

Using the rule, the limit here is

$\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0$