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3 years ago

HELP ME NO LINKS :( ........................

Mathematics

2 answers:

Reika [66]3 years ago

6 0

Answer:

y = x - 10

Step-by-step explanation:

10 less than x would be written as x - 10, and subtracting 10 from x would ensure y (the output) is always less than the number you put in for x

marishachu [46]3 years ago

5 0

$\huge{\mathbb{\tt { ANSWER↓}}}$

$\color{purple}{\tt{y=x−5}}$

$\huge{\mathbb{EXPLANATION↓}}$

You said "output" means y and "input" means x , so the only other thing you need to know is "is" means = (equals):

$\green{\boxed{\boxed{\sf{For \: This \: Part \: Just \: Click \: the \: file \: attatched \: (photo)}}}}$

$\color{orange}{\tt{Rewriting \: it \: yields:}}$

$\color{blue}{\tt=y=x−5}$

#CarryOnLearning

#LetsEnjoyTheSummer

<h3>→<em><u>XxKim02xX</u></em></h3>

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Gerald wants to buy a sweater that is $41.60 because it is on sale for 20% off

vovangra [49]

Answer: $33.28

Step-by-step explanation: 41.60x .20 = 8.32

41.60 - 8.32 = 33.28

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3 years ago

Help Please....................

Anon25 [30]

The sign would be positive because a negative x a negative is positive

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4 years ago

Read 2 more answers

How are they the same how are they different <br> y=2x +12 2x-y= -12

Anon25 [30]

Step-by-step explanation:

Let us move everything to the one side with y by itself to make it easier. the first equation we don't have to change. y=2x+12 and the second equation becomes -y=-2x-12 and we can divide both sides by negative to get rid of the negative on y so it becomes y=2x+12. So they are the same thing.

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3 years ago

Prove that the limit x tends to 1 (2x^4-6x^3+x^2+3)÷(x-1)=-8

elixir [45]

First note that if $x\neq1$ , you have

$\dfrac{2x^4-6x^3+x^2+3}{x-1}=2x^3-4x^2-3x-3$

Now, you're looking for $\delta>0$ such that for any $\varepsilon>0$ , you have

$|x-1|$

Note that you can divide through the left side of the $\varepsilon$ inequality by $x-1$ once more:

$\dfrac{2x^3-4x^2-3x+5}{x-1}=2x^3-2x-5\implies 2x^3-4x^2-3x+5=(x-1)(2x^2-2x-5)$

So it follows that you need to find an appropriate $\delta$ that will guarantee

$|(x-1)(2x^2-2x-5)|=|x-1||2x^2-2-5|$

For the moment, let's fix $\delta=1$ . Then by this assumption, we have

$|x-1|$

From this we get

$\implies0$
$\implies0$
$\implies0$
$\implies-5$
$\implies1$

where the upper bound is what we care about. With this assumption, we then get that

$|x-1||2x^2-2-5|$

which suggests that $\delta$ can be taken to be either the smaller of 1 or $\dfrac{\varepsilon}5$ , or $\delta=\min\left\{\dfrac{\varepsilon}5,1\right\}$ , to guarantee that the function gets arbitrarily close to -8.

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3 years ago

(x+y)^5 binom açılımı nedir ?

Aleksandr-060686 [28]

Answer:

Wat, hahahahahahahaha

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3 years ago