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3 years ago

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Twice a number and 5 more is 100

1 answer:

8090 [49]3 years ago

3 0

Hello there! Thank you for asking your question here at Brainly! I will be assisting you with answering this question today, and will be teaching you how to handle it on your own in the future.

First, let's take a look at our problem.
"Twice a number and 5 more is 100."
Based on this context, we are looking for a specific number to plug in.

To solve for this, I will rewrite the problem. As I do so, I will be writing (in parenthesis) the translation of an equation.

Twice a number (2x) and 5 more (+5) is 100 (=100).

Let's rewrite everything that's in parenthesis.
2x + 5 = 100

This is our equation.
To solve for this, we will need to do some basic algebra. Of course, I will teach you how to handle this on your own in future scenarios.

2x + 5 = 100
To solve for x, we need to isolate 2x, and then divide both sides by 2.

2x + 5 = 100
Subtract 5 from both sides to isolate 2x.
5 - 5 = 0
100 - 5 = 95

We now have the following equation:
2x = 95
Divide both sides by 2 to solve for x.

2x / 2 = x
95 / 2 = 47.5

x = 47.5

47.5 is your number.

I hope this helps!

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5 0

3 years ago

3x + 5 = x - 3<br> whats the answer?

tester [92]

Hello!

Here are the steps to your problem:

3x + 5 = x - 3

3x -x = -3 -5

2x = -8

x = -4

I hope this helps you! Have a lovely day!

-Mal

5 0

3 years ago

Read 2 more answers

Answer the following questions about the function whose derivative is f primef′(x)equals=x Superscript negative three fifths Ba

Elden [556K]

Answer:

(a) The critical points of f are x=0 and x=3.

(b)f is decreasing on $(0,3)$ and f is decreasing on $(3,\infty)$ .

(c) Therefore the local minimum of f is at x=3

Step-by-step explanation:

Given function is

$f'(x)= x^{-\frac35}(x-3)$

(a)

To find the critical point set f'(x)=0

$\therefore x^{-\frac35}(x-3)=0$

$\Rightarrow x=0,3$

The critical points of f are 0,3.

(b)

The interval are $(0,3)$ and $(3,\infty)$ .

To find the increasing or decreasing, taking two points one point from the interval (0,3) and another point $(3,\infty)$ .

Assume 1 and 4.

Now $f'(1)=(1)^{-\frac35}(1-3)$

and $f'(4)=(4)^{-\frac35}(4-3)>0$

Since 1∈ $(0,3)$ , f'(x)<0 and 4∈ $(3,\infty)$ , f'(x)>0

∴f is decreasing on $(0,3)$ and f is decreasing on $(3,\infty)$ .

(c)

$f'(x)= x^{-\frac35}(x-3)$

Differentiating with respect to x

$f''(x)=-\frac35x^{-\frac 85}(x-3)+x^{-\frac35}$

Now

$f''(0)=-\frac35(0)^{-\frac 85}(0-3)+(0)^{-\frac35}=0$

and

$f''(3)=-\frac35(3)^{-\frac 85}(3-3)+3^{-\frac35}$

$=0.517>0$

Since f''(x)>0 at x=3

Therefore the local minimum of f is at x=3

8 0

3 years ago

_+3=-9 pls help I will give branlyiest <br>

lions [1.4K]

Answer:

-12

Step-by-step explanation:

? + 3 = -9

Subtract 3 from -9

-9 - 3 = -12

? = -12

7 0

3 years ago