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4 years ago

A number decreased by two is at most four or at least nine

Mathematics

2 answers:

JulsSmile [24]4 years ago

6 0

$x-2 \leq 4$

or

$x-2 \geq 9$

DedPeter [7]4 years ago

5 0

You can also write the answer as

9≤x-2≤4

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Translate the following words to algebra symbols, using "n" as the variable. The product of six and a number is 24.

sveta [45]

So n will be the unknown, and the expression is:
6n = 24
from where we can find the number n:
n = 24/6
n = 4

7 0

3 years ago

Any help would be appreciated! <br> 7/9+ 8/9 as a mixed number

katovenus [111]

Answer:

15/9 :))

is your answerrr

7 0

3 years ago

Read 2 more answers

Can someone plzzzzzzzzzzzzzzzzzzzzzzzz help me

Marat540 [252]

9514 1404 393

Answer:

C. 168°

Step-by-step explanation:

The exterior angle at any vertex of an n-gon will be ...

360°/n

So a 30-gon will have exterior angles of ...

360°/30 = 12°

The adjacent interior angle is supplementary to this, so is ...

180° -12° = 168°

_____

<em>Using the reasonableness test</em>

You know that the interior angles of a square are 90°, and that regular polygons with more sides have larger interior angles. The most any regular polygon's interior angle can be is 180°. The only answer choice between 90° and 180° is the correct choice: 168°.

6 0

3 years ago

Can someone please help?

goldfiish [28.3K]

Answer:

a. max. 25c. min.13c

b. 2:00

Step-by-step explanation:

6 0

3 years ago

A) Use the limit definition of derivatives to find f’(x)

Ann [662]

$\text{Given that,}\\\\f(x) = \dfrac{ 1}{3x-2}\\\\\text{First principle of derivatives,}\\\\f'(x) = \lim \limits_{h \to 0} \dfrac{f(x+h) - f(x) }{ h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3(x+h) - 2} - \tfrac 1{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{1}{3x+3h -2} - \tfrac{1}{3x-2}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{3x-2-3x-3h+2}{(3x+3h-2)(3x-2)}}{h}\\\\\\$

$~~~~~~~= \lim \limits_{h \to 0} \dfrac{\tfrac{-3h}{(3x+3h-2)(3x-2)}}{h}\\\\\\~~~~~~~~= \lim \limits_{h \to 0} \dfrac{-3h}{h(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \lim \limits_{h \to 0} \dfrac{1}{(3x+3h-2)(3x-2)}\\\\\\~~~~~~~~=-3 \cdot \dfrac{1}{(3x+0-2)(3x-2)}\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)(3x-2)}\\\\\\~~~~~~~=-\dfrac{3}{(3x-2)^2}$

$\text{Given that,}~\\\\f(x) = \dfrac{1}{3x-2}\\\\\textbf{Power rule:}\\\\\dfrac{d}{dx}(x^n) = nx^{n-1}\\\\\textbf{Chain rule:}\\\\\dfrac{dy}{dx} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}\\\\\text{Now,}\\\\f'(x) = \dfrac{d}{dx} f(x)\\\\\\~~~~~~~~=\dfrac{d}{dx} \left( \dfrac 1{3x-2} \right)\\\\\\~~~~~~~~=\dfrac{d}{dx} (3x-2)^{-1}\\\\\\~~~~~~~~=-(3x-2)^{-1-1} \cdot \dfrac{d}{dx}(3x-2)\\\\\\~~~~~~~~=-(3x-2)^{-2} \cdot 3\\\\\\~~~~~~~~=-\dfrac{3}{(3x-2)^2}$

8 0

2 years ago