Nine times a number decreased by four

Mathematics

2 answers:

Wewaii [24]3 years ago

7 0

The answer for this question is 9x-4

PolarNik [594]3 years ago

5 0

Answer: 9x - 4

Step-by-step explanation:

We know that a number must be decreased by four. Let's break it up a little.

Nine times <em>some </em>number. We don't know what nine will be multiplied by, so we will just write a placeholder number: x. This number will be multiplied by nine. So it will turn into 9x.

We must decrease (subtract) from this by four. Subtracting 9x by four can be written as 9x - 4.

You might be interested in

❗️❗️Pleaseeee helpppp❗️❗️

inysia [295]

Oma made a mistake in step B, because she should have divided both sides by b instead of just moving it to the other side. The real next step would look like:
(2A)/b=h

6 0

3 years ago

PLEASE PLEASE HELP??!?

UNO [17]

In the case of linear equations, the graph will always be a line. In contrast, a nonlinear equation may look like a parabola if it is of degree 2, a curvy x-shape if it is of degree 3, or any curvy variation thereof. While linear equations are always straight, nonlinear equations often feature curves.

4 0

3 years ago

Rewrite 9cos 4x in terms of cos x.

rosijanka [135]

$\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
9cos(4x)\implies 9[8cos^4(x)-8cos^2(x)+1]
\\\\\\
72cos^4(x)-72cos^2(x)+9$

---------------------------------------------------------------------------

as far as the previous one on the 2tan(3x)

$\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\$

$\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[ \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[ \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
\\\\\\$

$\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
\\\\\\
2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}$

4 0

3 years ago

The depth of a lake is 1,400 meters.What is the depth in kilometers?

lina2011 [118]

1.4km 1000 meters is equal to 1 kilometer

8 0

3 years ago

Read 2 more answers

Help on math? Thanks

Flauer [41]

The answer to this is 58 in

3 0

3 years ago