All categories

3 years ago

El área de un rectángulo es 6 x al cuadrado más 2x en cm Determine su base y altura

Mathematics

1 answer:

Anettt [7]3 years ago

3 0

Answer:

Tres respuestas posibles son:

2 y 3x^2 + x

2x y 3x + 2

x y 6x + 2

y hay muchas mas.

Step-by-step explanation:

Hay muchas respuestas posibles.

6x^2 + 2x = 2(3x^2 + x) = 2x(3x + 2) = x(6x + 2)

You might be interested in

I will give u brainlist

masha68 [24]

Answer:

the y intercept is (0,-4) or just -4

Step-by-step explanation:

5 0

3 years ago

Find the graph of y=-3/4X-2

morpeh [17]

To graph the line $y=- \frac{3}{4}x-2$ we are going to take advantage of the fact that its y-intercept is -2 when x=0, so our first point is (0,-2). To find our second point, we are going to find the x-intercept; to do it, we are going to set y=0 and solve for x:
$y=- \frac{3}{4}x-2$
$0=- \frac{3}{4}x-2$
$2=- \frac{3}{4}x$
$x= \frac{2}{- \frac{3}{4} }$
$x=- \frac{8}{3}$
Now that we have our second point $(0,- \frac{8}{3} )$ , we just need to join our two points with a straight line.

We can conclude that the graph of <span>y=-3/4X-2 is:</span>

3 0

3 years ago

Solve for x. (16x + 23)°

Inga [223]

7 is the correct answer

6 0

3 years ago

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,