3x^2 - 5x + 12 5x^2 + x

MAXImum [283]

Answer:

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

Step-by-step explanation:

5 0

3 years ago

Pls help me with this :) .

san4es73 [151]

Answer:

35 is the answer thank u 35 is a real answer

3 0

3 years ago

F(x)=lx-2l-4 i need the range of this in interval notation,ASAP LIKE RIGHT THIS SECOND I WILL PUT YOU AS A BRAINLEST<3

kumpel [21]

Answer:

f = l + -2lx-1 + 3x-1

Step-by-step explanation:

8 0

3 years ago

usa el teorema de la altura para proponer como se podria construir un segmento cuya longitud sea media proporcional entre dos se

Mrac [35]

Usando el teorema de altura

El teorema de altura relaciona la altura (h) de un triángulo rectángulo (ver figura) y los catetos de dos triángulos que son semejantes al anterior ABC, al trazar la altura (h) sobre la hipotenusa. De manera que en todo triángulo rectángulo, la altura (h) relativa a la hipotenusa es la media geométrica de las dos proyecciones de los catetos sobre la hipotenusa (n y m). Es decir, se cumple que:

 $\frac{n}{h}= \frac{h}{m}$

Dado que el problema establece construir un segmento cuya longitud sea media proporcional entre dos segmentos de 4 y 9 cm, entonces, digamos que n = 4cm y m = 9cm tenmos que:

 $\frac{4}{h}= \frac{h}{9}$

De donde:

$h= \sqrt{(4)(9)}=\sqrt{36}=6cm$

¿Cómo se podria construir si los segmentos son de a cm y b cm?

Si los segmentos son de a y b cm entonces a y b son parámetros que pueden tomar cualquier valor positivo siempre que se cumpla que:

$h= \sqrt{ab}$

6 0

3 years ago

0.38, 1.52)

Elis [28]

Answer:

The best correct option is C

x = 1.79 to x = 3

Step-by-step explanation:

Making the sport off all little thing that crowed. They are still grouped into 7. Option C is theost appropriate.

5 0

3 years ago

3x^2 - 5x + 12 5x^2 + x - 11