5/4 (x-2)+2x-4=9 I need help with this please

2 answers:

4 0

1) distribute the 5/4 and subtract 9 from both sides

5/4x-5/2+2x-4-9=0

2)add/subtract common variables

3.25x-15.5=0

3) move the 15.5 to the other side

3.25x=15.5

4) divide both sides by 3.25 and you have x

x=4.77

marissa [1.9K]2 years ago

3 0

$\dfrac 54 (x-2) +2x -4 = 9\\\\\implies \dfrac{5x-10}{4} +2x = 9 +4 \\\\\\\implies \dfrac{5x -10 + 8x}4 = 13\\\\\implies 13x -10 = 13 \times 4 =52\\\\\implies 13x = 52 +10 = 62\\\\\implies x = \dfrac{62}{13}$

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2 years ago

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Ilia_Sergeevich [38]

I think it is coefficient variable.

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3 years ago

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2. El perimetro de un rectangulo es 64 pies. Su largo es 12 pies mas que 3 veces su ancho. Halla las dimensiones del rectangulo

Afina-wow [57]

Ancho= x
Largo= 3x+12 ("12 pies más (adición) que 3 veces (multiplicación) su ancho (x)")

<span>En el problema incluyen el perímetro (la suma de todos los lados de una figura). </span>En un rectángulo, los lados opuestos son congruentes, así que para calcular el valor de x, tenemos que sumar nuestras ecuaciones.

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x=40/8
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Luego lo único que queda hacer es substituir el valor de x en cada ecuación para calcular los valores que te piden.

4 0

3 years ago

If g(x)=4x^2-16 we’re shifted 9 united to the right and 1 down, what would the new equation be?

Inessa05 [86]

Shown in the graph

<h2>Explanation:</h2>

Using graph tools we can graph the function:

$g(x)=4x^2-16$

which is the red graph shown below. As you can see, this is a parabola. The rule for vertical and horizontal shifts is as follows:

$Let \ c \ be \ a \ positive \ real \ number. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:$

$\bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ h(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ h(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ right \ \mathbf{right}: \\ h(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ left \ \mathbf{left}: \\ h(x)=f(x+c)$