Can someone help me with this?

El perímetro de un rectángulo es de 54 pulgadas. Su longitud dos veces es ancho. Encontrar la longitud y anchura del hallazgo re

elixir [45]

Longitud = L Ancho = A
2A + 2L = 54
L = 2A Sustituir este valor por L en la primera ecuación:
2A + 2(2A) = 54
2A + 4A = 54
6A = 54
A = 9
También porque L = 2A,
L = 2(9) = 18

3 0

2 years ago

The area of a square is 72.25 inches squared. Find the length of the side and the perimeter.

Nadya [2.5K]

Answer:

8.5 inches and 34 inches

Step-by-step explanation:

Area of the square = 72.25

side² = 72.25

side = \sqrt{72.25}

Side = 8.5 in

Perimeter = 4 × side

Perimeter = 4 × 8.5 in

Perimeter = 34 in

6 0

3 years ago

Determine the equation of the circle graphed below. ( please help me )

Brums [2.3K]

Answer:

$Equation : \\(x+3)^2 + ( y -2)^2 = 36$

Step-by-step explanation:

For standard form the circle's equation we need the centre of the circle and the radius.

Step 1: Find the centre

If the centre is not given find the end points of the diameter

and then find the mid point.

Let the end points of the diameter be : ( - 3 , 8 ) and ( -3 , -4 )

The mid-point of the diameter is :

$Mid-point = (\frac{-3 + - 3}{2}, \frac{-4+8}{2}) = (-3, 2)$

Therefore, centre of the circle = ( -3 , 2 )

Step 2 : Find radius

$Radius = \frac{Diameter }{2}$

Diameter is the distance between the end points ( -3 , 8) and ( -3 , -4 )

That is ,

$Diameter = \sqrt{(-3-(-3))^2 + ( -4 -8)^2}\\$

$= \sqrt{(-3 + 3)^2 + (-12)^2}\\\\=\sqrt{0 + 144}\\\\=12$

Therefore ,

$Radius = \frac{12}{2} = 6$

Step 3 : Equation of the circle

Standard equation of the circle with centre ( h ,k )

and radius ,r is :

$(x - h)^2+(y -k)^2 = r^2$

Therefore, the equation of the circle with centre ( -3, 2)

and radius = 6 is :

$(x - (-3))^2 + (y - 2)^2 = 6^2\\\\(x + 3)^2 + (y - 2)^2 = 36$

5 0

2 years ago

Which equation represents the line shown on the graph?

Firlakuza [10]

Hello there,

To find out the equation you need to use slope intercept form...

y = mx + b

The only equation above that follows that form is...

y = 2x

Hope I helped!!

-Char

5 0

3 years ago

Read 2 more answers

The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g.

Dovator [93]

The statement "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g" is FALSE.

Domain is the values of x in the function represented by y=f(x), for which y exists.

THe given statement is "The domain of (fg)(x) consists of the numbers x that are in the domains of both f and g".

Now we assume the $g(x)=x+2$ and $f(x)=\frac{1}{x-6}$

So here since g(x) is a polynomial function so it exists for all real x.

$f(x)=\frac{1}{x-6}$ does not exists when $x=6$ , so the domain of f(x) is given by all real x except 6.

Now,

$(fg)(x)=f(g(x))=f(x+2)=\frac{1}{(x+2)-6}=\frac{1}{x-4}$

So now (fg)(x) does not exists when x=4, the domain of (fg)(x) consists of all real value of x except 4.

But domain of both f(x) and g(x) consists of the value x=4.

Hence the statement is not TRUE universarily.

Thus the given statement about the composition of function is FALSE.

Learn more about Domain here -

brainly.com/question/2264373

#SPJ10

3 0

1 year ago