Find the distance between the two points rounding to the nearest tenth (if necessary). (-1,8) and (8,5)

A rectangular room measures 15 feet by 20 feet. The ceiling is 9 feet high. What is the total area of the four walls? (to the ne

frozen [14]

Answer:

Answer:

825 feet of area of the wall required to be painted.

Explanation:

Given:

Length of the room = 15 feet

Width of the room =12 feet

Height of the room = 9-foot

Size of the door= 3 feet x 7 feet

To find:

The area of the walls that requirespainted=?

Solution:

We know that,

Area of walls that need paint = total surface area of room – area of door that doesn’t need paint…………(1)

Finding the Total area of the room:

room is cuboidal shape,

Then, its total surface area = 2(lb + bh + hl)

Where

h is height,

l is length,

b is width

Substituting the values,

Total surface area = 2(15 x 12 + 12 5 9 + 9 x 15)

Total surface area = 2(180+108+135)

Total surface area = 2( 423)

Total surface area = = 846 feet…………..(2)

Finding the area of the door:

Area of door = area of rectangle = length x width

Substituting the values,

area of door= 3 x 7 = 21 feet……………(3)

Substituting (2) and (3) in (1)

Area that is require paint = 846 – 21 = 825 feet

3 0

3 years ago

4{x + 8} +6 = 38 what is the value of x

zavuch27 [327]

The value of x is 0 because 4*8 is 32 +6 =38 and 4 * 0 is just 0

4 0

3 years ago

Read 2 more answers

Solve value of x. solve for x

user100 [1]

Answer:

6

Step-by-step explanation:

8x+6x+6 = 90

14x = 84

x = 6

3 0

2 years ago

The sum of the first 21 terms of the progression, - 18, - 15,-12.....is?

katovenus [111]

You can just use the tn formula to solve this tn= t1 + (n-1)d

8 0

2 years ago

Al factorizar el trinomio cuadrado perfecto, obtenemos el siguiente resultado: (que no se como resolver) xd algun pro que sepa r

PolarNik [594]

Answer:

$\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2$

Step-by-step explanation:

<u>Trinomio Cuadrado Perfecto</u>

El producto notable llamado cuadrado de un binomio se expresa como:

$(a-b)^2=a^2-2ab+b^2$

Si se tiene un trinomio, es posible convertirlo en un cuadrado perfecto si cumple con las condiciones impuestas en la fórmula:

* El primer término es un cuadrado perfecto

* El último término es un cuadrado perfecto

* El segundo término es el doble del proudcto de los dos términos del binomio.

Tenemos la expresión:

$\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8$

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

$\displaystyle a=\sqrt{\frac{100}{81}m^8p^{12}q^{16}z^2}$

$\displaystyle a=\frac{10}{9}m^4p^{6}q^{8}z$

Calculamos el valor de a como la raiz cuadrada del primer término del trinomio:

$\displaystyle b=\sqrt{\frac{1}{49}m^2p^2z^8$

$\displaystyle b=\frac{1}{7}mpz^4$

Nos cercioramos de que el término central es 2ab:

$\displaystyle 2ab=2\frac{10}{9}m^4p^{6}q^{8}z\frac{1}{7}mpz^4$

Operando:

$\displaystyle 2ab=\frac{20}{63}m^5p^7q^8z^5$

Una vez verificado, ahora podemos decir que:

$\displaystyle \frac{100}{81}m^8p^{12}q^{16}z^2-\frac{20}{63}m^5p^7q^8z^5+ \frac{1}{49}m^2p^2z^8=\left(\frac{10}{9}m^4p^{6}q^{8}z-\frac{1}{7}mpz^4\right)^2$

5 0

2 years ago

Find the distance between the two points rounding to the nearest tenth (if necessary). (-1,8) and (8,5)​

Find the distance between the two points rounding to the nearest tenth (if necessary). (-1,8) and (8,5)