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katen-ka-za [31]
4 years ago
12

(2 × 3) ^2 what's the answer to this problem help me out guys ​

Mathematics
1 answer:
Shtirlitz [24]4 years ago
7 0

(2*3)^2

do the parentheses because of using PEMDAS

P= Parentheses

E= Exponents

M= Multiplication

D= Division

A= Addition

S=Subtraction

(2*3)=6

6^2

6*6 ( write out 6 two times because of the exponent)

Answer: 6*6=36

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Use the equation for the surface area of a cylinder, SA = 2^r2 + 2arh, where r is
Alenkinab [10]

Answer:

<h2> 8in</h2>

Step-by-step explanation:

the given formula in the question is not correct

here is the correct formula

S. A. =2πr^2 +2πrh

given that S.A=207.24

radius r= 3in

207.24=2*3.142*(3)^2 +2*3.142*(3)h

207.24=56.556 +18.852h

207.24-56.556=18.852h

150.684=18.852h

h=150.684/18.852

h=7.99

Approximately the height is 8in

6 0
3 years ago
A cylinder has a height of 19 feet and a radius of 6 feet. What is the volume. Use 3.14 as pi and round the answer to the neares
Lapatulllka [165]

ANSWER:V≈2148.85.I hope this helps!!

6 0
3 years ago
HELP ASAP PLEASEEE!!!!!!!!!
Sergio [31]

Answer:

all the answer choices are correct

Step-by-step explanation:

reflections & rotations dont change any of the angle measures

8 0
2 years ago
Sea un cuadrado de 2 pulgadas de lado uniendo los puntos medios se obtiene otro cuadrado inscrito en el anterior si repetimos es
Ne4ueva [31]

Answer:

1) La serie geométrica formada es

4, 2, 1,..., ∞

2) La suma al infinito de las áreas de los cuadrados es 8 in.²

Step-by-step explanation:

1) El área del primer cuadrado, a₁ = 2² = 4 pulgadas²

El área del siguiente cuadrado, a₂ = (√ (1² + 1²)) ² = (√2) ² = 2 pulg²

El área del siguiente cuadrado, a₃ = ((√ (2) / 2) ² + (√ (2) / 2) ²) = 1 pulg²

Por lo tanto, la razón común, r = a₂ / a₁ = 2/4 = a₃ / a₂ = 1/2

Las áreas de los cuadrados progresivos forman una progresión geométrica como sigue;

4, 4×(1/2), 4 ×(1/2)²,...,4×(1/2)^{\infty}

De donde obtenemos la serie geométrica formada de la siguiente manera;

4, 2, 1,..., ∞

2) La suma de 'n' términos de una progresión geométrica hasta el infinito para -1 <r <1 se da como sigue;

S_{\infty} = \dfrac{a}{1 - r}

Por lo tanto, la suma de las áreas de los cuadrados hasta el infinito se obtiene sustituyendo los valores de 'a' y 'r' en la ecuación anterior de la siguiente manera;

La \ suma \ al \ infinito \ del \ cuadrado \ S_{\infty}  = \dfrac{4 \ in.^2}{1 - \dfrac{1}{2} } = \dfrac{4 \ in.^2}{\left(\dfrac{1}{2} \right)} = 2 \times 4 \ in.^2= 8 \ in.^2

La suma al infinito de las áreas de los cuadrados, S_{\infty} = 8 in.²

7 0
3 years ago
PLEASE HELP ME
Tju [1.3M]

Answer:

ummmm... what is the -4 -3 3 4 thing???

The other part makes sense. -324/-3=108 108/-3=-36 -36/-3=12 12/-3=-4 but -4/-3= 1.33333333333333333333

Step-by-step explanation:

/-3

6 0
3 years ago
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