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Rzqust [24]
2 years ago
14

Simplify 64 to the power of 0 first one to answer gets brainliest

Mathematics
1 answer:
AysviL [449]2 years ago
8 0

Answer:

A ; 1

Step-by-step explanation:

Anything to the power of 0 is equal to 1

Hope this helps

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5/9 of students at a school are girls. If there are 2367 students at the school, how many are boys?
Svet_ta [14]

the answer is 1052

5*2367 /9=1915

2367 -1315= 1052

5 0
3 years ago
Read 2 more answers
HCF of smallest 2 digit composite number and smallest 3 digit prime number is?​
posledela

Answer:

HCF = 1

Step-by-step explanation:

Smallest 2 digit composite number = 10

Smallest  3 digit prime number = 101

Factors of 10 = 1 , 2 , 5

Factors of 101 = 1 , 101

HCF = 1

6 0
2 years ago
Solve for X.<br><br> -3/2(x-2)=45/14<br><br> Please show work.
ICE Princess25 [194]

Answer:

-1/7

Step-by-step explanation:

-3/2(x-2)=45/14

x-2=(45/14)/(-3/2)

x-2=(45/14)(-2/3)

x-2=-90/42

simplify -90/42 into -15/7

x-2=-15/7

x=-15/7+2

x=-15/7+14/7

x=-1/7

7 0
3 years ago
Read 2 more answers
The sum of negative eighteen and a number is eleven. What is the number?
Oksanka [162]

Answer:

29

Step-by-step explanation:

-18 (Negative Eighteen) = A

29 (Positive Twenty Nine) = B

11 (Positive Eleven) = C


First, make A positive, and add C to it. You get B.

Now here is the equation:

-18 + (18 + 11)


-18 + (29)


11

5 0
3 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
2 years ago
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