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jok3333 [9.3K]
3 years ago
12

What is 20% of 41,844,000?

Mathematics
1 answer:
enot [183]3 years ago
4 0
<span>20% of 41,844,000 is 8,368,800</span>
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Mutiplying mixed numbers
Soloha48 [4]
You would first turn it into an improper fraction, to do this you leave the denominator(bottom number) then times it by your variable and add the numerator(top number)

Ex:

1 3/4x1/3= 7/12
7 0
3 years ago
A Family purchased tickets to the movies in spent a total of $40.75. The family purchase five tickets there was a $.90 processi
muminat

Answer:

$7.25

Step-by-step explanation:

Total amount spent on the ticket   = $40.75

Number of ticket purchased  = 5

Processing fee per ticket  = $0.90

Unknown:

Cost of each ticket = ?

Solution:

To solve this problem, we need to find the processing fee for the 5 tickets purchase;

      Processing fee = number ticket x processing fee per ticket

      Processing fee  = 5 x 0.9  = $4.5

Now, the real cost a ticket = $40.75  - $4.5  = $36.25

So, cost per ticket is;

        Cost  = \frac{36.25}{5}   = $7.25

7 0
2 years ago
Y = f(x + 3) + 3<br> Graph
-BARSIC- [3]

Answer:f in the chat

Step-by-step explanation:

4 0
2 years ago
Write the function that the balance after t years $35000 deposit that earns 9.2% annual interest compounded quarterly
Vlada [557]

Answer:A = P(1+r/n)nt

A = 3500(1+.092/4)4t

Step-by-step explanation:

8 0
2 years ago
Calcula y comprueba las ecuaciones: plisss lo necesito alguien me puede ayudar a) 2X = 6 b) 10 + Z = 20 c) P + 9 = 11 d) 3X + 8
vesna_86 [32]

Answer:

a) x=3

b) z=10

c) P= 2

d) X=7

e) U=1

Step-by-step explanation:

Resolver una ecuación consiste en hallar los valores de la variable que hacen cierta la igualdad.

a) 2x= 6

El coeficiente es el número junto a la variable. En este caso, el coeficiente es 2. Para eliminar este número en la expresión 2x, debido a que la variable x esta multiplicada por 2, deberás dividir ambos lados de la ecuación entre 2, debido a que la operación opuesta de la multiplicación es la división.

(2x)÷2=6÷2

x= 3

Comprobar la solución de una ecuación se hace al remplazar la variable en una ecuación con el valor de la solución. La solución debería satisfacer la ecuación cuando se ingresa en esta.

En este caso:

2*3= 6

6=6

b) 10 + z= 20

En este caso se debe sumar o restar la constante que se encuentra acompañando a la variable en ambos lados de la ecuación de manera de aislar el término de la variable. En este caso:

10 - 10 + z= 20 -10

z= 10

Comprobación:

10 + z=20

10 + 10=20

20=20

c) P + 9= 11

P +9 - 9= 11 -9

P=2

Comprobación:

2 + 9= 11

11=11

d) 3X + 8 = 29

En este caso, se suma o resta la constante en ambos lados de la ecuación y luego se elimina el coeficiente de la variable mediante la división o multiplicación. Esto es:

3X + 8 - 8= 29 - 8

3X= 21

3X ÷3= 21÷3

X=7

Comprobación:

3*7 + 8=29

21+8=29

29=29

e) 2U + 8= 10

2U + 8 - 8= 10 -8

2U= 2

2U ÷2= 2÷2

U=1

Comprobación:

2*1 + 8= 10

2 + 8= 10

10=10

8 0
2 years ago
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