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Nostrana [21]
3 years ago
6

the ceiling in jan's living room is 2.5 m high. she has a hanging lamp that hangs down 41 cm. her husbnd is 2 m tall. will he hi

t his head on the hanging lamp?
Mathematics
2 answers:
Alexus [3.1K]3 years ago
4 0

Answer:

No, his head will not hit the lamp.

Step-by-step explanation:

The height of ceiling in Jan's living room = 2.5 m

Height of the hanging lamp from the ceiling = 41 cm

                                                                         ≈ 0.41 meter

So distance between lamp and the ground = 2.5 - 0.41

                                                                        = 2.09 meter

If a person having height greater than 2.09 meter, will touch the lower edge of the lamp.

Since Jan's husband is 2(meter), tall lamp will not touch his head.

NARA [144]3 years ago
3 0

Hi! So It is impossible to say without knowing how long the lamp cord is. It also depends on what furniture is under the lamp: if there is a large coffee table directly under the lamp, for example, the husband is not going to be standing under the lamp!

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Answer:

2.82

Step-by-step explanation:

.15 x 18.80 = 2.82

8 0
3 years ago
En la división sintética sólo son tomados en cuenta los coeficientes del divisor?
Airida [17]

La división sintetica sirve para el cociente entre polinomios, la cual es particularmente util para cuando tenemos un numerador de grado alto y un denominador de grado 1.

Veremos que si, en la división sintética solo son tomados en cuenta los coeficientes del divisor.

Para explicarla, usaremos un ejemplo.

x^3 - 2x^2 + 5x + 3  será el <u>numerador</u>

x - 2 será el <u>denominador.</u>

Para poder aplicar la división sintetica, el maximo exponente en el <u>denominador </u>debe ser uno.

Así, nuestro cociente es:

\frac{x^3 - 2x^2 + 5x + 3}{x-2}

Procedemos a igualar el <u>denominador</u> a cero:

x - 2 = 0

x = 2

Ahora veremos los coeficientes del <u>numerador</u> (el <u>numerador </u>es el <u>divisor</u>).

Lo que debemos hacer es tomar cada coeficiente (comenzando por los que multiplican a los terminos con mayor exponente) y multiplicarlos por este número.

En este caso, el primer coeficiente es 1.

2*1 = 2

Ahora le sumamos esto al proximo coeficiente, que es -2:

-2 + 2 = 0

Multiplicamos esto por el valor de x, que es 2.

2*0 = 0

Ahora sumamos esto al proximo coeficiente, que es 5

5 + 0  = 5

Multiplicamos esto por el valor de x, que es 2:

5*2 = 10

Sumamos esto al ultimo coeficiente, que es 3.

10 + 3 = 13

Este ultimo valor es el residuo, y los primeros 3 números que obtuvimos son los coeficientes del cociente, entonces el cociente será:

1*x^2 + 0*x + 10 = x^2 + 10

Como vimos, solo usamos los coeficientes del numerador/divisor para realizar esta operación, por lo que si, solo los coeficientes del divisor son tomados en cuenta.

Si quieres aprender más, puedes leer:

brainly.com/question/10466403

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2 years ago
Can someone help me with this?
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Answer:

<h2>780</h2>

Step-by-step explanation:

What is the sum of the series 5^n from n=1 to 4.

5^1 + 5^2 + 5^3+ 5^4 =

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3 0
3 years ago
How many different ways can you make 82 cents using current u.s. currency
andrew11 [14]
 <span>You can probably just work it out. 

You need non-negative integer solutions to p+5n+10d+25q = 82. 

If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80. 


So this is the same as n + 2d + 5q ≤ 16 

So now you simply have to "crank out" the cases. 

Case q=0 [ n + 2d ≤ 16 ] 

Case (q=0,d=0) → n = 0 through 16 [17 possibilities] 
Case (q=0,d=1) → n = 0 through 14 [15 possibilities] 
... 
Case (q=0,d=7) → n = 0 through 2 [3 possibilities] 
Case (q=0,d=8) → n = 0 [1 possibility] 

Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81 

Case q=1 [ n + 2d ≤ 11 ] 
Case (q=1,d=0) → n = 0 through 11 [12] 
Case (q=1,d=1) → n = 0 through 9 [10] 
... 
Case (q=1,d=5) → n = 0 through 1 [2] 

Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42 

Case q=2 [ n + 2 ≤ 6 ] 
Case (q=2,d=0) → n = 0 through 6 [7] 
Case (q=2,d=1) → n = 0 through 4 [5] 
Case (q=2,d=2) → n = 0 through 2 [3] 
Case (q=2,d=3) → n = 0 [1] 

Total from case q=2: 1 + 3 + 5 + 7 = 16 

Case q=3 [ n + 2d ≤ 1 ] 
Here d must be 0, so there is only the case: 
Case (q=3,d=0) → n = 0 through 1 [2] 

So the case q=3 only has 2. 

Grand total: 2 + 16 + 42 + 81 = 141 </span>
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