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Lerok [7]
3 years ago
13

La temperatura mínima en una ciudad el día lunes fue de –2 ºC y la máxima fue de 7 ºC. ¿Cuál fue la variación de temperatura en

el día?
Mathematics
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

La variación de temperatura en el día es 9°C.

Step-by-step explanation:

Cuando tenemos dos valores, uno máximo M y uno mínimo m, la variación entre esos valores es la diferencia (la resta) entre el valor máximo M y el valor mínimo m

variación = M - m

En este caso sabemos que la temperatura mínima es -2°C

La temperatura máxima es 7°C

La variación sera entonces:

Variación = 7°C - (-2°C)

Recordar la regla de los signos:

(-)*(-) = (+)

Entonces:

Variación = 7°C - (-2°C) = 7°C + 2°C = 9°C

La variación de temperatura en el día es 9°C.

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Find the distance between points B and N.
Gekata [30.6K]

Answer:

5

Step-by-step explanation:

b is at 15

n is at 10

hence -15 - -10=5

positive 5 btw

5 0
2 years ago
Calculate the interest paid when 4920 is invested at 13% p.a simple interest for eight years​
Jlenok [28]

Answer:

I'll setup the problem and you can do the calculations

Step-by-step explanation:

The formula for simple interest:

I = P*r*t

I = interest

P = principle or amount invested

r = interest rate per period (period is a year)

t = number of periods

P = 4920

r = .13

t = 8

if you have questions, send a comment

5 0
3 years ago
Suppose you invested $10,000 part at 6% annual interest and the rest at 9% annual interest. If you received $684 in interest aft
victus00 [196]
Assuming simple interest (i.e. no compounding within first year), then
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At 9% interest = 10000*0.09 = $900

Two ways to find the ratio
method A. let x=proportion at 6%
then
600x+900(1-x)=684
Expand and solve
300x=900-684=216
x=216/300=0.72 or 72%
So 10000*0.72=7200 were invested at 6%
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method B: by proportions
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= 900-684 : 684-600
=216 : 84
= 18 : 7
Amount invested at 6% = 18/(18+7) * 10000 = 0.72*10000 = 7200
Amount invested at 8% = 7/(18+7)*10000=0.28*10000=2800
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