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2 years ago

¿Se puede afirmar que América es un conti- nente poco poblado? ¿Por qué?

Mathematics

1 answer:

notka56 [123]2 years ago

7 0

Answer:

El continente que se considera poco poblado debido a que su población ocupa menos de la mitad de su territorio es AMÉRICA.

Aunque no lo pareciera tenemos que el continente americano es considerado como poco poblado, pero esto es por la distribución de las sociedades que por falta de población.

La distribución de los pueblos americanos es bastante ortodoxa y esto ha hecho que las poblaciones se concentren en sitios y que ocupen menos territorio en el continente. Ademas, hay que tomar en cuenta que existe muchos territorios naturales que no se pueden poblar por ejemplo la amazonia.

Step-by-step explanation:

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His grandmother is sick, so Hassan is making a pot of bean soup to bring for dinner. The recipe calls for equal amounts of 3 kin

lions [1.4K]

Answer:

0.75 cup

Step-by-step explanation:

K + p + b = 2 1/4

K = p = b

Sence all beans are equal we can divide 2 1/4 by 3

2 1/4 / 3 = 3/4

7 0

2 years ago

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10

zvonat [6]

The approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

$FV=P\left(1-\dfrac{r}{100}\right)^n$

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is $n_1$ . Thus, by the above formula for the first car,

$0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}$

Take log both the sides as,

$\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85$

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is $n_2$ . Thus, by the above formula for the second car,

$0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}$

Take log both the sides as,

$\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14$

The difference in the ages of the two cars is,

$d=4.85-3.14\\d=1.71\rm years$

Thus, the approximate difference in the ages of the two cars, which depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

brainly.com/question/25297296

4 0

2 years ago

Help me please i didn’t pay attention in class i just fell asleep :D

Ghella [55]

3,000 feet i think since the formula is width *height *length so it should be correct.

3 0

3 years ago

Help me Solve 1/6 ÷ 2/6 please

pochemuha

$\frac{1}{6}$ ÷ $\frac{2}{6}$

First, apply this following rule: $a$ ÷ $\frac{b}{x} = a$ × $\frac{c}{b}$
$\frac{1}{6}$ × $\frac{6}{2}$
Second, apply this following rule: $\frac{a}{b}$ × $\frac{c}{d} = \frac{ac}{bd}$
$\frac{1x6}{6x2}$
Third, multiply 1 × 6 to get 6 and 6 × 2 to get 12.
$\frac{6}{12}$
Fourth, find the GCF of 6 and 12.
Factors of 6: 1, 2, 3, 6
Factors of 12: 1, 2, 3, 4, 6, 12
The GCF is 6.
Fifth, divide the numerator by the GCF.
$6$ ÷ $6 = 1$
Sixth, divide the denominator by the GCF.
$12$ ÷ $6 = 2$
Seventh, collect the new numerator and new denominator.
$\frac{1}{2}$

Answer as fraction: $\frac{1}{2}$
Answer as decimal: 0.5

8 0

3 years ago

Find the length of each side of a square plot having 172m.

Kisachek [45]

Is 172 the area or perimeter.

8 0

2 years ago