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3 years ago

Si 36 naranjas cuestan 48 soles ¿cuánto se pagará por tres decenas de naranjas

Geography

1 answer:

Reil [10]3 years ago

5 0

Answer:

40 soles

Explanation:

36 naranjas - 48 soles

1 naranja - 48/36 soles = 4/3 soles

3 naranjas - 4 soles

30 naranjas - 40 soles

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What is the value of X6? <br><br> Show the solution.

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Answer : The value of $x_6$ is $\sqrt{7}$ .

Explanation :

As we are given 6 right angled triangle in the given figure.

First we have to calculate the value of $x_1$ .

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Using Pythagoras theorem in triangle 2 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_2)^2=(1)^2+(X_1)^2$

$(x_2)^2=(1)^2+(\sqrt{2})^2$

$x_2=\sqrt{(1)^2+(\sqrt{2})^2}$

$x_2=\sqrt{3}$

Now we have to calculate the value of $x_3$ .

Using Pythagoras theorem in triangle 3 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_3)^2=(1)^2+(X_2)^2$

$(x_3)^2=(1)^2+(\sqrt{3})^2$

$x_3=\sqrt{(1)^2+(\sqrt{3})^2}$

$x_3=\sqrt{4}$

Now we have to calculate the value of $x_4$ .

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$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_4)^2=(1)^2+(X_3)^2$

$(x_4)^2=(1)^2+(\sqrt{4})^2$

$x_4=\sqrt{(1)^2+(\sqrt{4})^2}$

$x_4=\sqrt{5}$

Now we have to calculate the value of $x_5$ .

Using Pythagoras theorem in triangle 5 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_5)^2=(1)^2+(X_4)^2$

$(x_5)^2=(1)^2+(\sqrt{5})^2$

$x_5=\sqrt{(1)^2+(\sqrt{5})^2}$

$x_5=\sqrt{6}$

Now we have to calculate the value of $x_6$ .

Using Pythagoras theorem in triangle 6 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_6)^2=(1)^2+(X_5)^2$

$(x_6)^2=(1)^2+(\sqrt{6})^2$

$x_6=\sqrt{(1)^2+(\sqrt{6})^2}$

$x_6=\sqrt{7}$

Therefore, the value of $x_6$ is $\sqrt{7}$ .

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You are at 35 N and 175 W; you move to a new location which is 40 south and 60 west of your present position. What are the coord

evablogger [386]

Answer:

The new coordinates are 5S and 125E

Explanation:

Addition of latitudes:

If they are in the same direction, you just have to add the values.

For example, 5N + 10N = 15N.

You have to be careful that we don't have any latitude higher than 90. So, if the result of your sum is higher than 90, we subtract 180 by the result, and the position is in the other hemisphere.

For example

60N + 40N = 100N

100N is larger than 90, so not possible

Subtracting and changing hemisphere

180 - 100N = 80S

When they are in opposite directions, you can just say that one hemisphere is the positive one, and the other is the negative.

In this problem, i am going to say that the North hemisphere is the positive one and the South the negative.

So, currently at 35N moving 40S. The new position is

$P = 35 - 40 = -5$

Since the result is negative, this just means that we are in the Southern Hemisphere, so 5S

Adittion of Longitudes:

They work just like the latitudes, the difference is that we don't have any longitude higher than 180. So, if we have a result higher than 180, we subtract 360 by this value and change the hemisphere.

In this problem.

175W + 60W = 235, higher than 180, so:

360 - 235 = 125E

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4 years ago

Si 36 naranjas cuestan 48 soles ¿cuánto se pagará por tres decenas de naranjas​

Si 36 naranjas cuestan 48 soles ¿cuánto se pagará por tres decenas de naranjas