Wolfrich lived in portugal and brazil for a total period of 14 monthsin order to learn portuguese. He learned an average of 130 new words per month when he lived in protugal and an average of 150 new words per month when he lived in brazil. In total he learned 1920 new words. How long did wolfrich live in portugal and how long did he live in brazil?

Wolfrich live in Portugal 9 months and in Brazil 5 months.

Step-by-step explanation:

Given:

Wolfrich lived in Portugal and Brazil for a total period of 14 months in order to learn Portuguese.

He learned an average of 130 new words per month when he lived in portugal and an average of 150 new words per month when he lived in Brazil.

In total he learned 1920 new words.

Now, to find the months he live in Portugal and in Brazil.

Let the number of months he live in Portugal be

And let the number of months he lived in Brazil be

So, total number of months:

Now, the total words he learned:

Substituting the value of from equation (1):

Subtracting both sides by 1820 we get:

Dividing both sides by 20 we get:

The number of months he lived in Brazil = 5 months.

Now, substituting the value of in equation (1):

The number of months he lived in Portugal = 9 months.

Therefore, Wolfrich live in Portugal 9 months and in Brazil 5 months.

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)