Answer:
a. $14.50
b. $43.5
Step-by-step explanation:
a. $7.00 + $7.50 = $14.50
b. y = 14.5x x=3
y = $14.5 x 3 = $43.5
De acuerdo con un sistema de ecuaciones, tiene-se que los números son 31 y 84.
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- En el sistema de ecuaciones, tiene-se que los números son x e y.
- Suma de 115, o sea,
- <u>El número mayor es dos veces más 22 unidades que el otro</u>, o sea,
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Primero se encuenta el número menor, <u>reemplazando la segunda ecuación en la primera:</u>
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El número mayor es dado en <u>función de el menor</u>, o sea:
Los números son 31 y 84.
Otro problema resuelto por sistema de ecuaciones es dado en brainly.com/question/24637096
-- He must have at least one of each color in the case, so the first 3 of the 5 marbles in the case are blue-green-black.
Now the rest of the collection consists of
4 blue
4 green
2 black
and there's space for 2 more marbles in the case.
So the question really asks: "In how many ways can 2 marbles
be selected from 4 blue ones, 4 green ones, and 2 black ones ?"
-- Well, there are 10 marbles all together.
So the first one chosen can be any one of the 10,
and for each of those,
the second one can be any one of the remaining 9 .
Total number of ways to pick 2 out of the 10 = (10 x 9) = 90 ways.
-- BUT ... there are not nearly that many different combinations
to wind up with in the case.
The first of the two picks can be any one of the 3 colors,
and for each of those,
the second pick can also be any one of the 3 colors.
So there are actually only 9 distinguishable ways (ways that
you can tell apart) to pick the last two marbles.
See attachment for answer and math work.
