Enter the exponents for the variables of the GCF of the following monomials.

2 answers:

8 0

The smallest exponents of the x and y term are 2. The Greatest Common Factor ( G C F ) for 9 x² y² and 5 x² y³ is: x² y².

34kurt3 years ago

4 0

Answer:

The GCF of monomial $9x^2y^2$ and $5x^2y^3$ is $x^2y^2$

Step-by-step explanation:

Given monomial $9x^2y^2$ and $5x^2y^3$

We have to find the GCF of the given monomial.

GCF stands for greatest common factor that is a factor that divides the given numbers.

$9x^2y^2$ can be written in factors form as

$9x^2y^2=3 \cdot 3 \cdot x \cdot x \cdot y \cdot y$

And

$5x^2y^3$ can be written in factors form as

$5x^2y^3=5\cdot x \cdot x \cdot y \cdot y \cdot y$

Taking common factor, we have $x^2y^2$

Thus, the GCF of monomial $9x^2y^2$ and $5x^2y^3$ is $x^2y^2$

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A cruise line is planning its menu for the next trip. Vacationers like eating steak, lobster, and chicken. The cruise line has d

pychu [463]

Answer: They should order 160 steak, 80 lobster and 260 chicken. The total cost for the meals would be $4460

Step-by-step explanation:

The cruiser will have 400 passangers on board. By taking into account that the cruise line decided to buy 25% more of meals than bookings, we can conclude that in order to calculate the ammount of meals they will order, we have to take the number of passangers on board, 400, and add to it 25% more. So the number of meal orders will be 400 + (1/4 * 400) = 500.

Let's use the letter S to simbolize the ammount of steak they should buy, with L the ammount of lobster, and with C the ammount of chicken. The sum of those ammounts should be 500, so we deduce this formula

$S + L + C = 500$

we can replace C with 500 - S - L. Since at least forty percent of the passangers will have steak dinner and at least twenty percent will have lobster, we obtain this relations:

$S \geq 400*0.4 = 160$
$L \geq 400*0.2 = 80$

We note $f$ the total cost of the meals (in $). The value of $f$ will depend on the values of S and L. Using the information we are given about the cost of each dinner, and recalling that C was 500- S - L, we can represent $f$ by this formula

$f(S,L) = 10*S + 13*L + (500-S-L) * 7$

Which is equal to $10*S + 13*L + 3500 - 7*S - 7*L = 3500 + 3*S + 6*L$

We want to minimaze the value of $f$ , since 3*S + 6*L will get bigger as we use bigger values of S and L, then we will be in a good position if we take the values of S and L as low as possible.

With the restriction we obtained, the lowest values of S and L we can take are 160 and 80 respectively; so we take those values. In conclusion