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3 years ago

Sam bought 2 3/4 pounds of shrimp and 2 1/2 pounds of crabmeat. How much seafood did he buy?

Mathematics

2 answers:

Sedaia [141]3 years ago

7 0

5 1/4
thats a lot of seafood

marin [14]3 years ago

6 0

Sam bought 5 1/4 pounds of seafood.

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Answer:idk just here for the points

Step-by-step explanation:

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2 years ago

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Please help! I must match the numbers on the left with all appropriate number sets on the right . A number on the left may match

Flura [38]

Answer:

$\sqrt{5}$ = Irrational

6 = Natural, Whole, Integer, Rational

$\frac{1}{2}$ = Rational

-2 = Integers, Rational

Explanation:

See attached photo! (it is hard to explain in words haha)

Side note: All natural numbers are rational numbers, but not all rational numbers are natural numbers (this applies differently to all sections, like all integers are whole, but not all whole are integers, etc)

Side note #2: Whole numbers are just natural numbers AND zero

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2 years ago

PLZ HELP ASAP! ILL MARK BRAINLEST!

Sav [38]

Question 1.) Factor completely $12x^{4}+ 6x^{3}+18x^{2}$

Since $6$ divides all the three terms of the equation, so we take $6$ common out from the equation.

i.e., $6(2x^{4}+ 1x^{3}+3x^{2})$

Now, since $x^{2}$ divides all the three terms of the equation, so we take $x^{2}$ common out from the equation.

i.e., $6x^{2}(2x^{2}+ 1x+3)$

Now, this cannot be factored further as $2x^{2}+ 1x+3$ is a quadratic equation and taking out discriminant of it is:

$D = (1)^2 - 4(2)(3)\\ \ \ = 1- 24\\ \ \ = -23$

Since, the discriminant is less than zero. Therefore, the factored form of the given equation : $12x^{4}+ 6x^{3}+18x^{2}$ is : $6x^{2}(2x^{2}+ 1x+3)$

Question 2.) $7x^{3}y+ 14x^{2} y^{3} - 7x^{2}y^{2}$

Since $7$ divides all the three terms of the equation, so we take $7$ common out from the equation.

i.e., $7(x^{3}y+ 2x^{2} y^{3} - x^{2}y^{2})$

Now, since $x^{2}y$ divides all the three terms of the equation, so we take $x^{2}y$ common out from the equation.

i.e., $7x^{2}y(x+ 2y^{2} - y)$

Therefore, the factored form of the given equation : $7x^{3}y+ 14x^{2} y^{3} - 7x^{2}y^{2}$ is : $7x^{2}y(x+ 2y^{2} - y)$

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The Greatest Common Factor of two numbers is the biggest number which divides both the given numbers completely.

Now, for the expressions $x^{6}$ and $x^{3}$ :

Since, $x^{3}$ divides both the expressions completely and it is the biggest expression which divides both the expressions.

Therefore, $x^{3}$ is the Greatest Common Factor of $x^{6}$ and $x^{3}$

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Aleonysh [2.5K]

Answer:

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4π

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