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3 years ago

Which of the following represents the prime factortzation of 72?

Mathematics

1 answer:

rjkz [21]3 years ago

8 0

Answer:

2³ x 3²

Step-by-step explanation:

The prime factorization of a number is the process of breaking down a number to a set of prime numbers that their product will give the original number:

2 x 36

2 x 2 x 18

2 x 2 x 2 x 9

2 x 2 x 2 x 3 x 3

So;

The prime factorization is 2³ x 3²

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Six subtracted from a number equals 60.

Rama09 [41]

Answer:

Step-by-step explanation:

Is is wanting you to find the missing number, which is "x".

An equation for this is x-6=60

The is only one number that can be x.

So, I added 6 and 60, I got 66.

Than I subtracted 66 and 6, and got 60.

I know that 66 is the correct answer.

5 0

3 years ago

Read 2 more answers

Element X decays radioactively with a half life of 12 minutes. If there are 200

PolarNik [594]

$\textit{Amount for Exponential Decay using Half-Life} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{current amount}\dotfill &20\\ P=\textit{initial amount}\dotfill &200\\ t=\textit{elapsed time}\\ h=\textit{half-life}\dotfill &12 \end{cases} \\\\\\ 20=200\left( \frac{1}{2} \right)^{\frac{t}{12}}\implies \cfrac{20}{200}=\left( \frac{1}{2} \right)^{\frac{t}{12}}\implies \cfrac{1}{10}=\left( \frac{1}{2} \right)^{\frac{t}{12}}$

$\log\left( \cfrac{1}{10} \right)=\log\left[ \left( \frac{1}{2} \right)^{\frac{t}{12}} \right]\implies \log\left( \cfrac{1}{10} \right)=t\log\left[ \left( \sqrt[12]{\frac{1}{2}} \right) \right] \\\\\\ \cfrac{\log\left( \frac{1}{10} \right)}{\log\left[ \left( \sqrt[12]{\frac{1}{2}} \right) \right]}=t\implies \implies \stackrel{mins}{39.9}\approx t$

8 0

2 years ago

you have 3/5 of an apple pie. you divide the remaining pie into 5 equal slices. what fraction of the original pie is each slice?

9966 [12]

5/3.thats the fraction.

6 0

3 years ago

Read 2 more answers

Write the negation of each of the following logical expressions so that all negations immediately precede predicates. In some ca

zavuch27 [327]

Answer:

a. Negation of ∃x ∀y(P(x,y) → Q(x,y)) = ∀x ∃y P(x,y) ∧ ¬Q(x,y) ]

b. Negation of ∃x ∀y(P(x,y) → P(y,x)) = ∀x ∃y [ ¬P(x,y) ∨ ¬P(y,x) ] ∧ [P(x,y) ∨ P(y,x)]

c. Negation of ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y) = ∀x ∀y ¬ [P(x,y)] ∨ ∃x ∃y ¬ [Q(x,y)]

Step-by-step explanation:

a.∃x ∀y(P(x,y) → Q(x,y))

Negation = ¬ [ ∃x ∀y(P(x,y) → Q(x,y)) ]

= ∀x ¬ [ ∀y(P(x,y) → Q(x,y)) ]

= ∀x ∃y ¬ [ (P(x,y) → Q(x,y)) ]

= ∀x ∃y ¬ [ ¬P(x,y) ∨ Q(x,y) ]

= ∀x ∃y P(x,y) ∧ ¬Q(x,y) ]

Negation of ∃x ∀y(P(x,y) → Q(x,y)) = ∀x ∃y P(x,y) ∧ ¬Q(x,y) ]

b. ∃x ∀y(P(x,y) → P(y,x))

Negation = ¬ [ ∃x ∀y(P(x,y) → P(y,x)) ]

= ∀x ¬ [ ∀y(P(x,y) → P(y,x)) ]

= ∀x ∃y ¬ [ (P(x,y) → P(y,x)) ]

= ∀x ∃y ¬ [ ( P(x,y) ∧ P(y,x) ) ∨ ( ¬P(x,y) ∧ ¬P(y,x) )]

= ∀x ∃y ¬ [ P(x,y) ∧ P(y,x) ] ∧ ¬[ ¬P(x,y) ∧ ¬P(y,x) ]

= ∀x ∃y [ ¬P(x,y) ∨ ¬P(y,x) ] ∧ [ P(x,y) ∨ P(y,x) ]

∴ we get

Negation of ∃x ∀y(P(x,y) → P(y,x)) = ∀x ∃y [ ¬P(x,y) ∨ ¬P(y,x) ] ∧ [P(x,y) ∨ P(y,x)]

c. ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y)

Negation = ¬ [ ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y) ]

= ¬ [ ∃x ∃y P(x,y) ] ∨ ¬ [ ∀x ∀y Q(x,y) ]

= ∀x¬ [ ∃y P(x,y) ] ∨ ∃x ¬ [ ∀y Q(x,y) ]

= ∀x ∀y ¬ [ P(x,y) ] ∨ ∃x ∃y ¬ [ Q(x,y) ]

∴ we get

Negation of ∃x ∃y P(x,y) ∧ ∀x ∀y Q(x,y) = ∀x ∀y ¬ [ P(x,y) ] ∨ ∃x ∃y ¬ [ Q(x,y) ]

8 0

2 years ago

Take the sum of 6 and 7 and divide it by the sum of 10 and 3

DedPeter [7]

We can write the equation as: 6+7/10+3

Solve.

6+7/10+3

13/13

Therefore, the answer is 1.

Best of Luck!

6 0

3 years ago