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Bond [772]
4 years ago
11

Demostración de cada una de las conclusiones de los siguientes razonamientos Por medio de inferencias lógicas. Si los secuestrad

ores se cansan, se ponen nerviosos. Si los secuestradores están armados y se ponen nerviosos, la vida de los rehenes corre peligro. Los secuestradores están cansados y armados. Por consiguiente, la vida de los rehenes corre peligro.
Mathematics
1 answer:
Tatiana [17]4 years ago
7 0

Answer:.

Step-by-step explanation:Z

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Sara has two puppies, Sonny & Cher. Every month she purchases enough food to feed her puppies for the month. Sonny eats 8.25
Ipatiy [6.2K]

The amount of Cute Puppy food needed for the  year is 1140 cans.

<h3>Determining the total number of cans that the dogs eat in a year</h3>

The first step in determining the total number of cans needed in year is to calculate the sum of dog food consumed in a week.

8.25 + 15.5 = 23.75 cans

The second step is to determine the total cans the dogs consume in a month

23.75 x 4 = 95

The third second step is to determine the total cans the dogs consume in a year:

95 x 12 = 1140 cans

<h3>Cost of cute puppy food needed in a year</h3>

1140 x ($21 /40) = $598.50

To learn more about addition, please check: brainly.com/question/19628082

3 0
3 years ago
Emma is making a scale drawing of her farm using the scale 1cm = 2.5ft. In the drawing she drew a well with a diameter of 0.5 ce
Y_Kistochka [10]
Scale : 1 cm = 2.5 ft

1 cm = 2.5 ft
0.5 cm = 2.5 x 0.5 = 1.25 ft

Circumference = πD = π(1.25) = 3.93 ft

Answer: 3.93 ft
8 0
4 years ago
Can someone help me pls!!!
Cerrena [4.2K]

Answer:

First one: Function

Second one: not a function (a function cannot have two outputs)

Third one: Function

Last one: Not a function (doesn't pass vertical line test)

Step-by-step explanation:

Hope it helps!

6 0
3 years ago
Read 2 more answers
Which of the following expressions cannot be factored a. 16+4x b. 3x+7 c. 25x+30y d. 40x-30y
Alla [95]
C. 3x + 7 because you can factor out the other 3.
a. 16+4x=4(4+X)
b. 25x+30y=5(5x+6y)
d. 40x-30y = 5(8x-6y)
hope this helps!
6 0
3 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
2 years ago
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