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professor190 [17]
3 years ago
14

Write the equation of the line that passes through the points (−1,−6) and (−5,1). Put your answer in fully reduced point-slope f

orm, unless it is a vertical or horizontal line.
Mathematics
1 answer:
dem82 [27]3 years ago
7 0

Answer:you get two answers. y+6=-7/4(x+5) or y-1=-7/4(x+5)

Step-by-step explanation:

By the way the 7/4 is negative if you could not tell. You get this answer by finding the slope which is 1- -6/-5- -1 and get -7/4. Then you plug it into the formula y-y1=slope(x-x1). When you plug it in you should get that answer.

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6
Drupady [299]

Answer:

  • (a) 1/2
  • (b) 1/3

Step-by-step explanation:

(a) 3 of the 6 numbers on the die are even, so the probability that one of them will show is 3/6 = 1/2.

__

(b) 2 of the 6 numbers on the die are less than 3, so the probability that one of them will show is 2/6 = 1/3.

5 0
3 years ago
Can someone help me out
LiRa [457]

3/196

Step-by-step explanation:

multiply everything

5 0
3 years ago
The capacity at the local police 225 people
andreyandreev [35.5K]
If they were 19& people at the pool what is the percentage 16.9
7 0
3 years ago
Rewrite the following equation in slope-intercept form. 6x - 3y = 19​
tensa zangetsu [6.8K]

Answer:

y= 2x - 19/3

Step-by-step explanation:

6x-3y=19

3y=6x-19

y=2x- 19/3

slope: 2

y-intercept: -19/3

7 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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