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3 years ago

Parallelogram JKLM is shown on the coordinate plane below:

Mathematics

1 answer:

Bogdan [553]3 years ago

6 0

Answer:

B. K'(−6, −4); L'(−3, −3)

Step-by-step explanation:

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9. YOU BE THE TEACHER Your friend

Crank

Answer:

Step-by-step explanation:

Yes, it's reasonable.

What you are doing is solving the question by rounding. You come up with an answer. Suppose you loose the decimal somewhere and you get 0.36? Is that reasonable? Do you just write the answer in the provided blank and move on. What now?

You get it wrong?!!

But your estimate should be about 9/3 = 3. Now you look at your calculator with great misgivings, because it made a mistake. Did it or did you? Well ultimately you did, but you have to blame something. So the calculator takes the heat.

Who knows? Maybe the decimal doesn't work. It's stuck or something. In any event you should be aware that there's no way the answer could be 0.36 when you estimate it to be 3.

8 0

2 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

1 year ago

12 out of every 15

astraxan [27]

Answer: 348 students do not own a cell phone.

Step-by-step explanation:

12 out of 15 = 80%

80% of 435 = 348

4 0

3 years ago

What is the Pythagorean Theorem

Vlad [161]

Answer:

see below

Step-by-step explanation:

The Pythagorean theorem is for right triangles. It uses the two legs and the hypotenuse

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse

4 0

2 years ago

Find the coordinates of point P

Drupady [299]

Here is the formula for finding a partitioning point:
x=x1+k(x2-x1), y=y1+k(y2-y1)
k is the ratio of the segment from the beginning point to the partitioning : the whole segment. In this case, k=AP:AB=5/16
so x=1+(5/16)*(-2-1)=1/16
y=6+(5/16)(-3-6)=51/16
so the answer is (-1/16, 51/16)

Please double check my calculation by yourself.

refer to this website for the formula and how to find k:

"This ratio is called k, and is determined by writing the numerator over the sum of the numerator and the denominator of the original ratio."
https://cobbk12.blackboard.com/bbcswebdav/institution/eHigh%20School/Courses/CCVA%20CCGPS%20Coordina....

6 0

3 years ago