Find the equation for the tangent plane and the normal line at the point

Please help I’ll mark you as brainliest if correct!

WITCHER [35]

Answer:

576 cubic cm

Step-by-step explanation:

l×w×h :) 12 is l , 6 is w, 8 is h !

4 0

3 years ago

Amy is planning the seating arrangement for her wedding reception. Each round table can sit 10 guests. The head table can sit th

Zigmanuir [339]

236-6 because of head table

230 divided by 10 because of 10 per table

Answer: 23

3 0

2 years ago

BRAINLIEST!!!

Sonbull [250]

Answer:

$\sqrt{37}$

Step-by-step explanation:

The distance d between a point (m , n ) and a line in the form

Ax + By + C = 0 , is calculated as

d = $\frac{|Am+Bn+C|}{\sqrt{A^2+B^2} }$

Here (m, n ) = (6, 2 ) and

6x - y = - 3 ( add 3 to both sides )

6x - y + 3 = 0 → A = 6, B = - 1, C = 3

d = $\frac{|6(6)+-1(2)+3|}{\sqrt{6^2+(-1)^2} }$

= $\frac{|36-2+3|}{\sqrt{36+1} }$

= $\frac{|37|}{\sqrt{37} }$

= $\frac{37}{\sqrt{37} }$ × $\frac{\sqrt{37} }{\sqrt{37} }$

= $\frac{37\sqrt{37} }{37}$ ← cancel 37 on numerator/ denominator

= $\sqrt{37}$

7 0

3 years ago

Recall that a 6-bit string is a bit strings of length 6, and a bit string of weight 3, say, is one with exactly three 1's. How m

strojnjashka [21]

Answer:

1.. Total number of 6 bit strings is 64

2. Number of 6-bit strings with weight of 0 is 1

3. Number of 6-bit strings with weight of 1 is 6

4. Number of 6-bit strings with weight of 3 is 20

5. Number of 6-bit strings with weight of 5 is 6

6. Number of 6-bit strings with weight of 6 is 1

7. Number of 6-bit strings with weight of 7 is 0

Step-by-step explanation:

A bit string is a string that contains 0 and 1 only

1. Total number of 6 bit strings is 2^6 = 64

2. Number of 6 bit strings with weight 0 is 1

Explanation

Weight 0 means a string with no occurrence of 1

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 0 i.e. no occurrence of 1

6C0 = 6!/(6-0)!0!

6C0 = 6!/6!0!

6C0 = 1

Hence, the number of string with weight 0 (i.e. no occurrence of 1 ) is 1

3. Number of string with weight 1 is 6

Explanation

Weight 0 means a string with exactly 1 occurrence of '1'

Here, we are only interested in occurrence and not order of occurrence

We apply combination formula for this

nCr = n!/(n-r)!r!

n = 6 and r = 1

6C1 = 6!/(6-1)!1!

6C1 = 6!/5!1!

6C1 = 6

Hence, the number of string with weight 6

4. Number of string with weight 3 is 20

Explanation

n = 6 and r = 3

6C3 = 6!/(6-3)!3!

6C3 = 6!/3!3!

6C3 = 20

Hence, the number of string with weight 3 is 20

5. Number of string with weight 5 is 6

Explanation

n = 6 and r = 5

6C5 = 6!/(6-5)!5!

6C5 = 6!/1!5!

6C5 = 6

Hence, the number of string with weight 5 is 6

6. Number of string with weight 6 is 1

Explanation

n = 6 and r = 6

6C6 = 6!/(6-6)!6!

6C6 = 6!/0!6!

6C6 = 1

Hence, the number of string with weight 6 is 1

7. Number of string with weight 7 is 0

Weight of 7 means that a string that has 7 occurrence of 1

The total length of a 6 bit is 6

Since 6 is less than 7, there's no way a bit of weight 7 can occur.

So, the right answer for this is 0.

8 0

3 years ago

In a group of 50 people 28% of them has flue shots done. What is the probability to choose from this group 15 people that didn't

aliina [53]

Answer:

0.0025 = 0.25%

Step-by-step explanation:

First we need to find how many people don't have flue shots.

28% of 50 is equal to 0.28 * 50 = 14 people.

If 14 people have flue shots, we have 50 - 14 = 36 people that don't have flue shots.

Now, to solve this problem, the probability will be the cases where we have a group of people that don't have flue shots (that is, a combination of 36 choose 15) over the total cases (combination of 50 choose 15):

C(36,15) / C(50,15) = (36!/15!*21!) / (50!/15!*35!) = 0.0025 = 0.25%

8 0

3 years ago