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nataly862011 [7]
3 years ago
15

Which side lengths form a right triangle 4, 8, 12

Mathematics
1 answer:
Lostsunrise [7]3 years ago
7 0
(4^2+8^2)^.5 = 8.944
It's not 12.
So 4,8,12 can't be the side lengths of a right triangle.
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HELPPPP <br><br>which value is equal to 2✓28-5✓63?<br><br>a-37✓7<br>b-✓7<br>c-11✓7<br>d✓7​
IrinaVladis [17]

Answer:

c

Step-by-step explanation:

if I understand this right, then the expression is

2×sqrt(28) - 5×sqrt(63)

let's look for squared factors in the numbers under the square roots.

28 = 4×7 (4 is a squared number)

63 = 9×7 (9 is a squared number)

so,

2×sqrt(4×7) - 5×sqrt(9×7) =

= 2×2×sqrt(7) - 5×3×sqrt(7) = 4×sqrt(7) - 15×sqrt(7) =

= (4 - 15)×sqrt(7) = -11×sqrt(7)

7 0
1 year ago
A test requires that you answer first Part A and then either Part B or Part C. Part A consists of 4 true false questions, Part B
erma4kov [3.2K]

Answer:  374416

Step-by-step explanation:

Given : A test requires that you answer first Part A and then either Part B or Part C.

Part A consists of 4 true false questions, Part B consists of 6 multiple-choice questions with one correct answer out of five, and Part C consists of 5 multiple-choice questions with one correct answer out of six.

i.e. 2 ways to answer each question in Part A.

For 4 questions, Number of ways to answer Part A = 2^4

5 ways to answer each question in Part B.

For 6 questions, Number of ways to answer Part B = 5^6

6 ways to answer each question in Part C.

For 5 questions, Number of ways to answer Part C = 6^5

Now, the number of ways to  completed answer sheets are possible :_

2^4\times5^6+2^4\times6^5\\\\=2^4(5^6+6^5)\\\\=16(15625+7776)\\\\=16(23401)=374416

Hence, the number of ways to  completed answer sheets are possible = 374416

8 0
3 years ago
Can anyone pls help me to solve question 2 f and g and pls provide me a explanation I’m with that questions for three days
zvonat [6]

Answer:

  f)  a[n] = -(-2)^n +2^n

  g)  a[n] = (1/2)((-2)^-n +2^-n)

Step-by-step explanation:

Both of these problems are solved in the same way. The characteristic equation comes from ...

  a[n] -k²·a[n-2] = 0

Using a[n] = r^n, we have ...

  r^n -k²r^(n-2) = 0

  r^(n-2)(r² -k²) = 0

  r² -k² = 0

  r = ±k

  a[n] = p·(-k)^n +q·k^n . . . . . . for some constants p and q

We find p and q from the initial conditions.

__

f) k² = 4, so k = 2.

  a[0] = 0 = p + q

  a[1] = 4 = -2p +2q

Dividing the second equation by 2 and adding the first, we have ...

  2 = 2q

  q = 1

  p = -1

The solution is a[n] = -(-2)^n +2^n.

__

g) k² = 1/4, so k = 1/2.

  a[0] = 1 = p + q

  a[1] = 0 = -p/2 +q/2

Multiplying the first equation by 1/2 and adding the second, we get ...

  1/2 = q

  p = 1 -q = 1/2

Using k = 2^-1, we can write the solution as follows.

The solution is a[n] = (1/2)((-2)^-n +2^-n).

 

5 0
3 years ago
Plz need this asap ONLY ANSWER IF YOU KNOW IT will give brainliest
Natalija [7]

Answer:

<u>1</u> would be your y axis

Step-by-step explanation:

1. so the y axis is up and down

2. the x axis is horizontal side to side

3. the x axis is 0

4. the y axis is positive 1

5. So your anwser would be in the format (x,y)

6. It is (0,<u>1</u>)

7. It is positive

8. Plz name me brainliest I would really appreciate it if you need more help just ask me

Thank You

8 0
3 years ago
Show 2 different solutions to the task.
laila [671]

Answer with Step-by-step explanation:

1. We are given that an expression n^2+n

We have to prove that this expression is always is even for every integer.

There are two cases

1.n is odd integer

2.n is even integer

1.n is an odd positive integer

n square is also odd integer and n is odd .The sum of two odd integers is always even.

When is negative odd integer then n square is positive odd integer and n is negative odd integer.We know that difference of two odd integers is always even integer.Therefore, given expression is always even .

2.When n is even positive integer

Then n square is always positive even integer and n is positive integer .The sum of two even integers is always even.Hence, given expression is always even when n is even positive integer.

When n is negative even integer

n square is always positive even integer and n is even negative integer .The difference of two even integers is always even integer.

Hence, the given expression is always even for every integer.

2.By mathematical induction

Suppose n=1 then n= substituting in the given expression

1+1=2 =Even integer

Hence, it is true for n=1

Suppose it is true for n=k

then k^2+k is even integer

We shall prove that it is true for n=k+1

(k+1)^1+k+1

=k^1+2k+1+k+1

=k^2+k+2k+2

=Even +2(k+1)[/tex] because k^2+k is even

=Sum is even because sum even numbers is also even

Hence, the given expression is always even for every integer n.

3 0
3 years ago
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