All categories

3 years ago

Given a=5 b=12 c=13 5²+12²=13²

Mathematics

1 answer:

Advocard [28]3 years ago

4 0

Answer:

a^2+b^2=a(2)^1.5

Step-by-step explanation:

im not sure, just seems right

You might be interested in

Write the area A of a square as a function of its perimeter P. ...?

Svetradugi [14.3K]

The answer is A = P²/16

The perimeter P of a square is sum of its sides s: P = s + s + s + s = 4s
The area A of a square with side s is: A = s * s = s²

Step 1: Solve s from the formula for the perimeter.
Step 2: substitute s from the formula for the perimeter into the formula for the area.

Step 1:
P = 4s
s = P/4

Step 2:
A = s²
s = P/4
A = (P/4)²
A = P²/4²
A = P²/16

4 0

3 years ago

If you are buying at item at a store for a price of $17, and there is a 8.7% tax added on top, what would be the total amount du

anyanavicka [17]

Answer:

$18.48

Step-by-step explanation:

To find a percentage you just multiply the amount x the percentage in decimal form.

17 x 0.087 = 1.479 tax

Then you add that number too the original total.

17+1.479 = 18.479

Then you round to the nearest cent.

18.48

Total:

$18.48

8 0

3 years ago

Read 2 more answers

Suppose you got 3 out of every 4 questions

Andrej [43]

Answer:

16 questions

Step-by-step explanation:

12÷3/4

12×4/3

48/3=16

7 0

3 years ago

Read 2 more answers

Use matrices and elementary row to solve the following system:

LiRa [457]

I assume the first equation is supposed to be

$5x-3y+2z=13$

and not

$5x-3x+2x=4x=13$

As an augmented matrix, this system is given by

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\4&-2&4&12\end{array}\right]$

Multiply through row 3 by 1/2:

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\2&-1&2&6\end{array}\right]$

Add -1(row 2) to row 3:

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&5&5\end{array}\right]$

Multiply through row 3 by 1/5:

$\left[\begin{array}{ccc|c}5&-3&2&13\\2&-1&-3&1\\0&0&1&1\end{array}\right]$

Add -2(row 3) to row 1, and add 3(row 3) to row 2:

$\left[\begin{array}{ccc|c}5&-3&0&11\\2&-1&0&4\\0&0&1&1\end{array}\right]$

Add -3(row 2) to row 1:

$\left[\begin{array}{ccc|c}-1&0&0&-1\\2&-1&0&4\\0&0&1&1\end{array}\right]$

Multiply through row 1 by -1:

$\left[\begin{array}{ccc|c}1&0&0&1\\2&-1&0&4\\0&0&1&1\end{array}\right]$

Add -2(row 1) to row 2:

$\left[\begin{array}{ccc|c}1&0&0&1\\0&-1&0&2\\0&0&1&1\end{array}\right]$

Multipy through row 2 by -1:

$\left[\begin{array}{ccc|c}1&0&0&1\\0&1&0&-2\\0&0&1&1\end{array}\right]$

The solution to the system is then

$\boxed{x=1,y=-2,z=1}$

5 0

3 years ago

The statement is either true in all cases or false. If false, construct a specific example to show that the statement is not alw

natali 33 [55]

Answer: The correct option is (B).

True. The vector V₃ is a linear combination of V₁ and V₂, so at least one of the vectors in the set is a linear combination of the others and set is linearly dependent.

Step-by-step explanation:

Given that;

If V₁ ....... V₄ are in R⁴ and V₃ = 2V₁ + V₂ then {V₁, V₂, V₃, V₄} is linearly dependent.

Lets {V₁, V₂, V₃, V₄} are linearly dependent

Then there exist a scalars C₁, C₂, C₃, C₄

So that C₁V₁ + C₂V₂ + C₃V₃ + C₄V₄ = 0

where at least one of the Ci ≠ 0.

Take C₃ ≠ 0 then we have V₃ = (C₁V₁ + C₂V₂ + C₄V₄) / C₃

V₃ is a linear combination of V₁, V₂ and V₄}

that is

Given V₃ = 2V₁ + V₂

⇒ 2V₁ + V₂ - V₃ = 0

⇒ 2V₁ + V₂ + V₃ + 0V₃ = 0

Here, C₁ = 1, C₂ = 1, C₃ = -1 and C₄ = 0

So that {V₁, V₂, V₃, V₄} is linearly dependent.

therefore option B id the right answer.

- True. The vector V₃ is a linear combination of V₁ and V₂, so at least one of the vectors in the set is a linear combination of the others and set is linearly dependent.

4 0

3 years ago