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3 years ago

What are the steps to the Pythagorean theorem

Mathematics

1 answer:

Lorico [155]3 years ago

4 0

A squared + B squared = C squared

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which type of data would be best displayed in a box plot? HELP ME, GOD BLESS. I think it's between 1 and 2

kramer

1 because it represents a range

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3 years ago

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How many 1/4 - pound packages of cheese can the deli make with 12 - pound of cheese? Please help !!!

devlian [24]

Answer:

48 packages

Step-by-step explanation:

recipicate 1/4 which is 4 the multiply by 12

Hope this helps!!!

6 0

3 years ago

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After simplifying, how many terms does the expression 4y-6+y2-9 contain?

Alex73 [517]

Answer:

There are three terms in the simplified expression.

Step-by-step explanation:

We have to simplify the expression and have to count the number of terms that the expression has.

The expression is 4y - 6 + y² - 9

= 4y + y² - 6 - 9

= y² + 4y - 15

Therefore, there are three terms in the simplified expression, one for y² term, another is y term and the constant term. (Answer)

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3 years ago

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A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

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3 years ago

Number line , need help! Merry Christmas!

Kruka [31]

Answer:

It is A ( sister helped me if it does not work it is B) \

Have a great day and MERRY CHRISTMAS!!!

Step-by-step explanation:

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