Solve c x 13/5=2c-11/4

If ΔABC ≅ΔDEF, then what corresponding parts are congruent?

const2013 [10]

The part of the triangles which are congruent according to the description are; segment AB and segment DE.

<h3>Which parts of the triangles are congruent?</h3>

It follows from the task content that the two triangles ABC and DEF have been established as congruent. On this note, it can be established that by the congruence theorem that corresponding sides which are congruent and whose ratios equal to a constant ratio are segments AB and segment DE.

Read more on congruence theorem;

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6 0

2 years ago

Translate the following verbal statement into an algebraic expression. Use x for your variable.

mafiozo [28]

Answer:

Step-by-step explanation:

We can decipher this algebraic expression going step by step.

1) "Eight more", meaning +8 (we can add this on later).

2) The "quotient of a number (x) squared and eleven", a quotient by definition means the answer to a quantity divided by another. The quantity in this case being x² divided by 11 (so, x²/11).

3) We can combine both of these, getting x²/11 + 8.

Your expression is x²/11 + 8.

- Similarly, you could reposition this equation like 8 + x²/11 (due to the communicative property of addition). Both answers would be acceptable.

5 0

3 years ago

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

6 0

3 years ago

Which of the following is the solution to the equation - y = 24?

Naddik [55]

Answer:

The Answer to this equation Y = 32

Step-by-step explanation:

6 0

3 years ago

Can someone please help me with question 42?

Pavlova-9 [17]

These angles should add to 180 degrees because the angles are on a straight line so you should use the formula x=180-90-38. Therefore, x should be 52 degrees

8 0

3 years ago