Ask question

Ask question

All categories

3 years ago

9

Can you please help me with Questions #18-22 on this Math Worksheet

1 answer:

CaHeK987 [17]3 years ago

3 0

21. 4x+4y
22. 16 girls

You might be interested in

If mAB = 30° and mCD = 110°, find mLE.

umka21 [38]

Answer:

40

Step-by-step explanation:

Use the Outside Angles Theorem,

$\frac{110 - 30}{2} = 40$

7 0

3 years ago

Prove that the median to the base of an isosceles triangle is also: the altitude to the base.

EleoNora [17]

Hey there!

The base angles theorem converse states if two angles in a triangle are congruent, then the sides opposite those angles are also congruent. The Isosceles Triangle Theorem states that the perpendicular bisector of the base of an isosceles triangle is also the angle bisector of the vertex angle.

Let me know if you are still confused and need a better explination :)

Have a great day!

7 0

3 years ago

Read 2 more answers

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

Find the mode of this set of data 18,27,28,44,44,50,67

8_murik_8 [283]

Answer:

44

Step-by-step explanation:

the mode of a set of data is the value that occurs most often, in this case it is 44

8 0

3 years ago

Read 2 more answers

Help with this question thx

mario62 [17]

Answer:

3.5ft

Step-by-step explanation:

3 0

3 years ago