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oksano4ka [1.4K]

3 years ago

10

A courtyard in the shape of a right triangle is set in the middle of three square office buildings, with one office building alo

ng each side. Which statement is true about the three office buildings?

1 answer:

oksian1 [2.3K]3 years ago

3 0

Answer:

The area of Building C minus the area of Building B equals the area of Building A.

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In the diagram below of right triangle ACB, altitude CD is drawn to hypotenuse AB.

WITCHER [35]

I think the answer maybe will be 3)5 and the reason why Is because

8 0

2 years ago

jill bought a cd for $1000 that earns 4.2% apr and is compounded monthly. The cd matures in 2 years. How much will this cd be wo

OLEGan [10]

A = P(1+r/12)^12t, where A = amount after two years, P = Initial amount = $1000, r =apr = 4.2% = 0.042, t = period = 2 years

Then,

A = 1000 (1+0.042/12)^12*2 = $1087.47

7 0

3 years ago

Read 2 more answers

A meat market sold five 7/5 pounds of meatballs on Monday six 1/2 pounds of meat balls on Tuesday and 4 pounds of meatballs on W

Whitepunk [10]

Answer:

Add all of them together. So your answer is

61/10 or 6 and

Step-by-step:

7/5 =1 and 2/5 or 14/10

1/2= 5/10

4/1 or 4= 40/10

14/10 + 5/10 + 40/10 = 61/10

Simplified: 6 and 1/10

8 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

What is an algebraic expression for this word phrase?

snow_lady [41]

The first choice, 5+7n.

3 0

3 years ago