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3 years ago

Please help!!!

1 answer:

4 0

Mya took out money from the bank. When she went to get a bank statement she saw she had a $-2.00 in her account but she needed the number so she took out $30. So you would do $30+(-2.00)=(-32.00)

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Math question:

Vera_Pavlovna [14]

Answer: 102

Step-by-step explanation:

4 0

3 years ago

Find the area of the triangle.

Vsevolod [243]

Answer:

20 m²

Step-by-step explanation:

Formula (Area of triangle):

$\text{A (Triangle)} = \dfrac{1}{2} \times \text{Base} \times \text{Height}$

From the triangle, we can see that the base of the triangle is "7 + 3" m and the height of the triangle is "4" m.

$\implies \text{A (Triangle)} = \dfrac{1}{2} \times (7 + 3) \times (4)$

Simplify the right-hand-side as needed to evaluate the area.

$\implies \text{A (Triangle)} = \dfrac{1}{2} \times (10) \times (4)$

$\implies \text{A (Triangle)} = \dfrac{40}{2} = 20 \ \text{m}^{2}$

Therefore, the area of the provided triangle is 20 m².

Learn more about this topic: brainly.com/question/15442893

8 0

3 years ago

PLEASE HELP! BEEN WAITING FOR R0 MINS!!! WILL GIVE 20 POINTS!!!!

astraxan [27]

Answer:

the two minus symbols cancel out making it positive 5

Step-by-step explanation:

5 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

Eric spent $21.94, including sales tax, on 2 jerseys and 3 pairs of socks. The jerseys cost $6.75 each and the total sales tax w

kykrilka [37]

Answer:

7.72$ each

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers