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3 years ago

Which is the graph of y = 2.5x? A. B. C. D.

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

5 0

The answer is D. because first you got solve the problem.

y=2.5x
(turn y to 0)
0= 2.5x
÷2.5 ÷2.5
_________
0=x

So we got one answer (0,0)

y=2.5x
(turn x to 0)
y= 2.5(0)
y=2.5

(2.5, 0)

When you do (0,0) (2.5,0) on the graph it will look like D.)

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3x-13=74 solve for x

never [62]

Answer:

x=29

Step-by-step explanation:

3x-13=74

We want to solve for x, so we need to isolate it.

The first step is to add 13 to each side.

3x-13+13=74+13

3x = 87

Then we divide by 3 on each side

3x/3 = 87/3

x= 29

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3 years ago

Explain weather or not the two triangles below could lie on the same line. Be sure to use the slope of each triangle in your jus

KonstantinChe [14]

Answer:

No, these triangles cannot lie on the same line.

Step-by-step explanation:

For two triangles to lie on the same line they must have the same slope.

The slope of the bigger triangle is

$m=\frac{rise}{run} =-\frac{18}{10} =-1.8$

and the slope of the smaller triangle is

$m=\frac{rise}{run} =-\frac{2}{3} =-0.67$

slopes are negative because the triangles are leaning to the left.

We see that the slopes of the two triangles are not the same; therefore, they cannot lie on the same line.

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

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3 years ago

If f(t)=3t^2+11 find f(7)

tamaranim1 [39]

Answer:

f(7) = 158

Step-by-step explanation:

To evaluate f(7) substitute t = 7 into f(t)

f(7) = 3(7)² + 11 = (3 × 49) + 11 = 147 + 11 = 158

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3 years ago

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3 years ago