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3 years ago

Subtract 2x from 4x - 3

Mathematics

2 answers:

KatRina [158]3 years ago

8 0

Answer:

2x-3

Step-by-step explanation:

If you subtract 2x from 4x-3, you got 2x-3. This must be the answer you're looking for. Hope it helps!

velikii [3]3 years ago

6 0

2x -3
4x-2x= 2x
You can’t simplify anymore

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Which of the following could be the graph of 3x + 4y = 12?

Elan Coil [88]

Answer:

A graph that goes through (4, 0), (8, -3), (0, 3), and (-4, 6)

Step-by-step explanation:

Hello!

To find what this looks like it is easier to put it into slope-intercept form which is y = mx + b

To get our equation to look like that we have to get y by itself

3x + 4y = 12

Subtract 3x from both sides

4y = -3x + 12

Divide both sides by 4

$y =- \frac{3}{4}x + 3$

We now know the y-intercept is (0, 3)

Now we follow the slope to give us more points

More points are (4, 0), (8, -3), and (-4, 6)

The answer would be a graph that goes through those points.

Hope this helps!

6 0

3 years ago

8.) a round patio is 14m in diameter. What is the distance around the patio? A. 153.86m B. 21.98m C. 43.96m D. 615.44m

professor190 [17]

So by having a circle with a diameter of 14 metres it is a sake of calculating the circumference of the patio. This is simply done by (where d = 2r, d is the diameter, and r the radius)

$C = 2\pi r = 2\cdot \pi \cdot \left \frac{14}{2} \text{ m} \approx 43.98 \text{ m}$

3 0

4 years ago

Use the method of undetermined coefficients to solve the given nonhomogeneous system. x' = −1 5 −1 1 x + sin(t) −2 cos(t)

AlekseyPX

It looks like the system is

$x' = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} x + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}$

Compute the eigenvalues of the coefficient matrix.

$\begin{vmatrix} -1 - \lambda & 5 \\ -1 & 1 - \lambda \end{vmatrix} = \lambda^2 + 4 = 0 \implies \lambda = \pm2i$

For $\lambda = 2i$ , the corresponding eigenvector is $\eta=\begin{bmatrix}\eta_1&\eta_2\end{bmatrix}^\top$ such that

$\begin{bmatrix} -1 - 2i & 5 \\ -1 & 1 - 2i \end{bmatrix} \begin{bmatrix} \eta_1 \\ \eta_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Notice that the first row is 1 + 2i times the second row, so

$(1+2i) \eta_1 - 5\eta_2 = 0$

Let $\eta_1 = 1-2i$ ; then $\eta_2=1$ , so that

$\begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} = 2i \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix}$

The eigenvector corresponding to $\lambda=-2i$ is the complex conjugate of $\eta$ .

So, the characteristic solution to the homogeneous system is

$x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix}$

The characteristic solution contains $\cos(2t)$ and $\sin(2t)$ , both of which are linearly independent to $\cos(t)$ and $\sin(t)$ . So for the nonhomogeneous part, we consider the ansatz particular solution

$x = \cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}$

Differentiating this and substituting into the ODE system gives

$-\sin(t) \begin{bmatrix} a \\ b \end{bmatrix} + \cos(t) \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ -1 & 1 \end{bmatrix} \left(\cos(t) \begin{bmatrix} a \\ b \end{bmatrix} + \sin(t) \begin{bmatrix} c \\ d \end{bmatrix}\right) + \begin{bmatrix} \sin(t) \\ -2 \cos(t) \end{bmatrix}$

$\implies \begin{cases}a - 5c + d = 1 \\ b - c + d = 0 \\ 5a - b + c = 0 \\ a - b + d = -2 \end{cases} \implies a=\dfrac{11}{41}, b=\dfrac{38}{41}, c=-\dfrac{17}{41}, d=-\dfrac{55}{41}$

Then the general solution to the system is

$x = C_1 e^{2it} \begin{bmatrix} 1 - 2i \\ 1 \end{bmatrix} + C_2 e^{-2it} \begin{bmatrix} 1 + 2i \\ 1 \end{bmatrix} + \dfrac1{41} \cos(t) \begin{bmatrix} 11 \\ 38 \end{bmatrix} - \dfrac1{41} \sin(t) \begin{bmatrix} 17 \\ 55 \end{bmatrix}$

7 0

2 years ago

I need an answer please

photoshop1234 [79]

the answer is d im pretty sure

5 0

3 years ago

Read 2 more answers

A _ P

julsineya [31]

I think it ia c and there are 25 apples

8 0

3 years ago