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Gekata [30.6K]
3 years ago
15

Simplify each algebra expressions by combing like terms 5p + 2p + 6p​

Mathematics
2 answers:
xenn [34]3 years ago
5 0

Answer: 13p

Step-by-step explanation:

All three terms have the same variable. Therefore, you can add all three terms together into a singular term (below).

5p + 2p + 6p = 13p

wlad13 [49]3 years ago
3 0

Answer:

13 P

Step-by-step explanation:

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Please help I want to pass this semester
dusya [7]

Answer:

I think both angles are 95 degrees for question 14

Step-by-step explanation:

Because if you split the shape in half they make two triangles which is always the sum of 180 so all you do is add the two know angles together to get 128. Once you have that you would solve for your missing angles by subtracting 180 from 128 which gives you 95 degrees.

85+ m<b+ 43= 180 degrees goes for angle d to.

83+43=128

180-128=95

Now check work by plug in your answer.

85+95+43=180

180=180

making this a true statement.

4 0
2 years ago
7<br> Find mzx.<br> X (2x + 22)°<br> (9x – 15)<br> Z<br> Y<br> (4x + 11)<br> W
kakasveta [241]

Answer:

Step-by-step explanation:

7 0
2 years ago
Which of the following are polyhedra? Check all that apply.
KengaRu [80]
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7 0
2 years ago
Read 2 more answers
Solve the equation: 5 + 2(3 + 2x) = x + 3(x + 1)
Ann [662]

Answer:

No solutions

Step-by-step explanation:

In this question, you would be solving for x.

Solve:

5 + 2(3 + 2x) = x + 3(x + 1)

Use the distributive property.

5 + 6 + 4x = x + 3x + 3

Combine like terms.

11 + 4x = 4x + 3

Subtract 4x from both sides.

11 = 3

Since we don't have an "x value", there are no solutions.

"No solutions" would be your answer.

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}

6 0
3 years ago
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