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2 years ago

7

stephanie needs an object whose top view is a circle and whose left and front views are squares describe an object that will sat

isfy the conditions

1 answer:

kari74 [83]2 years ago

7 0

I believe it would be a cylinder

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Are these parallelograms why or why not

ale4655 [162]

Answer:

Yes, because they both have two sets of parallel lines.

Hope that helps!

Step-by-step explanation:

3 0

2 years ago

What is the square of 16

Morgarella [4.7K]

Answer:

Square root of 16 = 4

Square of 16 = 256

Hope this helps :)

Have a great day !

5INGH

Step-by-step explanation:

6 0

3 years ago

What happens to the mean of the data set {10, 2, 8, 9, 5, 2, 6} if the number 12 is added to the data set? A. The mean increases

Setler [38]

Answer:

C. The mean increases by 0.75.

Step-by-step explanation:

The mean of a data set is given by the sum of all its values, divided by the number of values.

Data set {10, 2, 8, 9, 5, 2, 6}

7 values

Sum: 10 + 2 + 8 + 9 + 5 + 2 + 6 = 42

Mean: 42/7 = 6

Adding number 12 to the data set

Now we have 8 values

The sum is 42 + 12 = 54

The mean is 54/8 = 6.75

This means that the mean increases by 0.75.

7 0

2 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

2 years ago

⬆️<br> ⬆️<br><br>Question<br><br><br>Please Help ☹

zysi [14]

Answer:

<em>See the Venn diagram attached</em>

<u>Numbers given:</u>

<em>Total</em>: 100
Either a cat of a fish: 70
Cat: 46
Cat and fish: 16

<u>Neither a cat or a fish:</u> 100 - 70 = 30

<u>Fish:</u> 70 - (46 + 16) = 70 - 62 = 8

<u>Required probabilities are:</u>

a) P(neither) = 30/100 = 0.3
b) P(f) = (16 + 8)/100 = 0.24
c) P(f not c) = 8/100 = 0.08

6 0

2 years ago