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lina2011 [118]
3 years ago
15

Which two events are independent?

Mathematics
2 answers:
swat323 years ago
8 0

Answer:

A and X

Step-by-step explanation:

kakasveta [241]3 years ago
6 0

Answer: ITSSSS AAAAAA

Step-by-step explanation:

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Answer please need belp ???
Stells [14]
Since a triangles angles add up to 180 you would add the other sides and then
180-161=19
So angle 1 is 19 degrees
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3 years ago
Which point is located at (-0.5, 0.75)?
olasank [31]

Answer:

<h2>Point Z</h2><h3>Step-by-step explanation:</h3><h3></h3>

When you go to -0.5, x, you go to the left 2 times, start at 0, then move to the left 2 times.

Points G and B are incorrect.

When you go to 0.75, you move up 3 times, start at 0, then move upwards 3 times.

Best of Luck to you.

If you have any questions, feel free to comment below.

Merry Christmas!

4 0
3 years ago
Seck<br> Simplify each expression.
VARVARA [1.3K]

Answer:

what do we have to simplify?

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3 years ago
Determine if these expression is equivalent or not.<br> 7+7x;7(x+1/7)<br> 2.5(3+x);2.5x+7.5
Helga [31]
7(x+\frac{1}{7})=7 \times x+7 \times \frac{1}{7}=7x+1 \not= 7+7x \\&#10;not \ equivalent \\ \\&#10;2.5(3+x)=2.5 \times 3+ 2.5 \times x=7.5+2.5x=2.5x+7.5 \\&#10;equivalent
4 0
3 years ago
Suppose f is a function of one variable that has a continuous second derivative. Show that for any constants a and b, the functi
salantis [7]

Answer:

We can find the solution using chain rule,

For instance, if u(x,y) = f(ax+by), then

u_{x}(x,y) = a f^{'} (ax+by)  which represents derivative of x with respect to x

and then derivative of u_{x}(x,y) with respect to y is

u_{xy}(x,y) = (ab)^2 (f^{''} (ax+by))^2,  

Now, the derivative of u_{x}(x,y) with respect to x, which is the second derivative, which is

u_{xx}(x,y) = a^2 f^{''} (ax+by)

and the derivative u_{y}(x,y) and   u_{yy}(x,y) are

u_{y}(x,y) = b f^{'} (ax+by),

u_{yy}(x,y) = b^2 f^{''} (ax+by)

Finally, the solution of PDE is

u_{xx}(x,y) u_{yy}(x,y) - u_{xy} ^2 (x,y)

= (a^2 f^{''} (ax+by)) (b^2 f^{''} (ax+by)) - (ab)^2 (f^{''} (ax+by))^2

= (ab)^2 (f^{''} (ax+by))^2) - (ab)^2 (f^{''} (ax+by))^2

= 0,

As the PDE is equal to 0, it means the function u(x,y) = f(ax+by) is the solution of the given PDE.

6 0
4 years ago
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