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timofeeve [1]
3 years ago
5

Expand (2x+y)(2x_y)​

Mathematics
1 answer:
OLga [1]3 years ago
7 0

Answer:

4 {x}^{2}  -  {y}^{2}

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Classify the triangle with side lengths 11 cm, 30 cm, and 39 cm as acute, right, or obtuse​
Ahat [919]

Answer:

This triangle is a obtuse traingle.

Step-by-step explanation:

First of all, it is impossible that this triangle is actute since the sides are different length.

Right is also not possible ( in this case ) because there are two pretty far sides.

30, 39

A obtuse triangle have a short side of 11 cm pointing top right and a side of 30 cm pointing directly to the right and the 39 cm side connecting the ends of the other sides.

                                        \                   Third side ( 39 cm , the one connecting

First side ( 11 cm )              \                   the ends of side 1 and 2)

                                            \______________

                                               Second side ( 30 cm )

<h3><u><em>Please rate this and please give brainliest. Thanks!!! </em></u></h3><h3><u><em>Appreciate it! : )</em></u></h3><h3><u><em></em></u></h3><h2><u><em>And always, </em></u></h2><h2><u><em> SIMPLIFY BANANAS          : )</em></u></h2>
5 0
3 years ago
What is the answer to this equation 5×.4x=10
oksano4ka [1.4K]

The answer to this equation is x=5.

6 0
3 years ago
Read 2 more answers
D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x
(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
b(1-16k)=30\implies b=\dfrac{30}{1-16k}

and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
8 0
3 years ago
Determine whether the given graph is a function or not
Alex777 [14]

Answer: yes is it a function

Step-by-step explanation:

A graph is a function when there is only one y value for each x value. You can also use the vertical line test. This example is a “quadratic function”.

7 0
3 years ago
If you remove my question you not big brain
guapka [62]

Answer:

Whats your question?

Step-by-step explanation:

5 0
3 years ago
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