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maria [59]
3 years ago
14

Simplify this expression : 2(y+z)

Mathematics
2 answers:
kap26 [50]3 years ago
6 0

Answer:2y + 2z

Step-by-step explanation: you need to multiply the number outside of the parenthesis with the one inside like.

dmitriy555 [2]3 years ago
5 0

The answer is 2y+2z

Use distributive property

2*y=2y

2*z=2z

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= 3,610.201  answer
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3 years ago
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ad-work [718]
It's not difficult to compute the values of A and B directly:

A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}
A=\tan^{-1}(\sin\theta)-\dfrac\pi4

B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt
B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}
B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2

Let's assume 0, so that |\csc\theta|=\csc\theta.

Now,

\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}
\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}
\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)
\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in A and B.
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svp [43]

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Step-by-step explanation:

4 0
3 years ago
Is 0.06 bigger than 1?
taurus [48]

Hi there!


Great question! 0.06 is actually SMALLER than 1.


Why?


Because 0.06 is almost half of the whole number. Half of 1 is 0.5, which is larger than 0.06.

 

Half is smaller than whole, so, therefore, 1 is larger.


Hope this helps!

Message me if you need anything else! I'd be happy to help! :D

8 0
3 years ago
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