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Semenov [28]
3 years ago
9

7 is ten times the value of what

Mathematics
1 answer:
MrRissso [65]3 years ago
8 0

Answer:

0.7

Step-by-step explanation:

0.7x10=7

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Kisachek [45]

Answer:

(+,-)

Step-by-step explanation:

quadrant 1 is (+,+)

quadrant 2 is (-,+)

quadrant 3 is (-,-)

quadrant 4 is (+.-)

7 0
3 years ago
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Que is on pic.i can't able to type in text.
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.
3 0
3 years ago
Each side of a square is lengthenedBy 4 inches. The area of the new large square is 25 square inches. Find the length of a side
melisa1 [442]

Answer:

1 inch.

Step-by-step explanation:

The length of a side of the new square = square root of 25 = 5 ins.

So a side of the original square = 5 - 4 = 1 inch.

7 0
4 years ago
Which if the following numbers is irrational?
Llana [10]

Answer:

B, the square root of 3

Step-by-step explanation:

8 0
3 years ago
What value for x will make the equation true?<br> -3 (7x - 5) = 87 + 3x
Gennadij [26K]

Answer:

x=-3

Step-by-step explanation:

follow the PEMDAS method then see if the 2 equation equal each other after plugging in the x.

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3 years ago
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