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3 years ago

Marcy played the piano for 45 minutes. She stopped playing at 4:15 p.m. At what time did she start playing the piano ?

1 answer:

4 0

Well, first off 1 hr =60 minutes. 60 min - 45 min = 15 minutes extra. what is one hour before 4:15? 3:15! Add the subtracted 15 minutes to get 3:30... Hope this helps

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What should be the first step to solve the equation?

katrin [286]

Answer:

Step-by-step explanation:2x + 9x - 9 = 16x + 16 - 5

= 11x - 9 = 16x + 11

11x - 16x = 11 + 9

5x = 20

x = 20/5

x = 4

Plz mark it as brainliest:)

7 0

3 years ago

Read 2 more answers

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

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Olegator [25]

Answer:

4.4 t = 13.2

t = 13.2 / 4.4 = 3

t = 3

Step-by-step explanation:

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