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3 years ago

The of -6 is 6, because it is 6 units from zero on the number line.

Mathematics

1 answer:

Greeley [361]3 years ago

7 0

Answer:

not answerable

Step-by-step explanation:

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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

2 years ago

Is it possible to have the same lower quartile as the minimum in a box plot?

Natalka [10]

No, bc the lower quartile should be in between the minimum and middle quartile(median) Hoped this helped!

6 0

3 years ago

Read 2 more answers

Which situation shows a constant rate of change?

charle [14.2K]

C is right I thinkkkk

8 0

3 years ago

Read 2 more answers

Please hurry i need help and im stuck!!! helpsosososos

sammy [17]

Answer:

Step-by-step explanation:

the x and y are being multiplied by 2 e.g. 1 × 2 is 2 and 2 × 2 is 4 so the constant is 2

7 0

3 years ago

Read 2 more answers

Q3. In a hydraulic lift, a 1400 N force is applied to a 0.5 m2 piston. Calculate the minimum surface area of the large piston to

blondinia [14]

Answer:

1.8m^2 approx

Step-by-step explanation:

Given data

P1= 1400N

A1=0.5m^2

P2=5000 N

A2=??

Let us apply the formula to calculate the Area A2

P1/A1= P2/A2

substitute

1400/0.5= 5000/A2

cross multiply

1400*A2= 5000*0.5

1400*A2= 2500

A2= 2500/1400

A2= 1.78

Hence the Area is 1.8m^2 approx

3 0

3 years ago