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3 years ago

How to find the square root of a negative 100

1 answer:

3 0

The thing is, you can't find the square root of a negative number, no matter what. Here's why:

Finding the square root of a number is essentially just finding which number, when squared, ends up to the original number. If you square a negative number, you will get a positive number as the outcome.

But the square root of positive 100 is 10, if that helps you.

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3c = 18 do I multiply 3 by 6

lilavasa [31]

Answer:

to find c you divide 18 by 3 to find that c = 6

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3 years ago

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HELP ASAP<br><br> Write the equation in standard form for the circle x2 + 32 = -y² + 12x.

vagabundo [1.1K]

Step-by-step explanation:

Move

to the right side of the equation because it does not contain a variable.

3x2−12x−2y2+16y=26

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2 years ago

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Graph: y < 1/3x+1/2<br> graph the point

Galina-37 [17]

Answer:

1/11

Step-by-step explanation:

First you have to times 1/3 and 1/2 and then you get 1/6. Then you add 1/3 and 1/2 and that's 1/5. then you add them both.

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3 years ago

This recipe makes 12 flapjacks.

wel

Answer:

Step-by-step explanation:

just times everything 3.5 as 42/12=3.5

490g margarine

420g sugar

350ml syrup

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175g raisins

4 0

3 years ago

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