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4 years ago

Name a semicircle for circle O.A) NQPB) MQNC) MPQD) RPQ

Mathematics

2 answers:

blagie [28]4 years ago

7 0

Answer:

Step-by-step explanation:

A semicircle is half the circle and is the region enclosed between a diameter and the arc subtending the diameter.

In this case MQN is a semicircle → B

Reika [66]4 years ago

3 0

i think the answer is MQN

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zalisa [80]

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Step-by-step explanation:

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3 years ago

What is the lowest common denominator of the following two fractions?

zloy xaker [14]

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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 $B$ .

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3 years ago

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just olya [345]

Answer:

The scale factor is 1/5

Step-by-step explanation:

Each of the lengths of the lengths of this triangle are divided by 5 (or multiplied by 1/5)

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3 years ago

Which of the following equations is written in standard form? A.-2X=13 B.3X+5Y=-15 C.1/ 2X+1/3Y=-15 D.-3X+13Y=25 - 3442595.

Inessa [10]

The answer is B. 3x + 5y = -1.

Hope this answer helped!! Please give me Brainliest Answer!!

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3 years ago