(a)(-5a) I have to find the product and simplify the answer. Can anyone please help me?

Hello! If someone could help me answer this it would be appreciated! <3

Leona [35]

Answers:

In the top row, the first and third polygons are regular polygons.
In the bottom row, only the first polygon is a regular polygon

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Explanation:

A regular polygon must have all sides of equal length. Also, all angles must be the same measure. This applies to the equilateral triangle in the upper left corner, and the square in the upper right corner. Each angle of the square is 90 degrees even though your teacher didn't use right angle markers.

The middle figure in the top row is not a regular polygon because the sides are not all the same length. Also, the angles are different.

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In the bottom row, only the first figure is a regular polygon for similar reasons mentioned earlier. The second figure is not a regular polygon because this figure is concave. Only convex polygons can be regular.

The third figure in the bottom row is not regular because one side is longer than the others.

3 0

2 years ago

To the nearest tenth, what is the area of a circle with circumference 35.6 centimeters?

AnnyKZ [126]

The answer is A. 100.9 cm2

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

If Jenny can read 7 books in 8 days, exactly how many days would it take her to read 20 books?

muminat

Answer:

Step-by-step explanation:

Given,

No. of days taken by Jenny to read 7 books

= 8

So,

No. of days taken by her to read 1 book

$= \frac{8}{7}$

Therefore,

No. of days taken by her to read 20 books

$= \frac{8}{7} × 20$

[On Simplification]

$= \frac{160}{7}$

[On dividing]

= 22.85.....

[On approximating]

= 23 days (approx. )

Hence,

Jenny will take 23 days to read 20 books (Ans)

3 0

3 years ago

Read 2 more answers

I have been stuck on the problem for so long. please help me:

Xelga [282]

Answer:

D-2

Step-by-step explanation:

You can use elimination to get the value of y. Move numbers to one side and x and ys to the other side. To use elimination, multiply the first equation with 2. You will get 2x+2y=24. add it with the equation below. y will be eliminated and you will get the value of x. X=10. Plug in x and get the y value. Y=2

6 0

3 years ago