Answer:
q = - 14 / 17
Step-by-step explanation:
9 - 17q = 23
-17q = 23 - 9
-17 q = 14
q = - 14 / 17
Answer:
The cost of hardbook is $9 and cost of paperback is $5.
Step-by-step explanation:
Let x be the price for hardback book and y be the price for paperback book.
Mr. Yamamoto purchased 42 new hardcover books and 64 new paperback books, which cost a total of $698.
We can write the equation as:

He also purchased 42 new hardcover books and 58 new paperback books, which cost a total of $668.
We can write the equation as:

We have to use the elimination method to solve the two equations.
We eliminate x by subtracting the two equations.

Thus, the cost of hardbook is $9 and cost of paperback is $5.
Answer:
-1
Step-by-step explanation:
We need to simplify the given expression . The given expression is ,
Here we can see that the power of the both exponent is same that is (2n+1) . Recall the property of exponents ,
Using this property , we have ,
This can be written as ,
Simplifying using ( a+b)(a-b) = a² - b² ,
Subtracting the numbers inside the brackets ,
Now we know that every odd number is in the form of 2n -1 , where n is any integer. Therefore , the <u>power is odd</u> .
Since the base is (-1) , for even power it is 1 and for odd power it is -1 . Therefore the final answer is ,
<u>Hence </u><u>the</u><u> </u><u>required</u><u> answer</u><u> is</u><u> </u><u>(</u><u>-</u><u>1</u><u>)</u><u> </u><u>.</u>
Check the picture below on the left-side.
we know the central angle of the "empty" area is 120°, however the legs coming from the center of the circle, namely the radius, are always 6, therefore the legs stemming from the 120° angle, are both 6, making that triangle an isosceles.
now, using the "inscribed angle" theorem, check the picture on the right-side, we know that the inscribed angle there, in red, is 30°, that means the intercepted arc is twice as much, thus 60°, and since arcs get their angle measurement from the central angle they're in, the central angle making up that arc is also 60°, as in the picture.
so, the shaded area is really just the area of that circle's "sector" with 60°, PLUS the area of the circle's "segment" with 120°.

![\bf \textit{area of a segment of a circle}\\\\ A_y=\cfrac{r^2}{2}\left[\cfrac{\pi \theta }{180}~-~sin(\theta ) \right] \begin{cases} r=radius\\ \theta =angle~in\\ \qquad degrees\\ ------\\ r=6\\ \theta =120 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20segment%20of%20a%20circle%7D%5C%5C%5C%5C%0AA_y%3D%5Ccfrac%7Br%5E2%7D%7B2%7D%5Cleft%5B%5Ccfrac%7B%5Cpi%20%5Ctheta%20%7D%7B180%7D~-~sin%28%5Ctheta%20%29%20%20%5Cright%5D%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0A%5Ctheta%20%3Dangle~in%5C%5C%0A%5Cqquad%20degrees%5C%5C%0A------%5C%5C%0Ar%3D6%5C%5C%0A%5Ctheta%20%3D120%0A%5Cend%7Bcases%7D)
Answer:
87777777777777777777777777
Step-by-step explanation: