Answer:
~ 80.75 x 10^7
Step-by-step explanation:
9 x 8 + 9 * .5 + .5 * 8 + .5 * . 5 = ~ 80.75 x 10 ^7
Step-by-step explanation:
![here \: given \: that \: total \: lenth = bd = 67 \\ then \: bd = bc + cd \\ so \: we \: can \: write \: that \\ 67 = 3x - 2 + 4x + 13 \\67 = 3x + 4x - 2 + 13 \\ 67 = 7x + 11 \\ 7x = 67 - 11 \\ 7x = 56 \\ x = \frac{56}{7} \\ x = 8 \\ thankyou](https://tex.z-dn.net/?f=here%20%5C%3A%20given%20%5C%3A%20that%20%5C%3A%20total%20%5C%3A%20lenth%20%3D%20bd%20%3D%2067%20%5C%5C%20then%20%5C%3A%20bd%20%3D%20bc%20%2B%20cd%20%5C%5C%20so%20%5C%3A%20we%20%5C%3A%20can%20%5C%3A%20write%20%5C%3A%20that%20%5C%5C%2067%20%3D%203x%20-%202%20%2B%204x%20%2B%2013%20%5C%5C67%20%3D%20%203x%20%2B%204x%20-%202%20%2B%2013%20%5C%5C%2067%20%3D%207x%20%2B%2011%20%5C%5C%207x%20%3D%2067%20-%2011%20%5C%5C%207x%20%3D%2056%20%5C%5C%20x%20%3D%20%20%5Cfrac%7B56%7D%7B7%7D%20%20%5C%5C%20x%20%3D%208%20%5C%5C%20thankyou)
In this problem, we are asked to determine the degree of the given expression 12X4 - 8X + 4X2 -3. To answer this, first, we need to arrange the mathematical expression in descending order with respect to its power such as the new arrangement will become 12x4 + 4x2 -8x -3. The degree is clearly visible and it is 4. Therefore, the answer to this problem is the letter "B" which is 4.
Now, we know that 90°< θ <180°, that simply means the angle θ is in the II quadrant, where sine is positive and cosine is negative.
![\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{13}}\impliedby \textit{now let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem}\\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20sin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B13%7D%7D%5Cimpliedby%20%5Ctextit%7Bnow%20let%27s%20find%20the%20%5Cunderline%7Badjacent%20side%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%5C%5C%5C%5C%0Ac%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ac%3Dhypotenuse%5C%5C%0Aa%3Dadjacent%5C%5C%0Ab%3Dopposite%5C%5C%0A%5Cend%7Bcases%7D)
Answer:
![\cos (a-b)=\cos a \cos b+\sin a \sin b](https://tex.z-dn.net/?f=%5Ccos%20%28a-b%29%3D%5Ccos%20a%20%5Ccos%20b%2B%5Csin%20a%20%5Csin%20b)
Step-by-step explanation:
Given : ![\cos (180^{\circ}-q)=-\cos q](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D-%5Ccos%20q)
We have to write which identity we will use to prove the given statement.
Consider ![\cos (180^{\circ}-q)=-\cos q](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D-%5Ccos%20q)
Take left hand side of given expression ![\cos (180^{\circ}-q)](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29)
We know
![\cos (a-b)=\cos a \cos b+\sin a \sin b](https://tex.z-dn.net/?f=%5Ccos%20%28a-b%29%3D%5Ccos%20a%20%5Ccos%20b%2B%5Csin%20a%20%5Csin%20b)
Comparing , we get, a= 180° and b = q
Substitute , we get,
![\cos (180^{\circ}-q)=\cos 180^{\circ} \cos (q)+\sin q \sin 180^{\circ}](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D%5Ccos%20180%5E%7B%5Ccirc%7D%20%20%5Ccos%20%28q%29%2B%5Csin%20q%20%5Csin%20180%5E%7B%5Ccirc%7D)
Also, we know
and ![\cos 180^{\circ}=-1](https://tex.z-dn.net/?f=%5Ccos%20180%5E%7B%5Ccirc%7D%3D-1)
Substitute, we get,
![\cos (180^{\circ}-q)=-1\cdot \cos (q)+\sin q \cdot 0](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D-1%5Ccdot%20%5Ccos%20%28q%29%2B%5Csin%20q%20%5Ccdot%200)
Simplify , we get,
![\cos (180^{\circ}-q)=-\cos (q)](https://tex.z-dn.net/?f=%5Ccos%20%28180%5E%7B%5Ccirc%7D-q%29%3D-%5Ccos%20%28q%29)
Hence, use difference identity to prove the given result.