Answer:
The process of calculating successive discounts of 8% and 10% on a $50 item is take 10% of $46.
Step-by-step explanation:
As given
successive discounts of 8% and 10% on a $50 item .
First find out for 8 % discount
8% is written in the decimal form
= 0.08
8 % of $50 item = 0.08 × 50
= $ 4
Price of item after 8% discount = 50 - 4
= $46
First find out for 10 % discount
10% is written in the decimal form
= 0.1
8 % of $48 item = 0.1× 46
= $4.6
Price of item after 8% discount = 46 - 4.6
= $41.4
Therefore in the successive discounts of 8% and 10% on a $50 item is $41.4 .
I assume you mean the product of mixed numbers,
3 1/2 × 3 1/2
If we write this as
(3 + 1/2) × (3 + 1/2) = (3 + 1/2)²
we can use the identity
(a + b)² = a² + 2ab + b²
so that
3 1/2 × 3 1/2 = 3² + (2 × 3 × 1/2) + (1/2)²
3 1/2 × 3 1/2 = 9 + 3 + 1/4
3 1/2 × 3 1/2 = 12 1/4
Alternatively, we can first write 3 1/2 as a mixed number:
3 + 1/2 = 6/2 + 1/2 = (6 + 1)/2 = 7/2
Then
3 1/2 × 3 1/2 = 7/2 × 7/2 = (7 × 7) / (2 × 2) = 49/4
and
49/4 = (48 + 1)/4 = ((4 × 12) + 1)/4 = 12 + 1/4
Answer:
m<2 = 4x - 26 = 4(30) - 26 = 94 degrees
m<3 = 3x + 4 = 3(30) + 4 = 94 degrees
Step-by-step explanation:
Congruent angles are equal, so m<2 = m<3
4x - 26 = 3x + 4
x = 30
m<2 = 4x - 26 = 4(30) - 26 = 94 degrees
m<3 = 3x + 4 = 3(30) + 4 = 94 degrees
Replace x with π/2 - x to get the equivalent integral
![\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-%5Cfrac%5Cpi2%7D%5E%7B%5Cfrac%5Cpi2%7D%20%5Ccos%28%5Ccot%28x%29%20-%20%5Ctan%28x%29%29%20%5C%2C%20dx)
but the integrand is even, so this is really just
![\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Ccos%28%5Ccot%28x%29%20-%20%5Ctan%28x%29%29%20%5C%2C%20dx)
Substitute x = 1/2 arccot(u/2), which transforms the integral to
![\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du](https://tex.z-dn.net/?f=%5Cdisplaystyle%202%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20%5Cfrac%7B%5Ccos%28u%29%7D%7Bu%5E2%2B4%7D%20%5C%2C%20du)
There are lots of ways to compute this. What I did was to consider the complex contour integral
![\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%5Cgamma%20%5Cfrac%7Be%5E%7Biz%7D%7D%7Bz%5E2%2B4%7D%20%5C%2C%20dz)
where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be
![\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%7C%5Cint_%7Bz%3DRe%5E%7Bi0%7D%7D%5E%7Bz%3DRe%5E%7Bi%5Cpi%7D%7D%20f%28z%29%20%5C%2C%20dz%5Cright%7C%20%5Cle%20%5Cfrac%7B%5Cpi%20R%7D%7B%7CR%5E2-4%7C%7D)
which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit
![\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-%5Cinfty%7D%5E%5Cinfty%20%5Cfrac%7B%5Ccos%28x%29%7D%7Bx%5E2%2B4%7D%20%5C%2C%20dx%20%3D%202%5Cpi%20i%20%7B%7D%20%5Cmathrm%7BRes%7D%5Cleft%28%5Cfrac%7Be%5E%7Biz%7D%7D%7Bz%5E2%2B4%7D%2Cz%3D2i%5Cright%29%20%3D%20%5Cfrac%5Cpi%7B2e%5E2%7D)
and it follows that
![\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%5Cpi%20%5Ccos%28%5Ccot%28x%29-%5Ctan%28x%29%29%20%5C%2C%20dx%20%3D%20%5Cboxed%7B%5Cfrac%5Cpi%7Be%5E2%7D%7D)
Circle A -- center(2, 0), radius 8 Circle A' -- center(-1, 5), radius 3