Answer:
The solution for the given algebraic equation is ![\frac{ - 2}{71}](https://tex.z-dn.net/?f=%5Cfrac%7B%20-%202%7D%7B71%7D)
Step-by-step explanation:
Given algebraic equation as :
( x - 8 ) + 1 = 2 ( 18 x - 1 )
Now solving the equation while opening the brackets
So,
× x -
× 8 + 1 = 2 × 18 x - 2
Or,
-
+ 1 = 36 x - 2
or,
- 4 + 1 = 36 x - 2
or,
- 3 = 36 x - 2
or,
- 36 x = - 2 + 3
Or,
- 36 x = 1
or,
= 1
or,
= 1
∴ 71 x = - 2
I.e x = ![\frac{ - 2}{71}](https://tex.z-dn.net/?f=%5Cfrac%7B%20-%202%7D%7B71%7D)
Hence The solution for the given algebraic equation is
Answer
Answer: Choice D
![\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%5C%20%27%28x%29%20%3D%202x%5Csqrt%7B1%2Bx%5E6%7D%5C%5C%5C%5C)
==========================================================
Explanation:
Let g(t) be the antiderivative of
. We don't need to find out what g(t) is exactly.
Recall by the fundamental theorem of calculus, we can say the following:
![\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7Ba%7D%5E%7Bb%7D%20g%27%28t%29dt%20%3D%20g%28b%29-g%28a%29)
This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).
So,
![\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint_%7B%5Cpi%7D%5E%7Bx%5E2%7D%5Csqrt%7B1%2Bt%5E3%7Ddt%5C%5C%5C%5C%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint_%7B%5Cpi%7D%5E%7Bx%5E2%7Dg%27%28t%29dt%5C%5C%5C%5C%20%5Cdisplaystyle%20F%28x%29%20%3D%20g%28x%5E2%29%20-%20g%28%5Cpi%29%5C%5C%5C%5C)
From here, we apply the derivative with respect to x to both sides. Note that the
portion is a constant, so ![g'(\pi) = 0](https://tex.z-dn.net/?f=g%27%28%5Cpi%29%20%3D%200)
![\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%28x%29%20%3D%20g%28x%5E2%29%20-%20g%28%5Cpi%29%5C%5C%5C%5C%20%5Cdisplaystyle%20F%20%5C%20%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bg%28x%5E2%29-g%28%5Cpi%29%5D%5C%5C%5C%5C%5Cdisplaystyle%20F%5C%20%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bg%28x%5E2%29%5D%20-%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bg%28%5Cpi%29%5D%5C%5C%5C%5C%20%5Cdisplaystyle%20F%5C%20%27%28x%29%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%5Bx%5E2%5D%2Ag%27%28x%5E2%29%20-%20g%27%28%5Cpi%29%20%5C%20%5Ctext%7B%20....%20chain%20rule%7D%5C%5C%5C%5C)
![\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F%5C%20%27%28x%29%20%3D%202x%2Ag%27%28x%5E2%29%20-%200%5C%5C%5C%5C%20%5Cdisplaystyle%20F%5C%20%27%28x%29%20%3D%202x%2Ag%27%28x%5E2%29%5C%5C%5C%5C%20%5Cdisplaystyle%20F%5C%20%27%28x%29%20%3D%202x%5Csqrt%7B1%2B%28x%5E2%29%5E3%7D%5C%5C%5C%5C%20%5Cdisplaystyle%20F%5C%20%27%28x%29%20%3D%20%5Cboldsymbol%7B2x%5Csqrt%7B1%2Bx%5E6%7D%7D%5C%5C%5C%5C)
Answer is choice D
Answer:
no complex zeros.
Step-by-step explanation:
The given equation is
![y=x^5](https://tex.z-dn.net/?f=y%3Dx%5E5)
To find the zeros equate y=0.
![x^5=0](https://tex.z-dn.net/?f=x%5E5%3D0)
We know that, if n>0, then
![x^n=0\Rightarrow x=0](https://tex.z-dn.net/?f=x%5En%3D0%5CRightarrow%20x%3D0)
So,
![x^5=0\Rightarrow x=0](https://tex.z-dn.net/?f=x%5E5%3D0%5CRightarrow%20x%3D0)
It means zero or root of the given equation is 0 with multiplicity 5.
Therefore, the given equation have no complex zeros.
Answer:
80%
Step-by-step explanation:
There are five separate points. One represents students who made more than 13 baskets. The other four represent students who made 13 or less baskets. 4/5 (four-fifths) is equal to 80%.
Whisper-20dB
Quiet Residence-30dB
Soft stereo in Residence-40dB
Average Speech-60dB
Cafeteria-80dB
Pneumatic Jackhammer-90dB
Loud crowd noise-100dB
Accelerating Bike-100dB
Rock concert-120dB
Jet Engine (75 ft away)-140dB
(1)Mode=100dB (since 100 is occurring the maximum no. of times, i.e. it has the highest frequency of 2)
(2)Mean=sum of observations/Total number of observations
=(20+30+40+60+80+90+100+100+120+140)/10
=770/10=10dB
(3)Median= 1/2[{n/2}th observation + {(n/2)+1}th observation], where, n=total no. of observations
So, 1/2[{10/2}th observation + {(10/2)+1}th observation]
=1/2[5th observation+6th observation]
=1/2[80+90] [Because:5th observation=80dB and 6th=90dB]
=1/2(170)
=85
Therefore, median=85dB