Answer: QP=11.18cm;
QP=11.18cm;
m<Q=42;m<R=48*
Step-by-step explanation:
Hope this helps and the 48 degree
Answer:
800,000+90,000+4,000+200+10+17= 894,217
Eight Hundred thousand+ Ninety thousand+ Four thousand+ Two hundred+ Ten+ Seven= 894,217
Step-by-step explanation:
(Image below)
Keep in mind that the first number in a coordinate is always on the x-axis (the horizontal line) and the second number is on the y-axis (the vertical line).
If the question asks us to plot (-3, 0), we need to find -3 on the horizontal line and find 0 on the vertical line.
Since 0 means that we go 0 units up, we can just plot the point at -3 on the x-axis.
Next, we need to plot (3, 5). First, let's find the first coordinate by locating 3 on the horizontal axis. Now, we need to find 5 up from that point, because our y-coordinate is 5.
Answer:
30,000 phone numbers.
Step-by-step explanation:
If the first three numbers are already defined then it means that only the other 4 digits remain to be defined.
The solution would be the number of numbers between 0000 and 9999, because the order here matters, since each number is different.
Therefore between these two numbers there are a total of 10,000 numbers, that is to say that for each of the three initial numbers there are 10,000 numbers, therefore in the town there are a total of 30,000 numbers.
Answer:
![f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....](https://tex.z-dn.net/?f=f%28x%29%3Dx%2B%5Cdfrac%7Bx%5E%7B3%7D%7D%7B3%7D%2B%5Cdfrac%7B2x%5E%7B4%7D%7D%7B3%7D....)
Approximate error = 0.4426
Step-by-step explanation:
f(x)=tanx, a=0
Maclaurin series formula used is given below
![f(x)=\sum_{n=0}^{\infty}\dfrac{f^{(n)}(0)x^{n}}{n!}=f(0)+f'(0)x+\dfrac{f''(0)}{2!}x^{2}+\dfrac{f'''(0)}{3!}x^{3}+....](https://tex.z-dn.net/?f=f%28x%29%3D%5Csum_%7Bn%3D0%7D%5E%7B%5Cinfty%7D%5Cdfrac%7Bf%5E%7B%28n%29%7D%280%29x%5E%7Bn%7D%7D%7Bn%21%7D%3Df%280%29%2Bf%27%280%29x%2B%5Cdfrac%7Bf%27%27%280%29%7D%7B2%21%7Dx%5E%7B2%7D%2B%5Cdfrac%7Bf%27%27%27%280%29%7D%7B3%21%7Dx%5E%7B3%7D%2B....)
f(x)=tanx
f(0)=tan0=0
![f'(x)=sec^{2}x\\f'(0)=sec^{2}0=1\\f''(x)=2sec^{2}xtanx\\f''(0)=2sec^{2}0tan0=0\\f'''(x)=-4sec^{2}x+6sec^{4}x\\f'''(0)=-4sec^{2}0+6sec^{4}0=-4+6=2\\](https://tex.z-dn.net/?f=f%27%28x%29%3Dsec%5E%7B2%7Dx%5C%5Cf%27%280%29%3Dsec%5E%7B2%7D0%3D1%5C%5Cf%27%27%28x%29%3D2sec%5E%7B2%7Dxtanx%5C%5Cf%27%27%280%29%3D2sec%5E%7B2%7D0tan0%3D0%5C%5Cf%27%27%27%28x%29%3D-4sec%5E%7B2%7Dx%2B6sec%5E%7B4%7Dx%5C%5Cf%27%27%27%280%29%3D-4sec%5E%7B2%7D0%2B6sec%5E%7B4%7D0%3D-4%2B6%3D2%5C%5C)
![f''''(x)=-8(2sec^{2}xtan^{2}x+sec^{4}x)+24(4sec^{4}xtan^{2}x)+sec^{6})\\f''''(0)=-8(0+1)+24(0+1)=-8+24=16\\](https://tex.z-dn.net/?f=f%27%27%27%27%28x%29%3D-8%282sec%5E%7B2%7Dxtan%5E%7B2%7Dx%2Bsec%5E%7B4%7Dx%29%2B24%284sec%5E%7B4%7Dxtan%5E%7B2%7Dx%29%2Bsec%5E%7B6%7D%29%5C%5Cf%27%27%27%27%280%29%3D-8%280%2B1%29%2B24%280%2B1%29%3D-8%2B24%3D16%5C%5C)
![f(x)=0+x+0+\dfrac{2x^{3}}{3!}+\dfrac{16x^{4}}{4!}\\](https://tex.z-dn.net/?f=f%28x%29%3D0%2Bx%2B0%2B%5Cdfrac%7B2x%5E%7B3%7D%7D%7B3%21%7D%2B%5Cdfrac%7B16x%5E%7B4%7D%7D%7B4%21%7D%5C%5C)
![f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}\\](https://tex.z-dn.net/?f=f%28x%29%3Dx%2B%5Cdfrac%7Bx%5E%7B3%7D%7D%7B3%7D%2B%5Cdfrac%7B2x%5E%7B4%7D%7D%7B3%7D%5C%5C)
Hence, the Taylor series for f(x)=tanx is given by
![f(x)=x+\dfrac{x^{3}}{3}+\dfrac{2x^{4}}{3}....](https://tex.z-dn.net/?f=f%28x%29%3Dx%2B%5Cdfrac%7Bx%5E%7B3%7D%7D%7B3%7D%2B%5Cdfrac%7B2x%5E%7B4%7D%7D%7B3%7D....)
Maclaurin series upper bound error formula used is given as
R_n(x)=|f(x)-T_n(x)|
R_3(x)=|tanx-T_3(x)|
![R_3(x)=|tanx-x-\dfrac{x^{3}}{3}-\dfrac{2x^{4}}{3}|](https://tex.z-dn.net/?f=R_3%28x%29%3D%7Ctanx-x-%5Cdfrac%7Bx%5E%7B3%7D%7D%7B3%7D-%5Cdfrac%7B2x%5E%7B4%7D%7D%7B3%7D%7C)
Plugging this value x=1
![R_3(x)=|tan(1)-1-\dfrac{1}{3}-\dfrac{2}{3}|\\](https://tex.z-dn.net/?f=R_3%28x%29%3D%7Ctan%281%29-1-%5Cdfrac%7B1%7D%7B3%7D-%5Cdfrac%7B2%7D%7B3%7D%7C%5C%5C)
R_3(x)=|1.5574-1-0.333-0.666|
R_3(x)=|-0.4426|=0.4426
Hence, upper bound on the error approximation
tan(1)=0.4426