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earnstyle [38]
3 years ago
9

Find the quadratic polynomial ax^2+bx+c which best fits the function f(x)=8^x at x=0, in the sense that g(0)=f(0), and f'(0)=g'(

0), and f''(0)=g''(0).
g(x)=?
Mathematics
1 answer:
tamaranim1 [39]3 years ago
7 0
F(0) = 8^0 = 1
f '(x) = 8^x * ln8; x= 0 => f '(0) = ln8
f ''(x) = 8^x (ln8)^2 ; x = 0=> f ''(0) = (ln8)^2

g(0) = a(0)^2 + b(0) + c = f(0) = 1 => c = 1

g '(x) = 2ax + b => g '(0) = 2a(0) + b = f '(0) = ln8 => b = ln(8)

g''(x) = 2a => g''(0) = 2a = f''(0) = (ln8)^2 => a = (ln8)^2 / 2

=> g(x) = [(ln8)^2 /2] x^2 + [ln(8)]x + 1

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so we're really looking for the equation of a line whose slope is 1/2 and passes through (-4 , 1)

(\stackrel{x_1}{-4}~,~\stackrel{y_1}{1})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{1}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{1}=\stackrel{m}{ \cfrac{1}{2}}(x-\stackrel{x_1}{(-4)}) \\\\\\ y-1=\cfrac{1}{2}(x+4)\implies y-1=\cfrac{1}{2}x+2\implies y=\cfrac{1}{2}x+3

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Find the missing side lengths. Leave your answers as radicals in simplest form. Please help!! Due today
Nataly_w [17]

Answer:

m = 10\sqrt 3

n = 10

Step-by-step explanation:

Required

Find m and n

Considering the given angle, we have:

\sin(60) = \frac{Opposite}{Hypotenuse}

This gives:

\sin(60) = \frac{m}{20}

Make m ths subject

m = 20 * \sin(60)

\sin(60) =\frac{\sqrt 3}{2}

So, we have:

m = 20 *\frac{\sqrt 3}{2}

m = 10\sqrt 3

Considering the given angle again, we have:

\cos(60) = \frac{Adjacent}{Hypotenuse}

This gives:

\cos(60) = \frac{n}{20}

Make n the subject

n = 20 * \cos(60)

\sin(60) =\frac{1}{2}

So, we have:

n = 20 *\frac{1}{2}

n = 10

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2 years ago
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ValentinkaMS [17]

Answer:

Use ≈ (approximately equal) sign as the scientific notation.

Step-by-step explanation:

(a) 0.001872 ≈ 0.0019

(b) 0.3411 ≈ 0.34

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*Zeros before non-zero numbers are not significant.

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Hope this helps!!

6 0
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Todea what did one pencil say to another mathematics algebra order of operations answer
Serjik [45]
Part 1

6- \frac{18-3^2}{4+(2-3)} =6- \frac{18-9}{4+(-1)}  \\  \\ =6- \frac{9}{4-1} =6- \frac{9}{3} =6-3 \\  \\ =3



Part 2

7-8\div4(3^2-1)=7-8\div4(9-1) \\  \\ =7-8\div4(8)=7-8\div32=7- \frac{1}{2} \\  \\ =6  \frac{1}{2} = \frac{13}{2}



Part 3

2-2(2-5)^2+3^2=2-2(-3)^2+9 \\  \\ =2-2(9)+9=2-18+9=-7



Part 4

(5^2-9\cdot3)^2-11=(25-27)^2-11 \\  \\ =(-2)^2-11=4-11=-7



Part 5

[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\  \\ =(1-5)\cdot4=-4\cdot4=-16



Part 6

12-11\cdot2+16\div8=12-22+2 \\  \\ =-8



Part 7

-5+1\cdot3-(7-2^3)=-5+3-(7-8) \\  \\ =-2-(-1)=-2+1=-1



Part 8

8+2\cdot3-14\div7=8+6-2 \\  \\ =12



Part 9

(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\  \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\  \\ =36+4=40



Part 10

7+(2-3^2)\cdot5=7+(2-9)\cdot5 \\  \\ =7+(-7)\cdot5=7+(-35)=7-35 \\  \\ =-28



Part 11

3-2[2^3+(-1)^3]=3-2[8+(-1)] \\  \\ =3-2(8-1)=3-2(7)=3-14 \\  \\ -11



Part 12

18+6\div3(2^2+5)=18+6\div3(4+5) \\  \\ =18+6\div3(9)=18+6\div27=18+ \frac{2}{9} \\  \\  =18 \frac{2}{9}
8 0
3 years ago
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