$\bf f(x)=ln(x-5)\qquad [6,8]\qquad \cfrac{df}{dx}=\cfrac{1}{x-5}\\\\
-----------------------------\\\\
\textit{mean value theorem}\qquad f'(c)=\cfrac{f(b)-f(a)}{b-a}\\\\
-----------------------------\\\\
f'(c)=\cfrac{1}{c-5}\qquad thus\quad \cfrac{1}{c-5}=\cfrac{f(8)-f(6)}{8-6}
\\\\\\
\cfrac{1}{c-5}=\cfrac{ln(3)-ln(1)}{2}\implies \cfrac{1}{c-5}=\cfrac{ln(3)-0}{2}
\\\\\\
\cfrac{2}{ln(3)}=c-5\implies \boxed{\cfrac{2}{ln(3)}+5=c}$

7 0

3 years ago

Paul deposits $4000 in a bank which pays a compound interest

yawa3891 [41]

Answer: He will have 3,200 in 5 years.

Step-by-step explanation: This is because 4% of $400 is 160, and if you multiply that by 5 you get 800, then you just subtract that from the total amount and you get your final answer.

5 0

3 years ago

Here are the questions:

yaroslaw [1]

Answer:

Q1) 0.46 Q2)0.2333

Step-by-step explanation:

Q1) P=(4+12+7)/50 =0.46

Q2) P=7/(4+7+10+9)

6 0

3 years ago

When solving the equation x2 – 8x - 7 = 0 by completing the square, which equation

IgorLugansk [536]

<h3>Answer:-</h3>

Lets Solve

$\\ \rm \longmapsto \: {x}^{2} - 8x - 7 = 0 \\ \\ \rm \longmapsto \: {x}^{2} - 8 {}^{}x = 7 \\ \\ \rm \longmapsto \: 4( {x}^{2} - 8x) = 4 \times 7 \\ \\ \rm \longmapsto \: {4x}^{2} - 32x = 28 \\ \\ \rm \longmapsto \: {(2x)}^{2} - 2 \times 2x \times 8 = 28 \\ \\ \rm \longmapsto \: (2x) {}^{2} - 2 \times 2x \times 8 + {8}^{2} = 28 + {8}^{2} \\ \\ \rm \longmapsto \: {(2x - 8)}^{2} = 64 + 28 \\ \\ \rm \longmapsto \: {(2x - 8)}^{2} = 92 \\ \\ \rm \longmapsto \: 2x - 8 = \underline{ + }9.3 \\ \\ \rm \longmapsto \:2x = 8 + 9.3 \: or \: 8 - 9.3 \\ \\ \rm \longmapsto \:2x = 17.3 \:or \: - 1.3 \\ \\ \rm \longmapsto \:2x = 17.3 \\ \\ \rm \longmapsto \:x = \frac{17.3}{2} \\ \\ \rm \longmapsto x = \:8.6$

Ignore negative value

4 0

3 years ago