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3 years ago

Find and simplify the following for f(x)=x(22-x), assuming h ≠ 0 in (C),a. f(x + h)b. f(x+h)- f(x)c. f(x+ h)- f(x)/h

Mathematics

1 answer:

Gre4nikov [31]3 years ago

4 0

Answer:

The answer is below

Step-by-step explanation:

Given that f(x)=x(22-x)

a) f(x)=x(22-x)

f(x + h) = (x + h)(22 - (x + h))

f(x + h) = (x + h)(22 - x - h)

f(x + h) = (22x - x² - xh + 22h - xh - h²)

f(x + h) = (-x² + 22x - h²+ 22h - 2xh)

b) f(x)=x(22-x) = 22x - x²

f(x + h) = (-x² + 22x - h²+ 22h - 2xh)

f(x+h)- f(x) = (-x² + 22x - h²+ 22h - 2xh) - (22x - x²)

f(x+h)- f(x) = -x² + 22x - h²+ 22h - 2xh - 22x + x²

f(x+h)- f(x) = -x² + x² + 22x - 22x - h²+ 22h - 2xh

f(x+h)- f(x) = - h²+ 22h - 2xh

c) f(x+h)- f(x) = - h²+ 22h - 2xh

$\frac{f(x+h)-f(x)}{h}=\frac{- h^2+ 22h - 2xh}{h}\\ \\\frac{f(x+h)-f(x)}{h}=\frac{- h^2}{h}+\frac{22h}{h}-\frac{2xh}{h}\\\\\frac{f(x+h)-f(x)}{h}=-h+22-2x$

$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}= \lim_{h \to 0} (-h+22-2x )=22-2x$

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2 years ago

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Answer:

$103.78\:\mathrm{ft}$

Step-by-step explanation:

We can form a right triangle where the distance between the ranger's current position and fire is the hypotenuse of the triangle. In a right triangle, the tangent of an angle is equal to its opposite side divided by the hypotenuse.

Therefore, we have:

$\tan 34^{\circ}=\frac{70}{x}$ , where $x$ is the distance between the base of the tower and the fire.

Solving, we get:

$x=\frac{70}{\tan 34^{\circ}}=103.779267796\approx \boxed{103.78\:\mathrm{ft}}$

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3 years ago

Jasmine can run 4 5/6 miles in 2/3 of an hour. how many miles can she run in one hour

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Jasmine can run in one hour is 7.25 miles.

<h3>What is the formula of unitary method?</h3>

The formula of the unitary method is to find the value of a single unit and then multiply the value of a single unit to the number of units to get the necessary value.

Given that,

Jasmine can run 4 $\frac{5}{6}$ miles in $\frac{2}{3}$ of an hour.

$\frac{2}{3}$ of an hour = 4 $\frac{5}{6}$ miles

= $\frac{29}{6}$ miles

By using unitary method we will find the value of an hour.

1 hour = $\frac{\frac{29}{6} }{\frac{2}{3} }$

= $\frac{29}{6}$ × $\frac{3}{2}$

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1 hour = 7.25 miles

Hence, Jasmine can run in one hour is 7.25 miles.

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1 year ago

A metal beam was brought from the outside cold into a machine shop where the temperature was held at 65degreesF. After 5 min, t

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Answer:

The beam initial temperature is 5 °F.

Step-by-step explanation:

If T(t) is the temperature of the beam after t minutes, then we know, by Newton’s Law of Cooling, that

$T(t)=T_a+(T_0-T_a)e^{-kt}$

where $T_a$ is the ambient temperature, $T_0$ is the initial temperature, $t$ is the time and $k$ is a constant yet to be determined.

The goal is to determine the initial temperature of the beam, which is to say $T_0$

We know that the ambient temperature is $T_a=65$ , so

$T(t)=65+(T_0-65)e^{-kt}$

We also know that when $t=5 \:min$ the temperature is $T(5)=35$ and when $t=10 \:min$ the temperature is $T(10)=50$ which gives:

$T(5)=65+(T_0-65)e^{k5}\\35=65+(T_0-65)e^{-k5}$

$T(10)=65+(T_0-65)e^{k10}\\50=65+(T_0-65)e^{-k10}$

Rearranging,

$35=65+(T_0-65)e^{-k5}\\35-65=(T_0-65)e^{-k5}\\-30=(T_0-65)e^{-k5}$

$50=65+(T_0-65)e^{-k10}\\50-65=(T_0-65)e^{-k10}\\-15=(T_0-65)e^{-k10}$

If we divide these two equations we get

$\frac{-30}{-15}=\frac{(T_0-65)e^{-k5}}{(T_0-65)e^{-k10}}$

$\frac{-30}{-15}=\frac{e^{-k5}}{e^{-k10}}\\2=e^{5k}\\\ln \left(2\right)=\ln \left(e^{5k}\right)\\\ln \left(2\right)=5k\ln \left(e\right)\\\ln \left(2\right)=5k\\k=\frac{\ln \left(2\right)}{5}$

Now, that we know the value of $k$ we can use it to find the initial temperature of the beam,

$35=65+(T_0-65)e^{-(\frac{\ln \left(2\right)}{5})5}\\\\65+\left(T_0-65\right)e^{-\left(\frac{\ln \left(2\right)}{5}\right)\cdot \:5}=35\\\\65+\frac{T_0-65}{e^{\ln \left(2\right)}}=35\\\\\frac{T_0-65}{e^{\ln \left(2\right)}}=-30\\\\\frac{\left(T_0-65\right)e^{\ln \left(2\right)}}{e^{\ln \left(2\right)}}=\left(-30\right)e^{\ln \left(2\right)}\\\\T_0=5$

so the beam started out at 5 °F.

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3 years ago

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The lateral surface area of the Pyramid will be 850547 square feet.

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The lateral surface area will be calculated as:-

A = $= l\sqrt{(\dfrac{w}{2})^2+h^2} + w\sqrt{(\dfrac{l}{2})^2+h^2}$

A = $= 750\sqrt{(\dfrac{750}{2})^2+450^2} + 750\sqrt{(\dfrac{750}{2})^2+450^2}$

A = 750 √321525 + 750 √321525

A = 150 √√321525

A = 1500 x 567.031

A = 850547 square feet

Therefore the lateral surface area of the Pyramid will be 850547 square feet.

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1 year ago

Find and simplify the following for ​f(x)=x(22-x), assuming h ≠ 0 in (C),a. f(x + h)b. f(x+h)- f(x)c. f(x+ h)- f(x)/h

Find and simplify the following for f(x)=x(22-x), assuming h ≠ 0 in (C),a. f(x + h)b. f(x+h)- f(x)c. f(x+ h)- f(x)/h