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2 years ago

7

Brad made a scale drawing of a rectangular room in his house. The actual length of the room is 12 4/5 ft. The scale used to make

the drawing was 1/4 in = 1 ft. What is the length, in inches, of the room in the drawing?
(show work pls)

1 answer:

zaharov [31]2 years ago

5 0

Answer:

3.2

Step-by-step explanation:

12 4/5 times 1/4= 3 1/5 or 3.2

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1. (5pts) Find the derivatives of the function using the definition of derivative.

andreyandreev [35.5K]

2.8.1

$f(x) = \dfrac4{\sqrt{3-x}}$

By definition of the derivative,

$f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}$

We have

$f(x+h) = \dfrac4{\sqrt{3-(x+h)}}$

and

$f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}$

Combine these fractions into one with a common denominator:

$f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}$

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

$f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}$

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

$\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}$

3.1.1.

$f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3$

Differentiate one term at a time:

• power rule

$\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4$

$\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}$

$\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}$

The last two terms are constant, so their derivatives are both zero.

So you end up with

$f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}$

8 0

2 years ago

Solve for x:<br> 4x + 10 = 34

VladimirAG [237]

Answer:

x=6

Step-by-step explanation:

4x+10=34

4x=24

x=6

7 0

3 years ago

Read 2 more answers

Using the completing-the-square method, rewrite f(x) = x2 − 8x + 3 in vertex form.

lys-0071 [83]

f(x) = x² - 8x + 3
y = x² - 8x + 3
- 3 - 3
y - 3 = x² - 8x + 16
y - 3 + 16 = x² - 8x + 16
y + 13 = x² - 8x + 16
y + 13 = (x - 4)²
- 13 - 13
y = (x - 4)² - 13
f(x) = (x - 4)² - 13

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3 years ago

Help, I need help with this math question as soon as possible!!

mina [271]

Answer:

Step-by-step explanation:

4x what = 98

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3 years ago

The distances of a port from 20 differ- ent locations form an AP. If the farthest distance is 300 km, and the nearest distance i

Mkey [24]

Answer:

Required difference = 270/19 km

Step-by-step explanation:

If we consider the AP of distances to be

a, a + r, a + 2r, ..., a + 19r,

where a is the distance of the nearest location from the port and r is the difference between any two successive location.

Given that,

a + 19r = 300 .....(1)

a = 30 ..... (2)

Using (2), from (1), we get

30 + 19r = 300

or, 19r = 300 - 30

or, 19r = 270

or, r = 270/19

Therefore the distance between any two successive location is 270/19 km.

5 0

3 years ago