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3 years ago

The rectangle below has an area of x^2-6x-7 square meters and a width of x-7 meters

Mathematics

1 answer:

Illusion [34]3 years ago

8 0

Complete Question:

The rectangle below has an area of x^2-6x-7 square meters and a width of x-7 meters. What expression represents the length of the rectangle?

Answer:

<h3>The expression represents the length of the rectangle is (x + 1) meter</h3>

Solution:

Given that,

$\text{Area of rectangle} = x^2 - 6x - 7 \\\\Width = x - 7$

TO FIND: LENGTH = ?

We know that,

$Area\ of\ rectangle = length \times width \\\\Therefore\\\\x^2 - 6x - 7 = length \times x - 7\\\\length = \frac{x^2 - 6x - 7 }{x - 7 }\\\\length = \frac{(x-7)(x + 1)}{x - 7}\\\\\text{cancel out x-7 from numerator and denominator} \\\\length = x + 1$

Thus expression represents the length of the rectangle is (x + 1) meter

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salantis [7]

Answer:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{1}{2}$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$

L'Hopital's Rule

Differentiation

Derivatives
Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

We are given the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)}$

When we directly plug in x = 0, we see that we would have an indeterminate form:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \frac{0}{0}$

This tells us we need to use L'Hoptial's Rule. Let's differentiate the limit:

$\displaystyle \lim_{x \to 0} \frac{\sqrt{cos(2x)} - \sqrt[3]{cos(3x)}}{sin(x^2)} = \displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)}$

Plugging in x = 0 again, we would get:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \frac{0}{0}$

Since we reached another indeterminate form, let's apply L'Hoptial's Rule again:

$\displaystyle \lim_{x \to 0} \frac{\frac{-sin(2x)}{\sqrt{cos(2x)}} + \frac{sin(3x)}{[cos(3x)]^{\frac{2}{3}}}}{2xcos(x^2)} = \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)}$

Substitute in x = 0 once more:

$\displaystyle \lim_{x \to 0} \frac{\frac{-[cos^2(2x) + 1]}{[cos(2x)]^{\frac{2}{3}}} + \frac{cos^2(3x) + 2}{[cos(3x)]^{\frac{5}{3}}}}{2cos(x^2) - 4x^2sin(x^2)} = \frac{1}{2}$

And we have our final answer.

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Limits

6 0

3 years ago

Consider the angle below.

Andrei [34K]

Answer:

x = 10 meter

y= 20 meter

shadw of a man = 10 meters

6 0

3 years ago

True or false? show how do you know: 57×2×5=10×57

gtnhenbr [62]

Answer:

true

Step-by-step explanation:

because If you multiply the 2 by 5 you will get 10

4 0

3 years ago

Use the associative property to identify which expression is equal to (10)(5x)(7). 1.(7)(5x)(10) 2. 10(5x + 7) 3. (5x)(10)(7) 4.

8090 [49]

Associative property is dealing with parentahsees and orders
it is
a(bc)=(ab)c and
a+(b+c)=(a+b)+c

so we want the multiplication one

given
(10)(5x)(7)
we can arragnge these a number of ways (we should use commutative property, but whatever)

first one works, the 7 was moved to front
2nd is wrong, that is a wierd version of the distributive
3rd is right, 5x was moved to the front
4th is wack, no bueno

answer is 1st and 3rd options

6 0

3 years ago

Julio earns $2 for 1/5 of an hour of work. How much does Julio earn in 1 hour of work?

Goshia [24]

Answer:

10 dollars

Step-by-step explanation:

he is racking in the money

6 0

3 years ago