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3 years ago

3.simplifyyyyyyyyyyyyyyyyyyy

Mathematics

2 answers:

Charra [1.4K]3 years ago

4 0

Answer:

C. 7

Step-by-step explanation:

garik1379 [7]3 years ago

4 0

Answer:

(3+√2)(3-√2) = 7

Step-by-step explanation:

We have given a expression.

(3+√2)(3-√2)

We have to simplify above expression.

We use difference formula to solve above expression.

(a-b)(a+b) = a²-b²

(3+√2)(3-√2) = (3)²-(√2)²

(3+√2)(3-√2) = 9-2

(3+√2)(3-√2) = 7 which is the answer.

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Jessie finish the first race in two minutes and seven seconds 10 finish 12 seconds faster than Jesse one finished eight seconds

Mandarinka [93]

20 seconds faster than Jesse so one minute and 57 seconds is my best guess.

3 0

3 years ago

Can somebody please help me. I would really appreciate it

laila [671]

90=6x+3x
90=9x
90/9=x
10=x

3x=3*10=30
6x=6*10=60
hope this helped

4 0

3 years ago

A ramp into a building forms a 6° angle with the ground. If the ramp is 8 feet long, how far away from the building is the entry

Sauron [17]

Answer:

7.96 ft

Step-by-step explanation:

Given;

Length of ramp L = 8 ft

Angle with the horizontal (ground) = 6°

Applying trigonometry;

With the length of ramp as the hypothenuse,

The horizontal distance d as the adjacent to angle 6°

Since we want to calculate the adjacent and we have the hypothenuse and the angle. We can apply cosine;

Cosθ = adjacent/hypothenuse

Substituting the values;

Cos6° = d/8

d = 8cos6°

d = 7.956175162946

d = 7.96 ft

The building is 7.96ft away from the entry point of the ramp.

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Clim_%7Bx%5Cto%20%5C%200%7D%20%5Cfrac%7B%5Csqrt%7Bcos2x%7D-%5Csqrt%5B3%5D%7Bcos3x%7D%20%7D%7

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

Crystal earns $5.50 per hour mowing lawns.how much does crystal earn if she works 3 hours and 45 minutes

Deffense [45]

Your question does not make sense unless you can estimate but the answer would be 20.625

Work:
You have to multiply 5.50 to 3.75 because 45 is 75 percent of 60

3 0

3 years ago