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3 years ago

For what value of y must QRST be a parallelogram? A: 1 B: 2 C: 3 D: 0.5

Mathematics

2 answers:

Reika [66]3 years ago

8 0

Answer:

Step-by-step explanation:

Diagonals of a parallelogram bisect each other.

3x = 3

x = 1

y = x

y = 1

laila [671]3 years ago

5 0

In a parallelogram, the parts in which each diagonal cuts the other are congruent.

This means that we must have

$\begin{cases}3=3x\\x=y\end{cases}$

From the first equation we can deduce x=1, and thus y=x=1.

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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

The minute hand of a clock is 3 inches long. How far does the tip of the minute hand move in 40 minutes?

Mkey [24]

Answer:

4π in ≈ 12.57 in

Step-by-step explanation:

One full revolution (2π radians) is made in 60 minutes. In 40 minutes, the hand moves through an angle θ of (40/60)(2π) = 4π/3 radians. The length of the arc is ...

s = rθ = (3 in)(4π/3) = 4π in ≈ 12.57 in

6 0

3 years ago

Please help me!!!

Vladimir79 [104]

Answer:

130

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

A 1 =7 and a_n=a_{n-1}+4a n =a n−1 +4

user100 [1]

Answer:

dhhshdhhdjejxj

Step-by-step explanation:

potanginaaaa

3 0

2 years ago

What is 6(5x-3) ??? I dont know what it is

aalyn [17]

Answer:

30x - 18

Step-by-step explanation:

6(5x - 3)

Distribute the 6 to both monomials.

6 * 5x = 30x

6 * -3 = -18

Combine both final monomials.

30x - 18

8 0

3 years ago