10. What is the equation of the line perpendicular to y= (1/2)x-2 that contains the point (2,5)

Which expression is equivalent to this expression? x(2x + 3) A) 2x2 + 3 B) 2x2 + 3x C) 2x2 - 3x Eliminate D) 3x + 3

ira [324]

Answer:

B

Step-by-step explanation:

x(2x + 3)

Expand the brackets.

x(2x)+x(3)

Multiply the terms.

2x² + 3x

The answer is 2x² + 3x.

3 0

3 years ago

Read 2 more answers

How do I solve this and what kind of formula is it?

soldier1979 [14.2K]

Answer:

160

Step-by-step explanation:

The measure of an arc is always twice the degree of the corresponding angle inside the circle. For example, the measure of arc DC is just double the measure of angle DOC; angle DOC is 44 degrees, so arc DC must be 88 degrees.

Angle COB is 80 degrees, so arc CB is 160 degrees.

5 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

I need help on this question plz

drek231 [11]

Given is triangle with base = 10 cm and perpendicular height = 3 cm.

We know the formula for area of triangle is given by :-

Area = $\frac{1}{2}$ × (Base) × (perpendicular height)

Area = $\frac{1}{2}$ × (10) × (3)

Area = $\frac{30}{2}$

Area = 15 cm²

So, final answer is 15 cm².

3 0

3 years ago

*simplify 5|-3+(-7)|=

Lena [83]

Answer:

-2l -3

Step-by-step explanation:

5l -3+ (-7)l

5l -3 -7l

5l -7l -3

-2l -3

6 0

3 years ago