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2 years ago

Select all the solutions to (x - 2)2 = -16.

Mathematics

1 answer:

slamgirl [31]2 years ago

4 0

Answer:

d and g

Step-by-step explanation:

Given

(x - 2)² = - 16 ( take the square root of both sides )

x - 2 = ± $\sqrt{-16}$ = ± $\sqrt{16(-1)}$ = ± 4i ( add 2 to both sides )

x = 2 ± 4i

Thus

x = 2 + 4i → (d)

x = 2 - 4i → (g)

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How do I solve the linear equation of 27=6(x)+4(y) over 20=2(x)+5(y)

larisa86 [58]

27 = 6x + 4y = 1 7/20 = 3 4/5
20 = 2x + 5y

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2 years ago

What is area? How do you find the area of a rectangle?

Savatey [412]

Answer:

The extent or measurement of a surface or piece of land is called area.

To find the area of a rectangle, we multiply the length of the rectangle by the width of the rectangle.

Step-by-step explanation:

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2 years ago

Read 2 more answers

The mass of a ball is 128 g. It has a density of 0.5 g/cm³.

liraira [26]

Answer:

The volume of the ball is 256 cm³

Step-by-step explanation:

we know that

The density is equal to the ratio of the mass by the volume

D=m/V

Solve for V

The volume is equal to the ratio of the mass by the density

V=m/D

In this problem we have

m=128 g

D=0.5 g/cm³

substitute

V=128/0.5=256 cm³

7 0

2 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

2 years ago

Millie went shopping and took £60 to spend. she spent 1/2 of her money on a pair of shoes, she then spent 2/3 of what she had le

Stels [109]

Total expenses;
1/2*60=30
2/3*30=20
Total = 50

Remainder;
60-50
£10

3 0

2 years ago