PLEASE PLEASE PLEASE HELP ME Which of the following is a perfect square trinomial?

The table shows the charges for using an internet service. Nicholas use the internet service for 16 hours and 40 Minutes last mo

yanalaym [24]

Answer:

A. Nicholas will have to pay less using plan A

B. He will pay US$ 4 less than plan B

Step-by-step explanation:

Let's compare how much Nicholas will pay on internet service in Plan A and plan B, after using it 16 hours and 40 minutes.

Plan A

Up to 10 hours = US$ 6

Every subsequent 1/2 hour: $1

10 hours + 14 (1/2 hour)

Nicholas will pay 6 + 14 * 1 = US$ 20

Plan B

Up to 12 hours = US$ 4

Every subsequent 1/2 hour: $2

12 hours + 10 (1/2 hour)

Nicholas will pay 4 + 2 * 10 = 24

7 0

2 years ago

Who made an error? The correct scale factor= ?

Ostrovityanka [42]

Answer:

reggie made an error

the correct scale factor is 2/3

Step-by-step explanation:

we want to get from A to A', B to B', and ultimately C to C'

to get there, we must multiply each value in each point by the scale factor.

let's start out with reggie's scale factor. he multiplies each value in C by 3/2 to get to C'. we can try this out with one point, e.g. A

for A: 3/2 * (-12, -6) = (-18, -9). this is not A' = (-8, -4)! thus, 3/2 cannot be the scale factor

now, onto hillary's scale factor of 2/3

for A: 2/3 * (-12, -6) = (-8, -4). this is A'! thus, hillary is correct and reggie made an error

the correct scale factor is thus hillary's: 2/3

4 0

1 year 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

-7-3 log₂ (-n + 10) = -16

SOVA2 [1]

Answer:

2

Step-by-step explanation:

-3 log₂ (-n + 10) = -16+7

-3 log₂ (-n + 10) = -9

log₂ (-n + 10) = -9/-3

log₂ (-n + 10) = 3

-n + 10 = 2^3

-n + 10 = 8

n=2

6 0

1 year ago

What number does 7 and 4 both fit into?

cestrela7 [59]

Easy tip: just multiply seven and four together!
7•4=28
So both seven and four fit into 28

7 0

3 years ago