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3 years ago

How can bargain-shopping help spending?

Mathematics

1 answer:

Over [174]3 years ago

5 0

Answer: It is shopping around for the best prices and saves money.

Step-by-step explanation: Bargain has the word "gain" in it. That's the best way I would look at it. Therefor I would cross out any other option that has to do with no benefit or the loss of something. Because we you bargain, you're getting multiple of something for a good price or maybe getting something that's on sale for a good price. For example: Buy 1 get 1 free items.

You might be interested in

What is £300 divided into a ratio of 3:7

3241004551 [841]

A) sum 3 + 7 =10
b) divide £300 ÷ 10 = £30
c) multiply:
3 x 30 = £90
7 x 30 = £210

Values are: £90 and £210

5 0

3 years ago

Read 2 more answers

ACFG is a parallelogram, If AX = 4y + 3 and XF = 2y + 7, find AX.

pashok25 [27]

Answer:

AX=XF

4y+3=2y+7

4y-2y=7-3

2y=4

y=2

then,

AX=4y+3

=4×2+3

=8+3

=11

3 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

How do I convert 4/5 to a percent

Gelneren [198K]

(4/5) x 100%
= 80%
=0.8

5 0

3 years ago

Suppose you received a job offer with a starting salary of $48,500 per year and a guaranteed raise of $1200 per year. Find the n

LuckyWell [14K]

Answer:

7 years

Step-by-step explanation:

This is going to be an arithmetic progression:

48500, 48500 + 1200, 48500 + 1200*2, ...

So the first term is

a₁ = 48500

and the common difference is

d = 1200

Sum of AP is:

S = (a₁ + aₙ)*n/2 = (2a₁ + (n - 1)d)*n/2

Substitute and solve for n:

(2*48500 + (n - 1)*1200)*n/2 = 321000
(97000 + 1200n - 1200)n = 642000
1200n² + 95800n - 642000 = 0
6n² + 479n - 3210 = 0

Solving we get a positive root of:

n ≈ 6.21 and it rounds up to 7 years

3 0

3 years ago