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2 years ago

Anthony buys a bag of oranges for $5.79. He pays with a $10 bill.

Mathematics

2 answers:

Lemur [1.5K]2 years ago

5 0

Answer:

4.21$

Step-by-step explanation:

Because his total bill is 5.79$

If he pays with a 10$ bill meaning he get's 4.21$ back.

To calculate this we could subtract 5.79 by 10

Add to 5.79 lentil 10

Subtraction would look like this.

10.00$

5.79$

______

Since 0 can't subtract nine he goes to his neighbor 0 for help but he can't help so his neighbor 0 also goes for help to his neighbor 0 but he still can't help so then he goes to 1 and puts a 1 in front of 0 to make it 10

And then 0 that turned in to 10 helps other 0 to go to a 9 and it turn into 10 and then that 0 helps that least 0 to turn into 10.

So what's 10 -9=1

9-7=2 so now we got .21

And last to get out final number what is 9 - 5=4 and finally makes 4.21 and that's our answer.

Hope this helps have a great day:)

balandron [24]2 years ago

3 0

Anthony receives $4.21 as change

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VashaNatasha [74]

Answer:

According what I can read, I have the following statements:

$\lim_{x \to 2} f(x) = 1$

$\lim_{x \to 2} g(x) = -4$

$\lim_{x \to 2} h(x) = 0$

a) Applying properties of limits

$\lim_{x \to 2} f(x) + 5g(x) = \lim_{x \to 2} f(x) + 5 \lim_{x \to 2} g(x) = 1 + 5*-4 = -19$

b) Applying properties of limits

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c) Applying properties of limits

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d) Applying properties of limits

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e) Applying properties of limits

$\lim_{x \to 2} g(x)*h(x) = \lim_{x \to 2} g(x)*\lim_{x \to 2} h(x) = -4*0 =0$

f) Applying properties of limits

$\lim_{x \to 2} g(x)*h(x)*f(x) = \lim_{x \to 2} g(x)*\lim_{x \to 2} h(x)*\lim_{x \to 2} f(x = -4*0*1 =0$

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3 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

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Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

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3 years ago

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Answer:

A set of parametric equations for the line y = 4x - 5 is;

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Step-by-step explanation:

To find a set of parametric equations for the line y = 4x - 5;

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We then substitute x = t in the original equation;

y = 4t - 5

Therefore, a set of parametric equations for the line y = 4x - 5 is;

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Answer:

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Step-by-step explanation:

Hope the above image help..

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