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3 years ago

Which method for calculating credit card balances is best for consumers who make large payments? A) Average daily balance method

Mathematics

1 answer:

Margarita [4]3 years ago

7 0

Answer:

D. Adjusted balance method

Step-by-step explanation:

a.p.e.x test

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Solve 6 and 7 please?

Charra [1.4K]

Hey!

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We know that (a = -3), (b = 2), (c = 5), and (d = -4)

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Question 6:

= 5ac - 2b

= 5(-3)(5) - 2(2)

= 5(-15) - 4

= -75 - 4

= -79

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Question 7:

= ab/d

= (-3)(2)/-4

= -6/-4

= 1 1/2

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Hope this helped, good luck!

6 0

3 years ago

Solve for x: |x| − 8 = −5

hjlf

X= 3 and x= -3 that should be the correct answer

3 0

3 years ago

The ratio of dark chocolate candies to white chocolate candies is 3 to 2. There are 18 pieces of dark chocolate. How many pieces

ch4aika [34]

there are 18 pieces of dark chocolate and the ratio of dark and white is 3:2 so we can divide 18 by 3.

18/3 = 6.

This means the ratio of dark chocolate to white as a 1:1 ratio would be 6:6 so we can multiply 6 by 2 to give us the total amount of pieces of white chocolate.

6*2 = 12.

Therefore there are 12 white chocolate pieces.

4 0

3 years ago

Whats the quotient of -8 divided by 2

malfutka [58]

Answer:

-4

Step-by-step explanation:

Dividing by two is the same as splitting in half. Half of -8 is -4

8 0

3 years ago

Read 2 more answers

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

3 years ago