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4 years ago

When paying off debts ,you should ___ .

Mathematics

2 answers:

shusha [124]4 years ago

7 0

B: because you will have to pay an little more when u have the money

kati45 [8]4 years ago

7 0

<h2>Answer:</h2>

B.pay as much as possible

Step-by-step explanation:

Debts are unpaid money that a person owes to its lenders. The lender can be any bank or any other financial institution or any person.

Long accumulated debts on banks, if not paid, can result in a person filing for bankruptcy. So, it is always good if you pay your debts on time.

Paying minimum or slightly more than minimum amount will result in paying more in the near future. It is possible that if you always pay minimum amount, you will not be able to clear debts anytime soon.

So, its best that if you have money, always pay as much as possible to get rid of your debt.

You might be interested in

What is the answer to 3(x+1) = 5+x

AleksandrR [38]

3 (x + 1) = 5 + x

First you must distribute the 3 to the values inside the parentheses:

3 (x + 1) = 5 + x

(3 * x) + (3 * 1) = 5 + x

3x + 3 = 5 + x

Now you must combine like terms. Like terms are numbers that have matching variables like 3x and x OR are numbers with out variables like 3 and 5.

3x - x = 5 - 3

2x = 2

Next, to completely isolate x, divide 2 to both sides. Since 2 is being multiplied by x, division (the opposite of multiplication) will cancel 2 out (in this case it will make 2 one) from the left side and bring it over to the right side.

2x / 2 = 2 / 2

1x = 1

Check:

3 (1 + 1) = 5 + 1

3 (2) = 5 + 1

6 = 6

^^^This is true, therefore:

x = 1

Hope this helped!

~Just a girl in love with Shawn Mendes

7 0

4 years ago

Read 2 more answers

Joey built a model sailboat that is similar to his real sailboat. The triangular sail on

lord [1]

Answer:

31.25

Step-by-step explanation:

25/12=2.083333

15 x 2.0833333=31.24999999995

round up to get 31.25

3 0

3 years ago

The measure of b is _______a0

postnew [5]

${c}^{2} = {a}^{2} + {b}^{2}$
${29}^{2} = {20}^{2} + {b}^{2}$
$b = \sqrt{ {29}^{2} - {20}^{2} }$
$b = \sqrt{441} \\ b = 21$

5 0

4 years ago

Read 2 more answers

Tobias won a $100 gift card at a local Fortnite tournament. He decided that he wanted to

Scrat [10]

Answer:

100-5.25x 100-5.25(15)= 22.25

Step-by-step explanation:

5 0

3 years ago

Relative extrema of f(x)=(x+3)/(x-2)

Salsk061 [2.6K]

Answer:

$\displaystyle f(x) = \frac{x + 3}{x - 2}$ has no relative extrema when the domain is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Step-by-step explanation:

Assume that the domain of $\displaystyle f(x) = \frac{x + 3}{x - 2}$ is $\mathbb{R} \backslash \lbrace 2 \rbrace$ (the set of all real numbers other than $2$ .)

Let $f^{\prime}(x)$ and $f^{\prime\prime}(x)$ denote the first and second derivative of this function at $x$ .

Since this domain is an open interval, $x = a$ is a relative extremum of this function if and only if $f^{\prime}(a) = 0$ and $f^{\prime\prime}(a) \ne 0$ .

Hence, if it could be shown that $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ , one could conclude that it is impossible for $\displaystyle f(x) = \frac{x + 3}{x - 2}$ to have any relative extrema over this domain- regardless of the value of $f^{\prime\prime}(x)$ .

$\displaystyle f(x) = \frac{x + 3}{x - 2} = (x + 3) \, (x - 2)^{-1}$ .

Apply the product rule and the power rule to find $f^{\prime}(x)$ .

$\begin{aligned}f^{\prime}(x) &= \frac{d}{dx} \left[ (x + 3) \, (x - 2)^{-1}\right] \\ &= \left(\frac{d}{dx}\, [(x + 3)]\right)\, (x - 2)^{-1} \\ &\quad\quad (x + 3)\, \left(\frac{d}{dx}\, [(x - 2)^{-1}]\right) \\ &= (x - 2)^{-1} \\ &\quad\quad+ (x + 3) \, \left[(-1)\, (x - 2)^{-2}\, \left(\frac{d}{dx}\, [(x - 2)]\right) \right] \\ &= \frac{1}{x - 2} + \frac{-(x+ 3)}{(x - 2)^{2}} \\ &= \frac{(x - 2) - (x + 3)}{(x - 2)^{2}} = \frac{-5}{(x - 2)^{2}}\end{aligned}$ .

In other words, $\displaystyle f^{\prime}(x) = \frac{-5}{(x - 2)^{2}}$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

Since the numerator of this fraction is a non-zero constant, $f^{\prime}(x) \ne 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ . (To be precise, $f^{\prime}(x) < 0$ for all $x \in \mathbb{R} \backslash \lbrace 2 \rbrace\!$ .)

Hence, regardless of the value of $f^{\prime\prime}(x)$ , the function $f(x)$ would have no relative extrema over the domain $x \in \mathbb{R} \backslash \lbrace 2 \rbrace$ .

7 0

3 years ago