All categories

3 years ago

Which of the following is not a credit card perk? A. Cash back B. Hotel Points C. apr D. Air Miles

Mathematics

1 answer:

Leona [35]3 years ago

5 0

Answer:

Step-by-step explanation:

ahahahahhahahahahahahahhahahannooooooooooooooooonoooo

You might be interested in

Help me and ill mark you brainliest

Harman [31]

Answer:

question 13 = D

Step-by-step explanation:

because you do some math, open a calculator and boom, you have that.

6 0

2 years ago

A number increased by 100% and decreased by 20% is x. what is the number ? NEEDED TODAY

vodka [1.7K]

Answer:

Let's assume the number be 100.

According to the problem , 100 is increased by 100%.

So, new number will be 100+100=200

Now 200 will decreased by 20%.

So, 20% of 200= 0.20*200= 40.

So, again the number has been changed into 200+40=240.

So, x=240

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

-pls answer all thanks!!! HAVE A BLESS DAY

blagie [28]

1. = 3
2.= 1
3.= 3/1= 3

5 0

2 years ago

Read 2 more answers

Evaluate -x^2-2x+5 when x=-10

kvv77 [185]

If I'm correct and the ^ stands for multiplication then the answer should be 5.

The picture is my work just ignore the fraction.

6 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

2 years ago