All categories

3 years ago

Can someone help me on this question

Mathematics

1 answer:

andre [41]3 years ago

5 0

Answer:

The answer is about 32,000.

Step-by-step explanation:

If you multiply 3.2 by 10, for 10 years, you get 32 percent. From there, if you multiply 25,000 by 0.32, you get 8000. If you add 25,000 to 8000, you get about 32,000.

You might be interested in

6(x-5)=18 please help me i need this really bad

slega [8]

The answer is 8
hopefully that helps you

8 0

3 years ago

Explain why it is almost impossible to end inflation and high unemployment at the same time PLEASE HELP ME I'M FALLING IN SOCIAL

tangare [24]

It is almost impossible to end inflation and high unemployment at the same time because as more people get money and unemployment goes down there will be high demands in good which will result in inflation.

3 0

2 years ago

Find all real solutions of this equation to answer

Lera25 [3.4K]

Answer:

(6-2x)(3-2x)x=40

Step-by-step explanation:

(-2x-2x) (6+3)x=40

8x=40

8x/8=40/8

x=5

6 0

3 years ago

Read 2 more answers

Janet has a square piece of cloth. She cuts 3 inches off of the length of the square and 3 inches off of the width. The area of

kolezko [41]

24
3+3=6
6×4=24
24 inches

8 0

3 years ago

Integrate 1 - x / x(x2 + 1) d x by partial fractions.

solniwko [45]

Answer:

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

Step-by-step explanation:

step 1:- by using partial fractions

$[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }$ ......(1)

step 2:-

solving on both sides

$1-x=A(x^{2} +1)+(Bx+C)x......(2)$

substitute x =0 value in equation (2)

1=A(1)+0

A=1

comparing x^2 co-efficient on both sides (in equation 2)

0 = A+B

0 = 1+B

B=-1

comparing x co-efficient on both sides (in equation 2)

-1 = C

step 3:-

substitute A,B,C values in equation (1)

now

$\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 } d x -\int\limits \frac{1}{x^{2}+1 } d x$

by using integration formulas

i) by using $\int\limits \frac{1}{x} d x =log x+c........(a)\\\int\limits \frac{f^{1}(x) }{f(x)} d x= log(f(x)+c\\$ .....(b)

$\int\limits tan^{-1}x dx =\frac{1}{1+x^{2} } +C$ .....(c)

step 4:-

by using above integration formulas (a,b,and c)

we get answer is

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

6 0

3 years ago