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egoroff_w [7]
3 years ago
10

Please answer it in two minutes

Mathematics
1 answer:
Setler79 [48]3 years ago
7 0

Answer:

0.9

I guess.

If yes

.. Follow me..

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Solve for x: 2x^2 + x - 4 = 0
ella [17]

Answer:

Hopefully this helps, sorry if it doesn't.

Step-by-step explanation:

7 0
3 years ago
Craig solicited donations for his school's jump-rope-a-thon. He collected $13 in fixed donations and pledges totaling $2 for eac
Nastasia [14]

Answer:

Step-by-step explanation:

18x2

6 0
2 years ago
What’s the unit rate for three dollars for 2 1/2 hours of work
Maru [420]

The unit rate for three dollars for 2 1/2 hours of work will be $1.2 using the concept of unitary method.

<h3>What is unitary method?</h3>

The unitary method is a technique that involves determining the value of a single unit and then calculating the value of the requisite number of units based on that value. The term unitary refers to a single or unique entity. As a result, the goal of this approach is to determine values in reference to a single unit. The unitary technique is a method for determining the value of any necessary quantity by first obtaining the value of the unit (one) quantity.

Here,

$3 for 2.5 hours of work,

1 hour will cost $x,

3/2.5=x/1

2.5x=3

x=$1.2

Using the unitary technique, the unit rate for three dollars for two and a half hours of labour will be $1.2.

To know more about unitary method,

brainly.com/question/22056199?referrer=searchResults

#SPJ1

4 0
1 year ago
Answer my question please
Viktor [21]
First number multiply by 3 for the next number subtract 10 and repeat process
6 0
3 years ago
SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).
Ann [662]

Answer:

Required series is:

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

Step-by-step explanation:

Given that

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

We know that:

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

Comparing (1) and (2)

                           f'(x)=-(tan^{-1}x) ---- (3)

Using power series expansion for tan^{-1}x

f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx

= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx

=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]

=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}

=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

as

                 tan^{-1}(0)=0 \implies C=0

Hence,

\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....

7 0
3 years ago
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