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Hitman42 [59]
3 years ago
6

4(8x+6)-2=22 Please help​

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
6 0

Answer:

0/32

Step-by-step explanation:

4(8x+6)-2=22

32x+24-2=22

32x+22=22

32x=22-22

32x=0

x= 0/32

VMariaS [17]3 years ago
4 0

Answer:

x=0

Step-by-step explanation:

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Step-by-step explanation:


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I have done the segment 1 pratice exam
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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)

<u>step 2:-</u>

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

<u>A=1</u>

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)

<u>-</u>1  =  C

<u>step 3:-</u>

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)

<u>step 4:-</u>

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
In August, 85% of the middle school students voted in a school election. The number of students who voted was 544. How many stud
Maksim231197 [3]

Answer:

640 students

Step-by-step explanation:

544 divided by 85%= 640

5 0
3 years ago
300 ml of pure alcohol is poured from a bottle containing 2 l of pure alcohol. Then, 300 ml of water is added into the bottle. A
Ede4ka [16]

Answer:

The present percentage of pure alcohol in the solution is 72.25% of pure alcohol

Step-by-step explanation:

The volume of pure alcohol poured from the 2 l bottle of pure alcohol = 300 ml of pure alcohol

The volume of water added into the bottle after pouring out the pure alcohol = 300 ml of water

The volume of diluted alcohol poured out of the bottle = 300 ml of diluted alcohol

The volume of water added into the bottle of diluted alcohol after pouring out the 300 ml of diluted alcohol = 300 ml of water

Step 1

After pouring the 300 ml of pure alcohol and adding 300 ml of water to the bottle, the percentage concentration, C%₁ is given as follows;

C%₁ = (Volume of pure alcohol)/(Total volume of the solution) × 100

The volume of pure alcohol in the bottle = 2 l - 300 ml = 1,700 ml

The total volume of the solution = The volume of pure alcohol in the bottle +  The volume of water added = 1,700 ml + 300 ml = 2,000 ml = 2 l

∴ C%₁ = (1,700 ml)/(2,000 ml) × 100 = 85% percent alcohol

Step 2

After pouring out 300 ml diluted alcohol from the 2,000 ml, 85% alcohol and adding 300 ml of water, we have;

Volume of 85% alcohol = 2,000 ml - 300 ml = 1,700 ml

The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol = 85/100 × 1,700 = 1,445 ml

The total volume of the diluted solution = The volume of the 85% alcohol in the solution + The volume of water added

∴ The total volume of the twice diluted solution = 1,700 ml + 300 ml = 2,000 ml

The present percentage of pure alcohol in the solution, C%₂ = (The volume of pure alcohol in the 1,700 ml, 85% diluted alcohol)/(The total volume of the diluted solution) × 100

∴ C%₂ = (1,445 ml)/(2,000 ml) × 100 = 72.25 %

The present percentage of pure alcohol in the solution, C%₂ = 72.25%

3 0
3 years ago
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