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3 years ago

What is the answer and work for these attached below

Mathematics

1 answer:

GrogVix [38]3 years ago

5 0

-4(-5-b)=1/3(b+16) Multiply both sides by 3 to get rid of the fraction
-12(-5-b)=b+16 distribute the -12 to get rid of the parenthesis
60+12b=b+16 get the b on the left side and non b values to the right side
11b=-44 solve for b
b=-44/11 simplify the fraction
b=-4

3/5(t+18)=-3(2-t) multiply both sides by 5/3 to get rid of thefraction
t+18=-5(2-t) distribute the -5 to get rid of the parenthises
t+18=-10+5t get the t to the left side and non t values to the right
-4t=-28 solve for t
t=7

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Someone please help:)

iogann1982 [59]

Answer:

2. Addition property of inequality. You're adding 6x to both sides.

3. 4x = 28

4. 4x/4 = 28/4 Justification: Division property of inequality

You're dividing both sides by 4.

5. x = 7

Hope that helps.

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2 years ago

Divide 2x^4 -5x^3+3-1 by 2x -1 <br> please help and please show the steps on how you did it

MaRussiya [10]

The answer below (I need 20 characters to respond so just ignore the words between the parenthesis on this response)

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3 years ago

Explain why each of the following integrals is improper. (a) 4 x x − 3 dx 3 Since the integral has an infinite interval of integ

erma4kov [3.2K]

Answer:

Since the integral has an infinite discontinuity, it is a Type 2 improper integral

Since the integral has an infinite interval of integration, it is a Type 1 improper integral

Since the integral has an infinite discontinuity, it is a Type 2 improper integral

Step-by-step explanation:

Considering a

$\int\limits^4_3 {\frac{x}{x- 3} } \, dx$

Looking at this we that at x = 3 this integral will be infinitely discontinuous

Considering b

$\int\limits^{\infty}_0 {\frac{1}{1 + x^3} } \, dx$

Looking at this integral we see that the interval is between $0 \ and \ \infty$ which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral

Considering c

$\int\limits^{\infty}_{- \infty} {x^2 e^{-x^2}} \, dx$

Looking at this integral we see that the interval is between $-\infty \ and \ \infty$ which means that the integral has an infinite interval of integration , hence it is a Type 1 improper integral

Considering d

$\int\limits^{\frac{\pi}{4} }_0 {cot (x)} \, dx$

Looking at the integral we see that at x = 0 cot (0) will be infinity hence the integral has an infinite discontinuity , so it is a Type 2 improper integral

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3 years ago

(2x to the 5th power) 3x to the 3/5th power

saul85 [17]

The expression simplifies to 6x^(28/5)
hope this helps :)

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2 years ago

The sum of two numbers is 41 . The smaller number is 11 less than the larger number . What are the numbers ?

AleksAgata [21]

26 and 15 is the answer!

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3 years ago