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IceJOKER [234]
4 years ago
6

3/4*5. u will get followed answer this

Mathematics
1 answer:
Alenkinab [10]4 years ago
7 0

\bf 5\cdot \cfrac{3}{4}\implies \cfrac{5\cdot 3}{4}\implies \cfrac{15}{4} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 15\div 4\implies \stackrel{quotient}{\boxed{3}}~~\stackrel{remainder}{3}~\hfill \cfrac{15}{4}\implies \boxed{3}\frac{3}{4}

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6th grade math help me pleaseeee
KIM [24]

Answer:

8.3

Step-by-step explanation:

Plug -1.9 into the equation.

-5-7(-1.9) = 8.3

Hope I helped!

6 0
3 years ago
What is 49 x 138 - 37 x 49
likoan [24]
49 x 138 = 6762
37 x 49 =1813
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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:

a

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

b

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

c

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

d

     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

     

7 0
3 years ago
..................................
coldgirl [10]

Step-by-step explanation:

What's Yo Question ?

3 0
3 years ago
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