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Nana76 [90]
2 years ago
10

Tell and write the correct answers as quickly as possible. (a)1÷2 part of a sum is Rs 10, (i)the sum= (ii)1÷4 part of the sum =

(b) part of the sum= 1÷3 part of a distance is 5km (i) the distance= (ii) 1÷5 part of the distance= I will give 100 points and mark at brainlist.​
Mathematics
1 answer:
Elden [556K]2 years ago
5 0

Answer:

I love you questions

Step-by-step explanation:

but I don't know the answer

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If the base of a right angle triangle is 8<br>cm and the hypotenuse is 17 cm find its<br>area.​
damaskus [11]
The area of the triangle is 60 cm
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Please help <br><br> Find the perimeter of AABC. Round to the nearest tenth
vova2212 [387]

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22.2

Step-by-step explanation:

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Suppose the diameter of a circle is \color{green}{4}4start color green, 4, end color green. what is its radius?
taurus [48]
By definition we have the relationship between the diameter of a circle and its radius is as follows:
 r = d / 2
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 the radius of the circle is 2 units.
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4 x 10^6 is how many times large then 1 x 10^4
Molodets [167]

Answer:

<h2>400</h2>

Step-by-step explanation:

4\times10^6\text{ is how many times large than}\ 1 \times 10^4?\\\\\dfrac{4\times10^6}{1\times10^4}=\dfrac{4}{1}\times\dfrac{10^6}{10^4}=4\times10^{6-4}=4\times10^2=4\times100=400

4 0
4 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
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