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Travka [436]
3 years ago
6

What is the best estimate of the circumference of a circle with a diameter of 47 mm: a) 141mm b) 139mm c) 121mm d) 100mm

Mathematics
1 answer:
Marina86 [1]3 years ago
7 0
Using the formulasC=2π r
d=2r
Solving for C:
C=π
d=π·47≈147.65485mm

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Solve.<br><br> x^2 - 49 = 0<br> A) ±49 <br> B) ±12 <br> C) ±7 <br> D) ±6
zlopas [31]
The answer is c
x^2 = 49
x = √49
√49 = ±7
3 0
3 years ago
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Solve the equation<br> -4(x - 26) = -200
Naddik [55]

Answer:

x=76

Step-by-step explanation:

-4(x - 26) = -200

x-26=50

x=50+26

x=76

I hope this helped you! If it did, please consider rating, pressing thanks, and giving my answer 'Brainliest.' Have a great day! :)

3 0
2 years ago
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The intersection of any two disjoint sets is a null set. Justify your answer.​
Ne4ueva [31]

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Yes The intersection of any two disjoint sets is a null set

if two sets do not have any common element  then their intersection does not have any element Therefore its intersection is a null set

Step-by-step explanation:

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3 years ago
All of the following values are part of the domain EXCEPT: <br> -1<br> -3<br> -4
artcher [175]
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4 0
2 years ago
Please help me with the below question.
VMariaS [17]

By letting

y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}

we get derivatives

y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}

y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0

Examine the lowest degree term \left(x^{r-1}\right), which gives rise to the indicial equation,

5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients c_k is

(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}

so that with r = 4/5, the coefficients are governed by

c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}

c) Starting with c_0=1, we find

c_1 = -\dfrac{c_0}5 = -\dfrac15

c_2 = -\dfrac{c_1}{10} = \dfrac1{50}

so that the first three terms of the solution are

\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}

4 0
2 years ago
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