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Alekssandra [29.7K]
4 years ago
13

Thanks for helping with calculus

Mathematics
1 answer:
allochka39001 [22]4 years ago
3 0

Integrating f(x) gives

g(x)=\displaystyle\int f(x)\,\mathrm dx=C+\sum_{n=0}^\infty\frac{a_n}{n+1}x^{n+1}

Notice that x=0 forces all terms in the sum to be 0, so that

g(0)=C=5

and hence

g(x)=\displaystyle5+\sum_{n=0}^\infty\frac{a_n}{n+1}x^{n+1}

making the first option correct.

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The ratio of the number of Mays necklaces to the number of Helen's necklace is 25:9. May has 100 necklaces. How many necklaces d
likoan [24]

Answer:

36

Step-by-step explanation:

25:9

25*4 = 100

9*4 = 36

6 0
3 years ago
The variance of a stock's returns can be calculated as the:
Tamiku [17]

Answer:

The correct option is b.

Step-by-step explanation:

The formula for standard deviation is

\sigma^2=\frac{\sum {(x-\overline{x})^2}}{n}

where, \overline{x} is mean of the data and n is number of observation.

The variance of a stock's returns can be calculated by the above formula.

Variance of stock's returns is the average value of squared deviations from the mean.

Therefore the correct option is b.

4 0
4 years ago
I need help on this asap, I'm giving 39 POINTS and mark BRAINLIEST the question is on the pic! thankss
shtirl [24]
Answer: I’m pretty sure it’s c also
3 0
3 years ago
Solve (round to two decimal places) <br><br> can someone please help me!
Degger [83]
Cos(40)= 21/ (hypotenuse)
hypotenuse(cos(40))= 21
hypotenuse = 21/ cos(40)
21/ cos40 = 27.41
4 0
3 years ago
In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an
Airida [17]

This is a little long, but it gets you there.

  1. ΔEBH ≅ ΔEBC . . . . HA theorem
  2. EH ≅ EC . . . . . . . . . CPCTC
  3. ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
  4. ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
  5. ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
  6. ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
  7. ΔDAC ≅ ΔDAG . . . HA theorem
  8. DC ≅ DG . . . . . . . . . CPCTC
  9. ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
  10. ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
  11. ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
  12. ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
  13. (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
  14. ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
  15. This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
5 0
3 years ago
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