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Sophie [7]
2 years ago
6

When calculating compound interest, the periodic interest rate is always less than the annual interest rate. True or false

Mathematics
1 answer:
uranmaximum [27]2 years ago
6 0

Answer: False


Step-by-step explanation: If your Compounding interest, the periodic interest is Sometimes less than the annual interest because the period of interest could be say 2 years

Hope this helps ☺  


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Help please And how would you get the answer need asap
Rashid [163]
No the 2 equations are not equal
the -1/10, -1/10 can't be subtracted from 1 3/10g
they are not like terms
the proper equation is
1/5g converts to 2/10g
2/10g - 1g + 1 3/10 g = 5/10g or 1/2g
1/2g - 2/10 is final equation
4 0
3 years ago
For f(x) = 2x – 3 find f(-1/4)
Firdavs [7]

9514 1404 393

Answer:

  -3 1/2

Step-by-step explanation:

Put the argument where the variable is and do the arithmetic.

  f(-1/4) = 2(-1/4) -3 = -1/2 -3

  f(-1/4) = -(3 1/2)

5 0
3 years ago
Given f (x ) = x^2 + 3x + 2 and g (x ) = x + 1, perform the indicated operations.
Nana76 [90]

Answer:

(i) (f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = x³ + 4·x² + 5·x + 2

Step-by-step explanation:

The given functions are;

f(x) = x² + 3·x + 2

g(x) = x + 1

(i) (f - g)(x) = f(x) - g(x)

∴ (f - g)(x) = x² + 3·x + 2 - (x + 1) = x² + 3·x + 2 - x - 1 = x² + 2·x + 1

(f - g)(x) = x² + 2·x + 1

(ii) (f + g)(x) = f(x) + g(x)

∴ (f + g)(x) = x² + 3·x + 2 + (x + 1) = x² + 3·x + 2 + x + 1 = x² + 4·x + 3

(f + g)(x) = x² + 4·x + 3

(iii) (f·g)(x) = f(x) × g(x)

∴ (f·g)(x) = (x² + 3·x + 2) × (x + 1) = x³ + 3·x² + 2·x + x² + 3·x + 2 = x³ + 4·x² + 5·x + 2

(f·g)(x) = x³ + 4·x² + 5·x + 2

7 0
3 years ago
There were 323232 volunteers to donate blood. Unfortunately, nnn of the volunteers did not meet the health requirements, so they
trapecia [35]

Answer:

Concept: Application of Real world problems

  • They counted 470470470 milliliters
  • You donate on average 470 ml per person
  • So 470470470÷ 470= 1,001,001 people.

4 0
3 years ago
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Travka [436]
Could you tell me what your question is exactly?
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3 years ago
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