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Burka [1]
3 years ago
6

Find the value of the 60th term of the following sequence: 33, 28, 23, 18

Mathematics
1 answer:
Otrada [13]3 years ago
4 0

Step-by-step explanation:

The first term of the AP i.e a = 33

and common difference i.e d = 28-33 = -5

So, 60th term = a +(n-1)d = a+ 59d

= 33 + 59( -5)

= 33- 295

= - 262

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19 is greater than the sum the number n and 3
masha68 [24]
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2 years ago
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Answer:

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6 0
3 years ago
4 + 6^2 + 9 ∙ 3 - 10
Korvikt [17]
4+6^2+9•3-10

PEMDAS
P=parentheses
E=exponent
m=multiplication
d=division
a=addition
s=subtraction

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6^2=36
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ANSWER=57







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3 0
3 years ago
Suppose that we don't have a formula for g(x) but we know that g(3) = −1 and g'(x) = x2 + 7 for all x.
Sliva [168]

Answer:

  • g(2.95) ≈ -1.8; g(3.05) ≈ -0.2
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Step-by-step explanation:

(a) The linear approximation of g(x) at x=b will be ...

  g(x) ≈ g'(b)(x -b) +g(b)

Using the given relations, this is ...

  g'(3) = 3² +7 = 16

  g(x) ≈ 16(x -3) -1

Then the points of interest are ...

  g(2.95) ≈ 16(2.95 -3) -1 = -1.8

  g(3.05) ≈ 16(3.05 -3) -1 = -0.2

__

(b) At x=3, the slope of the curve is increasing, so the tangent lies below the curve. The estimates are too small. (Matches description A.)

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