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Sergio [31]
3 years ago
13

the function f is defined by f(x)= 3x-4cos(2x+1), and it derivative is f'(x) = 3+ 8sin(2x+1). what are all the values of x that

satisfy the conclusion of the mean value theorem applied to f on the interval [-1,2].
Mathematics
1 answer:
olchik [2.2K]3 years ago
3 0

Answer:

hitgn vfpcik predlsxn[3edls

Step-by-step explanation:

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The measures of the angles of a triangle are ∠A = (2x)°, ∠B = (x+14)°, ∠C = (x−38)° Determine the value of x. Determine A B and
nexus9112 [7]

Answer:

Step-by-step explanation:

Measures of angles are,

m∠A = (2x)°

m∠B = (x + 14)°

m∠C = (x - 38)°

By triangle sum theorem,

m∠A + m∠B + m∠C = 180°

2x + (x + 14) + (x - 38) = 180

(2x + x + x) + (14 - 38) = 180

4x - 24 = 180

4x = 204

x = 51

m∠A = 2(51)° = 102°

m∠B = (51 + 14)° = 65°

m∠C = (51 - 38)° = 13°

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3 years ago
Convert 3.212121… into a fraction using variable
seraphim [82]

Answer:

In fraction it is

\frac{3212121}{1000000}

<h2>#NoNeedFriends</h2>
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2 years ago
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For the given values of n and d, find integers q and r such that n = dq + r and 0 ≤ r &lt; d. n = −67, d = 8
Firlakuza [10]

Answer:

q = 8

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Step-by-step explanation:

Given

n = dq + r

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n = 67 --- not -67

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Required

Find q and r

Substitute values for d and b

n = dq + r

67 = 8 * q + r

67 = 8q + r

Make q the subject

q = \frac{67 - r}{8}

0 \le r < d means that r is less than 8 but greater than or equal to 0

And r and q are integers.

Let r = 3

q = \frac{67 - 3}{8}

q = \frac{64}{8}

q = 8

No other true values of r and q can be gotten.

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Answer:

the answer to this problem would be 8.

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