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Setler79 [48]
2 years ago
6

I need help as fast as possible

Mathematics
1 answer:
aivan3 [116]2 years ago
4 0

Answer:

36°

Step-by-step explanation:

Given,

Measurement of <3 = 2x+8

Measurement of <4 = 6x+12

Since both are supplementary,

Therefore,

<3 + <4. = 180° [linear pair]

=> 2x + 8 + 6x + 12 = 180

=> 2x + 6x + 8 + 12 = 180

=> 8x + 20 = 180

=> 8x = 180 - 20 = 160

=  > x =  \frac{160}{8}

=> x = 20

Now as we have got the value of x,

According to the question measurement of <4 is,

<4 = 6×4 + 12 = 24 + 12 = <u>36</u><u>°</u><u> (Ans)</u>

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F(x) = x2 − x − ln(x) (a) find the interval on which f is increasing. (enter your answer using interval notation.) find the inte
Mekhanik [1.2K]

Answer:

(a) Decreasing on (0, 1) and increasing on (1, ∞)

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Step-by-step explanation:

ƒ(x) = x² - x – lnx

(a) Intervals in which ƒ(x) is increasing and decreasing.

Step 1. Find the zeros of the first derivative of the function

ƒ'(x) = 2x – 1 - 1/x = 0

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We reject the negative root, because the argument of lnx cannot be negative.

There is one zero at (1, 0). This is your critical point.

Step 2. Apply the first derivative test.

Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.

(1) x = ½

ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1

ƒ'(x) < 0 so the function is decreasing on (0, 1).

(2) x = 2

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ƒ'(x) > 0 so the function is increasing on (1, ∞).

(b) Local extremum

ƒ(x) is decreasing when x < 1 and increasing when x >1.

Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.

(c) Inflection point

(1) Set the second derivative equal to zero

ƒ''(x) = 2 + 2/x² = 0

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                   x² = -2

There is no inflection point.

(2). Concavity

Apply the second derivative test on either side of the extremum.

\begin{array}{lccc}\text{Test} & x < 1 & x = 1 & x > 1\\\text{Sign of f''} & + & 0 & +\\\text{Concavity} & \text{up} & &\text{up}\\\end{array}

The function is concave up on (0, ∞).

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3 years ago
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Firdavs [7]

Answer:

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Step-by-step explanation:

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