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2 years ago

PLEASE HELP FAST!!!

Mathematics

1 answer:

rewona [7]2 years ago

4 0

By trigonometric functions and law of cosines, the value of x associated with a missing angle in the geometric system is between 7.701 and 7.856.

<h3>How to find a missing variable associated to an angle by trigonometry</h3>

In this question we have a geometric system that includes a right triangle, whose missing angle is determined by the following trigonometric function:

sin (7 · x + 4) = 12/14

7 · x + 4 = sin⁻¹ (12/14)

7 · x + 4 ≈ 58.997°

7 · x = 54.997°

x ≈ 7.856

In addition, the geometric system also includes a obtuse-angle triangle and that angle can be also found by the law of the cosine:

7² = 8² + 6² - 2 · (8) · (6) · cos (7 · x + 4)

17/32 = cos (7 · x + 4)

7 · x + 4 = cos⁻¹ (17/32)

7 · x + 4 ≈ 57.910°

7 · x ≈ 53.910°

x ≈ 7.701

Hence, we conclude that the value of x associated with a missing angle in the geometric system is between 7.701 and 7.856.

To learn more on triangles: brainly.com/question/25813512

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Step-by-step explanation:

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3 years ago

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Step-by-step explanation:

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3 years ago

Consider functions of the form f(x)=a^x for various values of a. In particular, choose a sequence of values of a that converges

sleet_krkn [62]

Answer:

A. As "a"⇒e, the function f(x)=aˣ tends to be its derivative.

Step-by-step explanation:

A. To show the stretched relation between the fact that "a"⇒e and the derivatives of the function, let´s differentiate f(x) without a value for "a" (leaving it as a constant):

$f(x)=a^{x}\\ f'(x)=a^xln(a)$

The process will help us to understand what is happening, at first we rewrite the function:

$f(x)=a^x\\ f(x)=e^{ln(a^x)}\\ f(x)=e^{xln(a)}\\$

And then, we use the chain rule to differentiate:

$f'(x)=e^{xln(a)}ln(a)\\ f'(x)=a^xln(a)$

Notice the only difference between f(x) and its derivative is the new factor ln(a). But we know that ln(e)=1, this tell us that as "a"⇒e, ln(a)⇒1 (because ln(x) is a continuous function in (0,∞) ) and as a consequence f'(x)⇒f(x).

In the graph that is attached it´s shown that the functions follows this inequality (the segmented lines are the derivatives):

if a<e<b, then aˣln(a) < aˣ < eˣ < bˣ < bˣln(b) (and below we explain why this happen)

Considering that ln(a) is a growing function and ln(e)=1, we have:

if a<e<b, then ln(a)< 1 <ln(b)

if a<e, then aˣln(a)<aˣ

if e<b, then bˣ<bˣln(b)

And because eˣ is defined to be the same as its derivative, the cases above results in the following

if a<e<b, then aˣ < eˣ < bˣ (because this function is also a growing function as "a" and "b" gets closer to e)

if a<e, then aˣln(a)<aˣ<eˣ ( f'(x)<f(x) )

if e<b, then eˣ<bˣ<bˣln(b) ( f(x)<f'(x) )

but as "a"⇒e, the difference between f(x) and f'(x) begin to decrease until it gets zero (when a=e)

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4 years ago

For what interval is the value of (f – g)(x) negative?

mrs_skeptik [129]

Complete question:

Refer the attached diagram

Answer:

In reference to the attached figure, (-∞, 2) is the value where (f-g) (x) negative.

Step-by-step explanation:

From the attached figure, it shows that given data:

f (x) = x – 3

g (x) = - 0.5 x

To Find: At what interval the value of (f-g) (x) negative

So, first we need to calculate the (f-g) (x)

(f – g ) (x) = f (x) – g (x) = x-3 - (- 0.5 x)

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Now we are supposed to find the interval for which (f-g) (x) is negative.

⇒ (f - g) (x) = x - 3+ 0.5 x = 1.5 x – 3 < 0

⇒ 1.5 x – 3 < 0

⇒ 1.5 x < 3

⇒ $x$

⇒ x < 2

Thus for (f - g) (x) negative x must be less than 2. Thereby, the interval is (-∞, 2). Function is negative when graph line lies below the x - axis.

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