Solve for x. Round to the nearest tenth of a degree, if necessary.

Mathematics

1 answer:

kakasveta [241]3 years ago

6 0

Answer:

x ≈ 52.1°

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan x = $\frac{opposite}{adjacent}$ = $\frac{KL}{LM}$ = $\frac{36}{28}$ , then

x = $tan^{-1}$ ( $\frac{36}{28}$ ) ≈ 52.1° () to the nearest tenth )

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FInd the length of DB and CD. Please help!

Lerok [7]

Step-by-step explanation:

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= 63/12 + 8/12 = 71/12

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2 years ago

Find the critical points, domain endpoints, and local extreme values for the function

aliina [53]

Answer:

a) $x = -3$ , b) $x = 0$ , $x = -6$ , c) $x = 0$ , d) $x = -6$

Step-by-step explanation:

a) Let derive the function:

$f'(x) = \frac{10\cdot x \cdot (x+3)-5\cdot x^{2}}{25\cdot (x+3)^{2}}$

$f'(x)$ is undefined when denominator equates to zero. The critical point is:

$x = -3$

b) $f'(x) = 0$ when numerator equates to zero. That is:

$10\cdot x \cdot (x+3) - 5\cdot x^{2} = 0$

$10\cdot x^{2}+30\cdot x -5\cdot x^{2} = 0$

$5\cdot x^{2} + 30\cdot x = 0$

$5\cdot x \cdot (x+6) = 0$

This equation shows two critical points:

$x = 0$ , $x = -6$

c) The critical points found in point b) and the existence of a discontinuity in point a) lead to the conclusion of the existence local minima and maxima. By plotting the function, it is evident that $x = 0$ corresponds to a local maximum. (See Attachment)

d) By plotting the function, it is evident that $x = -6$ corresponds to a local minimum. (See Attachment)