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3 years ago

TRUE OR FALSE??

Mathematics

1 answer:

nordsb [41]3 years ago

3 0

Answer:

Step-by-step explanation:

A. True. Vertical angles have equal measure.

B. False. 12x-18=8x+10 -> x=7 -> the angles measure 66 degrees.

C. True. They're formed by a pair of intersecting lines.

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t^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b.

kondaur [170]

Answer:

a) $t = 1$ , b) $t = -2$ , c) Concave upward: $(-\infty, 1)$ . No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

$f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}$

By factorizing each component, the function is re-arranged as:

$f(t) = \frac{(t+2)^{2}}{(t-1)^{2}}$

There is no avoidable discontinuities. The only point where tangent line is vertical is $t = 1$ .

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is $t = -2$ .

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

$f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}$

$f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]$

$f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}$

$f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}$

The only critical point is:

$-6\cdot (t+2) = 0$

$t = -2$ (which coincides with the result of point b)

$f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]$

$f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}} \right]$

The value associated with the critical point is:

$f''(-2) = \frac{2}{9}$

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is $(-\infty, 1)$ . There is no absolute maximums and, therefore, there is no interval that is concave downward.

7 0

4 years ago

Television screens are measured along the diagonal. A certain 53-inch television is 45 inches wide. How tall is the television?

Kazeer [188]

By pythagoras
53^2 = h^2 + 45^2 where h is the height of the television.

h^2 = 53^2 - 45^2 = 784

h = sqrt 784 = 28 inches Answer

4 0

3 years ago

Read 2 more answers

Use the coordinate plane to plot the point (3,6). Then

Margarita [4]

Answer:

Start at the