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dimulka [17.4K]

3 years ago

6

Imagine two lines intersect. How can the properties of linear pairs and vertical angles help to determine the angle measures cre

ated by the intersecting lines? Explain. (PLEASE HURRY ITS TIMED!)

1 answer:

Blizzard [7]3 years ago

8 0

Answer:

A linear pair of angles is formed when two lines intersect. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. The measure of a straight angle is 180 degrees, so a linear pair of angles must add up to 180 degrees. Vertical angles are two angles opposite from each other and those two angles are then equal.

Step-by-step explanation:

For example, if you took your typical graph with four quadrants your lines would intersect. The linear pairs are equal to 180 and the vertical angles are equal.

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An analog signal with a bandwidth of 36MHz is sampled at a frequency (fs) of 36,000,000 samples per second during the ADC (analo

Elza [17]

Answer:

The signal would have experienced aliasing.

Step-by-step explanation:

Given that:

the bandwidth of the signal $f_m$ = 36MHz

= 36 × 10⁶ Hz

The sampling frequency $f_s$ = 36 × 10⁶ Hz

Suppose the sampling frequency is equivalent to the bandwidth of the signal, then aliasing will occur.

Therefore, according to the Nyquist criteria;

Nyquist criteria posit that if the sampling frequency is more above twice the maximum frequency to be sampled, a repeating waveform can be accurately reconstructed.

∴

By Nyquist criteria, for perfect reconstruction of an original signal, i.e. the received signal without aliasing effect;

Then,

$f_s \geq 2f_m$

∴

The signal would have experienced aliasing.

4 0

3 years ago

Can you define f(0, 0) = c for some c that extends f(x, y) to be continuous at (0, 0)? If so, for what value of c? If not, expla

Ahat [919]

(i) Yes. Simplify $f(x,y)$ .

$\displaystyle \frac{x^2 - x^2y^2 + y^2}{x^2 + y^2} = 1 - \frac{x^2y^2}{x^2 + y^2}$

Now compute the limit by converting to polar coordinates.

$\displaystyle \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = \lim_{r\to0} \frac{r^4 \cos^2(\theta) \sin^2(\theta)}{r^2} = 0$

This tells us

$\displaystyle \lim_{(x,y)\to(0,0)} f(x,y) = 1$

so we can define $f(0,0)=1$ to make the function continuous at the origin.

Alternatively, we have

$\dfrac{x^2y^2}{x^2+y^2} \le \dfrac{x^4 + 2x^2y^2 + y^4}{x^2 + y^2} = \dfrac{(x^2+y^2)^2}{x^2+y^2} = x^2 + y^2$

and

$\dfrac{x^2y^2}{x^2+y^2} \ge 0 \ge -x^2 - y^2$

Now,

$\displaystyle \lim_{(x,y)\to(0,0)} -(x^2+y^2) = 0$

$\displaystyle \lim_{(x,y)\to(0,0)} (x^2+y^2) = 0$

so by the squeeze theorem,

$\displaystyle 0 \le \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} \le 0 \implies \lim_{(x,y)\to(0,0)} \frac{x^2y^2}{x^2+y^2} = 0$

and $f(x,y)$ approaches 1 as we approach the origin.

(ii) No. Expand the fraction.

$\displaystyle \frac{x^2 + y^3}{xy} = \frac xy + \frac{y^2}x$

$f(0,y)$ and $f(x,0)$ are undefined, so there is no way to make $f(x,y)$ continuous at (0, 0).

(iii) No. Similarly,

$\dfrac{x^2 + y}y = \dfrac{x^2}y + 1$

is undefined when $y=0$ .

5 0

2 years ago

99 POINTS TO ANSWER AWARDING BRAINLIEST

kotegsom [21]

Hmm...this is difficult. The first thing I can think of is the relationship between age and time.

3 0

3 years ago

Read 2 more answers

Stanley noticed that he is both the 10th tallest and the 10th shortest student in his class. If everyone in the class is at a di

STALIN [3.7K]

the answer for this question is B.20

6 0

3 years ago

12, 3.1.52<br><br> If f(x) = x and g(x) = 2x³, find f[g(x)], and g[f(x)]

anastassius [24]

Given that $f(x) = x$ and $g(x) = 2x^3$ , we have

$f(g(x)) = f(2x^3) = 2x^3$

and

$g(f(x)) = g(x) = 2x^3$

5 0

2 years ago