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3 years ago

Which of the following sets represents M∪N?

Mathematics

1 answer:

Mnenie [13.5K]3 years ago

4 0

What are the following sets?

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This is converting to polar form, I need help and the answer too.

Mila [183]

Answer:

2 cis (7/6 pi)

Step-by-step explanation:

r = sqrt( a^2 + b^2)

r = sqrt (-sqrt(3) )^2 + (-1)^2)

= sqrt(3 +1)

= sqrt(4)

= 2

theta = arctan (b/a)

theta = arctan (-1/-sqrt(3))

theta = 30

but this is in the third quardrant -a and -b

so add 180

theta = 210 degrees

convert this to radians

210 * pi/180 = 210/180 * pi = 21/18 * pi = 7/6 * pi

r cis (theta)

2 cis (7/6 pi)

7 0

2 years ago

Help and explain Don’t use for points or I will take it back

KIM [24]

Step-by-step explanation:

step 1. x + 51 + 39 + 127 = 360 (1 revolution is 360°)

step 2. x + 90 + 127 = 360

step 3. x + 127 = 270. (subtract 90 on both sides)

step 4. x = 143° (subtract 127 on both sides)

4 0

3 years ago

What is the radius of a Ø6.75" circle? 3.375" or 3.750" or 4.375" or 6.750"

zysi [14]

Answer:4.375

Step-by-step explanation:

4 0

2 years ago

Help me pls i don’t know the answer

sleet_krkn [62]

Answer:

Step-by-step explanation:

$10 \log( \frac{106.8}{1 \times {10}^{ - 12} } ) \approx 140$

8 0

1 year ago

Read 2 more answers

Help !! Please I can’t find the answer

SVETLANKA909090 [29]

Answer:

$\large\boxed{r^2=(x+5)^2+(y-4)^2}$

Step-by-step explanation:

The equation of a circle:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

We have diameter endpoints.

Half the length of the diameter is the length of the radius.

The center of the diameter is the center of the circle.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the given points (-8, 2) and (-2, 6):

$d=\sqrt{(6-2)^2+(-2-(-8))^2}=\sqrt{4^2+6^2}=\sqrt{16+36}=\sqrt{52}$

The radius:

$r=\dfrac{d}{2}\to r=\dfrac{\sqrt{52}}{2}$

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$x=\dfrac{-8+(-2)}{2}=\dfrac{-10}{2}=-5\\\\y=\dfrac{2+6}{2}=\dfrac{8}{2}=4$

$(-5,\ 4)\to h=-5,\ k=4$

Finally:

$(x-(-5))^2+(y-4)^2=\left(\dfrac{\sqrt{52}}{2}\right)^2\\\\(x+5)^2+(y-4)^2=\dfrac{52}{4}\\\\(x+5)^2+(y-4)^2=13$

5 0

3 years ago