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3 years ago

A line segment that has one endpoint at the center of a circle and the other endpoint on the circle?

Mathematics

2 answers:

Umnica [9.8K]3 years ago

4 0

What's the actual question? That's the same as the radius of the circle if that's what you're asking

Igoryamba3 years ago

4 0

Answer:

This is true

Step-by-step explanation:

One of Elucid's 5 postulates

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What is 5.16 divided by 0.6 (pls show work)

IrinaVladis [17]

Answer:

8.6

Step-by-step explanation:

5.16 x 10 = 51.6 , 0.6 x 10 = 6, 6 divided by 51.6 = 8.6

6 0

1 year ago

Read 2 more answers

what is the lateral area of a square pyramid with a side length of 11.2 centimeters and slant height 20 centimeters

hammer [34]

Lateral sides consiste of 4 congruentisosceles triangle
lateral area

= 4 * 0.5 * 11.2 * 20
= 448 cm^2

4 0

3 years ago

An employer has a staff of eighty actuaries, ten of whom are student actuaries. A student actuary is allowed a total of ten week

sergiy2304 [10]

Answer:

0.1111

Step-by-step explanation:

From the given information;

Number of staffs in the actuary = 80

Out of the 80, 10 are students.

i.e.

P(student actuary) = 10/80 = 0.125

number of weeks in a year = 52

off time per year = 10/52 = 0.1923

P(at work || student actuary) = (50 -10/52)

= 42/52

= 0.8077

P(non student actuary) = (80 -10)/80

= 70 / 80

= 0.875

For a non-student, they are only eligible to 4 weeks off in a year

i.e.

P(at work | non student) = (52-4)/52

= 48/52

= 0.9231

∴

P(at work) = P(student actuary) × P(at work || student actuary) + P(non student actuary) × P(at work || non studnet actuary)

P(at work) = (0.125 × 0.8077) + ( 0.875 × 0.9231)

P(at work) = 0.1009625 + 0.8077125

P(at work) = 0.90868

Finally, the P(he is a student) = (P(student actuary) × P(at work || student actuary) ) ÷ P(at work)

P(he is a student) = (0.125 × 0.8077) ÷ 0.90868

P(he is a student) = 0.1009625 ÷ 0.90868

P(he is a student) = 0.1111

5 0

2 years ago

Solve the following equation:

Rama09 [41]

Complete the square.

$z^4 + z^2 - i\sqrt 3 = \left(z^2 + \dfrac12\right)^2 - \dfrac14 - i\sqrt3 = 0$

$\left(z^2 + \dfrac12\right)^2 = \dfrac{1 + 4\sqrt3\,i}4$

Use de Moivre's theorem to compute the square roots of the right side.

$w = \dfrac{1 + 4\sqrt3\,i}4 = \dfrac74 \exp\left(i \tan^{-1}(4\sqrt3)\right)$

$\implies w^{1/2} = \pm \dfrac{\sqrt7}2 \exp\left(\dfrac i2 \tan^{-1}(4\sqrt3)\right) = \pm \dfrac{2+\sqrt3\,i}2$

Now, taking square roots on both sides, we have

$z^2 + \dfrac12 = \pm w^{1/2}$

$z^2 = \dfrac{1+\sqrt3\,i}2 \text{ or } z^2 = -\dfrac{3+\sqrt3\,i}2$

Use de Moivre's theorem again to take square roots on both sides.

$w_1 = \dfrac{1+\sqrt3\,i}2 = \exp\left(i\dfrac\pi3\right)$

$\implies z = {w_1}^{1/2} = \pm \exp\left(i\dfrac\pi6\right) = \boxed{\pm \dfrac{\sqrt3 + i}2}$

$w_2 = -\dfrac{3+\sqrt3\,i}2 = \sqrt3 \, \exp\left(-i \dfrac{5\pi}6\right)$

$\implies z = {w_2}^{1/2} = \boxed{\pm \sqrt[4]{3} \, \exp\left(-i\dfrac{5\pi}{12}\right)}$

3 0

1 year ago

-1 3/5 divided by (-2/3) write as mixed number

hichkok12 [17]

Step-by-step explanation:

$- 1 \frac{3}{5} \div \frac{ - 2}{3}$

Convert the mixed numbers above to improper fractions. Thus:

$\frac{ - 8}{5} \div \frac{ - 2}{3}$

In order to simplify the fractions above, the division sign changes to multiplication sign thus turn the right hand side of the fraction inversely such that -2/3 to -3/2. Hence,

$\frac{ - 8}{5} \times \frac{ - 3}{2}$

$\frac{ - 8 \times - 3}{5 \times 2}$

$= \frac{24}{10}$

$= 2 \frac{4}{10}$

$= 2 \frac{2}{5}$

3 0

2 years ago