All categories

2 years ago

Which is true regarding chords and diameters of circles?

Mathematics

2 answers:

Gelneren [198K]2 years ago

6 0

Answer:

Both chords and diameters have two endpoints on a circle. Diameters must intersect the center of a circle.

Step-by-step explanation:

malfutka [58]2 years ago

4 0

Answer:

A - Both chords and diameters have two endpoints on a circle. Diameters must intersect the center of a circle.

Step-by-step explanation:

You might be interested in

The play took in $744 one night. The number of $8 adult tickets was 12 less than twice the number of $5 child tickets. How many

vfiekz [6]

Answer:

68 adult tickets and 40 child tickets were sold.

Step-by-step explanation:

Create a system of equations where a is the number of adult tickets sold and c is the number of child tickets sold:

8a + 5c = 744

a = 2c - 12

Solve by substitution by substituting the second equation into the first one:

8(2c - 12) + 5c = 744

16c - 96 + 5c = 744

21c - 96 = 744

21c = 840

c = 40

So, 40 child tickets were sold. Plug this into the second equation to solve for a:

a = 2c - 12

a = 2(40) - 12

a = 80 - 12

a = 68

So, 68 adult tickets and 40 child tickets were sold.

3 0

2 years ago

Find the laplace transform of the given function. (express your answer in terms of s. assume that s > 0.) f(t) = 0, t < 2

igomit [66]

$f(t)=\begin{cases}0&\text{for }t$

$\mathcal L_s\{f(t)\}=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt$
$=\displaystyle\int_2^\infty(t^2-4t+8)e^{-st}\,\mathrm dt=\frac{2e^{-2s}}{s^3}+\frac{4e^{-2s}}s$

4 0

3 years ago

Estimate 306% of 25 hkhiuiuuigkuuigigiiuuu

Snowcat [4.5K]

306% of 25

= (306/100) * 25

= 3.06 * 25

= 76.5

6 0

3 years ago

Anthony is playing a card game called Subzero, where the goal is to get the lowest possible score. In the first round, Anthony s

SVETLANKA909090 [29]

Answer -6

Step-by-step explanation: -12+-6 =-6

7 0

3 years ago

Read 2 more answers

10x +2y = 64

Musya8 [376]

Answer:

Step-by-step explanation:

$\left\{\begin{array}{ccc}10x+2y=64&\text{multiply both sides by 2}\\3x-4y=-36\end{array}\right\\\underline{+\left\{\begin{array}{ccc}20x+4y=128\\3x-4y=-36\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad23x=92\qquad\text{divide both sides by 2}\\.\qquad x=4\\\\\text{put the value of x to the first equation:}\\\\10(4)+2y=64\\40+2y=64\qquad\text{subtract 40 from both sides}\\2y=24\qquad\text{divide both sides by 2}\\y=12$

5 0

3 years ago