All categories

3 years ago

Brainliest question please help me please I need right answer please show your work

Mathematics

1 answer:

Oksana_A [137]3 years ago

7 0

Answer:

outside the circle

Step-by-step explanation:

diameter and radius are the main attractions in maths regarding circles.

You might be interested in

How to find the area only of a triangle

Cloud [144]

Area= height(base)/2

7 0

3 years ago

What is 1875 ÷ 125 (show ur work long divison)

Colt1911 [192]

Answer:

15 R7

Step-by-step explanation:

015

32/ 487

0

32

167

160

remainder 7

7 0

2 years ago

1. 232 is 8 times the sum of Mark's and his sister's ages. His sister is 21. Find Mark's age.

AURORKA [14]

Answer:

a. 8 years

b. $49.20

Step-by-step explanation:

The computation is shown below:

a. The mark age is

Given that

232 is 8 times of the sum of mark and his sister ages

And, his sister is 21

So,

Let us assume the mark age be x

now

8(x + 21) = 232

x + 21 = 29

x = 29 -21

x = 8

b. The original price of the socks is

As each pair get $3 off

And she bought 6 pairs for $31.20

= $31.20 + 6 × $3

= $31.20 + 18

= $49.20

3 0

3 years ago

I really need help with this one please

natka813 [3]

The distance from the sun is option 2 5.59 astronomical units.

Step-by-step explanation:

Step 1; To solve the question we need two variables. P which represents the number of years a planet takes to complete a revolution around the Sun. This is given as 13.2 years in the question so P = 13.2 years. The other variable is the distance between the planet and the sun in astronomical units. We need to determine the value of this using the given equation.

Step 2; So we have to calculate the value of 'a' in Kepler's equation. But the exponential power $\frac{3}{2}$ is on the variable we need to find so we multiply both the sides by an exponential power of $\frac{2}{3}$ to be able to calculate 'a'.

P = $a^{\frac{3}{2} }$ , $P^{\frac{2}{3} }$ = $a^{\frac{3}{2}*\frac{2}{3} }$ , $P^{\frac{2}{3} }$ = a, $13.2^{\frac{2}{3} }$ = a = 5.58533 astronomical units.