All categories

3 years ago

Whats 126/399 in its simplest form

1 answer:

4 0

A common factor of both 126 and 399 is 3 so we can divide both numbers by 3.

126 / 3 = 42
399 / 3 = 133

A common factor for both 42 and 133 is 7 so we can divide both numbers by 7.

42 / 7 = 6
133 / 7 = 19

19 is a prime number so we can't reduce it anymore, therefore 126/399 in its simplest form is 6/19

You might be interested in

Hey help me please (DON'T SUBMIT THAT ONE LINK THAT WILL HACK YOU! I WILL REPORT YOUR ACCOUNT

nikitadnepr [17]

Answer:

368m

Step-by-step explanation:

16 x 20 = 320

8 x 12 x 1/2 = 48

Add them together and get 368

6 0

2 years ago

What is the first step in solving: 14 + 2x - 3 = 43?

ch4aika [34]

D. Subtract 14

hope this helps !!!!

3 0

3 years ago

Express sin Q as a fraction in simplest terms.

Lesechka [4]

Answer:

Solution given

Sin Q=opposite/hypotenuse=28/29

28/29 is. a required answer.

3 0

2 years ago

During a period of 80 minutes, a music station played 8 minutes of commercials. What is the ratio of music they played to commer

igomit [66]

Answer:

Step-by-step explanation:

80 minutes.......8 minutes played commercials....that means (80 - 8) = 72 minutes played music

so the ratio of music to commercials is : 72/8 which reduces to 9/1...so the ratio is 9:1 <=== ur answer

** pay attention to the wording, because if it would have asked for the ratio of commercials to music, it would have been 8/72....ratio would be 1:9

8 0

3 years ago

A circle is centered at J(3, 3) and has a radius of 12.

stealth61 [152]

Answer:

$(-6,\, -5)$ is outside the circle of radius of $12$ centered at $(3,\, 3)$ .

Step-by-step explanation:

Let $J$ and $r$ denote the center and the radius of this circle, respectively. Let $F$ be a point in the plane.

Let $d(J,\, F)$ denote the Euclidean distance between point $J$ and point $F$ .

In other words, if $J$ is at $(x_j,\, y_j)$ while $F$ is at $(x_f,\, y_f)$ , then $\displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}$ .

Point $F$ would be inside this circle if $d(J,\, F) < r$ . (In other words, the distance between $F\!$ and the center of this circle is smaller than the radius of this circle.)

Point $F$ would be on this circle if $d(J,\, F) = r$ . (In other words, the distance between $F\!$ and the center of this circle is exactly equal to the radius of this circle.)

Point $F$ would be outside this circle if $d(J,\, F) > r$ . (In other words, the distance between $F\!$ and the center of this circle exceeds the radius of this circle.)

Calculate the actual distance between $J$ and $F$ :

$\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145} \end{aligned}$ .

On the other hand, notice that the radius of this circle, $r = 12 = \sqrt{144}$ , is smaller than $d(J,\, F)$ . Therefore, point $F$ would be outside this circle.

5 0

2 years ago