The circumference of a circle is 28 in.What is the diameter of the circle?

Responses

28 over pi, in.

14 over pi, in.

square root of 28 over pi end root, in.

14π−−√ in.

I think it is 28/pi but I would like to make sure

Answers

Answer 1

Answer:

The diameter οf the circle is 28/π inches οr apprοximately 8.89 inches (rοunded tο twο decimal places).

The fοrmula fοr the circumference (C) οf a circle is given by:

C = 2πr

where r is the radius οf the circle.

If the circumference οf the circle is 28 inches, we can sοlve fοr the radius by dividing bοth sides οf the equatiοn by 2π:

C/2π = r

Substituting the given value οf C = 28, we get:

r = 28/2π

r = 14/π

Finally, tο find the diameter (d) οf the circle, we multiply the radius by 2:

d = 2r

Substituting the value οf r = 14/π, we get:

d = 2(14/π) = 28/π

Therefοre, the diameter οf the circle is 28/π inches οr apprοximately 8.89 inches (rοunded tο twο decimal places).

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What is the equivalent of 27/5 in decimal form?

Leo dug up 6 4/5 pounds oftreasure on his first dig. Anne dug
up 4 1/2 pounds of treasure. How
much more treasure did Leo dig
up than Anne?

Answers

6 4/5 = 6.8
4 1/2 = 4.5

6.8-4.5= 2.3

Answer: 2 3/10 more pounds of treasure

Find the LCM and solve, it's very very urgent.

Answers

Answers:

1.1050

2.1200

3.1200

4.3360

5.1080

6.480

pleaseseetheattachedpictureforfullsolution..

Hopeit helps....

Goodluck on your assignment...

Answer this fast and I’ll make you Brainly!!

Answers

Answer:

8.5 or 4.25 i think pretty sure its 8.5 tho

Step-by-step explanation:

Help as quick and i will give the brainliest!!!

Answers

I believe both answers are Shannon.

I think it’s sally and Shannon

Find two unit vectors orthogonal to both given vectors. i j k, 4i k

Answers

The cross product of two vectors gives a third vector $\mathbf v$ that is orthogonal to the first two.

$\mathbf v=(\vec i+\vec j+\vec k)*(4\,\vec i+\vec k)=\begin{vmatrix}\vec i&\vec j&\vec k\n1&1&1\n4&0&1\end{vmatrix}=\vec i+3\,\vec j-4\,\vec k$

Normalize this vector by dividing it by its norm:

$(\mathbf v)/(\|\mathbf v\|)=(\vec i+3\,\vec j-4\,\vec k)/(√(1^2+3^2+(-4)^2))=\frac1{√(26)}(\vec i+3\,\vec j-4\vec k)$

To get another vector orthogonal to the first two, you can just change the sign and use $-\mathbf v$ .