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2 years ago

What is the answer plz

Mathematics

1 answer:

Fantom [35]2 years ago

8 0

Answer:

Sarah is correct.

Step-by-step explanation:

You need to find the diameter or radius whichever is used and you can do the equation backwards to find the radius or diameter. Once you find them then you will need to do the formula for circumference.

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Can someone please help me out

Paha777 [63]

Answer:

D. 20.

Step-by-step explanation:

-17 - -37 = 20.

These are all of the steps, to completely, get the correct answer to this question.

Hope this helps!!!

Kyle.

8 0

2 years ago

Select the correct answer. What is the solution to |x − 6| ≥ 5? A. 1 ≤ x ≤ 11 B. -11 ≤ x ≤ 1 C. x ≥ 11 or x ≤ 1 D. x ≥ 1 or x ≤

diamong [38]

The absolute value of a number a is written as |a| . And represents the distance between a and 0 on a number line.

So, Option C. x≥ 11 or x ≤ 1 is correct answer

Inequality given is,

|x − 6| ≥ 5

Taking positive case,

x − 6 ≥ 5

x − 6 + 6 ≥ 5 + 6

x ≥ 11

Taking negative case,

x − 6 ≤ - 5

x - 6 + 6 ≤ -5 + 6

x ≤ 1

So, solution of given inequality is x ≥ 11 or x ≤ 1

Learn more:

brainly.com/question/1779337

5 0

2 years ago

Sine, cosines and tangent of 30 45 and 60 degrees

bonufazy [111]

Answer:

Sine 30 =

$\frac{1}{2}$

Sine 45 =

$\sqrt{ \frac{2}{2} }$

Sine 60 =

$\sqrt{ \frac{3}{2} }$

Cos 30 =

$\sqrt{ \frac{3}{2} }$

Cos 45 =

$\sqrt{ \frac{2}{2} }$

Cos 60 =

$\frac{1}{2}$

Tan 30 =

$\sqrt{ \frac{3}{3} }$

Tan 45 =

$1$

Tan 60 =

$\sqrt{3}$

Step-by-step explanation:

I hope it helps

3 0

3 years ago

Please help me with my work

vaieri [72.5K]

,so first split them into separate shapes (refer to pic) then calculate each one
I started with the triangle
8×6=48 and 48÷2=24 so the triangles area its 24
next the rectangle
12×8=96
and I did the triangle shape cut out of the square
4×3=12 12÷2=6 so its area is 6
and the squares area is 8×8=64
but u need to subtract 6 from 64 because that area is missing
then you get 64-6=58
then add all the areas
58+24+96 to get your answer of 178m

8 0

3 years ago

The Continuing Education Division at the Ozark Community College offers a total of 30 courses each semester. The courses offered

dybincka [34]

Answer:

The college should offer 20 practical courses and 10 humanistic courses in order to maximize the revenue.

Step-by-step explanation:

Let x and y be

x= number of practical courses the college will offer

y= number of humanistic courses the college will offer

we have the following inequalities

x ≥ 10

y ≥ 10

x+y = 30

The only points (x,y) that satisfy all this inequalities are (10,20) and (20,10) (see picture attached).

On the other hand, the revenues would be given by

R = 1,500x + 1,000y

The maximum of R is attained in (x,y) = (20,10) and is

R = 300,000 + 100,000 = 400,000

and the college should be offering 20 practical courses and 10 humanistic ones.

If x = 21 then y must be 9, and then y does not satisfy y ≥ 10.

So you can only offer additional humanistic courses.

In n is an integer < 30

But if y=10+n then x must be 30-n and the revenue would be

R = 1,500(30-n) + 1,000(10+n) =

350,000 - 1,500n + 100,000+ 1,000n =

400,000 - 500n < 400,000

This result means in terms of offering additional courses, that is not worth doing it.

8 0

3 years ago