1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Lesechka [4]
3 years ago
15

The diameter of your bicycle wheel is 24 inches. How far will you move in one turn of your wheel?

Mathematics
1 answer:
Tju [1.3M]3 years ago
8 0

Answer:

24pi inches, or 75.3982237in

Step-by-step explanation:

circumference is diameter times pi

You might be interested in
Select the expression that represents the following statement multiply 6 by 2, and then subtract 4. 2 - 4x6
Helen [10]
The answer is the second one, 6x2 - 4
6 0
2 years ago
If f(x)=x^4+6, g(x)=x-2 and h(x)= sqrt (x), then f(g(h(x)))=
patriot [66]

Answer:

x^2 + 4x * (3 - sqrt(x)) - 2(5 + sqrt(x))

Step-by-step explanation:

Firstly let us split this up, we need to first work out what g(h(x)) is:

h(x) = Sqrt(x) so g(h(x)) = g(sqrt(x)) = sqrt(x) - 2

Now to work out f(g(h(x))) = f(sqrt(x) - 2) = (sqrt(x) - 2)^4 + 6

= (sqrt(x) - 2) * (sqrt(x) - 2) * (sqrt(x) - 2) * (sqrt(x) - 2) - 6

= (x - 2 * sqrt(x) + 4) * (x - 2 * sqrt(x) + 4) - 6

= x^2 - 2x * sqrt(x) + 4x - 2x * sqrt(x) + 4x - 8 * sqrt(x) + 4x - 8 * sqrt(x) + 16 - 6

= x^2 - 4x * sqrt(x)  + 12x - 16 * sqrt(x) + 10

= x^2 + 4x * (3 - sqrt(x)) - 2(5 + sqrt(x))

3 0
3 years ago
A 25-foot long ladder is propped against a wall at an angle of 18° with the wall. how high up the wall does the ladder reach? ro
Alecsey [184]

Answer

Find out the how high up the wall does the ladder reach .

To proof

let us assume that the height of the wall be x .

As given

A 25-foot long ladder is propped against a wall at an angle of 18° .

as shown in the diagram given below

By using the trignometric identity

cos\theta = \frac{Base}{Hypotenuse}

now

\theta = 18^{\circ}

Base = wall height = x

Hypotenuse = 25 foot

Put in the trignometric identity

cos18^{\circ} = \frac{x}{25}

cos18^{\circ} = \bar{0.9510565162}

x = 23.8 foot ( approx)

Therefore the height of the  ladder be 23.8 foot ( approx) .

8 0
3 years ago
Read 2 more answers
Complete the sentence to explain how to multiply 4.56x1000
s2008m [1.1K]
The answer is 4,560

4.56 X 1000
8 0
2 years ago
Solve x2 - 8x - 9 = 0. Rewrite the equation so that it is of the form x2 + bx = c.
barxatty [35]

Answer:

x=9,-1  and x^2+(-8)x=9

Explanation:

we have been given with the quadratic equation x^2-8x-9

we compare the given quadratic equation with general quadratic equation

general quadratic is ax^2+bx+c=0

from given quadratic equation a=1,b= -8,c= -9

substituting these values in the formula for discriminant D=b^{2}-4ac

D=(-8)^2-4(1)(-9)=100

Now, to find the value of x

Formula is x=\frac{-b\pm\sqrt{D}}{2a}

Now, substituting the values we will get

x=\frac{-(-8)\pm\sqrt{100}} {2}= \frac{8\pm10}{2}=9,-1

And rewritting the given equation by  shifting 9 to right hand side of the given equation and taking minus inside the bracket so as to convert it in the form of

x^2+bx=c

4 0
3 years ago
Read 2 more answers
Other questions:
  • The mean of the data is 18. What number is missing? 21,11,_,27,22,13,16
    13·1 answer
  • What is the value of u? A) 32° B) 34° C) 36° D) 68°
    10·1 answer
  • Can u help me with this question
    10·1 answer
  • Please guys help me out . Thanks
    7·1 answer
  • (5+9)+7 to associative
    12·1 answer
  • What is another way to name 1/4
    9·2 answers
  • This table shows how many sophomores and juniors attended two school events. What is the probability that the student attended t
    5·1 answer
  • Let X subscript 1 comma... comma X subscript 9 denote the Intelligence Quotient IQ) of a randomly selected individuals. Assume t
    8·1 answer
  • What is the solution to the inequality 1/4(3x-6) >2(7-x +1
    10·1 answer
  • I don’t understand how I get this answer
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!