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
xxMikexx [17]
3 years ago
10

A rectangle is placed around a semicircle as shown below. the length of the rectangle is 18yd . find the area of the shaded regi

on. use the value 3.14 for π , and do not round your answer. be sure to include the correct unit in your answer.
Mathematics
1 answer:
VladimirAG [237]3 years ago
7 0
The length of the rectangle is 18yd, therefore, the area of the shaded part is 56.52
You might be interested in
What is the area of the parallelogram?<br> 48 sqrt3 cm^2<br> 48 cm^2<br> 24 sqrt3 cm^2<br> 24 cm^2
Salsk061 [2.6K]
The area of a parallelogram is base times height. 

We can find the height by deriving a right triangle from the 60 degree angle and the 6 cm hypotenuse

sin(60°) = opp/hyp = x/6
√3/2 = x/6     Multiply both sides by 6
3√3 = x
x = 3√3

The height is 3 √3

Area = base x height
Area = 3√3 * 8
Area = 24 √3 cm²

Answer is C) 24 √3 cm²

3 0
3 years ago
Question: Solve it using the proportion method.​
Hatshy [7]
60% of what number is 348?

The answer is 580.
5 0
3 years ago
there are 36 roses and 27 carnations. anna is making flower arrangements using both flowers, if she made the maximum number of a
solniwko [45]
This is a GCF problem,
find the GCF of 36 and 27. To do this, list out the factors for each number.
EX 36- 1×36, 2×18, so on and so on. Then do this for 27. The GCF will be the greatest factor. In this case, that is 9.
So there would be 9 groups because that is the greatest common factor.
there would be 4 roses in each group because 4×9= 36.
There are 3 carnations in each group because 9×3 =27.
So 9 groups with 4 roses and 3 carnations.

4 0
3 years ago
Charles owns a catering business. The amount his 10 most recent clients spent is shown in the histogram below. Why is the graph
Alexxandr [17]
<span>The scales are different on the x-axis and the y-axis.
</span>Therefore, the correct answer choice is:
A. The x-axis scale shows the data is more clustered than it actually is.

5 0
3 years ago
Read 2 more answers
Which expressions are equivalent to when x0? Check all that apply.
Genrish500 [490]
We have that

\frac{(x+4)}{3} / \frac{6}{x} = \frac{x*(x+4)}{3*6} \\ \\ = \frac{( x^{2} +4x)}{18}

therefore

case a) 
\frac{(x+4)}{3} * \frac{x}{6}
Is equivalent

case b) 
\frac{6}{x} * \frac{(x+4)}{3}
Is not equivalent

case c) 
\frac{x}{6} * \frac{(x+4)}{3}
Is  equivalent

case d) 
\frac{(2 x^{2} +4x)}{6}
Is not equivalent

case e) 
\frac{(2 x^{2} +4x)}{18}
Is equivalent

Hence

the answer is

\frac{(x+4)}{3} * \frac{x}{6}

\frac{x}{6} * \frac{(x+4)}{3}

\frac{(2 x^{2} +4x)}{18}
3 0
3 years ago
Read 2 more answers
Other questions:
  • A caterer needs to buy 21 pounds of pasta to cater a wedding. At a local store, 8 pounds of pasta cost $12. How much will the ca
    12·1 answer
  • 2407 - -8 how do you do this type of problem
    10·1 answer
  • Solve for x... I dont know what to do with the square root
    6·1 answer
  • Find out the output, k, when the input, x, is -5. K=6x+100
    10·1 answer
  • What is the value of a+bc when a=4, b=6 and c =2? also please include how to do it step by step :)
    5·1 answer
  • Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x overbarequals103
    14·1 answer
  • At a restaurant, you buy a combo meal for $7.50. You also want an ice cream sundae for dessert for $3.00. You brought $10 in cas
    15·2 answers
  • What is <br> 12+5+3\2*10-5
    6·1 answer
  • How many solutions does the system of equations below have?<br> 2x – 3y = -9<br> 4х – бу = -18
    15·1 answer
  • A pot of soup, currently 16°C above room temperature, is left out to cool. If that temperature
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!