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
andreyandreev [35.5K]
3 years ago
6

Pls help bromeo’s

Mathematics
1 answer:
Viefleur [7K]3 years ago
3 0

Answer:

946.19 cm²

Step-by-step explanation:

Ok, we are going to split the shape into the semi-circle and the trapezoid:

<h3>Semi-Circle:</h3>

Here's the formula for the area circle: πr²

Because the semi-circle is just half of a circle, the area of a semi-circle :

(πr²) ÷2

Step 1: Input Information

The radius (r) is 1/2 the diameter (a) of the circle

Radius = 1/2(24) = 12

Now we have enough info to solve for the Area

Step 2: Solve for Area

A = (12²π) ÷2

A = (144π) ÷2

A = 72π

A ≈ 226.19

Area semi-circle is 226.19 cm²

<h3>Trapezoid</h3>

The formula for the area of a trapezoid is : h(b1 + b2) ÷ 2

The height (h) over here is c, or 24 cm

The base 1 (b1) is a, or 24 cm

The base 2 (b2) is b, 36 cm

Step 1: Input Information

Our new equation becomes:

A = c(a + b) ÷ 2

A = 24(24 + 36) ÷2

Step 2: Solve for area

A = 24(24 + 36) ÷ 2

A = 24(60) ÷ 2

A = 24(30)

A = 720

Area of the trapezoid is 720 cm²

<h3>Total Area:</h3>

Now, we just simply add the two values together

226.19 cm² + 720 cm² = 946.19 cm²

-Chetan K

You might be interested in
I need help with this equation it's confusing
enyata [817]

3 - 4x= -89 is the equation i hope it's right

5 0
3 years ago
Read 2 more answers
While organizing the magazines at the doctor's office, Camille put 12 magazines in the first pile, 15 magazines in the second pi
denpristay [2]

Answer:

27

Step-by-step explanation:

since you are adding 3 to the pile each time, on the 6th pile you would have 27 because 24 + 3 = 27

5 0
3 years ago
A woman 5 ft tall walks at the rate of 7.5 ft/sec away from a streetlight that is 10 ft above the ground. At what rate is the ti
lesya [120]
Height of the woman = 5 ft
Rate at which the woman is walking = 7.5 ft/sec
Let us assume the length of the shadow = s
Le us assume the <span>distance of the woman's feet from the base of the streetlight = x
</span>Then
s/5 = (s + x)/12
12s = 5s + 5x
7s = 5x
s = (5/7)x
Now let us differentiate with respect to t
ds/dt = (5/7)(dx/dt)
We already know that dx/dt = 7/2 ft/sec
Then
ds/dt = (5/7) * (7/2)
        = (5/2)
        = 2.5 ft/sec
From the above deduction, it can be easily concluded that the rate at which the tip of her shadow is moving is 2.5 ft/sec. 
8 0
3 years ago
You deposit money in an account that pays 5% interest compounded yearly. Find the balance after 5 years for the given initial am
mezya [45]
B I really believe you just got to read it carefully
5 0
3 years ago
Read 2 more answers
At school 161 students play at least one sport.This is 70% of the number of the student at the school. How many suenos are at th
kondor19780726 [428]
X=number total of student at the school.
70% of x=161
can suggest this equation:
(70/100)x=161
0.7x=161
x=161/0.7=230

Answer: the school has  230 students
7 0
3 years ago
Other questions:
  • 8/7 + 7/7 = ? Help please ?
    11·2 answers
  • Match each sequence to its appropriate recursively defined function
    12·1 answer
  • Please someone help
    7·1 answer
  • Yolanda makes wooden boxes for a crafts fair she makes a hundred boxes like the one showing and she wants to oaint all the outsi
    5·1 answer
  • F(x) = -4x + 1; Find x = -1
    14·1 answer
  • Helppp please will give brainliest ​
    14·1 answer
  • Topic integers<br> please solve this question <br> urgent
    14·1 answer
  • How many solutions does the following equation have? 2x+8=2(x+3)
    9·1 answer
  • Simplify the expression:<br> square root of p^10 where p&gt;0
    12·1 answer
  • Help ASAP, please!<br><br> What is the area of this quadrilateral?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!