The area of the geometry is the sum of the area of the triangle and the semicircle will be 49.12 square feet.
<h3>What is Geometry?</h3>
It deals with the size of geometry, region, and density of the different forms both 2D and 3D.
The geometry is the combination of the semicircle and right triangle.
Then the area of the geometry will be
Area = 1/2 x 6 x 8 + π/8 x 8²
Area = 4 x 6 + 3.14 x 8
Area = 24 + 25.12
Area = 49.12 square feet
More about the geometry link is given below.
brainly.com/question/7558603
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Answer:
<u>The original price of the belt was $ 59.32 and Jonathan paid $ 35 for it.</u>
Step-by-step explanation:
1. Let's review the information given to us to answer the question correctly:
Amount Jonathan paid for the belt = $ 35
Discount = 1/3 off the regular price
Coupon = additional 8% = 0.08
2. How much did the Gucci belt cost Jonathan? Show your work or explain how you know.
We assume the question is asking for the original price of the belt, then we make this calculation:
Amount Jonathan paid for the belt = (Original price - Discount) - Coupon
Replacing with the values we know:
Original price = x
35 = (x - 1/3x) - 0.08x
35 = 2/3x - 0.08x
35 = 0.67x - 0.08x
35 = 0.59x
x = 35/0.59
x = 59.32
<u>The original price of the belt was $ 59.32 and Jonathan paid $ 35 for it.</u>
Answer:
Line 1 is 1/3
Line 2 is 1/3 as well
Choice 1
Step-by-step explanation:
<u>Rise over run</u>
<em><u>Line 1:</u></em>


<u>Simplify:</u>
1/3
<em><u>Line 2:</u></em>


1/3
1 is Paralell to 2 because they have the same slope
I hope this helps! :D
Pls give thx and brainliest if i’m right!
we know that
<u>The triangle inequality theorem</u> states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side
so
Let
a,b,c------> the length sides of a triangle
The theorem states that three conditions must be met
<u>case 1)</u>

<u>case 2)</u>

<u>case3)</u>

therefore
<u>the answer is the option</u>
B. The sum of the lengths of any two sides of a triangle is greater than the length of the third side.