The Singh family ordered a large pizza with a diameter of 20 inches for dinner. If one of the members of the family ate one eighth of the pizza, how many square inches of pizza are remaining? Use 3.14 for π.274.75 square inches
39.25 square inches
314 square inches
157 square inches
Answers
274.75 square inches of pizza are remaining. Option A is the correct option.
What is π in math?
The ratio of a circle's diameter to its circumference, or "pi," is a mathematical constant that is roughly equal to 3.14159 (/pa/; also written as "pi"). Numerous mathematical and physics formulas contain the number. It is an irrational number, meaning that although fractions like 22/7 are frequently used to approximate it, it cannot be expressed exactly as a ratio of two integers.
Given that, the diameter of large pizza is 20 inches.
The radius of a circular shape is half of the diameter.
The radius of the pizza is (20 inches)/2 = 10 inches.
Area of a circular shape is ∏r² .
The area of the pizza is 3.14×10² = 314 in²
The total part of the pizza is considered as 1.
One-eighth of the pizza is eaten, and the remaining portion is (1-1/8) = 7/8.
The area of remaining pizza is
314 in² × (7/8)
= 274.75 square inches
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Related Questions
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M1Q11.) What percent of women in their 40s taking a screening mammography receive a false positive?
Select the correct answer. Solve the following equation by completing the square. 1/4x^2 + x + 1/4 = 0
Eric takes a train to his parent’s house 320 miles away. If the trip takes 4 hours, what is the unit rate of the train?
Answers
Answer:
The number of monthly memberships before the incentive was 100
Step-by-step explanation:
Remember that
The total memberships after the incentive is equal to the total memberships before the incentive
step 1
Find out the annual memberships after the incentive
Let
x -----> monthly memberships after the incentive
y -----> annual memberships after the incentive
we know that
-----> equation A
substitute the value of x in equation A
step 2
Find out the total memberships after the incentive
step 3
Find out the monthly members before the incentive
Let
x -----> monthly memberships before the incentive
y -----> annual memberships before the incentive
we know that
-----> equation A
-----> equation B
substitute equation A in equation B and solve for x
therefore
The number of monthly memberships before the incentive was 100
Final answer:
Initially, there were 100 monthly members at the community pool, before the incentives were offered.
Explanation:
We start with the information that the ratio of monthly to annual memberships was initially 10 to 3, and then became 5 to 8. After the incentive, we're told there are now 50 monthly members.
To solve this problem, we set up a proportion. Since each part of the new ratio equals 10 members (50 monthly members/5 parts = 10 members per part), we can infer that before the incentive there were 10*10=100 monthly members.
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(1, 6, 3), (-5, 3, 7)
Answers
Answer:
65
Step-by-step explanation:
Answers
Answer:
r < 3/2 or r < 1.5
Step-by-step explanation:
-8r - 9 > 21
-8r > -21 +9
-8r > -12
divide both sides by -8
= r < 3/2 or 1.5
9x2-5x = 94x-8
Answers
Answer:
x=26262626262626...
Step-by-step explanation:
Answers
Answer:
The graph is shown below.
=========================================================
Explanation:
Notice that if we multiplied the denominator (x-1) by 5, then we get 5(x-1) = 5x-5.
This is close to 5x-7, except we're off by 2 units.
In other words,
5x-7 = (5x-5)-2
since -7 = -5-2
Based on that, we can then say,
This answer can be reached through alternative methods of polynomial long division or synthetic division (two related yet slightly different methods).
-------------------------
Compare the equation to the form
We can see that
- a = -2
- h = 1
- k = 5
The vertical asymptote is x = 1, which is directly from the h = 1 value. If we tried plugging x = 1 into g(x), then we'll get a division by zero error. So this is why the vertical asymptote is located here.
The horizontal asymptote is y = 5, which is directly tied to the k = 5 value. As x gets infinitely large, then y = g(x) slowly approaches y = 5. We never actually arrive to this exact y value. Try plugging in g(x) = 5 and solving for x. You'll find that no solution for x exists.
The point (h,k) is the intersection of the horizontal and vertical asymptote. It's effectively the "center" of the hyperbola, so to speak.
The graph is shown below. Some points of interest on the hyperbola are
- (-1,6)
- (0,7) .... y intercept
- (1.4, 0) .... x intercept
- (2, 3)
- (3, 4)
Another thing to notice is that this function is always increasing. This means as we move from left to right, the function curve goes uphill.
What is the area of trapezoid?
Answers
Area = (1/2)(AB + DC)·h
We are given a relationship between AB and DC, so we can write
Area = (1/2)(AB + AB/4)·h = (5/8)AB·h
The given dimensions let us determine the area of ∆BCE to be
Area ∆BCE = (1/2)(5 cm)(12 cm) = 30 cm²
The total area of the trapezoid is also the sum of the areas ...
Area = Area ∆BCE + Area ∆ABE + Area ∆DCE
Since AE = 1/3(AD), the perpendicular distance from E to AB will be h/3. The areas of the two smaller triangles can be computed as
Area ∆ABE = (1/2)(AB)·h/3 = (1/6)AB·h
Area ∆DCE = (1/2)(DC)·(2/3)h = (1/2)(AB/4)·(2/3)h = (1/12)AB·h
Putting all of the above into the equation for the total area of the trapezoid, we have
Area = (5/8)AB·h = 30 cm² + (1/6)AB·h + (1/12)AB·h
(5/8 -1/6 -1/12)AB·h = 30 cm²
AB·h = (30 cm²)/(3/8) = 80 cm²
Then the area of the trapezoid is
Area = (5/8)AB·h = (5/8)·80 cm² = 50 cm²