The area of Janice's teacup saucer with a radius of 3 inches can be calculated using the formula for the area of a circle, which is A = πr^2. Using the value of π as 3.14, the saucer's area is approximately 28.26 square inches.
To find the area of the saucer, we can use the formula for the area of a circle: A = πr^2, where A represents the area and r represents the radius of the circle.
Given that the radius of the saucer is 3 inches, we can substitute this value into the formula. Using the approximation of π as 3.14, we calculate the area as follows:
A = 3.14 * (3 inches)^2
= 3.14 * 9 square inches
≈ 28.26 square inches
Therefore, the area of Janice's teacup saucer with a radius of 3 inches is approximately 28.26 square inches.
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You start at (-3, 5). You move left 1 unit and right 8 units. Where do you end?
The point where you end after moving is (4, 5)
How to determine the point where you endFrom the question, we have the following parameters that can be used in our computation:
Start = (-3, 5)
Direction: 1 unit left and 8 units right
using the above as a guide, we have the following:
New point = (-3, 5) + (-1 + 8, 0)
So, we have
New point = (-3 - 1 + 8, 5 + 0)
Evaluate
New point = (4, 5)
Hence, the point where you end is (4, 5)
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PLEASE HELP FAST!! WILL GIVE BRAINLIEST! The rectangle ABCD has diagonals that intersect at point O and ABD = 30. Find BC if AC = 16 in
The length of BC in the rectangle ABCD is 16 units.
To find the length of BC in the rectangle ABCD, we can use the properties of rectangles and the intersecting diagonals.
Let's consider the given information:
AC = 16 (given)
ABD = 30° (given)
In a rectangle, the diagonals are equal in length. Therefore, AO = CO and BO = DO.
Since ABD is a right triangle, we can use trigonometric ratios to find the length of AO. In triangle ABD, the angle ABD is 90°, and we know ABD = 30°. Therefore, the remaining angle BDA is 180° - 90° - 30° = 60°.
Using the trigonometric ratio for a right triangle:
sin(BDA) = AO / AB
sin(60°) = AO / AB
√3 / 2 = AO / AB
Since AB is the length of the diagonal of the rectangle, we can represent it as d:
√3 / 2 = AO / d
Now, we can find AO:
AO = (√3 / 2) * d
Since AO = CO and BO = DO, we can conclude that BO and CO also have lengths of (√3 / 2) * d.
Now, let's consider triangle AOC. We know that AC = 16, and AO = CO = (√3 / 2) * d. We can use the Pythagorean theorem to find OC:
OC^2 = AC^2 - AO^2
OC^2 = 16^2 - [(√3 / 2) * d]^2
OC^2 = 256 - (3/4) * d^2
OC = √(256 - (3/4) * d^2)
Similarly, in triangle BOC, we have BO = (√3 / 2) * d and OC = √(256 - (3/4) * d^2). We can again use the Pythagorean theorem to find BC:
BC^2 = BO^2 + OC^2
BC^2 = [ (√3 / 2) * d ]^2 + [ √(256 - (3/4) * d^2) ]^2
BC^2 = (3/4) * d^2 + 256 - (3/4) * d^2
BC^2 = 256
Taking the square root of both sides:
BC = √256 = 16
Therefore, BC = 16.
So, the length of BC in the rectangle ABCD is 16 units.
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