Q: The circumference of circle exceeds its diameter by 94 cm.Find the radius of the circle?

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the answer on A+ is 2.5 cm

Radius = 10Diameter = 20Circumference = pi D = 20 pi.50 degrees = (50 / 360) of the whole circle.Arc length = (50 / 360) of (20 pi) = (1,000 pi / 360) = 8.727 cm (rounded)

The radius length r of the inscribed circle equals to one half of the length side of the square, 10 cm. The area A of the inscribed circle: A = pir2 = 102pi ≈ 314 cm2 The radius length r of the circumscribed circle equals to one half of the length diagonal of the square. Since the diagonals of the square are congruent and perpendicular to each other, and bisect the angles of the square, we have sin 45⁰ = length of one half of the diagonal/length of the square side sin 45⁰ = r/20 cm r = (20 cm)(sin 45⁰) The area A of the circumscribed circle: A = pir2 = [(20 cm)(sin 45⁰)]2pi ≈ 628 cm2.

I know there is a equation to find the answer, but if youae like me someday you will forget it. So, I am going to use trig functions and a little work.Derive the side and angle of a triangleIn triangle xyz, yx equals 6cm, yz equals 7cm, angle y equals 78 degrees find the length of xz and the measure of angle x give answers to one decimal place?I know there is an equation to find the answer, but if you are like me, someday you will forget it. So, I am going to use trig functions and a little work.I wish I wiki would copy MS word drawings, but here we go.Draw a triangle with the base, YX = 6 cm.Label the left end of the line as Point Y.Starting at Point Y, draw a 7 cm line (YZ) up and right from the base YX making an angle of 78º angle with the base.Label the end point of this line as Point Z, so this is line YZThe angle XYZ should open to the right. The only reason I keep repeating this is so your picture looks like mine and you can follow my directions.Connect the points X and Z to complete the triangle.Draw an altitude straight down from point Z.Label the point where the altitude meets the base as point A.The altitude is line ZANow you have 2 right triangles.Look at triangle YZA (left one)Label the 78º angle.Sin θ = opposite ÷ hypotenuseSin 78º = ZA ÷ YZSin 78º = ZA ÷ 7ZA = 6.85 cm (Altitude)Cos θ = adjacent ÷ hypotenuseHypotenuse = YZ = 7 cmAdjacent = line YACos 78º = YA ÷ 7YA = 7 * Cos 78ºYA = 1.46 cmYX = 6 cmYX = YA + AX6 = 1.46 + AXAX = 4.54 cmNow we know the length of 2 sides of the right triangle on the right side, right Triangle ZXA.ZA = 6.85 cm (Altitude)AX = 4.54 cm (adjacent)Tan θ (ZXA) = opposite ÷ adjacentOpposite = ZA (altitude) = 6.85 cmAdjacent= AX = 4.54 cmTan θ (ZXA) = 6.85 ÷ 4.54 cm(ZXA) θ =56.5ºIn the right Triangle ZXA, line ZX is the hypotenuse, the line ZA(Altitude) is the opposite, and the angle between these two lines = 56.5º.Sin θ = opposite ÷ hypotenuseSin 56.5º = Altitude ÷ hypotenuseSin 56.5º = 6.85 ÷ hypotenuseHypotenuse = 6.85 ÷ Sin 56.5ºHypotenuse =8.22 cmZX = 8.22 cmfind the length of xz and the measure of angle x give answers to one decimal place?ZX = 8.22 cm, θ =56.5ºHere is the work from the beginning to the end.ZX = (7 * sin 78º) ÷ (7 * Cos 78º - 6) *-1, tan-1, sin, 1/x, * (7* sin 78º)

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the answer on A+ is 2.5 cm

Step One. Calculate the surface area of the inside of the vessel. Assume that the thickness of the vessel is negligible.Diameter is 20. Circumference is pi * diameter.C = 3.14159 * 20 = 62.8318 cmHeight = 14cm.Surface area = C * 14 = 879.6452 cm2Step Two. Calculate the cost of tinplating the surface area879.6452 * Re1 / 100 = CostCost = 8.796452 RE

Radius = 10Diameter = 20Circumference = pi D = 20 pi.50 degrees = (50 / 360) of the whole circle.Arc length = (50 / 360) of (20 pi) = (1,000 pi / 360) = 8.727 cm (rounded)

The radius length r of the inscribed circle equals to one half of the length side of the square, 10 cm. The area A of the inscribed circle: A = pir2 = 102pi ≈ 314 cm2 The radius length r of the circumscribed circle equals to one half of the length diagonal of the square. Since the diagonals of the square are congruent and perpendicular to each other, and bisect the angles of the square, we have sin 45⁰ = length of one half of the diagonal/length of the square side sin 45⁰ = r/20 cm r = (20 cm)(sin 45⁰) The area A of the circumscribed circle: A = pir2 = [(20 cm)(sin 45⁰)]2pi ≈ 628 cm2.

Assuming the load is 49 Newtons, the answer is force times distance, in other words 49 N times 0.001 metres, which is 0.049 Joules.

Please note that the diagonals cross one another at right angles, and bisect one another. You should draw a diagram to visualize this. From this center point, where the diagonals cross, you have a RIGHT TRIANGLE that involves half of one diagonal, half of the other diagonal, and one of the sides of the rhombus. This lets you use the Pythagorean Theorem to solve for the length of the side.

The cells in a bee hive are typically hexagonal in shape, forming a pattern known as a honeycomb. This shape allows for efficient use of space and optimal storage of honey, pollen, and eggs within the hive.

I know there is a equation to find the answer, but if youae like me someday you will forget it. So, I am going to use trig functions and a little work.Derive the side and angle of a triangleIn triangle xyz, yx equals 6cm, yz equals 7cm, angle y equals 78 degrees find the length of xz and the measure of angle x give answers to one decimal place?I know there is an equation to find the answer, but if you are like me, someday you will forget it. So, I am going to use trig functions and a little work.I wish I wiki would copy MS word drawings, but here we go.Draw a triangle with the base, YX = 6 cm.Label the left end of the line as Point Y.Starting at Point Y, draw a 7 cm line (YZ) up and right from the base YX making an angle of 78º angle with the base.Label the end point of this line as Point Z, so this is line YZThe angle XYZ should open to the right. The only reason I keep repeating this is so your picture looks like mine and you can follow my directions.Connect the points X and Z to complete the triangle.Draw an altitude straight down from point Z.Label the point where the altitude meets the base as point A.The altitude is line ZANow you have 2 right triangles.Look at triangle YZA (left one)Label the 78º angle.Sin θ = opposite ÷ hypotenuseSin 78º = ZA ÷ YZSin 78º = ZA ÷ 7ZA = 6.85 cm (Altitude)Cos θ = adjacent ÷ hypotenuseHypotenuse = YZ = 7 cmAdjacent = line YACos 78º = YA ÷ 7YA = 7 * Cos 78ºYA = 1.46 cmYX = 6 cmYX = YA + AX6 = 1.46 + AXAX = 4.54 cmNow we know the length of 2 sides of the right triangle on the right side, right Triangle ZXA.ZA = 6.85 cm (Altitude)AX = 4.54 cm (adjacent)Tan θ (ZXA) = opposite ÷ adjacentOpposite = ZA (altitude) = 6.85 cmAdjacent= AX = 4.54 cmTan θ (ZXA) = 6.85 ÷ 4.54 cm(ZXA) θ =56.5ºIn the right Triangle ZXA, line ZX is the hypotenuse, the line ZA(Altitude) is the opposite, and the angle between these two lines = 56.5º.Sin θ = opposite ÷ hypotenuseSin 56.5º = Altitude ÷ hypotenuseSin 56.5º = 6.85 ÷ hypotenuseHypotenuse = 6.85 ÷ Sin 56.5ºHypotenuse =8.22 cmZX = 8.22 cmfind the length of xz and the measure of angle x give answers to one decimal place?ZX = 8.22 cm, θ =56.5ºHere is the work from the beginning to the end.ZX = (7 * sin 78º) ÷ (7 * Cos 78º - 6) *-1, tan-1, sin, 1/x, * (7* sin 78º)