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emmasim [6.3K]
3 years ago
8

Two circles have radii of 4 units and 6 units. What is the radius of a circle whose area is equal to the sum of the areas of the

two given circles?
Mathematics
1 answer:
Bumek [7]3 years ago
3 0

Answer:

Step-by-step explanation:

We that the area of a circle is Pi times r^2 where r is the radius

Let A be the area of the first circle and S the area of the second one and T the total one

A=4^2× Pi

S=6^2× Pi

T = Pi( 4^2 +6^2)

= Pi ( 16+36) = Pi× 52

52 is the radius square

So r = root square 52 = 2 root square 13

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5 hours.

Step-by-step explanation:

5 hours * 60km = 300km in 5 hours.

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shtirl [24]

Answer:

y = 3

x = 1

y = 1.5

x = 2

Step-by-step explanation:

For the first x point, look at the graph where x is 0. When x is 0, y is 3. So the first one is 3. For the second one, there is a point on the graph where y is 4. Where y is 4, x is 1, so you the second answer is 1. For the third one, find the y point where x is -1. At the x value, -1, y is 1.5. For the last one, where y is 5, x is 2.

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Zigmanuir [339]

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a

Step-by-step explanation:

3 0
3 years ago
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Can someone please help!​
kozerog [31]

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fraction 4/12 or 1/3

decimal .333333333

percent 33%.333333

Step-by-step explanation:

6 0
3 years ago
42:28
gogolik [260]

Answer:

The statements about arcs and angles that are true in the figure are;

1) ∠EFD ≅ ∠EGD

2) \overline{ED}\cong \overline{FD}

3) mFD = 120°

Step-by-step explanation:

1) ∠ECD + ∠CEG + ∠CDG + ∠GDE = 360° (Sum of interior angle of a quadrilateral)

∠CEG = ∠CDG = 90° (Given)

∠GDE = 60° (Given)

∴ ∠ECD = 360° - (∠CEG + ∠CDG + ∠GDE)

∠ECD = 360° - (90° + 90° + 60°) = 120°

∠ECD = 2 × ∠EFD (Angle subtended is twice the angle subtended at the circumference)

120° = 2 × ∠EFD

∠EFD = 120°/2 = 60°

∠EFD ≅ ∠EGD

∠ECD = 120°

∠EGD = 60°

∴∠EGD ≠ ∠ECD

2) Given that arc mEF ≅ arc mFD

Therefore, ΔECF and ΔDCF are isosceles triangles having two sides (radii EC and CF in ΔECF and radii EF and CD in ΔDCF

∠ECF = mEF = mFD = ∠DCF (Given)

∴ ΔECF ≅ ΔDCF (Side Angle Side, SAS, rule of congruency)

\\ \overline{EF}\cong \overline{FD} (Corresponding Parts of Congruent Triangles are Congruent, CPCTC)

∠FED ≅ ∠FDE (base angles of isosceles triangle)

∠FED + ∠FDE + ∠EFD = 180° (sum of interior angles of a triangle)

∠FED + ∠FDE = 180° - ∠EFD = 180° - 60° = 120°

∠FED + ∠FDE = 120° = ∠FED + ∠FED (substitution)

2 × ∠FED  = 120°

∠FED = 120°/2 = 60° = ∠FDE

∴ ∠FED = ∠FDE = ∠EFD =  60°

ΔEFD  is an equilateral triangle as all interior angles are equal

\\ \overline{EF}\cong \overline{FD}\cong \overline{ED} (definition of equilateral triangle)

\overline{ED}\cong \overline{FD}

3) Having that ∠EFD = 60° and ∠CFE = ∠CFD (CPCTC)

Where, ∠EFD = ∠CFE + ∠CFD (Angle addition)

60° = ∠CFE + ∠CFD = ∠CFE + ∠CFE (substitution)

60° = 2 × ∠CFE

∠CFE =60°/2 = 30° = ∠CFD

\overline{CF}\cong \overline{CD} (radii of the same circle)

ΔFCD is an isosceles triangle (definition)

∠CFD ≅ ∠CDF (base angles of isosceles ΔFCD)

∠CFD + ∠CDF + ∠DCF = 180°

∠DCF = 180° - (∠CFD + ∠CDF) = 180° - (30° + 30°) = 120°

mFD = ∠DCF (definition)

mFD = 120°.

5 0
3 years ago
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