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arsen [322]
3 years ago
11

Select the correct answer.

Mathematics
1 answer:
Scilla [17]3 years ago
3 0

Answer:

Approximately (2, 1)

See image

Step-by-step explanation:

Rearrange each equation so it is ready to be graphed on an x-y-axis (see image) look for the point where the two lines cross. This is the solution for the system of equations.

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CALCULUS: Which of the following represents the volume of the solid formed by revolving the region bounded by the graphs of y =<
Mademuasel [1]

Answer:

π * [27,1]∫ (3-∛y)² dy

Step-by-step explanation:

6 0
3 years ago
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By the Triangle Inequality Theorem which set of side lengths could create a triangle?
Nonamiya [84]

Answer: B

Step-by-step explanation:

the sum of any two sides is bigger than the third side... so 5+9= 14, bigger than 6. 5+6=11, greater than 9. 9+6=15, bigger than 5.

4 0
3 years ago
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Can someone help me with this
Morgarella [4.7K]

Answer:

FG=30

Step-by-step explanation:

Since we know that Point G is on the Segment FH, it doesn't really matter where G is, but we can know for certain that:

FH=FG+GH

We are given that FH is 4x, GH is x, and FG is 2x+10. Substitute:

4x=(2x+10)+x

Solve for x. On the right, combine like terms:

4x=3x+10

Subtract 3x from both sides:

x=10

So, the value of x is 10.

To find the value of FG, substitute 10 into its x:

FG=2x+10\\FG=2(10)+10

Multiply:

FG=20+10

Add:

FG=30

And we're done!

6 0
3 years ago
John is younger than miraji for 15yrs after 4yrs the sum of their age will be 41yrs find the age of miraji at present
Margaret [11]
(x+4) + ((x-15)+4)= 41
2x-7= 41
2x= 48
x= 24
6 0
4 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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