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katrin2010 [14]
3 years ago
15

What it the ordered pair of 7x-2y=3

Mathematics
1 answer:
Genrish500 [490]3 years ago
3 0
In a system of equations with two variables, you generally need two equations to find a unique solution, and this system is no exception. One example of a solution to this problem is (1, 2), but (3, 9) and (5, 16) also work, and there are actually infinitely many solutions to this single equation.
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Two sides of a parallelogram are 93 feet and 80 feet. The measure of the angle between these sides is 18°. Find the area of the
azamat
<h3>Answer:  2299 square feet</h3>

Work Shown:

area = a*b*sin(C)

area = 93*80*sin(18)

area = 2299.086438

area = 2299 square feet

8 0
2 years ago
What is 21 inches to 9 feet in simplest form
Olenka [21]
2.41667, you can also find a converter online. 
7 0
4 years ago
How do you solve this don’t mind the (f) that’s just the problem number
zavuch27 [327]
Hello,
You need to evaluate the equation, so:
3xˆ2+14x+8 --> 17xˆ2+8. Hop this helps!

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3 0
3 years ago
In june 2005, Peter mailed a package from his local post office in Fayetteville North Carolina to a friend in Radford, virginia
den301095 [7]

The total price that Peter paid for the local parcel = $2.07

The rate per ounce at that time was = $0.23/ounce

Total weight of Peter's package = Total price paid by peter/ rate per ounce

Total weight of Peter's package = $2.07/$0.23 per ounce

We now have to divide 2.07/0.23 = 207/23 (Shifting both numerator and denominator by 2 decimal places) = 9

Total weight of Peter's package = 9 ounces



8 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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