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3 years ago

X ^ 2 + 4x - 7 = 0 Solve using the Quadratic Formula

1 answer:

8 0

There's a special case of the quadratic formula called the Shakespeare Quadratic Formula (2b or -2b) that applies when the linear coefficient is even.

$x^2-2Bx + C \textrm{ has zeros }x = B \pm\sqrt{B^2 - C}$

If you're planning on solving a lot of quadratic equations quickly, it's good to know.

Here we have

$x^2 + 4x - 7 = 0$

$x = -2 \pm \sqrt{2^2 - -7}$

$x = -2 \pm \sqrt{11}$

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Does anybody know this ??

Degger [83]

the correct Answer is:

3 0

3 years ago

An equation is graphed below. How many solutions does the equation have?

Dimas [21]

One as it intersects only on the Y axis

6 0

3 years ago

The number of adults going on a field trip was the largest prime number less than 50. How many adults were going on the field tr

Vilka [71]

Answer:

Step-by-step explanation:

Well 47 is the highest prime number that is smaller than 50

4 0

2 years ago

three adults and four children must pay $107. two adults and three children must pay $76. find the price of the adult ticks and

jenyasd209 [6]

Assigning "a" as cost of adult tickets and "c" the cost of children's tickets.

3a + 4c = 107
2a + 3c = 76

elimination method
2(3a + 4c = 107)
-3(2a + 3c = 76)

6a + 8c = 214
-6a - 9c = -228
----------------------
0 - c = - 14
c = 14

Then 2a + 3(14) = 76
2a = 76 - 42
2a = 34
a = 34/2
a = 17

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