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The factors of x² + 7x + 6 are?

Mathematics

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Step-by-step solution

18 steps
  1. The factors of x2+7x+6x^2 + 7x + 6 are:
  1. Identify the quadratic expression.

    The expression is x2+7x+6x^2 + 7x + 6.

  2. Find two numbers that multiply to the constant term (6) and add up to the coefficient of the middle term (7).

    The numbers are 1 and 6, because 1×6=61 \times 6 = 6 and 1+6=71 + 6 = 7.

  3. Write the factored form of the quadratic expression.

    x2+7x+6=(x+1)(x+6)x^2 + 7x + 6 = (x+1)(x+6)

    The factors are (x+1)(x+1) and (x+6)(x+6).

    The correct option is A.

    A) (x+1) and (x+6)\boxed{\text{A) } (x+1) \text{ and } (x+6)}
    1. The value of 6×27\sqrt{6} \times \sqrt{27} is:
  4. Combine the terms under a single square root.

    6×27=6×27\sqrt{6} \times \sqrt{27} = \sqrt{6 \times 27}
  5. Multiply the numbers inside the square root.

    6×27=1626 \times 27 = 162

    So, the expression becomes 162\sqrt{162}.

  6. Simplify the square root by finding perfect square factors of 162.

    162=81×2162 = 81 \times 2 162=92×2162 = 9^2 \times 2
  7. Extract the perfect square from the square root.

    162=92×2=92×2=92\sqrt{162} = \sqrt{9^2 \times 2} = \sqrt{9^2} \times \sqrt{2} = 9\sqrt{2}

    The value is 929\sqrt{2}.

    The correct option is B.

    B) 92\boxed{\text{B) } 9\sqrt{2}}
    1. Observe the triangles, the correct statement among the following is:
  8. Analyze the given triangles △ABC\triangle ABC and △QPR\triangle QPR.

    In △ABC\triangle ABC:
    Side AB=6AB = 6 cm
    Side BC=4BC = 4 cm
    Angle ∠B=90∘\angle B = 90^\circ (indicated by the square symbol)

    In △QPR\triangle QPR:
    Side QP=6QP = 6 cm
    Side PR=4PR = 4 cm
    Angle ∠P=90∘\angle P = 90^\circ (indicated by the square symbol)

  9. Compare corresponding sides and angles.

    We have:
    AB=QP=6AB = QP = 6 cm
    ∠B=∠P=90∘\angle B = \angle P = 90^\circ
    BC=PR=4BC = PR = 4 cm

  10. Apply the Side-Angle-Side (SAS) congruence criterion.

    Since two sides and the included angle of △ABC\triangle ABC are equal to two sides and the included angle of △QPR\triangle QPR, the triangles are congruent.
    The correspondence of vertices is A↔QA \leftrightarrow Q, B↔PB \leftrightarrow P, C↔RC \leftrightarrow R.

  11. Write the congruence statement.

    △ABC≅△QPR\triangle ABC \cong \triangle QPR

    The correct option is A.

    A) △ABC≅△QPR\boxed{\text{A) } \triangle ABC \cong \triangle QPR}
    1. If one side of an isosceles triangle is 5 cm & its equal sides measure 3 cm each, then half of its perimeter is:
  12. Identify the lengths of the sides of the isosceles triangle.

    The equal sides are 3 cm each. The third side is 5 cm.
    So, the side lengths are 3 cm, 3 cm, and 5 cm.

  13. Calculate the perimeter of the triangle.

    Perimeter P=sum of all sidesP = \text{sum of all sides}

    P=3 cm+3 cm+5 cmP = 3 \text{ cm} + 3 \text{ cm} + 5 \text{ cm} P=11 cmP = 11 \text{ cm}
  14. Calculate half of the perimeter.

    Half of perimeter=P2=11 cm2\text{Half of perimeter} = \frac{P}{2} = \frac{11 \text{ cm}}{2} Half of perimeter=5.5 cm\text{Half of perimeter} = 5.5 \text{ cm}

    The correct option is B.

    B) 5.5 cm\boxed{\text{B) } 5.5 \text{ cm}}
    1. Two angles measure (55∘+3a)(55^\circ + 3a) and (115∘−2a)(115^\circ - 2a) if each angle is a supplement of the other, then calculate the value of "a".
  15. Understand the definition of supplementary angles.

    If two angles are supplementary, their sum is 180∘180^\circ.

  16. Set up the equation using the given angle measures.

    (55∘+3a)+(115∘−2a)=180∘(55^\circ + 3a) + (115^\circ - 2a) = 180^\circ
  17. Combine like terms.

    55∘+115∘+3a−2a=180∘55^\circ + 115^\circ + 3a - 2a = 180^\circ 170∘+a=180∘170^\circ + a = 180^\circ
  18. Solve for aa.

    a=180∘−170∘a = 180^\circ - 170^\circ a=10∘a = 10^\circ

    The value of aa is 10∘10^\circ.

    a=10∘\boxed{a = 10^\circ}
    1. Write any two of Euclid's axioms.

    Here are two of Euclid's axioms:

    • Things which are equal to the same thing are equal to one another.
    • The whole is greater than the part.
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