We read your photo as:

Find the quadratic equation whose roots are x = -3 and x = (1)/(2).

Mathematics

ScanSolve AI Answer

Answer

B

Step-by-step solution

18 steps

Here are the solutions to the problems.

:
Find the quadratic equation whose roots are x=−3x = -3 and x=12x = \frac{1}{2}.

  1. Calculate the sum and product of the roots.

    Let the roots be r1=−3r_1 = -3 and r2=12r_2 = \frac{1}{2}.
    Sum of roots: S=r1+r2=−3+12=−62+12=−52S = r_1 + r_2 = -3 + \frac{1}{2} = -\frac{6}{2} + \frac{1}{2} = -\frac{5}{2}.
    Product of roots: P=r1×r2=(−3)×12=−32P = r_1 \times r_2 = (-3) \times \frac{1}{2} = -\frac{3}{2}.

  2. Form the quadratic equation using the formula x2−Sx+P=0x^2 - Sx + P = 0.

    x2−(−52)x+(−32)=0x^2 - \left(-\frac{5}{2}\right)x + \left(-\frac{3}{2}\right) = 0 x2+52x−32=0x^2 + \frac{5}{2}x - \frac{3}{2} = 0
  3. Multiply the entire equation by 2 to clear the denominators.

    2(x2)+2(52x)−2(32)=2(0)2(x^2) + 2\left(\frac{5}{2}x\right) - 2\left(\frac{3}{2}\right) = 2(0) 2x2+5x−3=02x^2 + 5x - 3 = 0

    The correct option is B.

    B. 2x2+5x−3=0\boxed{\text{B. } 2x^2 + 5x - 3 = 0}

    :
    Calculate bb in the figure below if B^=60∘\hat{B}=60^\circ, c=9c=9cm and a=11a=11cm.

  4. Identify the given values and the appropriate formula.

    We are given two sides (a=11a=11cm, c=9c=9cm) and the included angle (B^=60∘\hat{B}=60^\circ). We need to find the third side bb. The Cosine Rule is suitable here:

    b2=a2+c2−2accos⁡Bb^2 = a^2 + c^2 - 2ac \cos B
  5. Substitute the values into the Cosine Rule.

    b2=(11)2+(9)2−2(11)(9)cos⁡60∘b^2 = (11)^2 + (9)^2 - 2(11)(9) \cos 60^\circ

    Recall that cos⁡60∘=12\cos 60^\circ = \frac{1}{2}.

    b2=121+81−2(11)(9)(12)b^2 = 121 + 81 - 2(11)(9)\left(\frac{1}{2}\right) b2=202−(11)(9)b^2 = 202 - (11)(9) b2=202−99b^2 = 202 - 99 b2=103b^2 = 103
  6. Solve for bb.

    b=103b = \sqrt{103} b≈10.1488 cmb \approx 10.1488 \text{ cm}

    Rounding to two decimal places, b≈10.15b \approx 10.15 cm.
    The correct option is B.

    B. 10.15cm\boxed{\text{B. } 10.15\text{cm}}

    :
    Calculate the length of a chord of a circle of radius 5cm if the perpendicular distance from the center to the chord is 3cm.

  7. Visualize the geometry and identify the relevant right-angled triangle.

    Let the radius be r=5r=5cm. Let the perpendicular distance from the center to the chord be d=3d=3cm. Let half the length of the chord be xx. The radius, the perpendicular distance, and half the chord form a right-angled triangle, with the radius as the hypotenuse.

  8. Apply the Pythagorean theorem.

    x2+d2=r2x^2 + d^2 = r^2 x2+(3)2=(5)2x^2 + (3)^2 = (5)^2 x2+9=25x^2 + 9 = 25 x2=25−9x^2 = 25 - 9 x2=16x^2 = 16
  9. Solve for xx and then find the full length of the chord.

    x=16x = \sqrt{16} x=4 cmx = 4 \text{ cm}

    The length of the full chord is 2x2x.

    Chord length=2×4=8 cm\text{Chord length} = 2 \times 4 = 8 \text{ cm}

    The correct option is A.

    A. 8cm\boxed{\text{A. } 8\text{cm}}

    :
    A sector of a circle of 8cm has 90∘90^\circ at the centre. Calculate its area.

  10. Identify the given values.

    Radius r=8r = 8cm.
    Central angle θ=90∘\theta = 90^\circ.

  11. Use the formula for the area of a sector.

    The area of a sector is given by A=θ360∘×πr2A = \frac{\theta}{360^\circ} \times \pi r^2.

    A=90∘360∘×π(8cm)2A = \frac{90^\circ}{360^\circ} \times \pi (8\text{cm})^2 A=14×π(64cm2)A = \frac{1}{4} \times \pi (64\text{cm}^2) A=16π cm2A = 16\pi \text{ cm}^2
  12. Calculate the numerical value and round.

    Using π≈3.14159\pi \approx 3.14159:

    A=16×3.14159 cm2A = 16 \times 3.14159 \text{ cm}^2 A≈50.26544 cm2A \approx 50.26544 \text{ cm}^2

    Rounding to two decimal places, A≈50.27A \approx 50.27 cm2^2.
    The correct option is D.

    D. 50.27cm2\boxed{\text{D. } 50.27\text{cm}^2}

    :
    In the figure above, AD//GE, /BF/ = /CF/ and ∠FBC=55∘\angle FBC = 55^\circ. Find ∠BFC\angle BFC.

  13. Identify the type of triangle formed by B, F, and C.

    Given that /BF/ = /CF/, triangle BFC is an isosceles triangle.

  14. Use the property of isosceles triangles.

    In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, ∠FBC=∠FCB\angle FBC = \angle FCB.
    Given ∠FBC=55∘\angle FBC = 55^\circ, so ∠FCB=55∘\angle FCB = 55^\circ.

  15. Use the sum of angles in a triangle.

    The sum of angles in △BFC\triangle BFC is 180∘180^\circ.

    ∠BFC+∠FBC+∠FCB=180∘\angle BFC + \angle FBC + \angle FCB = 180^\circ ∠BFC+55∘+55∘=180∘\angle BFC + 55^\circ + 55^\circ = 180^\circ ∠BFC+110∘=180∘\angle BFC + 110^\circ = 180^\circ ∠BFC=180∘−110∘\angle BFC = 180^\circ - 110^\circ ∠BFC=70∘\angle BFC = 70^\circ

    The information AD//GE is not needed to solve for ∠BFC\angle BFC.
    The correct option is B.

    B. 70∘\boxed{\text{B. } 70^\circ}

    :
    The radii of two spheres are in the ratio 2:3. If the area of the smaller sphere is 32cm2^2, calculate the area of the bigger sphere.

  16. Set up the ratio of the radii.

    Let r1r_1 be the radius of the smaller sphere and r2r_2 be the radius of the bigger sphere.

    r1r2=23\frac{r_1}{r_2} = \frac{2}{3}
  17. Relate the ratio of radii to the ratio of surface areas.

    The surface area of a sphere is given by A=4πr2A = 4\pi r^2.
    Let A1A_1 be the area of the smaller sphere and A2A_2 be the area of the bigger sphere.
    The ratio of their areas is:

    A1A2=4πr124πr22=r12r22=(r1r2)2\frac{A_1}{A_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^2
  18. Substitute the given ratio and area to find the unknown area.

    A1A2=(23)2=49\frac{A_1}{A_2} = \left(\frac{2}{3}\right)^2 = \frac{4}{9}

    Given A1=32A_1 = 32 cm2^2.

    32A2=49\frac{32}{A_2} = \frac{4}{9}

    Cross-multiply to solve for A2A_2:

    4A2=32×94 A_2 = 32 \times 9 A2=32×94A_2 = \frac{32 \times 9}{4} A2=8×9A_2 = 8 \times 9 A2=72 cm2A_2 = 72 \text{ cm}^2

    The correct option is A.

    A. 72cm2\boxed{\text{A. } 72\text{cm}^2}
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp