Correct Answer: B. \(x^2-2x-15=0\)

Explanation

\(x^2-\)(sum of roots)\(x+\)(product of roots) = \(0\)

\(4x^2-4x-15=0\)

Divide through by 4

\(=x^2-x-\frac{15}{4}=0\)

\(=x^2-x+(-\frac{15}{4})=0\)

\(=x^2-(1)x+(-\frac{15}{4})=0\)

sum of roots =1

= m + (m + 4) = 1

=2m+4=1

=2m=-3

=m=-\(\frac{3}{2}\)

The equation whose roots are 2m and 2m+8

2m=2×-\(\frac{3}{2}=-3\)and \(2m+8=2×-\frac{3}{2}+8=5\)

\(=x^2-(-3+5)x+(-3)(5)=0\)

\(=x^2-2x+(-15)=0\)

\(∴x^2-2x-15=0\)

Correct Answer: A. \(\begin{pmatrix} -3 & 1 \end{pmatrix}\)

Correct Answer: B. -9

Correct Answer: D. q = pr^{\(\frac{1}{3}\)}

Explanation

p^{\(\frac{1}{3}\)} = \(\frac{3\sqrt{q}}{r}\)(cross multiply)

3\(\sqrt{q}\) = r x 3\(\frac{\sqrt{q}}{r}\)(cross multiply)

3\(\sqrt{q}\) = r x 3\(\sqrt{p}\) cube root both side

q = 3\(\sqrt{r}\) x p

q = r^{\(\frac{1}{3}\)}p = pr^{\(\frac{1}{3}\)}

Correct Answer: B. 6.8 m/s

Explanation

Since the bodies are in an opposite direction, one takes the negative velocity.

\(m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v\) (Momentum when the two bodies move in the same direction after collision)

\(3(-2) + 5(V) = (3 + 5)3.5\)

\(-6 + 5V = 28 \implies 5V = 34; V = 6.8 m/s\)

Explanation

(a) \(\frac{5}{8} of 2\frac{1}{2} - \frac{3}{4} \div \frac{3}{5}\)

= \((\frac{5}{8} \times \frac{5}{2}) - (\frac{3}{4} \div \frac{3}{5})\)

= \((\frac{25}{16}) - (\frac{5}{4})\)

= \(\frac{25 - 20}{16}\)

= \(\frac{5}{16}\)

(b)

\(V = \frac{1}{3} \times A \times h = \frac{1}{3} \pi r^{2} h\)

\(\therefore \frac{1}{3} \times 154 \times h = \frac{1}{3} \times \pi \times r^{2} \times h\)

Comparing the two equations,

\(154 = \frac{22}{7} \times r^{2}\)

\(r^{2} = \frac{154 \times 7}{22}\)

\(r^{2} = 49\)

\(\therefore r = 7 cm\).

Correct Answer: D. 20

Explanation

Recall how to find the mean: Mean/Average=Sum of all values/Number of values 30=(x)+(x+11)+(x?4)+(x?11)/4 4x?4=120 4x=124 x=31 Now, because the question asks you to find the smallest number, you need to find the value of x?11. x?11=31?11=20

Explanation

(a)

ST = QR ; PS = height of tower.

In \(\Delta\) PST,

\(\frac{72}{ST} = \tan 62°\)

\(ST = \frac{72}{\tan 62}\)

= 38.283m \(\approxeq\) 38.3m

In \(\Delta\) PQR, \(\frac{h}{QR} = \tan 51°\)

\(h = QR \times \tan 51°\)

= \(38.3 \times 1.235\)

= \(47.3m\)

QS = RT (height of the building)

QS = PS - PQ

= 72 - 47.3 = 24.7m

(b)

(i) \(\Delta POR = \Delta QOR\) (isosceles triangle)

\(\frac{PT}{30} = \sin 60\)

\(PT = 30 \sin 60 = 30 \times \frac{\sqrt{3}}{2}\)

= \(15\sqrt{3} cm\)

PT = TR

PR = 2PT

= \(2 \times 15\sqrt{3} = 30\sqrt{3} cm\)

= \(30 \times 1.732 = 51.96 cm\)

\(\approxeq 52 cm\)

(ii) Length of minor arc, PQR = \(\frac{\theta}{360} \times 2\pi r\)

= \(\frac{120}{360} \times 2 \times 3.142 \times 30\)

= \(62.84 cm\)

(iii) Perimeter of shaded portion = chord PR + Arc PQR

= 51.96 + 62.84

= 114.8 cm \(\approxeq\) 115cm.

Correct Answer: D. 4.62 x 10^-2

Correct Answer: B. 14.2m