amygdaleon305

The formula of CI is 

                \(A=P(1+ {r\over n})^{nt}\)

where A = final amount

           P = principal amount

           r = rate

           t = time (in years) 

           n = no. of times compounded

So, here P = 10000, r = 10% = 0.1, n = 2, t = 1 year. 

Now, simply put the values

\(A= 10000(1+ {0.1\over 2})^2\)

\(A=$11,025\)

~Hope you got it. 

No that's not right measure of arc is given in degrees or radian but that doesn't mean you can simply subtract it from 360 to calculate the length of larger arc. The formula for length of arc is 

                    length of arc \(= \theta × {\pi r\over 180}\)

where \(\theta\) is the angle subtended by respective arc at the centre and r is the radius. 

So, we need the radius to calculate..

\({x+2 \over x^2-x-2} = {x+2 \over (x-2)(x+1)} = {A\over x-2}+{B\over x+1}\)

Multiplying by (x - 2)(x + 1) both sides, 

\(A(x+1)+B(x-2)=x+2\)

Put x = -1  ⇒ -3B = 1

                      B = -1/3

Put x = 2  ⇒  3A = 4

                     A = 4/3 

Thus the answer is \(({4\over 3},{-1\over 3})\). 

Let the set of values of r = {4,5} and k = {2,3,4,5,6,7} and let E be the event that  \(r\over k\)  is an integer. 

⇒Total no. of outcomes = 12 

No. of favourable outcomes = 2 

∴ \(P(E) = {2\over 12} = {1\over 6}\)

Digits are 1, 2, 3, 4 and 5. 

⇒no. of 1-digit odd no.'s  = 3

           [Repetition of digits not allowed] 

No. of 2-digit odd numbers

          one's place has = 3 possibilities 

          Then, 10's place has = 4 possibilities

⇒ no. of 2-digit odd numbers = 4 × 3  = 12 

No. of 3-digit odd numbers

           one's place has = 3 possibilities 

           10's place has = 4 possibilities 

           then, 100's place has = 5 - 2 = 3 possibilities 

 ⇒ no. of 3-digit odd numbers = 3 × 4 × 3 = 36 

No. of 4-digit odd numbers

           one's place has = 3 possibilities

           10's place has = 4 possibilities

           100's place has = 3 possibilities

           then, 1000's place has = 2 possibilities

⇒ no. of 4-digit odd numbers = 3 × 4 × 3 × 2 = 72

No. of 5-digit odd numbers

           one's place has = 3 possibilities

           10's place has = 4 possibilities

           100's place has = 3 possibilities

           1000's place has = 2 possibilities

           then, 10000's place has = 1 possibility

⇒ no. of 5-digit odd numbers = 3 × 4 × 3 × 2 × 1 = 72

∴ No. of odd numbers = 72 + 72 + 36 + 12 + 3 = 195

ABCDEF is a regular hexagon and PQR is an equilateral triangle.

In ABCDEF, 

area(△AOB) = area(△BOC) = area(△COD) = area(△DOE) = area(△EOF) = area(△FOA) 

⇒area(ABCDEF) = area(△AOB + △BOC + △COD + △DOE + △EOF + △FOA) 

                            \(= {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}+ {\sqrt3 \over 4}\)

                            \(=6× {\sqrt3 \over 4}\)

                            \(={3\sqrt3 \over 2}\) sq. units

⇒area(△PQR) \(={\sqrt3 \over 4}×4\)

                        \(=\sqrt3\) sq. units

\({area(ABCDEF)\over area(△PQR)} = {3\over 2} = 3:2\)

Let n(F) choose football, n(T) in tennis, n(R) in running, n(B) boys and n(G) girls

• n(F∪T∪R) = 115
• n(B) = 54    ⇒n(G) = 115 - 54 = 61
• n(F) = 52    ⇒n(F∩G) = 25    so    n(F∩B) = 52 - 25 = 27
• n(T) = 43
• n(R∩G) = 10

No. of girls in tennis

              n(T∩G) = n(G) - [n(F∩G) + n(R∩G)] = 61 - (25 + 10) = 26

No. of boys in tennis

              n(T∩B) = n(T) - n(T∩G) =  43 - 26 = 17 

∴ Remaining no. of boys in running

              n(R∩B) = n(B) - [n(F∩B) - n(T∩B)] = 54 - (27 + 17) = 10

⇒n(R) = n(R∩G) + n(R∩B) = 10 + 10 = 20

Probability of choosing running 

            \(P(R) = {20\over 115} = {4\over 23}=0.17\)

Idk but how to solve unless the radius is given? Circle LOM is irrelevant to the question. 

What's the question?

\(x+y=10\)

\(x^2+y^2+2xy=100\)

\(2xy=100-52\)

\(2xy = 48\)

\(xy=24\)

⇒\(x+{24\over x}=10\)

\(x^2-10x+24=0\)

\(x=6,4\)

1st Case : when x = 6 ⇒y = 4 

2nd Case : when x = 4 ⇒y = 6 

∴ All pairs of solutions are (6,4) and (4,6)