$1. \: MC \: is \: tangent \: to \: (B)\Rightarrow \widehat{BCM}=90° \\ 2. \: MC \: and \: MZ \: are \: tangent \: to \: (B)\Rightarrow MC = MZ\Leftrightarrow 5x - 9 = x + 7\Leftrightarrow x = 4 \\ 3. \: BM = \sqrt{ {BC}^{2} + {MC}^{2} } = \sqrt{ {5}^{2} + {12}^{2} } = 13 \\ \Rightarrow EM = BM - BE = BM - BC = 13 - 5 = 8 \\ 4. \: \tan(60°) = \frac{MC}{BC} = \frac{13 \sqrt{3} }{ BC}\Leftrightarrow BC = \frac{13 \sqrt{3} }{\tan(60°)} = \frac{13 \sqrt{3} }{ \sqrt{3} } = 13 \\ 5. \: MC = \sqrt{ {BM}^{2} - {BC}^{2} } = \sqrt{ {20}^{2} - {12}^{2} } = 16 \\ 6. \:BZ = \sqrt{ {BM}^{2} - {MZ}^{2} } = \sqrt{ {25}^{2} - {20}^{2} } = 15 \\ \Rightarrow XE=BX+BE=BZ+BZ=15+15=30$

7 0

3 years ago

Find the smallest number which when divided by 12,15,18&27 leaves as remainder 8,11,14,23 respectively.

Free_Kalibri [48]

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{536}}}}}$

Step-by-step explanation:

Solution :

Here, 12 - 8 = 4

15 - 11 = 4

18 - 14 = 4

27 - 23 = 4

Thus, every divisor is greater than its remainder by 4. So, the required smallest number is the difference of the L.C.M of the given number and 4

Finding the L.C.M

First of find the prime factors of each numbers

12 = 2 × 2 × 3

15 = 3 × 5

18 = 3 × 3 × 3

27 = 3 × 3 × 3

Take out the common prime factors : 3 , 3 and 3

Also take out the other remaining prime factors : 2 , 2 and 5

Now, Multiply those all prime factors and obtain L.C.M

L.C.M = Common factors × Remaining factors

= 3 × 3 × 3 × 2 × 2 × 5

= 540

L.C.M of 12 , 15 , 18 and 27 = 540

So, The required smallest number = 540 - 4

= 536

Hope I helped!

Best regards!!

3 0

3 years ago

I need help on this please help me tnx lolZ

sammy [17]

Mhm guess on it ! I would choose 1 5/9

5 0

2 years ago