Im running out of titles for these things but answer the question from below

natulia [17]

Answer:

$\frac{6+\sqrt{27} }{4-\sqrt{3} } = \frac{r+s\sqrt{3} }{13} \\We\ may\ multiply\ the\ numerator\ and\ denominator\ with\ (4+\sqrt{3} ).\\Hence,\\\frac{(6+\sqrt{27})(4+\sqrt{3}) }{13} = \frac{r+s\sqrt{3} }{13} \\Hence,\\24+6\sqrt{3} +4\sqrt{27} +9= r+s\sqrt{3}\\24+6\sqrt{3} +4*3\sqrt{3} } +9= r+s\sqrt{3}\\24+6\sqrt{3} +12\sqrt{3}+9 = r+s\sqrt{3}\\33+18\sqrt{3} = r+s\sqrt{3}\\Hence,\\r=33, s=18$

5 0

3 years ago

Helppppppppppppppppppppppppppppppppp

liq [111]

Answer:

The answer is B

you should still invest in a calculator (:

4 0

2 years ago

Verify the identity. 4 csc 2x = 2 csc2x tan x

vlada-n [284]

Step-by-step explanation:

$4csc(2x) = 2csc^2(x) tan(x)$

We start with Left hand side

We know that csc(x) = 1/ sin(x)

So csc(2x) is replaced by 1/sin(2x)

$4 \frac{1}{sin(2x)}$

Also we use identity

sin(2x) = 2 sin(x) cos(x)

$4 \frac{1}{2sin(x)cos(x)}$

4 divide by 2 is 2

Now we multiply top and bottom by sin(x) because we need tan(x) in our answer

$2\frac{1*sin(x)}{sin(x)cos(x)*sin(x)}$

$2\frac{sin(x)}{sin^2(x)cos(x)}$

$2\frac{1}{sin^2(x)} \frac{sin(x)}{cos(x)}$

We know that sinx/ cosx = tan(x)

Also 1/ sin(x)= csc(x)

so it becomes 2csc^2(x) tan(x) , Right hand side

Hence verified

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%283%5Csqrt%7B6%7D%29%283%5Csqrt%7B10%7D" id="TexFormula1" title="(3\sqrt{6})(3\sqrt{10}" alt=

mote1985 [20]

18sqrt(15)

This has to be at least 20 characters long so yeaaaa

6 0

3 years ago

2067 Supp Q.No. 2a Find the sum of all the natural numbers between 1 and 100 which are divisible by 5. Ans: 1050

Alborosie

5

Answer:

1050

Step-by-step explanation:

Natural Numbers are positive whole numbers. They aren't negative, decimals, fractions. We can just divide 5 into 100 to find how many natural numbers go up to 100 and just add them but that is just to much.

There is a easier method.

E.g: Natural Numbers that are divisible by a Nth Number. is the same as adding the Nth Numbers to a multiple of that Nth Term. For example, let say we need to find numbers divisible by 2. We know that 4 is divisible by 2 because 4/2=2. We can add the Nth numbers which is 2 to 4. 4+2=6. And 6 is divisible by 2 because 6/2=3. We can call this a arithmetic series. A series which has a pattern of adding a common difference

Back to the problem, we can use the sum of arithmetic series formula,

$y = x( \frac{z {}^{1} + {z}^{n} }{2} )$

Where x is the number of terms in our sequence. Z1 is the fist term of our series. ZN is our last term. And y is the sum of all of the terms

The first term is 5, the numbers of terms being added is 20 because 100/5=20. The last term is 100.

$y = 20( \frac{5 + 100}{2} )$

$y = 20( \frac{105}{2} )$

$y = 1050$

5 0

3 years ago