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Where are the POINTS................

This is the ANSWER. L HOPE it will HELP you.

Your ANSWER is 10803

0.0555555555555555555555555⅝555

So what is the QUESTION here

48.6 will be the ANSWER

4221 should be the ANSWER

Here is the ANSWER for you!!

Cube ROOT of 765 upto 3 DECIMAL places is 9.145

5/3 -7/6 +(1/1×3/3 -2/3)
5/3 -7/6 + (3/3-2/3)
5/3-7/6 +1/2
5/3-7/6 +1/2 × 3/3
5/3 -7/6 + 3/6
5/3 -10/6
5/3× 2/2 -10/6
10/6 - 10/6
=0
pls MARK as BRAINLIST

The number of positive INTEGER which DIVIDE 2^5 i.e 32 is 6
they are 1,2,4,8,16,32.

1.11085 is the VALUE

In CROSS multiplication we MULTIPLY the numerator of each (or one) side by the DENOMINATOR of the other side.

Heyyy DEAR..
No. of letters in 'Mathematics '=11
No. of TIMES 'h' occurs in mathematics =1
No. of times 'S'in mathematics =1
PROBABILITY of getting 'H'= No of event happen/total no. of EVENTS =1/11
No. of times 'S'in mathematics =1/11
Thus, the probability of getting H and S is 1/11
According to me this the answer........
Hope it will help u

Let's TAKE first term be a. The 5th term =ar^4 =p and 8th term =ar^7=q , 11TH term=ar^10 =s. p×s=ar^4(ar^10)=a^2r^14. q^2=ar^7(ar^7)=a^2r^14. HENCE proved where r is the ratio.

What PAIR? PLEASE re-post FULL QUESTION.

2√3 + √3 + 6 + √3 - 8/√2 + √2

= 4√3 + 6 - 4√2 + √2

= 4√3 + 6 - 3√2

hope it HELP u

plz MARK as a BRAINLIEST & follow me

In a TRIANGLE ABC angle C =3 angle B = 2 (angle A+ angle B). What are the three angles?

C = 2(A+B)

C = 3B, or

2(A+B) = 3B, or

2A+2B=3B, or

B=2A.

C =2(A+2A) = 6A

A+2A+6A = 180 or 9A = 180.

Or A = 20 deg. B = 40 deg, C = 120 deg.

20.904 it's CORRECT ANSWER

let total BOYS be X

3/4 of x = 18

x = 24

total boys = 24

total boys is 2/3 of total STUDENTS

let total students be y

2/3 of y = 24

y = 36

total students = 36

girls = total students - boys

= 36 - 24

= 12

150coin are TWENTY psise

Given: b = a + C and the quadratic equation has equal roots
Here,
b2–4ac = 0 (∵ The quadratic equation has equal roots)
⇒ b2–4ac = 0
⇒ (a + c)2–4ac = 0 (∵ b = a + c)
⇒ a2 + 2AC + c2–4ac = 0
⇒ a2–2ac + c2 = 0
⇒ (a–c)2 = 0
Applying sq.rt on both sides
⇒ √(a–c)2 = √0
⇒ a–c = 0
⇒ a = c
∴ a = c

18.002 KG .............................

in my SENCE 12578 will be the COMPOUND INTREST

Let's the no of students in A and B be 4x and 3x. new average =(students in classA)(avg of A)+(students in class B)(avg of B)/TOTAL no of students =
4x(35)+3x(63)/4x+3x= 140x+189x/7x=329x/7x=47.

The first prime minister of India is Jawaharlal NEHRU PLEASE MARK me as brain LIST

If
6.25 =>2000
100=>X

Total RUPPES he spend at WHEAT = 500
and 30 KG at RS = 420
total ruppes = 500+420
= 920
total rupees for each kg=26.5
precent 26.5/100÷920=243 each

please mark brainlist

I HOPE this will HELP .

The EQUIVALENT RATIONAL NUMBER is -45/2

(x+1/x)(x-1/x)(x^2+1/x^2)(x^4+1/x^4)
{x^2-1/x^2}{x^2+1/x^2}{x^4+1/x^4}
.
.
x^8-1/x^8

ONE because all POINTS LIE on same LIME

Option B is correct hope this will help PLZ MARK as brainlist

Math]secx + TANX[/math]

[math]secx = \dfrac{1}{cosx}[/math]

And we know that

[math]cosx = \dfrac{1-TAN^2\frac{x}{2}}{1+tan^2\frac{x}{2}}[/math]

So,

[math]secx = \dfrac{1+tan^2\frac{x}{2}}{1-tan^2\frac{x}{2}}[/math]

ALSO

[math]tanx = \dfrac{2tan\frac{x}{2}}{1-tan^2\frac{x}{2}}[/math]

On adding the 2 equations we get

[math]secx + tanx = \dfrac{1+2tan\frac{x}{2}+tan^2\frac{x}{2}}{1-tan^2\frac{x}{2}}[/math]

LHS = [math]\dfrac{(1+tan\frac{x}{2})^2}{(1-tan\frac{x}{2})(1+tan{x}{2})}[/math]

Cancelling one of the terms of 1+tan(x/2) we get

LHS = [math]\dfrac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}[/math]

LHS = [math]tan (\frac{\pi}{4} + \frac{x}{2})[/math]

So we see that we get a TERM of [math]\frac{\pi}{4}[/math] and not [math]\frac{\pi}{8}[/math]