What is the arithmetic square root of the root 2500

What is the arithmetic square root of the root 2500

5 times root 2

Given that a is the integral part of Radix 17 and B-1 is the arithmetic square root of absolute value of 2, find the root a + B

A is the integral part of Radix 17, that is, a = 4
B-1 is the arithmetic square root of the absolute value of 2, that is, B-1 = root 2, that is, B = 1 + root 2
A + B = 4 + 1 + 2 = 3 + 2

It is known that x, y are reciprocal to each other, C and B are opposite to each other, the absolute value of a is 3, and the arithmetic square root of Z is 5 Find the square of C - the square of D + XY + the square root of Z of A

The arithmetic square root of ∵ Z is 5 ᙽ z = 25 ∵ the absolute value of a is 3 ᙽ a = ± 3
Solution 1; (square of C-D + XY + Z of a)
The original formula = (C-B) (c + b) + 1 + 25 / 3 or the original formula = (C-B) (c + b) + 25 / 1-3
=0 + 1 + 25 / 3 or = 0 + 1-3 / 25
=Either ± {2 √ 21)} or not true
Solution 2; (the square of C - the square of D + XY) + the square root of Z of A
Original formula = 0 + 1 + √ 25 / 3 (no negative value under the root sign)
=1 + 3 (5 √ 3) or 1-3 (5 √ 3)
=3 {(5 √ 3) + 1} or {3 - (5 √ 3)} of 3

Let m and n be real numbers, and let n = root sign (M2-4) + root sign (4-m2) + 2-m-2 to find the value of root sign (MN) M2 means m to the power of 2, followed by both N = {root (M2-4) + Radix (4-m2) + 2} m-2

n=[√(m^2-4)+√(4-m^2)+2]/(m-2)
m^2-4≥0,4-m^2≥0
m=±2
And denominator m-2 ≠ 0
So m = - 2
So n = 2 / (- 2-2) = - 1 / 2
mn=1
√(mn)=1

Let x, y be real numbers, y= x−4+ 4 − x + 2, find the value of 3x + 4Y

According to the meaning of the title, we can get
x−4≥0
4−x≥0 ，
The solution is x = 4,
When x = 4, y = 2,
∴3x+4y=3×4+4×2=20．

Given that x, y are real numbers, the root sign is 3x + 4, then + y squared - 6y + 9 = 0. If axy-3x = y, then the value of real number a is

3x+4=0
y^2-6y+9=0
x=-4/3
Y=3
a*(-4/3)*3-3*(-4/3)=3
a=1/4

Given that X and y are real numbers, y = x square - 4 + √ 4x square + 1 divided by X - 2, try to find the value of 3x + 4Y I want now, write well, clear, I will add points! Given that X and y are real numbers, y = (under the root sign √ x square - 4) + (√ 4x Square) + 1 divided by X-2, try to find the value of 3x + 4Y.

From y = √ (x? - 4) + √ (4-x?) + 1 / (X-2), (√ 4x square should be √ (4-x})

Given that a and B are real numbers, and the root sign a-4-3, root sign 8-2a = B + 3, find the arithmetic square root of a-b

The root sign must be greater than or equal to 0, so a = 4, so B + 3 = 0, that is, B = - 3,
So the arithmetic square root of A-B = root 7

The three inner angles a, B and C of △ ABC are a, B and C respectively. Asinasinasinb + bcos ^ 2A = root 2A. If C ^ 2 = B2 ^ + root 3A ^ 2, find B

bsinA=asinBasinAsinB+bcos^2A=√2absin²a+bcos²A=√2ab=√2ab^2=2a^2c^2=b^2+√3a^2=2a^2+√3a^2c=a√[2+√3]=a√[(4+2√3)/2]=a(√3+1)/√2cosB=(a^2+c^2-b^2)/(2ac)=(a^2+2a^2+√3a^2-2a^2)/[2a*(√3+1...

The three inner angles a, B and C of the triangle ABC are a, B, C, asinasinasinb + bcos2a = radical 2A 1. Find B of A 2. If C 2 = B 2 + root 3A 2, find B

(1) According to the sine theorem a = 2rsina, B = 2rsinb, where R is the diameter of circumscribed circle, we can get 2rsinasinb + 2rsinb (COSA) ^ 2 = √ 2 * 2rsina [(Sina) ^ 2 + (COSA) ^ 2] SINB = √ 2sinasinb / Sina = √ 2 (II) from cosine theorem and C2 = B2 + √ 3a2, CoSb = 1 + √ 3