Given that the arithmetic square root of 2a-b is 3, the square root of 3A + B-1 is plus or minus 4, and C is the integral part of root 13. Find the square root of a + 2b-c

Given that the arithmetic square root of 2a-b is 3, the square root of 3A + B-1 is plus or minus 4, and C is the integral part of root 13. Find the square root of a + 2b-c

The arithmetic square root of 2a-b is 3
2a-b=3²=9 (1)
The square root of 3A + B-1 is plus or minus 4
3a+b-1=(±4)²=16 (2)
(1)+(2)
5a=25
A=5
b=2a-9=1
C is the integer part of root 13,
So C = 3
So a + 2b-c = 5 + 2-3 = 4
So the square root of a + 2b-c = - 2 and 2

It is known that the square root of root 7-2a is positive and negative. The arithmetic square root of 3,4a + B-11 is 7. Find the cube root of ab

√(7-2a)=9;
7-2a=81;
2a=-74;
a=-37;
4a+b-11=49;
b=49+11+148=208;
Cube root of AB = -19.74
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It is known that a = 1-B degree √ A is the arithmetic square root of a, B = 4A + B degree √ B + 1 is the cube root of B + 1. Find the square root and cube root of a + B

Because 1-B degree √ A is the arithmetic square root of a, so 1-B is equal to 2. Because 4A + B degree √ B + 1 is the cube root of B + 1, 4A + B is equal to 3. The equation system can be obtained as follows: 1-B = 24a + B = 3, a = 1, B = - 1, so a = √ 1 = 1, B = 3 √ (- 1 + 1) = 3 √ 0 = 0A + B = 1 + 0 = 1 because (± 1) 2 = 1, 1

If the square root of a positive number is a minus 1 and a plus 3 Find the square root of X + y + Z with the square of (x-1) + the absolute value y + 2 plus the root sign Z-3 = 0

Root 0.49 - root 0.81 - root 4 / 9
=0.7-0.9-2/3
=-1/5-2/3
=-13/15
25-36% of the root sign - 3 / 4 of the absolute value
=5/12-1/9-3/4
=（5*3-4-27）/36
=-4/9
If the square root of a positive number is a minus 1 and a plus 3
a-1+a+3=0
a=-1
a-1=-2
Positive number = 2 ^ 2 = 4
The square of (x-1) + absolute value y + 2 plus root sign Z-3 = 0
x-1=y=z-3=0
X=1
Y=0
Z=3
The square root of X + y + Z = ± 2
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Find the square root of the following numbers: 100, 16, 90.25, root sign 36 (- 4), the square root - 2 and the absolute value root of 1 / 4

√100=10
√9/16=3/4
√0.25√36=√1.5
√（-4）²=4
√|-9/4|=3/2
√81=9

Given that the integer part of Radix 15 is a, B-1 is the square root of 9, and the absolute value of A-B = B-A, find a + B

8729833462074168851792653997824 The integer part of Radix 15 = ± 3 and ∵ the integer part of Radix 15 is a (known)

If α is the second quadrant angle, what is the value of (1-cos? α) - cos α × root (1-sin? α) equal to?

Sin α × root sign (1-cos 2 α) - cos α × root sign (1-sin 2 α)
=Sin α × root sign (sin 2 α) - cos α × root sign (COS 2 α)
=sin²α-cosα(-cosα)
=sin²α+cos²α
=1

It is known that α and β are all acute angles, and cos α = 2 times the root 5 / 5, cos β = 10 / 10?

Because α and β are all acute angles, there are:
Cos α = 2 √ 5 / 5, sin α = √ 5 / 5
Cos β = √ 10 / 10, sin β = 3 √ 10 / 10
sin(α-β)=sinαcosβ-cosαsinβ
=√5/5x√10/10-2√5/5x3√10/10
=-√2/2
That is, α - β = - 45 degrees

It is known that α, β are acute angles and cos α = 1 10，cosβ=1 Then the value of α + β is equal to 5___ .

α. β is an acute angle and cos α = 1
10，cos β=1
So sin α = 3
10，sinβ=2
Five
cos（α+β）=cosαcosβ-sinαsinβ=1
Five
2-6
Five
2=-
Two
Two
The value of α + β is equal to 3 π
Four
So the answer is: 3 π
Four

If cos (a-b) = 5 / 5 under the root sign, cos2a = 10 / 10 under the root sign, and a and B are acute angles and a is less than B, then the value of a + B is

So cos (a + b) = cos [2A - (a-b)] = √ 10 / 10 · √ 5 / 5 ＋ (- 2 √ 2 / 5) · 3 √ 10 / 10 = if you calculate again, it is too difficult to input on the computer