What is the ninth power of 99 plus the tenth power of 1 / 99

What is the ninth power of 99 plus the tenth power of 1 / 99

It's a multiplication, not a plus
99^9*(1/99)^10
=99^9*(1/99)^9*1/99
=(99*1/99)^9*1/99
=1^9*1/99
=1*1/99
=1 / 99 Where the formula A ^ n * B ^ n = (AB) ^ n is used

A power of 3 = 5, B power of 3 = 10 9 times of A-B power is equal to what?

3 ^ a = 5 3 ^ B = 10, so 3 ^ (a-b) = 5 / 10 = 1 / 2, so 9 ^ (a-b) = [3 ^ (a-b)] ^ 2 = 1 / 2 = 1 / 4 = 0.25

How many meters is 5 × 10 to the negative 9th power

The answer is 5 * 10 minus 18 meters

1 + 3 + 3's square + 3's 3rd power + 3's 4th power + 3's 5th power + 3's 6th power + 3's 7th power + 3's 8th power + 3's 9th power + 3's 10th power is equal to

1+3+3²+3³+.+3^10
=(3^11-1)/2
=88573

If the x power of 10 is equal to 20, and the Y power of 10 is equal to the - 1 power of 5, try to find the value of 2x power of 3 divided by the Y power of 9

(10 ^ x) / (10 ^ y) = 10 ^ (x / y) = 20 ^ (1 / 5} = 100 = 10 ^ 2, so (x / y) = 2
3^(2x)÷9^y=9^(x/y)=9^2=81

Given cos2a = 3 / 5, what is the value of sin ^ 4 + cos ^ 4? I know to mention sin ^ 2 + cos ^ 2, but the answer is 17 / 25

cos2a=3/5
2(cosa)^2-1=1-2(sina)^2=3/5
(cosa)^2=4/5,(sina)^2=1/5
sin^4+cos^4=[(sina)^2+(cosa)^2]^2-2(sina)^2*(cosa)*2
=1-2*1/5*4/5
=17/25

It is known that Cos2 θ = Two 3, then the value of SiN4 θ + Cos4 θ is______ .

sin4θ+cos4θ=（sin2θ+cos2θ）2-2sin2θcos2θ=1-1
2sin22θ=1-1
2（1-cos22θ）=11
Eighteen
So the answer is: 11
Eighteen

If Cos2 alpha = the root of 3, then the value of the fourth power of sin alpha + the fourth power of COS alpha is?

sin^4a+cos^4a=(sin^2a+cos^2a)^2-2(sina*cosa)^2
=1-1/2*sin^2 2a
Because cos2a = root 2 / 3
So cos ^ 2 2A = 2 / 9
sin^2a=7/9
Substituting the original formula = 11 / 18

Mathematics, using cosa to express SiN4 power - Sin square a + cos square a ——

I just want you to take it for nothing else
This is very simple. Do you know (Sina) ^ 2 + (COSA) ^ 2 = 1?
Just make a simple replacement
The original formula = [1 - (COSA) ^ 2] ^ 2 - [1 - (COSA) ^ 2] + (COSA) ^ 2
=(COSA) ^ 4 is the fourth power of cosa

Cos α + cos α squared = 1, find the maximum value of sin α + sin α squared + sin α quartic + sin α sixth power

cosa+cos²a=1
cosa=1-cos²a
cosa=sin²a
sina+sin²a+sin^4a+sin^6a
=sina+cosa+sin^4a(1+sin²a)
=sina+cosa+(cos²a)(1+cosa)
=cos³a+cos²a+cosa+sina
=cosa(1-sin²a)+cos²a+cosa+sina
=cosa(1-cosa)+cos²a+cosa+sina
=2cosa+sina
=5sin (a + b) SINB = 5 / 5 under Radix CoSb = 5 / 5 under Radix