Given 2x + 5y-3 = 0, find the value of (4 ^ x) * (32 ^ y)

Given 2x + 5y-3 = 0, find the value of (4 ^ x) * (32 ^ y)

2x+5y-3=0
2x+5y=3
(4^x)*(32^y)
=2^2x*2^5y
=2^(2x+5y)
=2^3
=8
2x+5y=3
Original formula = (2 ^ 2) ^ x * (2 ^ 5) ^ y
^2x*2^5y
=2^(2x+5y)
=2^3
=8
On the inequality X & # 178; - ax-6a ＜ 0 of X, the solution set is {x| m ＜ x ＜ n}, and N-M ≤ 5, find the range of A
(n-m) 2 ≤ 25, Mn is two solutions of the equation x2-ax-6a = 0, N + M = a, Mn = - 6a, that is, (M + n) 2-4mn ≤ 25, A2 + 24a ≤ 25, - 25 ≤ a ≤ 1
The focal point of a parabola is f (- 1,0), and its quasilinear equation is x-2y + 2 = 0
Make a vertical line of x-2y + 2 = 0 through F
X-2y + 2 = 0, the slope is 1 / 2
So the slope of the perpendicular is - 2
So y-0 = - 2 (x + 1)
y=2x+2
The intersection of x-2y + 2 = 0 is a (- 2 / 3,2 / 3)
So the vertex is the midpoint of AF
So it's (- 5 / 6,1 / 3)
2x-5y-4 = 0, find the x power of 4 and the Y power of 32
2x-5y-4=0
2x-5y=4
4^x÷32^y
=(2^2)^x÷(2^5)^y
=2^2x÷2^5y
=2^(2x-5y)
=2^4
=16
2x-5y=4
4^x÷32^y
=(2^2)^x÷(2^5)^y
=2^2x÷2^5y
=2^(2x-5y)
=2^4
=16
X of 4 = 2x of 2, y of 32 = 5Y of 2,
The (2x-5y) power of the original formula = 2
=The fourth power of 2
=16
If the inequality (A-2) x ^ (a + 2) - 1 ＜ 5 about X is a system of linear equations with one variable, the solution set of inequality (2a-b) x + 3a-4b ＜ 0 about X is x ＞ 4 / 9
Find a, B
That is, the degree of X is 1
So a + 2 = 1
a=-1
So (- 2-B) x-3-4b-3-4b
Divide both sides by B + 2
Solution set x > 4 / 9
So (- 3-4b) / (B + 2) = 4 / 9
-27-36b=4b+8
40b=-35
b=-7/8
a=-1,b=7/8
It is known that the equation of the parabola is 4x-y = 0. The focal coordinates and the standard equation of the parabola are obtained
4x-y & # 178; = 0, that is, the standard equation is
Y & # 178; = 4x according to the standard equation of parabola y & # 178; = 2px
We can get 2p = 4, so p = 2
The focus of the parabola is (P / 2,0), so the focus of the parabola is (1,0)
The standard equation is y ^ 2 = 4x
Focus coordinates (1,0): pro, how to write the process
Given 2x + 5y-3 = 0, the value of 4 ^ x * 32 ^ y is?
2x+5y-3=0
2x+5y=3
therefore
4^x*32^y
=2^2x * 2^5y
=2^(2x+5y)
=2³
=8
4^x*32^y=2^(2x)*2^(5y)=2^(2x+5y)=8
2x+5Y-3=0
So, 2x = 5Y
4^x=(2^2)^x=2^2x
32^y=(2^5)^y=2^5y
So: 4 ^ X32 ^ y = 2 ^ 2x 2 ^ 5Y = 2 ^ (2x + 5Y) = 2 ^ 3 = 8
Let (2a-b) x + 3a-4b0 be an inequality about X
The two symbols are not the same. The problem should be changed to: (2a-b) x + 3a-4b ≤ 0 solution set is x ≤ 4 / 9 or (2a-b) x + 3a-4b < 0 solution set is x < 4 / 9. The solution is as follows: (2a-b) x < 4b-3a because the solution set is x < 4 / 9, so (4b-3a) / (2a-b) = 4 / 9 and (2a
It is known that the equation of the parabola is 4x-y = 0. The focal coordinates and the Quasilinear equation of the parabola are obtained
Given that the equation of the parabola is 4x-y & # 178; = 0, the focal coordinates and the Quasilinear equation of the parabola are obtained
Y & # 178; = 4x; 2p = 4, P = 2, so the focal point F (1,0); collimator: x = - 1
4x-y & # 178; = 0, that is, the standard equation is
Y & # 178; = 4x according to the standard equation of parabola y & # 178; = 2px
We can get 2p = 4, so p = 2
The focus of the parabola is (P / 2,0), so the focus of the parabola is (1,0)
It is pointed out that the power series (x) = 1 / 2 is convergent
The best way to solve this problem is to use binomial expansion and click to enlarge