Given the exponential function f (x) = a ＾ x, and f (- 2 / 3) = 4 ＾ (1 / 3), find the value of F (- 1 / 2)

Given the exponential function f (x) = a ＾ x, and f (- 2 / 3) = 4 ＾ (1 / 3), find the value of F (- 1 / 2)

f(-2/3)=a^(-2/3)=4^(1/3)
-2/3lga=1/3lg4
lga=-1/2*lg4=-lg2=lg1/2
a=1/2
So: F (x) = (1 / 2) ^ x
f(-1/2)=(1/2)^(-1/2)=√2

If the image of exponential function f (x) passes through the point (- 1,1 / 2), try to find the value of F [f (1)]

Let f (x) = a ^ x, pass through the point (- 1,1 / 2), and substitute it to get a = 2
f(x)=2^x
∴f(1)=2,f[f(1)]=f(2)=4

If the image of exponential function f (x) passes through point (1,2), then f (- 1) =?

If the image of exponential function f (x) passes through point (1,2), then f (- 1) = 1 / 2

1 (x²+3x-3)(x²+3x+4)-8=
2 (x-1)x(x+1)(x+2)-24=
3 2x²+xy-y²-4x+5y-6=

(x²+3x-3)(x²+3x+4)-8
=(x^2+3x)^2＋(x^2+3x)－20
=(x^2+3x-4)(x^2+3x＋5)
=(x-1)(x+4)(x^2+3x＋5)
x(x + 1)(x - 1)(x + 2) - 24
= x(x^2 - 1)(x + 2) - 24
= x(x^3 + 2x^2 - x - 2) - 24
= x^4 + 2x^3 - x^2 - 2x - 24
= (x^4 - 2x^3) + 4x^3 - x^2 - 2x - 24
= x^3(x - 2) + 4x^3 - 8x^2 + 7x^2 - 2x - 24
= (x^3 + 4x^2)(x - 2) + (7x^2 - 14x) + (12x - 24)
= (x^3 + 4x^2 + 7x + 12)(x - 2)
= [(x^3 + 3x^2) + (x^2 + 3x) + (4x + 12)](x - 2)
= (x^2 + x + 4)(x + 3)(x - 2)
2x²+xy-y²-4x+5y-6
=2x^2+(y-4)x+(-y^2+5y-6)
=2x^2+(y-4)x+[-(y^2-5y+6)]
=2x^2+(y-4)x+[-(y-2)(y-3)]
=[1x+(y-3)][2x+(-y+2)]
=(x+y-3)(2x-y+2)

The solutions X and y of the equations 4x-3y = k, 2x + 5Y = 12D are opposite to each other, and K is obtained

Because X and y are opposite to each other
So y = - X
2x+5y=2x-5x=-3x=12
x=-4,y=4
k=4x-3y=-16-12=-28

How about 8 / 17 x 11 / 16 + 5 / 17 x 1 / 2?
It should be calculated in a simple way

8/17x11/16+5/17x1/2
=1/17x11/2+5/17x1/2
=5/17x11/10+5/17x1/2
=5/17x(11/10+1/2)
=5/17x16/10
=8 / 17 (8 / 17)

Given that a and B are matrices of order 4, if AB + 2B = 0, R (b) = 2 and determinant a + e = a-2e = 0, (1) find the eigenvalue of a; (2) prove that a can be diagonalized; (3) calculate determinant a + 3E. This is a complete topic. According to the topic, we can get two eigenvalues - 1 and 2. According to ab = - 2b, then use a β 1 = - 2 β 1 a β 2 = - 2 β 2 a β 3 = - 2 β 3 a β 4 = - 2 β 4. These are part of the teacher's solution, Then from the above, we can get that - 2 is the eigenvector of a, and then according to R (b) = 2, we can know that - 2 is the double root. I only know that R (2) = 2, we can know that there are two linearly independent vector groups. Can we get that a has four linearly independent eigenvectors according to the two linearly independent vector groups?

A β 1 = - 2 β 1 a β 2 = - 2 β 2 a β 3 = - 2 β 3 a β 4 = - 2 β 4, where β I, I = 1,2,3,4 are the four column vectors of B respectively. According to the equation: - 2 is an eigenvalue of A. since R (B) = 2, we can know that the rank of β I, I = 1,2,3,4 is also 2

Dry bulb temperature 23 wet bulb temperature 17 relative humidity

The temperature difference is 6 degrees
The dry bulb temperature was 23 & nbsp;
The relative humidity is 0.52

[(2 ^ x + 3 ^ x + 6 ^ x) / 3] ^ 1 / SiNx when x approaches the limit of 0

[(2 ^ x + 3 ^ x + 6 ^ x) / 3] ^ 1 / SiNx = [1 + (2 ^ x + 3 ^ x + 6 ^ x-3) / 3] ^ [3 / (2 ^ x + 3 ^ x + 6 ^ x-3) * (1 / SiNx) * (2 ^ x + 3 ^ x + 6 ^ x-3) / 3] original formula = e ^ LIM (2 ^ x + 3 ^ x + 6 ^ x-3) / 3sinx = e ^ LIM (2 ^ xln2 + 3 ^ xln3 + 6 ^ xln6) / 3cosx = e ^ [(LN2 + Ln3 + ln6) / 3] = & # 179; √ 36

Simple calculation: 8 / 7 * 75% + 0.75 * 6 / 7-3 / 4

8/7*75％+0.75*6/7-3/4
=8／7×3／4＋3／4×6／7－3/4
=3／4×(8／7＋6／7－1)
=3／4×1
=3／4