Solve the equation: 5 / 16 x = 3 / 8

Solve the equation: 5 / 16 x = 3 / 8

5x/16=3/8
8×5x=16×3
40x=48
x=6/5

It is known that the ellipse X & # 178 / / A & # 178; + Y & # 178 / / A & # 178; - 1 = 1 and the line y = X-1 intersect at two points ab,
Take AB as the diameter of the circle through the left focus of the ellipse, find the value of a & #

Let a (x1, Y11), B (X2, Y22) let a (x1, Y11, let B (X2, Y22) substitute y = X-1 for y (x = x-1) let a (x (X-2, Y22) let a (x (x1, Y11, Y11) let B (X2, Y22) take y = X-1, and let y = x-x-1 be substituted into x \\35\35\\; (x-1) \\\\\\\\\\\\\\\\\\\\\\\\\\\(A & # 178; - 2) = 0x1 + x2 = - 2 x1x2 = 2 - A & # 1

A question on the principle of computer organization
It is known that the efficiency of the cache / main memory system is 85%, the average access time is 60ns, and the cache is 4 times faster than the main memory?

H = hit rate; TA = average access time; E = efficiency;
TM = main memory period, TC = cache access time;
e=Tc/Ta=0.85
Ta=h*Tc＋(1-h)Tm=60ns
Tm=4Tc
The results show that: TC = 51ns; TM = 204ns; H = 16 / 17;

What is (2a + b) * (2a + 3b) - 3A * 2B equal to

(2a+b)*(2a+3b)-3a*2b
=4a²+6ab+2ab+3b²-6ab
=4a²+3b²+2ab

Let function y = (x ^ 2-1) / (x-1), then x = 1 is the break point of function ()

The limit of X from greater than 1 to 1 is equal to the limit of X from less than 1 to 1

The kindergarten bought 150 apples and divided them into classes according to the number of students

150 ÷ (20 + 25 + 30), = 150 ÷ 75, = 2 (pieces), 20 × 2 = 40 (pieces), 25 × 2 = 50 (pieces), 30 × 2 = 60 (pieces), thus the above table can be completed, as shown in the following figure:

Equilateral △ OAB in the plane rectangular coordinate system, given the point a (2,0), rotate △ OAB a ° clockwise around the point O (0 < a < 360)
When a = 30, calculate the area of the overlap part (the shaded part in Figure 2) between △ OAB and △ oa1b1;
When the ordinates of A1 and B1 are the same, calculate the value of A;
When 60 ＜ a ＜ 180, let the line A1B1 and Ba intersect at point P, and the lengths of PA and PB1 are the two real roots of the equation x2 MX + M = 0, then the coordinates of point P at this time can be obtained

（1） B（1,√3）
(2) Because ∠ bob1 = 30 degree
So ob1 ⊥ AB, am = 2 - √ 3
Then QM = √ 3 × am = 2 √ 3-3
The overlap area is s △ aon-s △ AMQ = 6-3 √ 3
(3) Because the ordinates of A1 and B1 are the same
So A1B1 ‖ X axis
If A1B1 is below the x-axis, a = 120 degree
If A1B1 is above the x-axis, a = 300 degree
(4) Make or ⊥ A1B1 in R and ot ⊥ AB in t
From △ ORP ≌ △ OTP (HL), we can get ∠ OPA1 = ∠ OPA
And ∠ oa1p = ∠ OAP = 120 ° OA = OA1
Then △ oa1p ≌ △ OAP (AAS)
So PA = PA1
PB1=PA1+A1B1=PA+2
In the equation x2-mx + M = 0, two satisfy X1 + x2 = m, x1 × x2 = M
Then X1 + x2 = x1 × X2, so x2 = X1 / x1-1
Let PA = x, then PB1 = x + 2
Let X be X1 and X + 2 be x2
Then x + 2 = x / X-1 gives X1 = √ 2, X2 = - √ 2 (rounding off)
So PA = √ 2
Then p (4 + √ 2 / 2, - √ 6 / 2)

Solve the equation: 3 / 100 divided by x = 3 / 10

The answer is one in 10

Given the two-dimensional vector group A1, A2, A3, A4, what is R (A1, A2, A3, A4) at most?

maxr(a1,a2,a3,a4) =2

What are the last three digits of the 10234 power of 93

93^10234=(100-7)^10234=100^10234-C(10234,10233)*100^10233*7+C(10234,10232)*100^10232*7^2-...-C(10234,3)*100^3*7^10231+C(10234,2)*100^2*7^10232-C(10234,1)*100*7^10233+7^10234
It is easy to know that only the last two terms in the above expansion contribute to the last three digits of the result, namely
【-C(10234,1)*100*7^10233+7^10234】
Simplified to - 1023400 * 7 ^ 10233 + 7 ^ 10234
Because 10234 is a multiple of 4, and the last three digits of the multiple power of 7 are to the power of 20
007 049 343 401 、807 649 543 801、 607 249 743 201、 407 849 943 601、 207 449 143 001
So the last four digits of [- 1023400 * 7 ^ 10233] are * 300;
The last three digits of 7 ^ 10234 are 849
So the last three digits of the 10234 power of 93 are (* 849 - * 300) = 549