Given that the center of the ellipse is at the origin, a focal point is f (- 23,0), and the length of the major axis is twice that of the minor axis, the standard equation of the ellipse is___ ．

Given that the center of the ellipse is at the origin, a focal point is f (- 23,0), and the length of the major axis is twice that of the minor axis, the standard equation of the ellipse is___ ．

It is known that a = 2B, C = 23a2-b2 = C2  B2 = 4a2 = 16F (- 23, 0) ℅ x216 + y24 = 1 is the required; so the answer is: x216 + y24 = 1

The center of the ellipse is at the origin, a focus is on the X axis, and a + C = 7, a-c = 1. Try to find the standard equation of the ellipse

First get a = 4, C = 3, then get B ^ 2 = a ^ 2-C ^ 2 = 1, and then get the equation x ^ 2 / 16 + y ^ 2 = 1

In the pyramid o-abcd, the bottom surface ABCD is rhombic, OA ⊥ plane ABCD, e is the midpoint of OA, f is the midpoint of BC

It is proved that: (1) the ⊥ OA ⊥ plane ABCD, BD ⊂ plane ABCD, so OA ⊥ BD, ⊥ ABCD is rhombus, ∩ AC ⊥ BD, OA ∩ AC = a, ∩ BD ⊥ plane OAC, and ⊂ BD ⊂ plane OBD, ⊂ plane bd0 ⊥ plane ACO. (2) take the midpoint m of OD, connect km and cm, then me ∥ ad, me = 12ad, ∩ ABCD is rhombus, ∩ ad ∥ BC, ad = BC, ∥ f is the midpoint of BC, ∥ CF ∥ ad, CF = 12ad, ∥ me ∥ CF, Me = cf.. Quadrilateral EFCM is a parallelogram, EF cm, EF plane OCD

The image of the known function f (x) = - x ^ 3 + ax ^ 2 + BX is shown in the figure. The area bounded by the tangent of F (x) to the origin on the X axis is one twelfth
The other point is the tangent point. What is the value of a?
In fact, it doesn't matter if you don't look at that picture

F (x) = - x (x ^ 2-ax-b) one intersection point is 0, the other intersection point is negative, so the zero point of negative number is double zero point, that is, x ^ 2-ax-b = 0 has double root of negative number, a 0 = - x ^ 4 / 4 + ax ^ 3 / 3 + BX ^ 2 / 2 = - A ^ 4 / 64-A ^ 4 / 24 + Ba ^ 2 / 8 = - A ^ 4 / 64-A ^ 4 / 24-a ^ 4 / 32 = - 17a ^ 4 / 192 area is one twelfth, there are: 17a ^ 4

How to subtract the time of two columns in MySQL?
For example, there are two columns of data
A B
2002-08-17 23:22:23 2002-08-17 23:22:30
2002-08-18 07:48:52 2002-08-18 07:49:57
2002-08-18 11:05:39 2002-08-18 11:11:27
Then what we need to do is to subtract the time in front from the time in the back
Then add all the differences
This is what I wrote in PHP
$sql5=mysql_ query("select SEC_ TO_ TIME(UNIX_ TIMESTAMP(JSTime)- UNIX_ TIMESTAMP(JTTime)) from log_ call where State='0'");
But in the end, only the first line of data is output, and the rest is not output

--The result is the total number of seconds of the difference select sum (Unix)_ timestamp(B)-unix_ Time stamp (a)) as seconds from TB xxx:xx The form of: XX select sec_ to_ time(sum(unix_ timestamp(B)-unix_ timestamp(A))) as tifrom tb ...

Derivative of y = SiNx + 3 / X

Y'=sinx'+【3(X-1）】’
=cosx-3（x-2）
(in the solution, we don't know how to make negative one square and negative two square, so we write them as - 1 and - 2)

Are there any oral arithmetic problems in grade three?
It's multiplication and division. It takes 200 channels
Give as many as you have, but it's better to have 200 courses. If you have more, you can increase the score

4.8+2.3= 5.6÷7= 30×(200+3)= 4.7+6.2=
8.5÷5= 24÷4+56÷4= 3.7–0.07= 27÷0.01=
4.6–4.6÷4.6= 0.62+0.38= 1.25×4= 0.45÷0.09×5=
3.72+2.28= 5.2×0= 8.7–1.9–3.1= 1.73+3.2=
8.7÷0.29= 45÷0.5÷2= 10.3+5.2= ÷0.8=
1.5+4.2+0.5= 1.03–0.9= 1.12×100= (5.25–4)×8=
12.4+0.37= 0.04×25= 0.8+0.2–0.8+0.2= 4.83–2.53=
1.25×0.8= 4×3.7×0.25= 1.7–0.9= 3.6+4=
3.82+2.5+6.18= 0.29–0.18= 1–0.02= 21.6–3.8–7.2=
6+7.8= 3.9–2.8= 4.9+0.1–4.9+0.1= 1–0.62=
7÷0.5= 3.6×0.2×1= 2.73–0.63= 1÷4=
7.5×5.3÷5.3= 4–0.9= 6.3×101= 25×8×125×4=
0.9×6= 0.1×0.01= 0.7×16–16×0.3= 28.6÷1000=
1.8÷0.9= 1.7×25×4= 1.4×0.2= 5÷2.5=
2.7–1.8–0.2= 2.4÷0.6= 0.36÷3= 1.8×6–1.8=
4.2÷0.3= 10–4.3= 2.8×9+2.8= 60×0.05=
1.7×3= 0.32×0.2×5= 7×0.08= 2.8÷0.04=
10–0.8×10= 1.6×5= 0.37×20= 4.8÷0.6+4.8÷0.4=
4.5×0.2= 4.5+0.55= 8.5–0.24–1.76= 0.81÷0.9=
6.8–0.08= 9×1.25×8= 0.25×28= 0.72÷0.12=

It is known that the solution of the system of equations 2x-3y = 1-A - 3A + 2Y = 7 + 5A satisfies the inequality x + y < 0, and the value range of a is

From - 3A + 2Y = 7 + 5a, we get that
y＝﹙7＋8a﹚/2
If y = (7 + 8a) / 2 generations 2x-3y = 1-A, then
2x－3×﹙7＋8a﹚/2＝1-a
2x－﹙21/2＋12a﹚＝1-a
2x＝1-a ＋21/2＋12a
2x＝23/2＋11a
x＝23/4＋﹙11/2﹚a
∵x+y＜0
∴23/4＋﹙11/2﹚a＋﹙7＋8a﹚/2＜0
The solution is a ＜ - 37 / 38

Fill the nine numbers L ~ 9 in the following formula respectively to make the equation hold: (each number can only be used once) □ □ × □ = □ × □ = 3634

158*23=46*79

If the center is at the origin, an asymptote equation is y = x, and the distance between two vertices is 2, then the standard equation of hyperbola is

If y = x, then a = B. the vertex distance is 2, then a = b = 1, C = radical 2, e = radical 2
The Quasilinear equation is x = ± A & # 178 / C or y = ± A & # 178 / C
The Quasilinear equation is x = ± (radical 2) / 2, or y = ± (radical 2) / 2