The number a is three fifths of the number B, the number a is 300, what do you mean by the solution of the equation, five times four fifths Four times four and three out of seven

The number a is three fifths of the number B, the number a is 300, what do you mean by the solution of the equation, five times four fifths Four times four and three out of seven

The number a is three fifths of the number B, and the number a is 300. Use the equation to solve,
Let B be X
x×3/5=300;
x=300×5÷3;
x=500;
Five times four out of five means what is four out of five of five
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5 times of a number is 12.3 times more than 2 times of it

5x-2x=12.3
x=4.1

BD and CE are the bisectors of the inner angle of the triangle ABC. AF is perpendicular to CE and Ag is perpendicular to BD. the perpendicular foot is the point F and D. It is proved that FG is parallel to AC

Because. BD bisector angle ABC, Ag is perpendicular to BD and G, so. Angle abd = angle CBD, angle AGB = angle Ngb = 90 degrees, because. BG = BG, so. Triangle ABG is equal to triangle nbg, so. Ag = ng, G is the midpoint of an, similarly. F is the middle point of am, so. FG is parallel to BC

If f (x) = AF (x) + BG (x) + 2 and f (- 2) = 5, then f (2)=___ ．

Let H (x) = f (x) - 2 = AF (x) + BG (x), since both f (x) and G (x) are odd functions defined on R, so the function H (- x) = AF (- x) + BG (- x) = - AF (x) - BG (x) = - H (x), so the function H (x) is odd. From F (- 2) = 5, we can get H (- 2) = f (- 2) - 2 = 5-2 = 3, so h (- 2) = - H (2) = 3, then H (2) = - 3, f (2) - 2 = - 3, we can get f (2) = - 1, So the answer is: - 1

On the mathematical problems of vector
Let points a (2,0), B (4,2), if point P is on line AB, and the module of vector AB = 2 times the module of vector AP, the coordinates of point P are obtained
I have answers (3, 1) and (1, - 1), but I don't know how to get them

∵ point P is on line AB, and the module of vector AB is twice the module of vector AP,
Ψ vector AB = 2 vector AP or vector AB = - 2 vector AP
(1) When vector AB = 2 vector AP, vector AP + vector Pb = 2 vector AP
Ψ vector AP = vector PB
The point P is the midpoint of the line AB, and a (2,0), B (4,2)
The coordinate of point P is (3,1)
(2) When vector AB = - 2 vector AP, vector AP + vector Pb = - 2 vector AP
3 vector AP = - vector PB
Ψ vector AP = (- 1 / 3) vector PB
Let P (x, y) be the ratio of the point P to the vector AB (- 1 / 3)
Also a (2,0), B (4,2)
From the fixed point formula of the line segment, we get
x=[2+（-1/3）×4]/（1-1/3）=1
y=[0+（-1/3）×2]/（1-1/3）=-1
The coordinates of point P are (1, - 1)
In conclusion, the coordinates of point P are (3,1) or (1, - 1)

As shown in the figure, in △ ABC, ∠ a = 70 °. If ⊙ o cuts the three sides of △ ABC to get the same chord length, then the degree of ∠ BOC is ()
A. 160°B. 135°C. 125°D. 110°

In ∵ △ ABC, ∵ o cuts the three sides of ∵ ABC to get the same chord length, and the distance from ∵ o to the three sides of triangle is equal, that is, O is the heart of ∵ ABC, ∵ 1 = ∵ 2, ∵ 3 = ∵ 4, ∵ 1 + ∵ 3 = 12 (180 ° - a) = 12 (180 ° - 70 °) = 55 °, ∵ BOC = 180 ° - (∵ 1 + ∵ 3) = 180 ° - 55 ° = 125 °. So C is selected

Write an application problem, so that the unknown x satisfies the fractional equation 20 / 3 + x = 15 / X
Yeah, yeah. There's a little way to go. (6y+12/y^2+4y+4)-(y^2-4/y^2-4y+4)+(y^2/y^2-4)=0

1. B walks 20 / 3 kilometers more than a per hour, and the product of two people's walking speed is 15 kilometers per hour
2.(6y+12)/(y^2+4y+4)-(y^2-4)/(y^2-4y+4)+y^2/(y^2-4)
=6(y+2)/(y+2)^2-(y+2)(y-2)/(y-2)^2+y^2/[(y+2)(y-2)]
=6/(y+2)-(y+2)/(y-2)+y^2/[(y+2)(y-2)]
=[6(y-2)-(y+2)^2+y^2]/[(y+2)(y-2)]
=(2y-16)/[(y+2)(y-2)]
=(2y-16)/(y^2-4)

If △ ABC is inscribed in a circle with o as the center and 1 as the radius, and 3oa vector + 4ob vector + 5oC vector = 0 vector, then OC vector · AB vector =?
A 6/5 B -6/5 C 1/5 D -1/5

It is known that OA ^ 2 = ob ^ 2 = OC ^ 2 = 1
3oa vector + 4ob vector + 5oC vector = 0 vector,
So 3oa vector = - 4ob vector - 5oC vector,
Square: 9 OA ^ 2 = 16 ob ^ 2 + 25 OC ^ 2 + 40 OC · ob,
9=16+25 +40 OC·OB,
OC·OB=-4/5.
3oa vector + 4ob vector + 5oC vector = 0 vector,
So 4ob vector = - 3oa vector - 5oC vector,
Square: 16 ob ^ 2 = 9 OA ^ 2 + 25 OC ^ 2 + 30 OC · OA,
16=9+25 +30 OC·OA,
OC·OA=-3/5.
Ψ OC vector · AB vector = = OC · (ob - OA)
= OC·OB- OC·OA=-1/5.
Choose D

As shown in the figure, point m is the midpoint of the right side AC of the isosceles teaching triangle ABC, ad ⊥ BM at e, ad intersects BC at D, FC ∥ ab ∠ ABM = 25 degrees, find the degree of ∠ DMB

The results show that the triangle ADC is equal to the triangle BMD, so DC = MC, triangle DMC is isosceles right triangle, ∠ DMC = 45 degrees, so ∠ DMB = 45 + 65 = 110 degrees

How to solve the equation, 1 question 5 multiplied by 1.3 minus 2x equals 5.5 (x =) (x =) 2 question 2.8x divided by 25 = 1.6 (x =) (x =) (x =)=

5 * 1.3-2x = 5.56.5-2x = 5.52x = 1x = 0.52 2.8x divided by 25 = 1.62.8x = 1.6 * 25 =