1. X-180 + 310 = 350 2.5x-5 = 3.35 3.13x divided by 2 = 104 4.2.5x + 210 = 360 how to solve the equation? It's ten o'clock tonight. Come on

1. X-180 + 310 = 350 2.5x-5 = 3.35 3.13x divided by 2 = 104 4.2.5x + 210 = 360 how to solve the equation? It's ten o'clock tonight. Come on


1. X-180 + 310 = 350 = > x + 130 = 350 = > x = 350-130 = 2202.5x-5 = 3.35 = > 5x = 3.35 + 5 = 8.35 = x = 8.35/5 = 1.673.13x divided by 2 = 104 = > 13X = 104 * 2 = 208 = > x = 208 / 13 = 164.2.5x + 210 = 360 = > 2.5x = 360-210 = 150 = > x = 150 / 2.5 = 60



3x8 + 8x = 360 / 4-5x to solve the equation


The first possibility: the first step: 3x8 = 24, the second step: 360 / 4 = 90, the third step: you can get the equation as: 24 + 8x = 90-5x, the fourth step: you can move the terms on both sides, 8x + 5x = 90-24, the fifth step: you can get 13X = 66 after adding and subtracting on on both sides, the last step is x = 66 / 13, the second possibility: if your topic is 3x (8 + 8x)



As shown in the figure, in △ ABC, ab = 15, BC = 14, s △ ABC = 84 are known


(1) Let a be ad ⊥ BC at point D. ∵ s △ ABC = 12bc · ad = 84, ∵ 12 × 14 × ad = 84, ∵ ad = 12. Let B be ab = 15, ∵ BD = AB2 − ad2 = 152 − 122 = 9. ∵ CD = 14-9 = 5. In RT △ ADC, AC = ad2 + DC2 = 122 + 52 = 13, ∵ Tanc = ADDC = 125 (2) let B be ⊥ AC at point E. ∵ s △ ABC = 12ac · EB = 84, ∵ be = 16813, ∵ sin ∠ BAC = BEAB = 1681315 = 168195 = 5665



Do, spell, your, how, name, you


How do you spell your name?



The two right sides of a right triangle are a and B, the length of the hypotenuse is C, and the height of the hypotenuse is h


Answer: right triangle
In a right triangle, we have:
A ^ 2 + B ^ 2 = C ^ 2 (Pythagorean string theorem) (1)
At the same time, 1 / 2Ab = 1 / 2CH = > AB = CH (two calculation methods of area) (2)
In the new triangle:
(c+h)^2= c^2+2ch+h^2 (3)
(a+b)^2+h^2=a^2+b^2+2ab+h^2 (4)
By introducing (1) and (2) into (4), we get the following results:
c^2+2ch+h^2 (5)
The comparison between (3) and (5) is as follows:
(a+b)^2+h^2=(c+h)^2
It shows that the triangle composed of (a + b), H ^ 2 and (c + H) is right triangle, where C + H is hypotenuse



Antonym (corresponding word)
below___________ wrong_____________ tall_____________ fat______________
come____________ those_____________ over_____________ good_____________
ask_____________ sell______________ now______________ new______________
cool____________ wet_______________ many_____________ much_____________
put on__________ get on____________ forget___________ outside__________
free____________ clever____________ late_____________ always___________
slow____________ clean_____________ messy____________ after____________
hungry__________ pretty____________ quickly__________ upstairs_________
left____________ interesting_______ exepensive_______ eniptv___________
west____________ thick_____________ light____________ poor_____________
different_______ sick______________ last_____________ difficult________
safe____________ weak______________


above,right,short,skinny
go,these,below,bad
answer,buy,before,old
warm,dry,few,less
take off,get off,remember,inside
busy,stupid,early,never
fast,dirty,clean,before
full,ugly,slowly,downstairs
right,boring,cheap,empty?--> full
east,thin,dark,rich,
same,healthy,first,easy
dangerous,strong



Solving triangle problem ~ involving sine and cosine theorem (Mathematics of grade one in senior high school)
11. In known triangle ABC, Sina (CoSb + COSC) = sinbsinc
Prove that this triangle is a right triangle
The problem comes from p19.11, mathematics compulsory 5 of PEP B edition


sinA=(sinB+sinC)/(cosB+cosC)
sin(B+C)=(sinB+sinC)/(cosB+cosC)
sinBcosC+cosBsinC=(sinB+sinC)/(cosB+cosC)
sinBcosBcosC+sinB(cosC)^2+(cosB)^2sinC+cosBsinCcosC=sinB+sinC
sinBcosBcosC+cosBsinCcosC=sinB-sinB(cosC)^2+sinC-(cosB)^2sinC
sinBcosBcosC+cosBsinCcosC=sinB(sinC)^2+(sinB)^2sinC
cosBcosC(sinB+sinC)=sinBsinC(sinB+sinC)
(cosBcosC-sinBsinC)(sinB+sinC)=0
cos(B+C)(sinB+sinC)=0
sinB+sinC≠0
So cos (B + C) = 0
B + C = 90 degrees, right triangle



When to use the ing form and the to + original form for sentences that begin with verbs in English


To solve this problem, he spends two hours on it... To solve this problem, he spends two hours on it... To solve this problem, he spends two hours on it



How does the education of perceptual set affect the development and education of children's number concept


It is to let children perceive the set and its elements without teaching them the set terms, so that they can have a perceptual understanding of what is set and elements, and learn to compare the number of elements in the set with corresponding methods, The idea of set, subset and the relationship among them is permeated in the content and method of early childhood mathematics education. The introduction of perceptual set education is the result of the reform of mathematics education in the past two or three decades, because set is not only a basic concept of modern mathematics, but also an important content of early childhood mathematics education, The main contents of perceptual set education are: 1) the teaching of object classification. Let children master different classification methods, understand and master the words about classification. Class contains subclass, and class is greater than subclass; ② The teaching of distinguishing "1" and "many" is the teaching content of small class. Let children distinguish "1" and "many" and understand the relationship between them, infiltrate the idea of set composed of elements; ③ compare the teaching of two groups of objects equal and unequal. It belongs to the teaching content of perceptual set before learning to distinguish "1" and "many" in small class. The main purpose is to make children learn to correspond, We should use the method of one-to-one correspondence to compare the number of two groups of objects (no more than 5), and be able to understand the vocabulary of "as much", "not as much", "more" and "less". We should mainly use the method of comparison for teaching, such as overlapping method and juxtaposition method. We should not only carry out the perceptual set education in the above teaching, but also infiltrate the idea of set in the teaching of number and quantity



Matlab complex comparison size
x=12+i*5;
y=5-i*13;
(1)result=x>y;
(2)result=abs(x)>abs(y)
What is the result and why? (mainly why)


In MATLAB, x > y is equivalent to real (x) > real (y), that is, only the real part is compared
So the result of result = x > y is 1
The result of result = ABS (x) > ABS (y) is 0