Boshlang’ich, oraliq va yakuniy nazorat ishlar to’plami I bosqich talabalari uchun boshlang’ich, oraliq va yakuniy nazorat ishlar

Javob: y= Javob: y=7^x

3. Sodda logarifmik tenglamani yeching.

a) log₄ x=5 b) log₃(x+2)=4

x=4⁵ x+2=3⁴

x=1024 x+2=81

log₄ 1024=5 x=79

log₃(79+2)=4

Javob: x=1024 Javob: x=79

4. Masala.

AB,AC va AD to’g’ri chiziqlar juft-juft perpendikulyar. Bunda AB=3sm,

BC=7sm, AD=1,5 sm bo’lsa, CD kesmani toping.

Berilgan: C Yechish:

A BACAD Pifagor teoremasidan foydalanib

AB=3sm yechamiz: c²=a²+b².

BC=7sm

A D=1,5 D B

CD=?

A

Oraliq nazorat ish – 2.

I-variant.

1-misol.

To’rtta nuqta berilgan : A(0;1;-1), B(1;-1;2), C(3;1;0), D(2;-3;1). AB va

CD vektorlar orasidagi φ burchakning kosinusini toping.

Berilgan nuqtalardan va vektorlarni yasaymiz:

=> a₁=1-0=1 (1;-2;3) => b₁=2-3=-1 (-1;-4;1)

a₂ =-1-1=-2 b₂=-3-1=-4

a₃=2-(-1)=3 b₃=1-0=1

Ikki vektor orasidagi burchak cos φ= formulaga asosan topamiz.

cos φ=