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1. Multiply: 5/7 x 7/6 = N *
1 point
a. 5/6
b. 4/6
c. 1 5/6
d. 3/5
2. What is the product of 3/4multiplied by 2/3? *
1 point
a. 5/3
b. 3/12
c. 1/2
d. 6/10
3. Multiply 7/2 and 4/5? *
1 point
a. 2 2/5
b. 2 4/5
c. 1 2/5
d. 1 4/5
Other:
4.What is the product of 2/8 and 16/5? *
1 point
a. 4/5
b. 32/5
c. 5/16
d. 5/4
5.What is 2/3 of 5 1/3? *
1 point
a. 8
b. 3 5/9
c. 5 1/9
d. 18/6
6. Sam gathers 25 kilos of mangoes. Only 4/5 of it are ripe and will be used for mango float and buko pandan salad. How many kilos of mangoes are ripe? *
1 point
a. 18 1/2
b. 20
c. 21 1/5
d. 22
7.Multiply:3/15 x 6/9= N *
1 point
a. 12/15
b. 5/15
c. 3/15
d. 2/15


Sagot :

Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming pool in 4 hours. How many hours will it take 4 large and 4 small pumps to fill the swimming pool.(We assume that all large pumps are similar and all small pumps are also similar.)

Find all sides of a right triangle whose perimeter is equal to 60 cm and its area is equal to 150 square cm.
A circle of center (-3 , -2) passes through the points (0 , -6) and (a , 0). Find a.
Find the equation of the tangent at ( 0 , 2) to the circle with equation
(x + 2)2 + (y + 1)2 = 13

An examination consists of three parts. In part A, a student must answer 2 of 3 questions. In part B, a student must answer 6 of 8 questions and in part C, a student must answer all questions. How many choices of questions does the student have?

Solve for x
x2 - 3|x - 2| - 4x = - 6

The right triangle ABC shown below is inscribed inside a parabola. Point B is also the maximum point of the parabola (vertex) and point C is the x intercept of the parabola. If the equation of the parabola is given by y = -x2 + 4x + C, find C so that the area of the triangle ABC is equal to 32 square units.


∫ (x³cos(x/2)+(1/2))√(4-x²) dx [-2,2]

Distribute

∫ x³√(4-x²)cos(x/2) + (1/2)√(4-x²) dx [-2,2]

Now
Determine the function of

f(x) = x³√(4-x²)cos(x/2)
Is an odd or a function

f(-x) = -x³√(4-x²)cos(x/2)

Therefore the function is odd so automatically the answer for

∫ x³√(4-x²)cos(x/2) dx [-2,2] = 0

While for

(1/2)∫ √(4-x²) dx[-2,2]

f(-x) = √(4-x²)
So the function is even therefore

(1/2)•2 ∫ √(4-x²) dx [0,2]

∫ √(4-x²) dx [0,2]

x = 2sinθ
dx = 2cosθ dθ

2 ∫ √(4-4sin²θ)cosθ dθ

4 ∫ cos²θ dθ
2 ∫ 1+cos(2θ) dθ
2(θ+(1/2)sin(2θ))|[0,2]
2θ+sin(2θ)|[0,2]
2arcsin(x/2)+(1/2)x√(4-x²)|[0,2]

2arcsin(1)+0
= 2(π/2)

= π




#ᴄᴀʀʀʏᴏɴʟᴇᴀʀɴɪɴɢ

1.) 5/6

2.) 1/2

3.) 14/5

4.) 4/5

5.) 2/5

6.)24 1/5

7.) 2/15