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Josue Mat160-Test1-2021

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MAT- 160: Test 1

Professor S. Kha Total Points (100) Time- 90 minutes

Name of The Student:……………………………………………………..


Q.1)
a. Simplify:

√3 56 a16 b8 2a5b2 3√ 7ab2


b. Solve for x. x = √ 1 ─ 4 x ─5 x=-2

5
c. √
4

2m
Rationalize the denominator. 4√40m3

2m
Q.2) Perform the following below as indicated:
1 1 17 1
a. Solve for p. 2
(p+1) ─ = + (2p ─5) 1/6=1/6
3 12 4
1
b. Solve for b 2. A = 2 h (b 1 +b 2) 2A/b1+b2

Q.3) Determine the domain of each relation and determine whether each
relation describes y/f(x) as a function of x.
a. x= y 3 (−∞,∞)
2
x +x ─ 2
b. y= 2 [-1,2]
x ─ x─2

1 3
c. f(x) = x+7 + x+9
[-9,-7]

1
Q.4) a. Find the equation of the line containing the points
(─4 8) and (3, 5). Write your answer in slope intercept form

as well as standard form. y=-3/7x+44/7 - y=-(4)

(7)x+8

b. Graph the linear equation 8x ─ 4y = 12 by finding x and y


intercepts and one other point. -3,1.5 -1,1

c. Find the equation of the line perpendicular to ─3x +y=2


containing the point (9, 4). Write the equation in standard
form as well as point -slope form. y-4=1/3 (x-9), y=x/3+1/3

Q.5) Let f(x) = ─4x+7 and g(x) = x² +9x ─2 find the function values
a. f (h ─7)f=4h
b. g (t² -3t +7) g=t4+9t2-6t+12
c. The longer leg of a right triangle is 7 cm more than the shorter
leg.The length of the hypotenuse is 3cm more than twice the length of the
shorter leg. Find the length of the hypotenuse. 13cm
Q.6) Perform as indicated below.
2
a. If g (x) = 3 x + 1 find x so that g (x) = 5
−3 10 −3
8x y
b. Simplify: ( −3
) 125x12/8y39
20 x y

Q.7) Factor completely if possible:


a. 54q² ─144q +42 ,6 (3q-1)x(3q-7)

2
b.. solve for t : 15t³+40t =70t² t=0, t=2/3 or t=4

Q.8)

a. Write the rational expression in lowest form.


r ³ ─ 3 r ²+2r ─ 6
21 ─ 7 r
- r2+2/7
−1 ─1
W ─V
b. Simplify. ─ 2 +¿V vw-w2/2v+w2
−1

2W ¿

Q.9) a. Simplify.
2
1+
d−3
d-1/5d-d2
−2 d
─d
d−3

b. Solve for n.
3 n 2
n+4
= n+6
─ 2
n +10 n+24
n=-5 or n=4

Q.10) Perform the indicated operation:


1 x 4
a. x+ y + 2
x ─y
2 ─
2x ─ 2 y
-x2 -3xy+3y3 +x3/x3-x2y2-xy2+y4
2
x +2 xy + y ² 7 x +7 y
b. Divide: 3 y−12 ÷ xy−4 x +3 y−12

3
x3+2xy+xy2+3x2-15y+3y2-21x/3xy+9y-12x-36

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