BARU
Ukuran huruf
Lembar kerjaPRETEST (Week 4)
Total soal: 15
Worksheet time: 7mins
Which of the following is a quadratic inequality?
x−1+x>7
y2+x=z
p2+p≤ 5
a+b=c
Which of the following is NOT a quadratic inequality?
x2>2x−5
2x2−x≥7
x2+2x<5
2x−2−x ≤ 7
Which of the following quadratic inequality is in the standard form?
y2≥2y+6
y2−2y<6
y2+2y+6>0
y(y+2)−6≤0
Which quadratic inequality is the same as:
x(x+2)>1?x2+2x−1≥0
x2+2x−1>0
x2+2>1
2x2+2x−1>0
The length of a rectangle is 5 units more than its width(w). If its area is greater than 50 sq. units, which of these represents the given as a quadratic inequality?
w(w+5)>50
w(w+5)≥50
w(w+50)<5
w(w+50)≤5
Which of the following is NOT a solution to the interval notation
0≤x≤3?4
3
2
1
Which interval notation stands for the graph of inequality shown above?
−3<x<4
−3≤x≤4
−4<x<3
−4≤x≤3
Which of the following is the graph of
x2+x−6≤0?Which graph illustrates the inequality
(x−1)(x+2)>0?Which inequality is represented by the graph on the number line above?
x2+x−12>0
x2+x−12<0
x2+x−12≥0
x2+x−12≤0
Jessica is printing a tarpaulin for a birthday party banner. The length of the tarpaulin is 3 feet more than its width(w) and its area is at least 40 square feet.
Which inequality represents the problem?
w(w−3)≥40
w(w+3)≥40
3(w+3)≥40
3(w+40)≥0
Jessica is printing a tarpaulin for a birthday party banner. The length of the tarpaulin is 3 feet more than its width(w) and its area is at least 40 square feet.
What is the solution set of this problem?
w≥5
w≥6
w≤5
w≤8
Jessica is printing a tarpaulin for a birthday party banner. The length of the tarpaulin is 3 feet more than its width(w) and its area is at least 40 square feet.
Which of the following is a possible dimension of the tarpaulin?
l=10ft.,w=5ft.
l=9ft.,w=5ft.
l=8ft.,w=5ft.
l=7ft.,w=5ft.
A ball is thrown vertically upward with initial velocity of 96 dm./sec. The distance s (in decimeters) of the ball from the ground after time t seconds is s(t)=96t−16t2 . For what time is the ball more than 128 dm. above the ground?
Which inequality represents the problem above?
96t−16t2>128
−16t2+96t<128
96t−16t2≥128
−16t2+96t≤128
A ball is thrown vertically upward with initial velocity of 96 dm./sec. The distance s (in decimeters) of the ball from the ground after time t seconds is s(t)=96t−16t2. For what time is the ball more than 128 dm. above the ground?
For what time is the ball more than 128 dm. above the ground?
from 2 - 4 sec.
from 3 - 5 sec.
bet. 3 sec. and 5 sec.
bet. 2 sec. and 4 sec.
