the Point (-3,-4) lies on the graph of y+ f(x). This corresponding point that must lie on the graph of
y= l f(x)l is
a. (3,4)
b. (3,-4)
c. (-3,4)
d. (-3,-4)
Posted 7 years ago by Ruby1 in Math 20-1 | 1 answers
#absolute values
if r and s are numbers on a number line, the best representation of how much farther one value is from zero than the other value is
A. (r+s)
B. (r-s)
C. (r) + (s)
D. (r) - (s)
() represent the absolute value sign
Posted 7 years ago by Ruby1 in Math 20-1 | 0 answers
#absolute values
the Equation $\left|x^2-px+5\right|=7$|x2−px+5|=7 has real solutions
A. only when p2 > or equal to 48
B. Only when p2 < 48
C. for any p-values
D. for no values of p
Posted 7 years ago by Ruby1 in Math 20-1 | 0 answers
#absolute values
If f(x)= x2+4x+4, then the domain of y= 1/f(x) is?
Posted 7 years ago by Veronica in Math 20-1 | 5 answers
#absolute values
when an object is dropped from the top of a 75 ft tall building, the object will be h feet above the ground after t seconds,where t = $\frac{1}{4}\sqrt{75-h,}h\le75$14 √75−h,h≤75. How far above the ground will the object be after 1 s? Later verify answers.
Posted 7 years ago by gurleenm in Math 20-1 | 1 answers
#absolute values
when an object is dropped from the top of a 75 ft tall building, the object will be h feet above the ground after t seconds,where t = $\frac{1}{4}\sqrt{75-h,}h\le75$14 √75−h,h≤75. How far above the ground will the object be after 1 s? Later verify answers.
Posted 7 years ago by gurleenm in Math 20-1 | 1 answers
#radicals
the Equation $\left|x^2-px+5\right|=7$|x2−px+5|=7 has real solutions
A. only when p2 > or equal to 48
B. Only when p2 < 48
C. for any p-values
D. for no values of p
Posted 7 years ago by gurleenm in Math 20-1 | 2 answers
#absolute values
use the graph of y= 1/ f(x) to graph y= f(x)
Posted 7 years ago by gurleenm in Math 20-1 | 3 answers
#absolute values
Sketch the graph y= 1/3x-6, outlining the steps used.
Posted 7 years ago by Veronica in Math 20-1 | 2 answers
#absolute values
# 14 in the pre calculus textbook in Alberta
suppose a set of 10 beads were laid in a line where each successive bead had a diameter that was 3/4 of the diameter of the previous bead. If the first bead had a diameter of 24mm, determine the total length of the line of the beads. Express your answer to the nearest millimetre.
Posted 7 years ago by Sm10002 in Math 20-1 | 2 answers
#geomentric series