The weight ($w$) of the soccer ball used by the age group 8 to 12, vary from 373 grams to 403 grams, inclusive. Write an absolute value inequality describing the weights of soccer balls that are outside the acceptable range.
a.
| $w$ - 403 | < 373
b.
| $w$ - 388 | < 15
c.
| $w$ - 388 | > 15
d.
$w$ < 373 or $w$ > 403
Solution:
w be the weight of a soccer ball, then 373 ≤ w ≤ 403 [Weight of a soccer ball vary from 373 grams to 403 grams, inclusive.]
w < 373 or w > 403
[Outside the acceptable range.]
w < 388 - 15 or w > 388 + 15
[Average of 373 and 403 is 388.]
w - 388 < - 15 or w - 388 > 15
[Rearrange the terms.]
- (w - 388) > 15 or (w - 388) > 15
[If a < - b then - a > b.]
| w - 388 | > 15 [Use: if - x > a or x > a, then | x | > a.]
Correct answer : (3)
2.
Earth is not a perfect sphere, but nearly so, with the approximate surface area 510926783 km^{2}. Find the earth's diameter approximately. (Use $\pi $ = 3.14)
a.
7198 km
b.
12756 km
c.
6378 km
d.
3189 km
Solution:
Surface area of a sphere, S = 4πr^{2}, where r is the radius of the sphere.
d = 2r = 2s4π = sπ
d = 5109267833.14≈ 12756 km.
So, the diameter of the Earth approximately is 12756 km.
Correct answer : (2)
3.
The diameter of a tennis ball is between 6.34 cm and 6.68 cm. Write an absolute value inequality that describes the diameter ($d$) range of a tennis ball.
a.
6.34 < $d$ < 6.68
b.
| $d$ - 6.51 | > 0.17
c.
| $d$ - 6.51 | < 0.17
d.
| $d$ - 6.68 | < 6.34
Solution:
Let d be the diameter of a tennis ball, then 6.34 < d < 6.68 [Diameter of a tennis ball vary from 6.34 cm to 6.68 cm.]
6.34 - 6.51 < d - 6.51 < 6.68 - 6.51 [Average of 6.34 and 6.68 is 6.51.]
- 0.17 < d - 6.51 < 0.17 [Simplify.]
| d - 6.51 | < 0.17 [If - a < ( x - k ) < a, then | x - k | < a.]
Correct answer : (3)
4.
A rectangular gift box with volume (V) 968 cm^{3} and a height of $h$ cm has a square base. Write an equation to find the length of one side of the base and find the value of it for $h$ = 8 cm.
a.
$\sqrt{\frac{\mathrm{V}}{h}}$ ; 11 cm
b.
$\sqrt{\mathrm{V}h}$ ; 88 cm
c.
$\frac{\sqrt{\mathrm{V}}}{h}$ ; 6.2 cm
d.
$\sqrt{\frac{\mathrm{V}}{h}}$ ; 121 cm
Solution:
Volume of the rectangular gift box, V = base area × height
Let x be the base length of a rectangular gift box.
V = x^{2}h⇒x^{2} = Vh [Base area = x × x = x^{2}.]
So, x = Vh
x = 9688 = 11
[Replace V = 968 cm^{3} and h = 8 cm.]
So, the length of the one side of the base of a rectangular gift box is 11 cm.
Correct answer : (1)
5.
The manufacturer of a particular equipment for fitness, claims that the reduction in weight ($w$) resulting by using this equipment satisfies the inequality | $w$ - 18 | ≤ 10. Find the maximum weight in pounds reduced by using that equipment as claimed by the manufacturer.
a.
18 pounds
b.
28 pounds
c.
10 pounds
d.
180 pounds
Solution:
| w - 18 | ≤ 10 [Weight reduction inequality.]
- 10 ≤ (w - 18) ≤ 10 [If | x | ≤ a then - a ≤ x ≤ a.]
- 10 + 18 ≤ w ≤ 10 + 18 [Add 18 to each term of the inequality.]
8 ≤ w ≤ 28 [Simplify.]
So, the maximum weight reduced by using the equipment is 28 pounds.
Correct answer : (2)
6.
Find the equation of the inverse of $y$ = 2$x$.
a.
$y$ = - 2$x$
b.
$y$ = $\frac{1}{2}$$x$
c.
$x$ = $\frac{1}{2}$$y$
d.
$y$ = - $\frac{1}{2}$$x$
Solution:
y = 2x
x = 2y [Interchange x and y.]
y = x2 [Solve for y.]
So, the equation of the inverse is y = 1 / 2x.
Correct answer : (2)
7.
Find the equation of the inverse of $y$ = $x$ - 2.
a.
$x$ = $y$ + 2
b.
$x$ = $\sqrt{y}$ - 2
c.
$y$ = $x$ - 2
d.
$y$ = $x$ + 2
Solution:
y = x - 2
x = y - 2 [Interchange x and y.]
y = x + 2 [Solve for y.]
So, the equation of the inverse is y = x + 2.
Correct answer : (4)
8.
Find the equation of the inverse of $f$($x$) = $\frac{x+2}{3}$. [Hint: Replace $f$($x$) with $y$]
a.
$y$ = - 2$x$ + 3
b.
$y$ = 3$x$ - 2
c.
$x$ = $\frac{\sqrt{y+2}}{3}$
d.
Inverse does not exist
Solution:
f(x) = x+23
y = x+23 [Replace f(x) by y.]
x = y+23 [Interchange x and y.]
y = 3x - 2 [Solve for y.]
So, the equation of the inverse is y = 3x - 2.
Correct answer : (2)
9.
Find the equation of the inverse of $y$ = $\frac{1}{7}$$x$^{2}.
a.
$y$ = ± $\sqrt{7x}$
b.
$x$ = $\frac{1}{7}$$y$
c.
$y$ = - $\sqrt{7}$$x$
d.
$y$ = $\sqrt{7}$$x$
Solution:
y = 1 / 7x^{2}
x = 1 / 7y^{2} [Interchange x and y.]
7x = y^{2} [Multiply throughout by 7.]
y = ± 7x [Solve for y.]
So, the equation of the inverse is y = ± 7x.
Correct answer : (1)
10.
Find the equation of the inverse of $y$ = $x$^{2} - 6.