MATHEMATICS HIGH SCHOOL

A factory produces engraved gold disks. The cost of the disks is directlyproportional to the cube of the radius of the disk.
A disk with a radius of cm costs US dollars(USD).
(a) Find an equation which links and . [3]
(b) Find, to the nearest USD, the cost of the disk that has a radius of
cm.

Answers

Answer 1
Answer:

Answer:

(a) To find an equation that links the cost of the disk and its radius, we can use the given information that the cost is directly proportional to the cube of the radius. Let's denote the cost of the disk as C and the radius as r.

According to the given information, we can write:

C ∝ r^3

Since we are looking for an equation, we need to introduce a constant of proportionality. Let's call this constant k. Therefore, we can rewrite the equation as:

C = k * r^3

Now, we need to find the value of k. We are given that a disk with a radius of cm costs USD . Substituting these values into our equation, we get:

= k * (^3)

Simplifying further:

= k *

Now, we can solve for k by dividing both sides of the equation by :

k =

Therefore, our equation linking the cost of the disk (C) and its radius (r) is:

C = * r^3

(b) To find the cost of a disk with a radius of cm, we can substitute this value into our equation from part (a). Let's denote this cost as C1 and the radius as r1.

C1 = * (r1)^3

Substituting r1 = cm into the equation:

C1 = * (^3)

Calculating this expression will give us the cost of the disk to the nearest USD.

The answer to part (b) cannot be provided in this format as it requires specific numerical values for and r1. Please provide those values so that I can calculate and provide you with an accurate answer.

Step-by-step explanation:

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6th grade math help me plzz

Answers

Answer:

A-B really simple.

Step-by-step explanation:

Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth. x^2 + 3 = -4x

A. 1,3
B. -1,-3
C. 1,-3
D. -1, 3

Answers

Hello,

x²+4x+3=0
x=(-4-2)/2=-3  or x=(-4+2)/2=-1


Answer B (-1,-3)

Eli plans to repaint some classroom bookcases. He has 12 galnt. All of the bookcases are the same size and each requires 3/2 gallon of paint. How many bookcases will he be able to paint?

Answers

Answer:

8 bookcases

Step-by-step explanation:

12 / (3/2) = (12*2)/3 = 8

the number of people who own computers has increased by 23.2 percent annually since 1990. if half a million people owned a computer in 1990, predict how many people will own a coumpter in 2015

Answers

There would be at least 92 million people who will own a computer in 2015.

I don't know how to make a function out of a percentage, that is why I used excel to compute annual population.

YEAR     POPULATION   Increase of previous years population
1990         500,000 
1991         616,000         1.232
1992         758,912         1.232
1993         934,978         1.232
1994     1,151,894         1.232
1995     1,419,134         1.232
1996     1,748,373         1.232
1997     2,153,996         1.232
1998     2,653,723         1.232
1999     3,269,387         1.232
2000     4,027,885         1.232
2001     4,962,354         1.232
2002     6,113,620         1.232
2003     7,531,980         1.232
2004     9,279,400         1.232
2005   11,432,221         1.232
2006   14,084,496         1.232
2007   17,352,099         1.232
2008   21,377,786         1.232
2009   26,337,433         1.232
2010   32,447,718         1.232
2011   39,975,588         1.232
2012   49,249,925         1.232
2013   60,675,908         1.232
2014   74,752,718         1.232
2015   92,095,349        

Solve the system of equations using cramer's rule -x+y-3z=-4 3x-2y+8z=14 2x-2y+5z=7

Answers

System of Equations
-1x + 1y - 3z = -4 \n3x - 2y + 8z = 14 \n2x - 2y + 5z = 7

Coefficient Matrix's Determinant
D = \left[\begin{array}{ccc}-1&1&-3\n3&-2&8\n2&-2&5\end{array}\right]

Answer Column
\left[\begin{array}{ccc}-4\n14\n7\end{array}\right]

Dx: Coefficient Determinant with Answer-Column values in X-Column
D_(x) = \left[\begin{array}{ccc}-4&1&-3\n14&-2&8\n7&-2&5\end{array}\right]

Dy: Coefficient Determinant with Answer-Column Values in Y-Column
D_(y) = \left[\begin{array}{ccc}-1&-4&-3\n3&14&8\n2&7&5\end{array}\right]

Dz: Coefficient Determinant with Answer-Column Values in Z-Column
D_(z) = \left[\begin{array}{ccc}-1&1&-4\n3&-2&14\n2&-2&7\end{array}\right]

Evaluating each Determinant
D= \left[\begin{array}{ccc}-1&1&-3\n3&-2&8\n2&-2&5\end{array}\right] \nD = (-1 * (-2) * 5) + (1 * 8 * 2) + (-3 * 3 * (-2)) - (2 * (-2) * (-3)) - (-2 * 8 * (-1)) - (5 * 3 * 1) \nD = (10) + (16) + (18) - (12) - (16) - (15) \nD = 10 + 16 + 18 - 12 - 16 - 15 \nD = 26 + 18 - 12 - 16 - 15 \nD = 44 - 12 - 16 - 15 \nD = 32 - 16 - 15 \nD = 16 - 15 \nD = 1

D_(x) = \left[\begin{array}{ccc}-4&1&-3\n14&-2&8\n7&-2&5\end{array}\right] \nD_(x) = (-4 * (-2) * 5) + (1 * 8 * 7) + (-3 * 14 * (-2)) - (7 * (-2) * (-3)) - (-2 * 8 * (-4)) - (5 * 14 * 1)) \nD_(x) = (40) + (56) + (84) - (42) - (64) - (70) \nD_(x) = 40 + 56 + 84 - 42 - 64 - 70 \nD_(x) = 96 + 84 - 42 - 64 - 70 \nD_(x) = 180 - 42 - 64 - 70 \nD_(x) = 138 - 64 - 70 \nD_(x) = 74 - 70 \nD_(x) = 4

D_(y) = \left[\begin{array}{ccc}-1&-4&-3\n3&14&8\n2&7&5\end{array}\right] \nD_(y) = (-1 * 14 * 5) + (-4 * 8 * 2) + (-3 * 3 * 7) - (2 * 14 * (-3)) - (7 * 8 * (-1)) * (5 * 3 * (-4)) \nD_(y) = (-70)+ (-64) + (-63) - (-84) - (-56) - (-60) \nD_(y) = -70 - 64 - 63 + 84 + 56 + 60 \nD_(y) = -134 - 63 + 84 + 56 + 60 \nD_(y) = -197 + 84 + 56 + 60 \nD_(y) = -113 + 56 + 60 \nD_(y) = -57 + 60 \nD_(y) = 3

D_(z) = \left[\begin{array}{ccc}-1&1&-4\n3&-2&14\n2&-2&7\end{array}\right] \nD_(z) = (-1 * (-2) * 7) + (1 * 14 * 2) + (-4 * 3 * (-2)) - (2 * (-2) * (-4)) - (-2 * 14 * (-1)) - (7 * 3 * 1) \nD_(z) = (14) + (28) + (24) - (16) - (28) - (21) \nD_(z) = 14 + 28 + 24 - 16 - 28 - 24 \nD_(z) = 42 + 24 - 16 - 28 - 21 \nD_(z) = 66 - 16 - 28 - 21 \nD_(z) = 50 - 28 - 21 \nD_(z) = 22 - 21 \nD_(z) = 1

x = (D_(x))/(D) = (4)/(1) = 4 \ny = (D_(y))/(D) = (3)/(1) = 3 \nz = (D_(x))/(D) = (1)/(1) = 1 \n(x, y, z) = (4, 3, 1)

Find the distance between points P(1, 2) and Q(4, 8) to the nearest tenth. 11.2 5 6.7 9

Answers

distance = sqrt (x2 - x1)^2 + (y2 - y1)^2
(1,2)...x1 and y1
(4,8)...x2 and y2
now we sub
distance = sqrt (4 - 1)^2 + (8 - 2)^2
distance = sqrt (3^2 + 6^2)
distance = sqrt (9 + 36)
distance = sqrt (45)
distance = 6.7 <===
PQ=√(x2-x1)²+(y2-y1)²
     =√(4-1)²+(8-2)²
     =√(3)²+(6)²
     =√9+36
     =√45
     =3√5
     =3*2.236
     =6.708
     =6.79.
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