MATHEMATICS COLLEGE

During a game of cards, Megan divided the deck of cards into 4 equal groups. She then placed (3 ofher cards in the center of the table. If Megan now has
5 cards in her hand, how many cards were in the
entire deck?

Answers

Answer 1

Answer:

There were originally 3 cards in the entire deck.

Let's denote the number of cards in the entire deck as $D$.

Megan divided the deck into 4 equal groups, so each group has $(D)/(4)$ cards.

She placed 3 cards in the center, and now she has 5 cards in her hand.

So, the equation representing this situation is:

$(D)/(4) - 3 + 5 = D$

Now, let's solve for $D$:

$(D)/(4) + 2 = D$

Multiply both sides by 4 to get rid of the fraction:

$D + 8 = 4D$

Subtract $D$ from both sides:

$8 = 3D$

Divide both sides by 3:

$D = (8)/(3)$

However, the number of cards in a deck must be a whole number, so we round up:

$D = 3$

Therefore, there were originally 3 cards in the entire deck.

Answer 2

Answer:

Answer: 32

Step-by-step explanation: If she had 8 in her had and placed 3 cards down she would have 5 and if there were 8 cards in each group then, 8x4=32

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Which of the following is false regarding number sets

Of (x - 12)(3x + 4)?

Answers

Answer:

3x^2 - 32x - 48

Step-by-step explanation:

3x^2-36x+4x-48

3x^2 - 32x -48

Rewrite the expression without using a negative exponent.12x^−4
Simplify your answer as much as possible.

Answers

12x^-4 =
12 * x^-4 =
12 * 1/(x^4) =
12 / (x^4)

12x^-4
To make the exponent positive you have to bring the term x^-4 down so your answer would be= 12/x^4

What is the solution to –4|–2x + 6| = –24?x = 0x = 0 or x = –6x = 0 or x = 6no solution

Answers

The correct answers are the x = 6 and x = 0.

In order to find this, we can follow the order of operations.

–4|–2x + 6| = –24

Divide by -4

|–2x + 6| = 6

Since it is now equal to a absolute value, we have to solve for the positive and negative versions of the final answer. Let's start with the positive.

-2x + 6 = 6

-2x = 0

x = 0

Now for the negative

-2x - 6 = -6

-2x = -12

x = 6

These would be your two answers.

Answer:

Step-by-step explanation:

A package of 6 pairs of insulated socks costs $34.14. What is the unit price of the pairs of socks?

Answers

Answer:

$\boxed {\tt \$5.69 \ per \ pair}$

Step-by-step explanation:

We want to find the unit price for the insulated socks, or the price per pair of insulated socks. To find the unit price, we must divide the cost by the pairs of socks.

$unit \ price =(cost)/(pairs)$

We know that it costs $34.14 for 6 pairs of socks.

$unit \ price =(\$34.14)/(6 \ pairs)$

Divide.

$unit \ price = \$5.69 / pair$

The unit price is $5.69 per pair of insulated socks.

Answer:

$5.69

Step-by-step explanation:

34.14/6

=5.69

The unit price of the pairs of socks are $5.69.

Please help thanks!!

Answers

Answer:

y = -2x + 4

Explanation:

Pick two points from the line: (1, 2), (0, 4)

Slope intercept form: y = mx + b

Find 'm': 2 - 4 = -2

Find 'b': 4

Insert numbers into equation: y = -2x + 4

Answer:

y= -2x + 4

Step-by-step explanation:

Find the radius and height of a cylindrical soda can with a volume of 256cm^3 that minimize the surface area.B: Compare your answer in part A to a real soda can, which has a volume of 256cm^3, a radius of 2.8 cm, and a height of 10.7 cm, to conclude that real soda cans do not seem to have an optimal design. Then use the fact that real soda cans have a double thickness in their top and bottom surfaces to find the radius and height that minimizes the surface area of a real can (the surface area of the top and bottom are now twice their values in part A.

B: New radius=?

New height=?

Answers

Answer:

A) Radius: 3.44 cm.

Height: 6.88 cm.

B) Radius: 2.73 cm.

Height: 10.92 cm.

Step-by-step explanation:

We have to solve a optimization problem with constraints. The surface area has to be minimized, restrained to a fixed volumen.

a) We can express the volume of the soda can as:

$V=\pi r^2h=256$

This is the constraint.

The function we want to minimize is the surface, and it can be expressed as:

$S=2\pi rh+2\pi r^2$

To solve this, we can express h in function of r:

$V=\pi r^2h=256\n\nh=(256)/(\pi r^2)$

And replace it in the surface equation

$S=2\pi rh+2\pi r^2=2\pi r((256)/(\pi r^2))+2\pi r^2=(512)/(r) +2\pi r^2$

To optimize the function, we derive and equal to zero

$(dS)/(dr)=512*(-1)*r^(-2)+4\pi r=0\n\n(-512)/(r^2)+4\pi r=0\n\nr^3=(512)/(4\pi) \n\nr=\sqrt[3]{(512)/(4\pi) } =\sqrt[3]{40.74 }=3.44$

The radius that minimizes the surface is r=3.44 cm.

The height is then

$h=(256)/(\pi r^2)=(256)/(\pi (3.44)^2)=6.88$

The height that minimizes the surface is h=6.88 cm.

b) The new equation for the real surface is:

$S=2\pi rh+2*(2\pi r^2)=2\pi rh+4\pi r^2$

We derive and equal to zero

$(dS)/(dr)=512*(-1)*r^(-2)+8\pi r=0\n\n(-512)/(r^2)+8\pi r=0\n\nr^3=(512)/(8\pi) \n\nr=\sqrt[3]{(512)/(8\pi)}=\sqrt[3]{20.37}=2.73$

The radius that minimizes the real surface is r=2.73 cm.

The height is then

$h=(256)/(\pi r^2)=(256)/(\pi (2.73)^2)=10.92$

The height that minimizes the real surface is h=10.92 cm.

Final answer:

The minimal surface area for a cylindrical can of 256cm^3 is achieved with radius 3.03 cm and height 8.9 cm under uniform thickness, and radius 3.383 cm and height 7.14 cm with double thickness at top and bottom. Real cans deviate slightly from these dimensions possibly due to practicality.

Explanation:

For a cylinder with given volume, the surface area A, radius r, and height h are related by the formula A = 2πrh + 2πr^2 (if the thickness is uniform) or A = 3πrh + 2πr^2 (if the top and bottom are double thickness). By taking the derivative of A w.r.t r and setting it to zero, we can find the optimal values that minimize A.

For a volume of 256 cm^3, this gives us r = 3.03 cm and h = 8.9 cm with uniform thickness, and r = 3.383 cm and h = 7.14 cm with double thickness at the top and bottom. Comparing these optimal dimensions to a real soda can (r = 2.8 cm, h = 10.7 cm), we see that the real can has similar but not exactly optimal dimensions. This may be due to practical considerations like stability and ease of holding the can.