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Crazy boy [7]
3 years ago
8

What is the shape of a cross section that is parallel to the base of a cube?

Mathematics
1 answer:
babymother [125]3 years ago
7 0
Im pretty sure it’s a hexagon
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Find f (x )if f apostrophe apostrophe (x )equals x to the power of negative 2 end exponent comma space x greater than 0 comma sp
viva [34]

Step-by-step explanation:

Given g'(x) = g^{-2}, x> 0\ g(1) = 0 \ and \ g(2) = 0

To get g(x), we will have to integrate g'(x)

g(x)  = \int\limits {x^{-2}} \, dx\\ g(x)  = \frac{x^{-2+1}}{-2+1} + C\\ g(x) = -x^{-1} + C\\g(x) = -x^{-1} + C\\

If g(1) = 0, this means at x = 1, g(x) = 0

0 = -1⁻¹ + C

C= 1

Substitute C = 1 into the function

g(x) = -x⁻¹ + 1

If g(2) = 0, this means at x = 2, g(x) = 0

0 = -2⁻¹ + C

C= 2⁻¹

C =  1/2

Substitute C = 2 into the function

g(x) = -x⁻¹ + 1/2

3 0
3 years ago
The number of loaves of bread purchased and the total cost of the bread in dollars can be modeled by the equation c = 3. 5b. Whi
Allisa [31]

You can use the fact that number of breads purchased cannot be negative since a customer either buys them or not and usually do not sell to the shopkeeper.(if somehow they end up selling to shop owner, then yes that will go in negative, but we'll assume it is wrong in most cases as generally shop owners are there to sell stuffs).

The third table of values matches the equation and includes only viable solutions.

<h3>What is a viable solution here?</h3>

It is talking about those solutions which are seen in real world. As stated above, a customer either buys the bread or not, thus number of breads sold will be either positive or 0(in case of no selling). Thus, we cannot have number of breads as negative.

Such solutions which are correct in the real world context here are called here as viable solutions.

<h3>Checking one by one all the tables for them being matched with table and viability</h3>

For first table, the number of breads are in negative, thus it is not going to have viable solution.

For second table, we have:

b = 0 thus c = 3.5b = 3.5 times 0 = 0 which is correctly given in second column.

b = 0.5, thus c = 3.5b = 3.5 times 0.5 =1.75 which is correctly given.

b = 1, thus c= 3.5 times 1 = 3.5 which is correctly given

b = 2001.5 thus c = 3.5 times 2001.5  = 7005.25 which is not correctly given, thus wrong.

For third table, we have:

b = 0, thus c = 3.5 \times 0 = 0, correctly given in second column.

b = 3, thus c = 3.5 \times 3 = 10.5, correctly given.

b = 6, thus c = 3.5 \times 6 = 21, correctly given.

b = 9, thus c = 3.5 \times 9 = 31.5, correctly given.

Thus, the third table of values matches the equation and includes only viable solutions.

Learn more about purchasing to cost relation here:
brainly.com/question/13727919

8 0
3 years ago
in the diagram below, line segment AB is drawn. Point A has been mapped to image point A' by the use of a translation.​
maksim [4K]
Sawyer you please please call me eueuew
7 0
3 years ago
What is the relative maximum value of the function f(x) = –x3 + 6x2 – 9x – 1?
Vlad [161]

Answer:

-1.

Step-by-step explanation:

First find the derivative of f(x):

f'(x) = -3x^2 + 12x - 9 = 0   for a maximum or minimum.

-3(x^2 - 4x + 3) = 0

(x - 1)(x - 3) = 0

x = 1, 3.

To find which gives a relative maximum we find the second derivative:

f"(x) = -6x + 12

When x = 1  f"(x) = 6,    positive.

when x = 3, f"(x) = -6,  negative.

So x = 3 gives a maximum value of f(x).

f(3) = -(3)^3  + 6*(3)^2  - 9(3) - 1 = -1 (answer).

7 0
3 years ago
Anyone know how to do graphs
PSYCHO15rus [73]

Answer:

yeah sure I know.........

8 0
3 years ago
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