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Vladimir [108]
2 years ago
9

Presents of equal size are in a boxes each with dimensions of 10 inches length, 4 inches width, and 2 inches height. How many pr

esents can be gift wrapped if you have a roll of 600 square inches of wrapping paper? Show all work
Mathematics
1 answer:
Maurinko [17]2 years ago
4 0

Answer:

You can make 7.5 gifts presents (or just 7)

Step-by-step explanation:

First I calculated the area of 1 box, which is 80 inches squared. Then I just divided the 600 square inches with the 80 inches. 600/80= 7.5. Therefore, you can make 7 presents (since you cannot wrap half of a present (0.5))

Hope this helps !

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Please help me :( I don’t understand at all :(
DENIUS [597]

Answer:

1/16

Step-by-step explanation:

f(x) = 2^x

Let x = -4

f(-4) = 2 ^ -4

We know that a^-b = 1/a^b

      = 1/2^4

     = 1/16

8 0
3 years ago
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<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20x%20%7B%7D%5E%7B2%7D%20%20-%205" id="TexFormula1" title="f(x) = x {}^{2} - 5
faust18 [17]

Answer:

The vertex is (0, -5) and the axis of symmetry is x = 0.

Step-by-step explanation:

In order to find the axis of symmetry, we look for the x value of the vertex. We can find this using the equation -b/2a, in which a is equal to the coefficient of the x^2 term and b is the coefficient of the x term.

x = -b/2a

x = 0/2(1)

x = 0/2

x = 0

So the axis of symmetry is 0 as is the x value of the vertex. Now we can find the y value by plugging in the x value into the equation.

f(x) = x^2 - 5

f(x) = 0^2 - 5

f(x) = 0 -5

f(x) = -5

This gives us a vertex of (0,-5)

7 0
3 years ago
Which is an equation of the line that passes through (–1, –5) and (–3, –7)
mestny [16]

Answer: y=x-4

Step-by-step explanation: The y value is exactly x minus four.

4 0
3 years ago
Paul is making loaves of raisin bread to sell at a fundraising event. The recipe calls for 1\3 cup of raisins for each loaf, and
azamat
39 divided by 4 equals 9 batches with 3/12 raisins left over
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3 years ago
Can someone please help me with this question?!? I am so confused and I don't know how to answer it.
lbvjy [14]

9514 1404 393

Answer:

  a. f(0) = 1

  b. DNE (does not exist)

  c. DNE

  d. lim = 3

Step-by-step explanation:

The function exists at a point if it is defined there. The function is defined anywhere on the solid line and at solid dots. It is not defined at open circles. So, the function is defined everywhere except (2, 3), which has an open circle.

The open circle at (0, 4) prevents the function from being doubly-defined at x=0, since it is already defined to be 1 at x=0.

This discussion tells you ...

  f(0) = 1

 f(2) does not exist. There is a "hole" in the function definition there.

__

The function has a limit at a point if approaching from the left and approaching from the right have you approaching that same point.

Consider the point (1, 2). The graph is a solid line through that point. Approaching from values less than x=1, we get to the same point (1, 2) as when we approach from values greater than x=1.

Similarly, consider the point (2, 3). Approaching from values of x less than 2, we get to the same point (2, 3) as when we approach from x-values greater than 2. The limit at x=2 is 3. The only difference from the previous case is that the function is not actually defined to be that value there.

__

Now consider what happens at x=0. When we approach from the left, we approach the point (0, 4). When we approach from the right, we approach the point (0, 1). These are different points. Because they are different coming from the left and from the right, we say "the limit as x→0 does not exist."

__

In summary, ...

  a) f(0) = 1

  b) lim x → 0 does not exist

  c) f(2) does not exist

  d) lim x → 2 = 3

_____

<em>Additional comment</em>

The significance of the function not being defined at a point where the limit exists, (2, 3), is that <em>the function is not continuous there</em>. This kind of discontinuity is called "removable", because we could make the function continuous at x=2 by defining f(2) = 3 (that is, "filling the hole").

6 0
2 years ago
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