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liubo4ka [24]
3 years ago
8

(15 points) make a table and graph the function y = x + 3

Mathematics
2 answers:
DaniilM [7]3 years ago
8 0
Its the second to last picture where
-1 = 2
0 = 3
1 = 4
Serhud [2]3 years ago
3 0
Third one is your answer
x   -2  -1   0   1    2
y   1    2    3   4   5
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Give two different meanings for each of the phrases below. Use a dictionary to look up the meanings of individual words in each
Nata [24]
A cold wave - a series of cold days in terms of weather; a physical wave that is given by someone with a "cold" demeanor.

A fair day- a day with nice, mild weather; making a modest amount of money in a days work.

hope those help.
8 0
3 years ago
Write an expression with a negative exponent that has a value between 0 and 1:2
Naddika [18.5K]
Edit to reflect comments:

3^-1=1/3
4^-1=1/4

Flip the fraction over basically i.e. 4/1---->1/4
7 0
3 years ago
Let g(x)=5x-1 and h(x)=X^2-1<br> Solve:<br> g(h(x))=74
AURORKA [14]

Answer:

Let's replace the h(x) function in g(x) and then use 74 as a result on the axis y. The correct answer is 4 .

4 0
3 years ago
Prove that it is impossible to dissect a cube into finitely many cubes, no two of which are the same size.
solniwko [45]

explanation:

The sides of a cube are squares, and they are covered by the respective sides of the cubes covering that side of the big cube. If we can show that a sqaure cannot be descomposed in squares of different sides, then we are done.

We cover the bottom side of that square with the bottom side of smaller squares. Above each square there is at least one square. Those squares have different heights, and they can have more or less (but not equal) height than the square they have below.

There is one square, lets call it A, that has minimum height among the squares that cover the bottom line, a bigger sqaure cannot fit above A because it would overlap with A's neighbours, so the selected square, lets call it B, should have less height than A itself.

There should be a 'hole' between B and at least one of A's neighbours, this hole is a rectangle with height equal to B's height. Since we cant use squares of similar sizes, we need at least 2 squares covering the 'hole', or a big sqaure that will form another hole above B, making this problem inifnite. If we use 2 or more squares, those sqaures height's combined should be at least equal than the height of B. Lets call C the small square that is next to B and above A in the 'hole'. C has even less height than B (otherwise, C would form the 'hole' above B as we described before). There are 2 possibilities:

  • C has similar size than the difference between A and B
  • C has smaller size than the difference between A and B

If the second case would be true, next to C and above A there should be another 'hole', making this problem infinite. Assuming the first case is true, then C would fit perfectly above A and between B and A's neighborhood.  Leaving a small rectangle above it that was part of the original hole.

That small rectangle has base length similar than the sides of C, so it cant be covered by a single square. The small sqaure you would use to cover that rectangle that is above to C and next to B, lets call it D, would leave another 'hole' above C and between D and A's neighborhood.

As you can see, this problem recursively forces you to use smaller and smaller squares, to a never end. You cant cover a sqaure with a finite number of squares and, as a result, you cant cover a cube with finite cubes.

3 0
3 years ago
Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
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