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sdas [7]
3 years ago
12

Which of the following is an equation of a line perpendicular to the equation y=-2/3x-4

Mathematics
1 answer:
Bond [772]3 years ago
3 0

Answer:

B

Step-by-step explanation:

Perpendicular lines have related slopes. The slopes are negative reciprocals of each other. The slope of y=-2/3-4 is -2/3. So the perpendicular line will have slope 3/2. Y=3/2+1 is the solution.

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Tamiku [17]
Sure you giving points?
5 0
2 years ago
82, 62, 95, 81, 89, 51, 72, 56, 97, 98, 79, 85 order from least to greatest find the The median of the lower half of the data an
Studentka2010 [4]

Answer:

The median of the lower half of the data was determined to be 67 while the median of the upper half of the data was determined to be 92.

Step-by-step explanation:

<u>Step 1:  Order from least to greatest</u>

51, 56, 62, 72, 79, 81, 82, 85, 89, 95, 97, 98

<u>Step 2:  Determine the median of the lower half of the data</u>

There are 12 numbers which means that there is going to be 6 numbers on both sides of the data.  So using the first 6 numbers we will be able to determine the median of the lower half of the data.

51, 56, 62, 72, 79, 81

Since there is an even amount of numbers, that means that we will need to find the middle between the 3rd and 4th number.  We can do that by adding them up together and dividing by 2.

(62 + 72) / 2 = 67

Step 3:  <u>Determine the median of the upper half of the data</u>

So using the first 6 numbers we will be able to determine the median of the upper half of the data.

82, 85, 89, 95, 97, 98

Since there is an even amount of numbers, that means that we will need to find the middle between the 3rd and 4th number.  We can do that by adding them up together and dividing by 2.

(89 + 95) / 2 = 92

Answer: The median of the lower half of the data was determined to be 67 while the median of the upper half of the data was determined to be 92.

5 0
2 years ago
Number 34 please guys!!!!!!
GenaCL600 [577]
The answer is obviously three.

3 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
A metal bar has the D of 7.9 g/mL. If the volume of the bar is 500mL, what is the mass in kg?​
morpeh [17]

Answer:

It will be 0.5kg cuz kg and liters are same

8 0
3 years ago
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