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GREYUIT [131]
3 years ago
13

How many solutions exist for the given equation?

Mathematics
1 answer:
Tcecarenko [31]3 years ago
5 0

Answer:

infinitely many

Step-by-step explanation:

Given

12x + 1 = 3(4x + 1) - 2 ← distribute parenthesis and simplify

12x + 1 = 12x + 3 - 2, that is

12x + 1 = 12x + 1

The expressions on both sides are equal.

This indicates that any value of x makes the equation true.

Thus there are infinitely many solutions

You might be interested in
Which line segment could be mid segment of abc? 21 points!!!!
Pavel [41]

Answer:

DF

Step-by-step explanation:

the mid segment is a line that connects the midpoints.

4 0
2 years ago
Read 2 more answers
Factorization of:
zubka84 [21]

Answer:

Step-by-step explanation:

2800: 2 * 2 * 2 * 2 * 5 * 5 * 7

75: 3 * 5 * 5

168: 2 * 2* 2 * 3 * 7

It's not really multiplication. It's more division.

Try 2800 as a sample. What you are trying to do is break this down into primes. The first prime is 2

2800/2 = 1400

1400 / 2 = 700

700 / 2 = 350

350 / 2 = 175. That's the end of what the 2s can do.

175 / 5 = 35

35/ 5 = 7 7 is a prime. You are done. Now run up the ladder.

2800: 2 * 2 * 2 * 2 * 5 * 5 * 7

75 is not an even number. It has no 2s. Go to 3

75 / 3 = 25.

25 / 5 = 5

That's the end

75: 3 * 5 * 5

Your calculator can be of great help. The rule is keep factoring until you get a decimal remainder. Move on to the next prime. Stop when the last division gives you a prime.

3 0
3 years ago
HELP ME pleaseeeee ill mark you as brainiest if I get two answers
sergiy2304 [10]

                                        Question 9

Given the segment XY with the endpoints X and Y

Given that the ray NM is the segment bisector XY

so

NM divides the segment XY into two equal parts

XM = MY

given

XM = 3x+1

MY = 8x-24

so substituting XM = 3x+1 and MY = 8x-24 in the equation

XM = MY

3x+1 = 8x-24

8x-3x = 1+24

5x = 25

divide both sides by 5

5x/5 = 25/5

x = 5

so the value of x = 5

As the length of the segment XY is:

Length of segment XY = XM + MY

                                = 3x+1 + 8x-24

                                = 11x - 23

substituting x = 5

                               = 11(5) - 23

                               = 55 - 23

                               = 32

Therefore,

The length of the segment = 32 units

                                        Question 10)

Given the segment XY with the endpoints X and Y

Given that the line n is the segment bisector XY

so

The line divides the segment XY into two equal parts at M

XM = MY

given

XM = 5x+8

MY = 9x+12

so substituting XM = 5x+8 and MY = 9x+12 in the equation

XM = MY

5x+8 = 9x+12

9x-5x = 8-12

4x = -4

divide both sides by 4

4x/4 = -4/4

x = -1

so the value of x = -1

As the length of the segment XY is:

Length of segment XY = XM + MY

                                = 5x+8 + 9x+12

                                = 14x + 20

substituting x = 1

                               = 14(-1) + 20

                               = -14+20

                               = 6

Therefore,

The length of the segment XY = 6 units

8 0
2 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
Can somebody please help me
Nataly [62]

Answer:

The probability of drawing the first pink ball is \frac{3}{10}.

Because you drew that first pink ball, there are two pink balls left and the total number of balls left is nine.

Hence, the probability of drawing another pink ball after drawing the first pink ball would be:

\frac{3}{10} *\frac{2}{9} =\frac{1}{15}

5 0
2 years ago
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