1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Fittoniya [83]
2 years ago
9

I need to find and write the equation for the problem.

Mathematics
2 answers:
Julli [10]2 years ago
7 0
0.8p - 50 < = 150
0.8p < = 150 + 50
0.8p < = 200
p < = 200/0.8
p < = 250

the reason I set it up this way is because when it is 20% off, u r actually paying 80% of the original price (p)....80% of the original price is written as 0.8p...then u subtract ur 50 dollar discount coupon...- 50.....and if all she can spend is 150....it would be less then or equal to 150. So the most she can spend on the phone is 250

Firdavs [7]2 years ago
7 0
(p-20/100p)-50 =80/100p-50 =4/5p-50
You might be interested in
Solve the following compound inequality 4
CaHeK987 [17]

Answer:

0 < x < 8

Step-by-step explanation:

Given

4 < x + 4 < 12

Required

Solve

4 < x + 4 < 12

Split:

4 < x + 4  and  x + 4< 12

Collect Like Terms

4 - 4 < x and x < 12- 4

0 and x < 8

Combine

0 < x < 8

7 0
3 years ago
The difference of 9 times a number and 2 is 67. Write an equation that can be used to find the unknown number?
pochemuha
9x - 2 = 67
67 + 2 = 69
69 × 9 = 621
621 = x
3 0
3 years ago
A pair of basketball shoes was originally priced at $80, but was marked up 47.5%. What was the retail
Lelu [443]
47.5% in decimal form is 0.475.

Since the price was increased, we add 1 to 0.475 to get 1.475.

80 * 1.475 = 118

Another way to solve this is to multiply 80 by 0.475, which is 38, then add it to 80.

The retail price of the shoes is $118.
6 0
3 years ago
Eighty degrees is Drawn how
alexdok [17]
It is a little less than 90 degrees, which is a right angle, so it will be more acute
8 0
3 years ago
Read 2 more answers
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
Other questions:
  • What number has 2 more tens than 40 and the same number of ones as 18
    8·1 answer
  • Question 8.am I correct I got B
    12·1 answer
  • To copy an angle using only a compass and a straightedge, begin by marking the vertex of the new angle. Then draw a ray from the
    5·2 answers
  • a bowl contains 80 M&amp;M candies If 15 are orange what fraction of the candies is NOT orange. Answer reduced to lowest terms
    5·2 answers
  • The diameter of a circle is two and 1/2 inches what is the length of the radius of the circle
    12·1 answer
  • (1) Cos AcosecA=cot A​
    13·1 answer
  • Maranda had 4 lbs. 6 oz. of clay she divided it into 10 equal parts how heavy was each part
    10·2 answers
  • What is the slope of 2x-4y=5
    9·2 answers
  • 20 PTS PLEASE HELP ASAP!
    15·1 answer
  • Write the coordinates of the vertices after a translation 4 units left.
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!