skibidibrain901

To find the portion of the class that did not study, we can set up an equation using the given information: Let x be the proportion of students who studied and (1 - x) be the proportion of students who did not study. The equation can be set up as follows: \(78x + (1 - x) 54 = 70\) . Solving this equation, we get:

1. \(78x + 54 - 54x = 70\)
2. \(24x + 54 = 70\)
3. \(24x = 16\)
4. \(x = 16 / 24\)
5. \(x = 2 / 3\)

Therefore, the portion of the class that did not study is \(1 - 2/3 = 1/3\).

Answer:

n = 1

Explanation:

    You want the largest positive integer n such that n³ divides 15.

Divisors

The positive integer divisors of 15 are ...

  1, 3, 5, 15

The only one of these that is a cube is 1 = 1³.

  n = 1

 Hope u found this helpful!

or you could just use Eulers thereom to get the same thing

Sorry, my first response was wrong because i read the question incorrectly. Here's the corrected version:

Let's break down the problem step by step: 

1. On a normal day, the mathematician solves t×p problems.

2. On the day he drinks coffee, he solves (t−3)×(4p+5) problems.

3. According to the problem, he solves twice as many problems as he would on a normal day: (t−3)×(4p+5)=2×(t×p)

4. Now, we can solve the equations to find the value of t and p:

 \((t−3)×(4p+5)=2×t×p\)  

\(4tp−12p+5t−15=2tp\)

\( 2tp−5t−12p+15=0 \)

5. The solution to this equation will give us the values of t and p, which we can then use to calculate how many problems the mathematician solves on the day he drinks coffee.

To find the total weight of the berries after two days, we can follow these steps: 1. Initially, the box of berries weighed 200 kg, and 99% of this weight was water. This means the weight of water in the berries was 0.99 * 200 kg = 198 kg. 2. Since the remaining 1% of the weight was the actual berry material, the weight of the berries themselves was 0.01 * 200 kg = 2 kg. 3. After two days in the sun, the water content decreased to 90%. This means the weight of water in the berries after two days was 0.9 * total weight of the berries. 4. The weight of the berries remained the same at 2 kg, so the weight of water after two days is 0.9 * total weight - 2 kg. 5. We know that the weight of water after two days was originally 198 kg. So, we can set up the equation: 0.9 * total weight - 2 kg = 198 kg. 6. Solving this equation will give us the total weight of the berries after two days. Let's solve the equation: 0.9×total weight−2=198 Adding 2 to both sides: 0.9 × total weight = 200 Dividing by 0.9: total weight = 0.9200​ ≈ 222.22 kg Therefore, the total weight of the berries after two days would be approximately 222.22 kg.

Let's assume that the mathematician works for x hours a day and can solve y problems per hour. Also, the mathematician drinks some coffee and discovers that he can now solve z problems per hour. So, the mathematician works for n hours that day. We are given that:x*y = number of problems solved in a dayz * n = number of problems solved on the day he drank coffee

Then, we can write the equations:x*y = n * 2*z (he still solves twice as many problems as he would in a normal day)andx = n (he only works for n hours that day)Now, we need to simplify these equations to solve for the number of problems solved on the day he drank coffee. Here is how to do it:\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)Since x, y, n, and z are all positive integers, we can say that the expression 2*n*z/x is also a positive integer. Therefore, we can write:\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)where k is a positive integer.

Finally, the number of problems solved on the day he drank coffee is:  y = 2k Therefore, the answer is that the mathematician solved 2k problems on the day he drank coffee.

Answer: 2000 problems

 Hope u found this helpful!

To find the area of trapezoid EFGH, we can use the fact that the area of a trapezoid is equal to the average of the lengths of the two bases multiplied by the height. Given that P is the midpoint of side EH, we can use the areas of triangles PEF and PGH to find the height of trapezoid EFGH. 1. Since P is the midpoint of EH, we know that the height of trapezoid EFGH is equal to the distance between EF and GH. 2. The area of triangle PEF is given as $18.Let′sdenotethebaseoftrianglePEFasx.Therefore,theheightoftrianglePEFisalsoh,wherehisthedistancebetween\overline{EF}and\overline{GH}.3.Usingtheformulafortheareaofatriangle,wehave\frac{1}{2} \times x \times h = 18.Thus,x \times h = 36.4.Similarly,theareaoftrianglePGHisgivenas36, and let's denote the base of triangle PGH as y. Therefore, y×h=72. 5. Now, the average of the bases of trapezoid EFGH is 21​(x+y). Multiplying this by the height h, we get the area of trapezoid EFGH as 21​(x+y)×h=21​(xh+yh)=21​(36+72)=54. Therefore, the area of trapezoid EFGH is $54$ square units.

Answer: 54 square units

(Sorry for the misformatted writing; i was on an ipad)