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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!

Let B = (0,0)

Let C = (32,0)

Let A  be the apex

The line through AB has the equation

y = (3/5)x

The line through CA has the equation

y = (-1/4) (x - 32)  =  (-1/4)x + (-1/4)

To find the x coordiante of A, set the lines equal

(3/5)x = (-1/4)x + 8

(3/5 + 1/4)x  = 8

(17/20) x = 8

x = 160 / 17

y = (3/5) (160/17)  =  480/85   = 96/17 =  the height of the triangle

[ABC ]  =  (1/2) (32) (96/17)  = 1536/17 ≈  90.35

xz = x + z                    yz = 2 ( y + z)

xz - x = z                     yz = 2y + 2z

x ( z  -1) = z                yz - 2y  = 2z

x = z / ( z -1)                  y ( z - 2)   =2z

                                    y = (2z) / (z -2)

xy = x + y

z/ (z-1)* (2z) ( z -2)  =   z/(z -1) + (2z) (z -2)

[2z^2] / [ (z-1)(z -2)] =   [  z (z -2) +  2z ( z -1)] / [ (z -1) ( z -2) ]

2z^2  = z^2 - 2z + 2z^2 - 2z

2z^2  = 3z^2 - 4z

z^2 - 4z  =  0

z ( z - 4)  = 0 

Reject z  = 0

z =  4

              A

       3           5

B           D            C 

              7

Law of Cosines

7^2 = 3^2 + 5^2  - 2(3*5)cos ( BAC)

[7^2 - 3^2 - 5^2 ] / [ -2 * 3 *5] =  cos (BAC)  =  -1/2

arccos (-1/2)  = BAC =  120°

So  cos (BAD)  =   cos (120 / 2) =  cos 60 =  1/2

We can use simple conversions and calculations to solve this problem. 

We first find the total area needed, then divide it by 400, before doing conversions. 

Thus, we have

\( [ 15 * 60 *1 / 400] = 9/4\)

Converting to meters, we have

\(\frac{9/4}{(0.3048^2)} \approx 25\)

so our answer is just 25. 

Thanks! :)

First, let's find the distance Will travels before Grace even starts rowing. 

Since Will has a 45 minute head start, we have the equation

\((45 min)(50 m/min) = 2250 m \)

This means that Will and Grace must cover

\((2800 m) – (2250 m) = 550 m \) together. 

Let's say the two row t minutes. Since we add their speeds together, we have

\( (50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875 \)

This rounds to about 6 minutes and 52 seconds. 

In total, the two took 51 minutes and 32 seconds to meet each other . So, the two will meet at \(2:51:52 pm \)

So 2:51:52 is our answer. 

Thanks! :)

Let's first note that the quadratic with roots 1 and -1 can be written in the form of

\((x-1)(x+1)=0\)

This is in the form of what we call factored form. Now, expanding this, we have

\(x(x+1) - 1(x+1)\)

Distributing the values in, we have

\(x^2+x-x-1\)

\(x^2-1\)

Thus, we have that the first blank is 0 and the second blank is -1. 

Thanks! :)

First, let's combine all our like terms and move all the terms to oneside. We get

\(6x^{2}-21x-31=0\)

Now, we use the quadratic equation. We get that

\(x=\frac{-({-21})\pm \sqrt{{-21})^{2}-4\cdot {6}({-31})}}{2\cdot {6}}\)

Simplifying furher. We get 2 values for x We get

\(x=\frac{\sqrt{1185}+21}{12}\\ x=\frac{-\sqrt{1185}+21}{12}\)

Clearly, the first value of x is greater in this case, so we have

\(x=\frac{\sqrt{1185}+21}{12}\) as our final anwer. 

Thanks! :)

(a  + 2) / ( a + 1)  = b /3     cross-multiply

3(a + 2) = b (a + 1)

3a + 6  =  ab + b

3a - ab - 1b  + 6  = 0       multiply   the coefficients on a,b = (3)(-1) = -3  and add to both sides

3a - ab -1b + 6  - 3 = -3       

3a - ab - 1b + 3   =  -3

(-a -1)(b -3)  =  -3

- (a + 1) ( b -3)  = -3

(a + 1) ( b -3) =  3

a     b

0     6

2     4

-2    0

-4    2

Oh! I think you have a typo there :). It should be "plsplspls help me cheat on my homework." Good luck!

What an obvious homework cheater.

But don't worry, this forum panders to dishonest and lazy students, so your questions will be answered very soon!

1.   Some possibilites are

        98 , 89

        97,  79

        96,  69....etc.

2. 

D = Dad's age now     G = Grogg's age now

 D - 7  = 9( G - 7)   →   D - 7  = 9G - 63  →  9G - D  = 56     (1)

      D - 4 = 5(G - 4)  →   D - 4 = 5G -20  → 5G - D  =  16    (2)

Multiply (1) by 5   and (2)  by -9

45G - 5D = 280

-45G + 9D = -144      add these

4D  = 136

D = 136 / 4  =   34

(n + 2) / n   *  (n + 3)/ ( n + 1) * ( n + 4) / (n + 2)  = 10     simplify

1/n * (n + 3) / (n + 1) * (n + 4) =  10

(n + 3) (n + 4)  =  10 ( n)(n + 1)

n^2 + 7n + 12 = 10n^2 + 10n

9n^2 + 3n - 12 = 0 

3n^2 + n - 4  =  0

(3n + 4) (n -1 )  = 0

n = 1      n  =  -4/3

Only n = 1  is good

Value of  first fraction  =  3 / 1  =  3

The common ratio is  .036 / .4 =  9/100

The sum is

.4 / (1 - 9/100)  =  (4/10) / ( 91/100)  =  40 /  91

Slope =  [ 17 - 8 ] / [ -4 -5 ] = 9/-9  = -1

Equation of line using (5,8)

y = -1 (x -5) + 8

y = -x + 13

x + y  = 13      divide through by 13

x  / 13  + y /13  = 1

a = 13

Let  d  be the distance between  the center of the circle and UV

We can  form a right triangle  with legs of d , (UV/2)   and a hypotenuse of r

So  we have

d^2  + (UV/2)^2  = r^3

d^2  + 32^2  = r^2      (1)

Likewise  let  the distance between the center of the circle and  YZ = d + 2 + 2 = d + 4

And similar to  the first case we have

(d + 4)^2 + (YZ / 2)^2  = r^2

(d + 4)^2 + 24^2 = r^2    (2)

Set (1)  = (2)

d^2 + 32^2  = (d + 4)^2 + 24^2       simplify

d^2 + 1024 = d^2 + 8d + 16 + 576

1024 - 16 - 576  =  8d

432  = 8d

d =432 / 8   =  54

And

d^2 + 32^2  = r^2

54^2 + 32^2 = r^2

3940  = r^2

So  we have another right triangle  such that the distance from the center of the circle to WX = d + 2

So

(d + 2)^2  + ( WX / 2)^2  = r^2

(54 + 2)^2  +  WX^2 / 4  = 3940

56^2  + WX^2 / 4  =3940

3136 + WX^2 /  4  = 3940

WX^2 =  4 ( 3940 - 3136)

WX^2  = 4 * 804

WX^2  = 3216

WX = sqrt (3216)   ≈  56.7

This makes sense.....WX should  be > 48  but < 64

Simplify as

x^2 + nx + 316 < 0

Set as an equality 

We will have real solutions whenever the discriminant is  ≥ 0

So

n^2 - 4(1)(316) ≥ 0

n^2 - 1264 ≥ 0

n^2 ≥ 1264             take both roots

n ≥ sqrt (1264)  → ≈ 35.6   or    n  ≤  -sqrt (1264) → ≈  -35.6

Thus....we have real solutions when  n  = ( inf , ≈ -35.6] U [ ≈ 35.6 , inf )

So  the integers for n that  make this  have NO real solutions  in x are the integers between these two values

n = [ -35 , 35 ]  =    2(35) + 1 =   71  integers

55 *35 * 30  = 57750 cm^3

62 L  =  62000 cm^3

So

62000 - 57750  = 4250 cm^3  overflows  =  4.25 L  =  4L +  250 ml

The product of a set of distinct positive integers is $630$.   

If one of the numbers is 42, what are the other two numbers?    

630 divided by 42 is 15.  15 has only two factors.   

Therefore the three integers are 42 and 5 and 3.   

edited to add:  15 has only two factors other than 1 and itself.   

_.

WX= 6sqrt6 is impossible because WX is bigger than YZ and YZ = 22