All Answers 358460 Answers

H==[1/16  +  1/18  + 1/20] / 2 ==0.0840277778

S ==4 * Sqrt[ H * (H - 1/16) * (H - 1/18) * (H - 1/20)]==0.00529539678

Area of the triangle==1 / 0.00529539678 ==188.84326 u^2

There are 9 students in total. In your answer, in case 1 there are 6 students, in case 2 7 students, and in case 3 there are 6 students.

If you had uploaded the picture of the rectangle, then I would give details of the solution. Since you haven't, this is all you will get:

The width of the recatngle=5

The length of the rectangle =18

Area of the rectangle=5  x  18 ==90 u^2

The areas of the 3 given triangles = 12 + 15 + 27 =54 u^2

Therefore, the area of triangle FBE = 90  -  54 =36 u^2

There are three cases to consider:

Case 1: Each group has 2 students. Then there are (29​)=36 ways to form the groups.

Case 2: One of the groups has 3 students and the other two groups have 2 students each. Then there are (39​)=84 ways to choose which student is in the group of 3, and then 3 ways to order the 3 students in that group.

Case 3: One of the groups has 4 students and the other two groups have 1 student each. Then there are (49​)=126 ways to choose which student is in the group of 4, and then 4 ways to order the 4 students in that group.

Summing the number of ways in each case, we get a total of 36+84+126=246​ ways to form the groups.

I figured it out, the answer is 471 for anyone who is wondering. The legs are x and y, so you get xy/2 = 3x + 3y. xy = 6x + 6y, so xy - 6x - 6y = 0. 

(x - 6)(y - 6) - 36 = 0, so (x - 6)(y - 6) = 36. Then, just calculate all possible values of x and y to find the sum of all areas as 471

Let $f(x) = x^3(ax^2+bx+c)$. Then $f(x)$ is divisible by $x^3$. We want to find $a$, $b$, and $c$ such that $f(x) + 3 = x^3(ax^2+bx+c)+3$ is divisible by $(x+1)^3$. Note that $(x+1)^3 = x^3 + 3x^2 + 3x + 1$. Thus, we want to find $a$, $b$, and $c$ such that:

f(x)+3=x3(ax2+bx+c)+3=(x3+3x2+3x+1)q(x)

for some polynomial $q(x)$. Equivalently, we want to find $a$, $b$, and $c$ such that:

ax5+bx4+cx3+3=x5q(x)+3x4q(x)+3x3q(x)+q(x)

for some polynomial $q(x)$. Comparing coefficients of like powers of $x$, we get the following system of equations:

\(\begin{align*} a &= q(x) \ b &= 3q(x) \ c &= 3q(x) \ 3 &= q(x) \end{align*}\)

Thus, $q(x) = 3$, and $a = 3$, $b = 9$, $c = 9$. Therefore, the polynomial $f(x) = \boxed{3x^5 + 9x^4 + 9x^3}$ satisfies the given conditions.

Cost of a pear: $5.30 * 2 - $8 = $10.60 - $8 = $2.60. 

Cost of an orange: $2.60 * 1/2 + $7 = $1.30 + $7 = $8.30. 

That should be all you need to solve the problem. 

conditions are first set, for example, the size of the field is 27 * 22 (x * y), n=3 step either up or to the right, d=4 diagonal, and so here is the winning strategy for the one who goes second

Is the Queen able to move any number of squares,     

as is the case on a chessboard?   

If so, your winning strategy would be to move first,  

then just move her all the way to the other corner.  

_.

A nonagon has angles that measure 140°, 140°, 130°, 140°, 140°, 145°, 155°, 140°, and a. What is a?    

You would calculate the sum of all the angles,   

then subtract the total of the angles you know.   

The sum of the internal angles of a polygon is  S = (n – 2)(180^o)    

where n is the number of sides of the polygon.  

                                                             S  =  (9 – 2)(180^o)   

                                                             S  =  7 • 180^o   

                                                             S  =  1260^o   

The total of the known angles is                    1130^o   

if neither of us made a mistake.  

You'd better check my addition.   

Subtract                                                 a  =  130^o     

_.

Find the coefficient of y^4 in the expansion of (2y-5+3y^2)^6.   

That's way too much multiplication to do by hand.  

Fortunately, a useful tool has been provided for us, at   

https://www.symbolab.com/solver/expand-calculator

Per the "expand" tool on that page, the coefficient of y⁴ is 375.  

But don't take my word for it, go there and check it for yourself.  :-)  

Is it considered cheating, to use a tool to avoid the drudgery of endless

multiplications?  I don't consider it cheating to use any tool that has been  

made available.  The abacus, the slide rule, the calculator, the computer.      

_.

Find the ordered triple (p,q,r) that satisfies the following system:  

p - 2q = 3  

q - 2r = -2 + q  

p + r = 9 + p  

Let's just number those equations, for identification:  

(1)  p – 2q  =  3  

(2)  q – 2r  =  –2 + q  

(3)  p + r  =  9 + p  

Per equation (2), r = 1  

Per equation (3), r = 9  

It can't be both at the same time, so the system has no solution.  

^.

We will have the same amount of money if I give my brother $5. Instead, if he offers me $25, I'll have twice as much cash as he does.  

Well........ if he only offers $25 nothing changes.  

He would have to actually hand over the $25.  

But I think I know what you mean.  

Let M stand for My original amount  

Let B stand for Brother's original amount   

                                                             (M – 5)  =  (B + 5)   

                                                             (M + 25)  =  (2)(B – 25)  

                                                              M – 5  =  B + 5   gives us  M = B + 10    (1)  

                                                              M + 25  =  2B – 50                                 (2)  

Substitute the value of M from (1)   

into the equation (2)                             B + 10 + 25  =  2B – 50   

Combine like terms                                          85  =  B    

Brother has 85 dollars before they start swapping moneys.   

My amount is 95 dollars   

Check answer  

If I give my brother 5, then I'll have 90 and so will he, so we have the same.  

If my brother gives me 25, then I'll have 120 and my brother will have 60.  

_.

a=listfor(n, 1, 40, (2* n* (sqrt(2* n + 1) - sqrt(2* n - 1));print a, sum a

(1.464101615, 2.01606868, 2.458100001, 2.833989511, 3.166247904, 3.467117821, 3.74404899, 4.001956471, 4.244279723, 4.473535028, 4.691628224, 4.90004344, 5.09996299, 5.292346764, 5.477986671, 5.657545079, 5.831582643, 6.000578899, 6.164947788, 6.325049561, 6.481200049, 6.633677961, 6.782730723, 6.928579181, 7.071421427, 7.211435958, 7.348784282, 7.483613098, 7.616056131, 7.746235682, 7.874263952, 8.000244167, 8.124271556, 8.246434191, 8.366813718, 8.485485994, 8.602521647, 8.717986566, 8.831942334, 8.944446615)>>Sum== 242.809263

Hi Ron,

LOL. I’m familiar with that location. It’s the same place that some of the forum’s homework-mooching students shove the detailed-answer-solutions presented by competent answerers.    

Submitted for your education and disapproval, Web2.0calc’s Math Forum, formally one of the repositories of knowledge for the bright and brilliant-minded, and also the drab and undistinguished, dusty-minded dolts who have propensity for falling into uncovered manholes while reading math solutions on their phones; now, a forum cluttered with suppositories waiting for the bone-heads, who set all educational topics back 3000 years as their brain dies decades before death claims the rest of them.

Why suppositories? Because shoving it up their ass is the only way to get the knowledge into them. Of course, as soon as they take a shit they lose that knowledge, but a copy remains on their homework.

To find the sum of f(1) + f(2) + ... + f(40), we can simplify each term and then calculate the sum.

Let's simplify f(n) for any positive integer n:
f(n) = 4n / (sqrt(2n - 1) + sqrt(2n + 1))

To simplify the expression, we can multiply both the numerator and the denominator by the conjugate of the denominator, which is (sqrt(2n - 1) - sqrt(2n + 1)):

f(n) = (4n / (sqrt(2n - 1) + sqrt(2n + 1))) * ((sqrt(2n - 1) - sqrt(2n + 1)) / (sqrt(2n - 1) - sqrt(2n + 1)))

Simplifying further, we get:

f(n) = (4n * (sqrt(2n - 1) - sqrt(2n + 1))) / ((2n - 1) - (2n + 1))
= (4n * (sqrt(2n - 1) - sqrt(2n + 1))) / (-2)

Now we can calculate the sum of f(1) + f(2) + ... + f(40):

sum = f(1) + f(2) + ... + f(40)
= (41(sqrt(21 - 1) - sqrt(21 + 1)))/(-2) + (42(sqrt(22 - 1) - sqrt(22 + 1)))/(-2) + ... + (440(sqrt(240 - 1) - sqrt(240 + 1)))/(-2)
= -2 * (1*(sqrt(21 - 1) - sqrt(21 + 1)) + 2*(sqrt(22 - 1) - sqrt(22 + 1)) + ... + 40*(sqrt(240 - 1) - sqrt(240 + 1)))

Now we can simplify each term in the parentheses and calculate the sum:

sum = -2 * (1*(sqrt(21 - 1) - sqrt(21 + 1)) + 2*(sqrt(22 - 1) - sqrt(22 + 1)) + ... + 40*(sqrt(240 - 1) - sqrt(240 + 1)))
= -2 * (1*(sqrt(1) - sqrt(3)) + 2*(sqrt(3) - sqrt(5)) + ... + 40*(sqrt(79) - sqrt(81)))

a+b=14,   

a^3+b^3=812, solve for a, b

use substitution to get:

a = 7 - sqrt(3)    and     b = 7 + sqrt(3)

a = 7 + sqrt(3)   and     b = 7 - sqrt(3)

That is incorrect!  You are given only the altitudes of the triangle and no side is known.

Let's make two variables, y(your money) and b(brother's money).

The first sentence creates the equation y-5 = b+5

Rearranging this equation gives me y=b+10.

The second sentence creates the equation y+25 = 2(b-25).

I can plug in y from the first equation, giving me (b+10)+25=2(b-25), which simplifying gives b+35=2b-50, which b=85, so your brother has 85 dollars. 

We can check by plugging b in, getting the y value, then plugging it into the other equation. 

y-5=85+5 so y=95

Plugging y=95 and b=85 into the second equation, I get 95+25=2(85-25) which is 120=120, which is true, meaning that I solved everything correct.

So your brother currently has 85. 

The area of a triangle is given by the formula:

area = (1/2)bh

where b is the base of the triangle and h is the height of the triangle.

In this case, we are given the heights of the triangle, so we can simply plug them into the formula to find the area.

area = (1/2)(H(a) + H(b) + H(c))

= (1/2)(16 + 18 + 20)

= 136

Therefore, the area of triangle ABC is 136.

Your calculations are generally correct.
And the specified financial calculator does not work, maybe you should look for another one

It is not possible to anwser this question because 50 > 40 yet you are arriving later.... please clarify / change your numbers to fit the problem accordingly 

Hi guest! There is a similar problem already on this website here is the link good luck  : 

https://web2.0calc.com/questions/please-help-me_130