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3 years ago

A total of

2 answers:

6 0

Since there were three times as many student tickets sold, you can write the equation C = 3A C being children tickets and A being adult tickets

to get the total number of tickets sold you add children and adults so the equation would be C + A = 496

this is a substitution problem. where you take what C equals in the first equation, and plug it into C for the second equation

(3A) + A = 496
4A = 496
A = 124
so you now know that 124 adult tickets where sold.

Airida [17]3 years ago

3 0

A+s=496

s=496-a

and we are also told that:

s=3a

Since s=s we can say:

3a=496-a add a to both sides

4a=496 divide both sides by 4

a=124

So 124 adult tickets were sold.

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Answer: 12s^2

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3 years ago

Write the following equations in slope intercept form

Rus_ich [418]

Answer:

problem 1 2x+5y= 0

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3 years ago

Please, I need help in this ??

nignag [31]

Answer:

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

Step-by-step explanation:

$\int\frac{x^{4}}{x^{4} -1}dx$

Adding and Subtracting 1 to the Numerator

$\int\frac{x^{4} - 1 + 1}{x^{4} -1}dx$

Dividing Numerator seperately by $x^{4} - 1$

$\int 1 + \frac{1}{x^{4}-1 }\, dx$

Here integral of 1 is x +c1 (where c1 is constant of integration

$x + c1 + \int\frac{1}{(x-1)(x+1)(x^{2}+1)}\, dx$ ----------------------------------(1)

We apply method of partial fractions to perform the integral

$\frac{1}{(x-1)(x+1)(x^{2}+1)}$ = $\frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{x^{2} + 1}$ ------------------------------------------(2)

$\frac{1}{(x-1)(x+1)(x^{2}+1)} = \frac{A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)}{(x-1)(x+1)(x^{2} +1)}$

1 = $A(x+1)(x^{2} +1) + B(x-1)(x^{2} +1) + C(x-1)(x+1)$ -------------------------(3)

Substitute x= 1 , -1 , i in equation (3)

1 = A(1+1)(1+1)

A = $\frac{1}{4}$

1 = B(-1-1)(1+1)

B = $-\frac{1}{4}$

1 = C(i-1)(i+1)

C = $-\frac{1}{2}$

Substituting A, B, C in equation (2)

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\int\frac{1}{4(x-1)} - \frac{1}{4(x+1)} -\frac{1}{2(x^{2}+1) }$

On integration

Here $\int \frac{1}{x}dx = lnx and \int\frac{1}{x^{2}+1 } dx = arctanx$

$\int\frac{x^{4}}{x^{4} -1}dx$ = $\frac{1}{4} ln(x-1)$ - $\frac{1}{4} ln(x+1)$ - $\frac{1}{2} arctanx$ + c2---------------------------------------(4)

Substitute equation (4) back in equation (1) we get

$x + c1 + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1) - \frac{1}{2} arctanx + c2$

Here c1 + c2 can be added to another and written as c

Therefore,

$\int\frac{x^{4}}{x^{4} -1}dx = x + \frac{1}{4} ln(x-1) - \frac{1}{4} ln(x+1)-\frac{1}{2} arctanx + c$

4 0

4 years ago

5 -2x + 6y= -38 ? 3x – 4y = 32 •(-4, - 5) •(-5, 4) •(1, – 6) (4, - 5)

Svet_ta [14]

Answer:

Option D (4, -5)

Step-by-step explanation:

This question can be solved by various methods. I will be using the hit and trial method. I will plug in all the options in the both the given equations and see if they balance simultaneously.

Checking Option 1 by plugging (-4, -5) in the first equation:

-2(-4) + 6(-5) = -38 implies 8 - 30 = -38 (not true).

Checking Option 2 by plugging (-5, 4) in the first equation:

-2(-5) + 6(4) = -38 implies 10 + 24 = -38 (not true).

Checking Option 3 by plugging (1, -6) in the second equation:

3(1) - 4(-6) = 32 implies 3 + 24 = 32 (not true).

Since all the options except Option 4 have been ruled out, therefore, (4,-5) is the correct answer!!!

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3 years ago

Write two inequalities to compare -5and -3

AleksandrR [38]

-3 > -5
-5 < -3
hope it helps, sorry if I'm wrong

6 0

3 years ago