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Tatiana [17]
3 years ago
15

The table below shows the numbers of vanilla,

Mathematics
1 answer:
Naya [18.7K]3 years ago
8 0

Answer:

marcia must have sold at least 22 chocolate cones.

Step-by-step explanation:

Jan sold a total of 65 cones (20+15+30). Tomas sold a total of 68 cones (20+18+30).

If Marcia sold more cones than Jan or Tomas, she had to have sold more than 68 cones in total. 30+17 = 47. I subtracted 68-47, to see how many chocolate cones she should have sold to match the number of cones Tomas sold, which is 21 chocolate cones. Therefore, to surpass Tomas in the total number of cones sold, she had to have sold at least 22 chocolate cones.

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egoroff_w [7]

Answer:

It will take 1.47 hours for the horse and clown to collide

Step-by-step explanation:

They are driving toward each other, in opposite directions.

So to find their speed of approximation, we add their speeds.

19 mph + 15 mph = 34 mph.

This means that each hour, they are 34 miles closer.

How long it takes for them to find each other if they are 50 miles apart?

1 hour - 34 miles

x hours - 50 miles

34x = 50

x = \frac{50}{34}

x = 1.47

It will take 1.47 hours for the horse and clown to collide

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the area of a projected picture on a movie screen varies directly as the square of the distance from the projector to the screen
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Ivenika [448]
Since the length of this rectangle is 13 units and the width is 5 units, the area can be found this way:
A=l*w
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3 years ago
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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

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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

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\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

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