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mr Goodwill [35]
3 years ago
15

PLEASE HELP ASAP! Question 12. Thank you

Mathematics
1 answer:
a_sh-v [17]3 years ago
8 0

Answer:

Step-by-step explanation:

In the first column, put days at the top and pounds at the bottom.

So Day 1 = 75 pounds, Day 2=150 pounds, Day 3 = 225 pounds, etc..

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Set up, but do not evaluate, the integral that represents the length of the curve given by x = 1 + 3t^2, y = 4 + 2t^3 over the i
kherson [118]

L

=

∫

t

f

t

i

√

(

d

x

d

t

)

2

+

(

d

y

d

t

)

2

d

t

. Since  

x

and  

y

are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function:  

L

=

∫

b

a

√

1

+

(

d

y

d

x

)

2

d

x

. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:

d

x

d

t

=

3

−

3

t

2

d

y

d

t

=

6

t

And we substitute these into the integral:

L

=

∫

√

3

0

√

(

3

−

3

t

2

)

2

+

(

6

t

)

2

d

t

And solve:

=

∫

√

3

0

√

9

−

18

t

2

+

9

t

4

+

36

t

2

d

t

=

∫

√

3

0

√

9

+

18

t

2

+

9

t

4

d

t

=

∫

√

3

0

√

(

3

+

3

t

2

)

2

d

t

=

∫

√

3

0

(

3

+

3

t

2

)

d

t

=

3

t

+

t

3

∣

∣

√

3

0

=

3

√

3

+

3

√

3

=6The arclength of a parametric curve can be found using the formula:  

L

=

∫

t

f

t

i

√

(

d

x

d

t

)

2

+

(

d

y

d

t

)

2

d

t

. Since  

x

and  

y

are perpendicular, it's not difficult to see why this computes the arclength.

It isn't very different from the arclength of a regular function:  

L

=

∫

b

a

√

1

+

(

d

y

d

x

)

2

d

x

. If you need the derivation of the parametric formula, please ask it as a separate question.

We find the 2 derivatives:

d

x

d

t

=

3

−

3

t

2

d

y

d

t

=

6

t

And we substitute these into the integral:

L

=

∫

√

3

0

√

(

3

−

3

t

2

)

2

+

(

6

t

)

2

d

t

And solve:

=

∫

√

3

0

√

9

−

18

t

2

+

9

t

4

+

36

t

2

d

t

=

∫

√

3

0

√

9

+

18

t

2

+

9

t

4

d

t

=

∫

√

3

0

√

(

3

+

3

t

2

)

2

d

t

=

∫

√

3

0

(

3

+

3

t

2

)

d

t

=

3

t

+

t

3

∣

∣

√

3

0

=

3

√

3

+

3

√

3

=

6

√

3

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.    

Be aware that arclength usually has a difficult function to integrate. Most integrable functions look like the above where a binomial is squared and adding the two terms will flip the sign of the binomial.

8 0
3 years ago
2(x-3)=5(x-3)+10<br><br><br> Need urgent help
ArbitrLikvidat [17]

Step-by-step explanation:

2(x-3)=5(x-3)+10

=> 2x - 6 = 5x - 15 + 10

=> -6 + 15 -10 = 5x - 2x

=> 5x -2x = 15 - 6 - 10

=> 3x = 15 - 16

=> 3x = -1

=  > x =  \frac{ - 1}{3} (ans)

4 0
3 years ago
Write the first five terms of the geometric sequence in which a1=64 and the common ratio is 5/4
natita [175]

Remember that finding terms in a geometric sequence is done by multiplying the previous term by a common ratio r. For example, we can say:

a_2 = a_1 r

a_3 = a_2 r = (a_1 r)r = a_1 r^2


We have a_1 = 64. To find a_2, let's multiply this term by \frac{5}{4}:

a_2 = 64 \cdot \frac{5}{4} = 80


Now, let's use this to find all of our other terms:

a_3 = 80 \cdot \frac{5}{4} = 100

a_4 = 100 \cdot \frac{5}{4} = 125

a_5 = 125 \cdot \frac{5}{4} = \frac{625}{4}


Thus, our terms are 64, 80, 100, 125, and (625/4).

8 0
3 years ago
Find X : log2+ logx=1
Dominik [7]
log2+logx=1;\ D:x\in\mathbb{R^+}\\\\log(2\ \cdot\ x)=log10\iff2x=10\ \ \ /:2\\\\x=5\in D\\\\Solution:x=5.
8 0
3 years ago
Read 2 more answers
The corner where the states of Utah, New Mexico, and Colorado meet is called the Four Corners.
kogti [31]
If Utah is angle a and Colorado is angle b and Arizona is angle c and New Mexico is angle d Angle a and d are vertical angles angle b and c are vertical angles 
5 0
3 years ago
Read 2 more answers
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