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Eddi Din [679]
2 years ago
9

Joan saved $2.34 on an item that is usually $7.89. What percent did she save

Mathematics
1 answer:
xxTIMURxx [149]2 years ago
5 0
(2.34/ 7.89) times 100 = ˜29.65%
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PLZ<br> HELP<br> ILL<br> GIVD<br> MORE<br> IF<br> ITS<br> RIGHT
ira [324]

Answer:

1707 I think if I did it right

7 0
3 years ago
9,732,005 rounded to the nearest million
lys-0071 [83]
10,000,00 is the correct answer! Since 7 is greater then 5 you would round up to 10 which = to 10,000,000
8 0
3 years ago
Read 2 more answers
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
a store sells 14.5- ounces cans of tomatoes for $0.89 each. if the store sells 35 cans one day, how many total ounces of tomatoe
solniwko [45]

Answer:

451.675

Step-by-step explanation:

14.5x35=507.5

507.5x0.89=451.675

4 0
3 years ago
2. Draw the image of RST under the dilation with scale factor 2/3 and center of dilation (1,-1). Label the image RST .
professor190 [17]

Answer:

From the graph: we have the coordinates of RST i.e,

R = (2,1) , S = (2,-2) , T = (-1,-2)

Also, it is given the scale factor \frac{2}{3} and center of dilation C (1,-1)

The mapping rule for the center of dilation applied for the triangle as shown below:

(x, y) \rightarrow (\frac{2}{3}(x-1)+1, \frac{2}{3}(y+1)-1)

or

(x, y) \rightarrow (\frac{2}{3}x -\frac{2}{3}+1 , \frac{2}{3}y+\frac{2}{3}-1)

or

(x, y) \rightarrow (\frac{2}{3}x+\frac{1}{3} , \frac{2}{3}y-\frac{1}{3} )

Now,  

for R = (2,1)  

the image R' = (\frac{2}{3}(2)+\frac{1}{3} , \frac{2}{3}(1)-\frac{1}{3} ) or

(\frac{4}{3}+\frac{1}{3} , \frac{2}{3}-\frac{1}{3} )

⇒ R' = (\frac{5}{3} , \frac{1}{3})  

For S = (2, -2) ,

the image S'=  (\frac{2}{3}(2)+\frac{1}{3} , \frac{2}{3}(-2)-\frac{1}{3} ) or

(\frac{4}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )

⇒ S' = (\frac{5}{3} , -\frac{5}{3})

and For T = (-1, -2)

The image T' =  (\frac{2}{3}(-1)+\frac{1}{3} , \frac{2}{3}(-2)-\frac{1}{3} ) or

(\frac{-2}{3}+\frac{1}{3} , \frac{-4}{3}-\frac{1}{3} )

⇒ T' = (\frac{-1}{3} , \frac{-5}{3})

Now, label the image of RST on the graph as shown below in the attachment:

5 0
2 years ago
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