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Arte-miy333 [17]
2 years ago
12

Please help this is a REALLY BIG test please guysss

Mathematics
1 answer:
Brums [2.3K]2 years ago
5 0

Answer:

A

Step-by-step explanation:

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This table shows some values of an exponential function. x у 01.25 1 2.50 25 3 10 What is the function?​
Helen [10]

Answer:

y= 1.25(2)^{x}

Step-by-step explanation:

y= 1.25(2)^{x}

For x=0

y= 1.25(2)^{0} = 1.25

For x=1

y= 1.25(2)^{1} = 1.25(2)=2.5

For x=2

y= 1.25(2)^{2} = 1.25(4)=5

For x=3

y= 1.25(2)^{3} = 1.25(8)=10

7 0
2 years ago
Can someone help me with this. Will Mark brainliest.
HACTEHA [7]

Answer: True!

Step-by-step explanation: Rhombuses are like squares with an 4 equivalent sides!

hope this helped have a nice day ! :)

4 0
2 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
Choose correct response
Illusion [34]

Answer:

D

Step-by-step explanation:

11.1/0.01=11.1*100=1110

7 0
2 years ago
Solve for q.<br> a=qb+r<br> What is q equal to?<br> q=a−r−b<br> q=a−r/b<br> q=a/b −r<br> q=a/rb
Mademuasel [1]

The given equation is a=qb+r

We need to solve the equation for q.

<u>Value of q:</u>

The value of q can be determined by solving the equation a=qb+r for q.

Thus, subtracting both sides of the equation by r, we get;

a-r=qb

Now, dividing both sides of the equation by b, we have;

\frac{a-r}{b}=\frac{qb}{b}

Simplifying the terms, we get;

\frac{a-r}{b}=q

Therefore, the value of q is q=\frac{a-r}{b}

Hence, Option B is the correct answer.

7 0
3 years ago
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