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3 years ago

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Which of the following rules represents a dilation? A. (x, y) → (x − 3, y + 5) A. , (x, y) → (x − 3, y + 5) , B. (x, y) → (−y, x

) B. , (x, y) → (−y, x) , C. (x, y) → (0.5x, 0.5y) C. , (x, y) → (0.5x, 0.5y) , D. (x, y) → (−x, y) D. , (x, y) → (−x, y) ,

1 answer:

murzikaleks [220]3 years ago

7 0

Answer:

35

Step-by-step explanation:

mutilply the 3 by the 9 subtract dez uts

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Jean and Clint stack boxes at a warehouse. Jean stacks 50 boxes per hour. Clint stacks 60 boxes per hour. Which of the following

Valentin [98]

Represent the total number of boxes that both jean and clint stack in h hours,

total number of boxes=(50xh)+(60xh)
you want the total number of boxes for both of them together. so the amount they can stack times h or the hours. so if h=5 you would do 50x5 plus+ 60x5 which would give you 250 and 300. add the two you get 550 boxes for 5 hours total.

5 0

3 years ago

I need help with part "C" and "D"

dmitriy555 [2]

3x²cos( x³ ) and 3sin²( x ) cos( x ) are the derivatives of the composite functions f(x) = sin(x³) and f(x) = sin³(x) respectively.

<h3>What are the derivative of f(x) = sin(x³) and f(x) = sin³(x)?</h3>

Chain rule simply shows how to find the derivative of a composite function. It states that;

d/dx[f(g(x))] = f'(g(x))g'(x)

Given the data in the question;

f(x) = sin(x³) = ?

f(x) = sin³(x) = ?

First, we find the derivate of the composite function f(x) = sin(x³) using chain rule.

d/dx[f(g(x))] = f'(g(x))g'(x)

f(x) = sin(x)

g(x) = x³

Apply chain rule, set u as x³

d/du[ sin( u )] d/dx[ x³ ]

cos( u ) d/dx[ x³ ]

cos( x³ ) d/dx[ x³ ]

Now, differentiate using power rule.

d/dx[ xⁿ ] is nxⁿ⁻¹

cos( x³ ) d/dx[ x³ ]

In our case, n = 3

cos( x³ ) ( 3x² )

Reorder the factors

3x²cos( x³ )

Next, we find the derivative of f(x) = sin³(x)

d/dx[f(g(x))] = f'(g(x))g'(x)

f( x ) = x³

g( x ) = sin( x )

Apply chain rule, set u as sin( x )

d/du[ u³ ] d/dx[ sin( x )]

Now, differentiate using power rule.

d/dx[ xⁿ ] is nxⁿ⁻¹

d/du[ u³ ] d/dx[ sin( x )]

3u² d/dx[ sin( x )]

Replace the u with sin( x )

3sin²(x) d/dx[ sin( x )]
Derivative of sin x with respect to x is cos (x)

3sin²( x ) cos( x )

Therefore, the derivatives of the functions are 3x²cos( x³ ) and 3sin²( x ) cos( x ).

Learn more about chain rule here: brainly.com/question/2285262

#SPJ1

4 0

1 year ago

Find two consecutive whole numbers that 3 lies between.<br><br><br> Please help

Aleksandr-060686 [28]

Answer:

1 and 2?

Step-by-step explanation:

6 0

3 years ago

Which number is the numerator in the fraction 1/2

Paraphin [41]

1

a numerator in a fraction is the top number

5 0

3 years ago

2,17,82,257,626,1297 next one please ?

In-s [12.5K]

The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule $n^4+1$ . The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the <em>n</em>-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

Let $b_n$ denote the <em>n</em>-th term in the sequence of forward differences of $\{a_n\}$ , defined by

$b_n=a_{n+1}-a_n$

for <em>n</em> ≥ 1. That is, $\{b_n\}$ is the sequence with

$b_1=a_2-a_1=17-2=15$

$b_2=a_3-a_2=82-17=65$

$b_3=a_4-a_3=175$

$b_4=a_5-a_4=369$

$b_5=a_6-a_5=671$

and so on.

Next, let $c_n$ denote the <em>n</em>-th term of the differences of $\{b_n\}$ , i.e. for <em>n</em> ≥ 1,

$c_n=b_{n+1}-b_n$

so that

$c_1=b_2-b_1=65-15=50$

$c_2=110$

$c_3=194$

$c_4=302$

etc.

Again: let $d_n$ denote the <em>n</em>-th difference of $\{c_n\}$ :

$d_n=c_{n+1}-c_n$

$d_1=c_2-c_1=60$

$d_2=84$

$d_3=108$

etc.

One more time: let $e_n$ denote the <em>n</em>-th difference of $\{d_n\}$ :

$e_n=d_{n+1}-d_n$

$e_1=d_2-d_1=24$

$e_2=24$

etc.

The fact that these last differences are constant is a good sign that $e_n=24$ for all <em>n</em> ≥ 1. Assuming this, we would see that $\{d_n\}$ is an arithmetic sequence given recursively by

$\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}$

and we can easily find the explicit rule:

$d_2=d_1+24$

$d_3=d_2+24=d_1+24\cdot2$

$d_4=d_3+24=d_1+24\cdot3$

and so on, up to

$d_n=d_1+24(n-1)$

$d_n=24n+36$

Use the same strategy to find a closed form for $\{c_n\}$ , then for $\{b_n\}$ , and finally $\{a_n\}$ .

$\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}$

$c_2=c_1+24\cdot1+36$

$c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2$

$c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3$

and so on, up to

$c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)$

Recall the formula for the sum of consecutive integers:

$1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$

$\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)$

$\implies c_n=12n^2+24n+14$

$\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}$

$b_2=b_1+12\cdot1^2+24\cdot1+14$

$b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2$

$b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3$

and so on, up to

$b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)$

Recall the formula for the sum of squares of consecutive integers:

$1^2+2^2+3^2+\cdots+n^2=\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$

$\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)$

$\implies b_n=4n^3+6n^2+4n+1$

$\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}$

$a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1$

$a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2$

$a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3$

$\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1$

$\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$

$\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)$

$\implies a_n=n^4+1$

4 0

3 years ago