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4 years ago

$(60) = 21–42 + 10.7 V ON f(x) O

Mathematics

1 answer:

Varvara68 [4.7K]4 years ago

3 0

Your answer is ON your welcome!

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Typical misalignments between product design and operations include A. technology, reward systems, and infrastructure. B. techno

Lady bird [3.3K]

Answer:

(A) technology, reward systems, and infrastructure.

Explanation:

Misalignments can occur in technology, infrastructure, and reward systems, and also in marketing. A company is misaligned when people pursue goals and agendas that are incongruent with each other and do not combine to effectively advance a single purpose.

In fact, in marketing, misalignment is the scourge of the modern marketing department, and it’s no surprise why there is so many channels to manage with no way to predict the customer journey. Highly fragmented roles and teams executing on compressed time cycles. Intense pressure to drive business impact, while staff are drowning in disconnected data and unclear strategy.

4 0

4 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 n-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

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

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

for n ≥ 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 n-th term of the differences of $\{b_n\}$ , i.e. for n ≥ 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 n-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 n-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 n ≥ 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

Give the place value of the 5 in the given number. 63,475 a. ones c. hundreds b. tens d. thousands

ivolga24 [154]

Ones because it is the last place value

5 0

3 years ago

John borrowed $ from bank for 3 years at a rate of 10%

garik1379 [7]

<h3>Given:</h3>

P= $12500
R= 10%
T= 3 years

P= Principal amount
R= Rate of interest
T= Time period

The simple interest
The total amount paid

<h3>Solution:</h3>

$\large\boxed{Formula: I= \frac{PRT}{100}}$

First, we'll have to multiply, principal amount (12500), rate (10) and time period (3).

$12500×10×3$

$=375000$

Now, we'll have to divide the amount (375000) by 100.

$\frac{375000}{100}$

I=$3750

Now, we can find the total amount paid.

$\large\boxed{A=SI+P}$

Let's substitute according to the formula.

$A=3750+12500$

A=$16250

Therefore, simple interest is $3750 and $16250 was paid in total.

5 0

2 years ago

Plz help Plz help Plz help Plz help Plz help Plz help Plz help Plz help Plz help Plz help Plz help Plz help Plz help Plz help Pl

erastovalidia [21]

Answer:

1. KLP + PLM = 180 degrees (straight line)

2. 3x + angle PLM = 180 degrees

3. angle PLM = 180 - 3x

4. PMN = P + PLM (Exterior angle)

5. 2x + 72 = x + 180 - 3x

6. x = 27

Step-by-step explanation:

1. Notice that angle KLP + angle PLM is a straight line, so KLP + PLM = 180 degrees (straight line)

2. angle KLP = 3x, so

3x + angle PLM = 180 degrees

3. angle PLM = 180 - 3x

4. PMN = P + PLM (Exterior angle)

5. 2x + 72 = x + 180 - 3x

6. 5 gives 4x = 108, so x = 27

6 0

3 years ago

Read 2 more answers