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3 years ago

4th term of (4x-y)^9

1 answer:

3 0

Now, let's do the same as we did for the previous one here.

$\bf (4x-y)^9\implies 
\begin{array}{llll}
term&coefficient&value\\
-----&-----&-----\\
1&+1&(4x)^9(-y)^0\\
2&+9&(4x)^8(-y)^1\\
3&+36&(4x)^7(-y)^2\\
4&+84&(4x)^6(-y)^3
\end{array}$

notice again, how did we get 84 for the 4th element's coefficient? well 36 * 7 / 3. and so on. And you can just expand it from there.

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Tyrone used 15 centimeters of tape to wrap 5 presents. How many presents did Tyrone wrap if he used 42 centimeters of tape? Assu

Elan Coil [88]

Answer: 14 presents

Step-by-step explanation:

5 presents 42 cm tape

——————I-———————.

15 cm tape 1

= (5x42) /15 = 14 presents

4 0

3 years ago

The vector (a) is a multiple of the vector (2i +3j) and (b) is a multiple of (2i+5j) The sum (a+b) is a multiple of the vector (

kow [346]

Answer:

$\|a\| = 5\sqrt{13}$ .

$\|b\| = 3\sqrt{29}$ .

Step-by-step explanation:

Let $m$ , $n$ , and $k$ be scalars such that:

$\displaystyle a = m\, (2\, \vec{i} + 3\, \vec{j}) = m\, \begin{bmatrix}2 \\ 3\end{bmatrix}$ .

$\displaystyle b = n\, (2\, \vec{i} + 5\, \vec{j}) = n\, \begin{bmatrix}2 \\ 5\end{bmatrix}$ .

$\displaystyle (a + b) = k\, (8\, \vec{i} + 15\, \vec{j}) = k\, \begin{bmatrix}8 \\ 15\end{bmatrix}$ .

The question states that $\| a + b \| = 34$ . In other words:

$k\, \sqrt{8^{2} + 15^{2}} = 34$ .

$k^{2} \, (8^{2} + 15^{2}) = 34^{2}$ .

$289\, k^{2} = 34^{2}$ .

Make use of the fact that $289 = 17^{2}$ whereas $34 = 2 \times 17$ .

$\begin{aligned}17^{2}\, k^{2} &= 34^{2}\\ &= (2 \times 17)^{2} \\ &= 2^{2} \times 17^{2} \end{aligned}$ .

$k^{2} = 2^{2}$ .

The question also states that the scalar multiple here is positive. Hence, $k = 2$ .

Therefore:

$\begin{aligned} (a + b) &= k\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 2\, (8\, \vec{i} + 15\, \vec{j}) \\ &= 16\, \vec{i} + 30\, \vec{j}\\ &= \begin{bmatrix}16 \\ 30 \end{bmatrix}\end{aligned}$ .

$(a + b)$ could also be expressed in terms of $m$ and $n$ :

$\begin{aligned} a + b &= m\, (2\, \vec{i} + 3\, \vec{j}) + n\, (2\, \vec{i} + 5\, \vec{j}) \\ &= (2\, m + 2\, n) \, \vec{i} + (3\, m + 5\, n) \, \vec{j} \end{aligned}$ .

$\begin{aligned} a + b &= m\, \begin{bmatrix}2\\ 3 \end{bmatrix} + n\, \begin{bmatrix} 2\\ 5 \end{bmatrix} \\ &= \begin{bmatrix}2\, m + 2\, n \\ 3\, m + 5\, n\end{bmatrix}\end{aligned}$ .

Equate the two expressions and solve for $m$ and $n$ :

$\begin{cases}2\, m + 2\, n = 16 \\ 3\, m + 5\, n = 30\end{cases}$ .

$\begin{cases}m = 5 \\ n = 3\end{cases}$ .

Hence:

$\begin{aligned} \| a \| &= \| m\, (2\, \vec{i} + 3\, \vec{j})\| \\ &= m\, \| (2\, \vec{i} + 3\, \vec{j}) \| \\ &= 5\, \sqrt{2^{2} + 3^{2}} = 5 \sqrt{13}\end{aligned}$ .

$\begin{aligned} \| b \| &= \| n\, (2\, \vec{i} + 5\, \vec{j})\| \\ &= n\, \| (2\, \vec{i} + 5\, \vec{j}) \| \\ &= 3\, \sqrt{2^{2} + 5^{2}} = 3 \sqrt{29}\end{aligned}$ .

6 0

3 years ago

Six times a number is greater than 20 more than that number. What are the possible values of that number?

seropon [69]

It would be easier to translate this into mathematical terms first.
Let n = the unknown number

6n > n + 20

Subtracting both sides by n to simplify:
6n - n > n + 20 - n
5n > 20

Dividing both sides by 5:
5n/5 > 20/5
n > 4

Among the choices, the correct one is the second choice.

4 0

3 years ago

−8x+6−2=−12 what is x?

zubka84 [21]

Answer: x=2

Step-by-step explanation:

have a good day! :)

plz give me brainliest

4 0

3 years ago

Read 2 more answers

PLEASE HELP!!!

ioda

Answer:

Step-by-step explanation:

5 0

3 years ago