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3 years ago

What is reciprocal of 6 and 13

Mathematics

1 answer:

LuckyWell [14K]3 years ago

5 0

Answer:

Step-by-step explanation:

1/6 and 1/13

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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:

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

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

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3 years ago

A circle has its Center at (-2, 5) and a radius of 4 units. What is the equation of the circle?

Artist 52 [7]

$\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{-2}{ h},\stackrel{5}{ k})\qquad \qquad 
radius=\stackrel{4}{ r}
\\\\\\\
[x-(-2)]^2+[y-5]^2=4^2\implies (x+2)^2+(y-5)^2=16$

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3 years ago

Question 1 (Essay Worth 10 points)

Shalnov [3]

✿————✦————✿————✦————✿

The answer is: <u>2(k2−4k)(2c+5)</u>

✿————✦————✿————✦————✿

Step:

* Consider 2ck2+5k2−8ck−20k. Do the grouping 2ck2+5k2−8ck−20k=(2ck2+5k2) +(−8ck−20k), and factor out k2 in the first and −4k in the second group.

* Factor out the common term 2c+5 by using the distributive property.

* Rewrite the complete factored expression.

✿————✦————✿————✦————✿

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3 years ago

156787 can go into 4 how many times???????

kiruha [24]

39196.75 is the answer to your problem

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3 years ago

9n+21−3n−24 simplify

Thepotemich [5.8K]

Alright First you combine the like terms.
9n - 3n = 6n
21 - 24 = -3

Rewrite the simplified equation
6n - 3 = 0

Solve for n
6n = 3
n = 3/6
n = 1/2

Ask if you need more help! :)

6 0

3 years ago

What is reciprocal of 6 and 13 ​

What is reciprocal of 6 and 13