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3 years ago

Please help this is due today!

Mathematics

1 answer:

Temka [501]3 years ago

4 0

Answer:

For ABC A= (-3,3) B= (0.2) and C= (-3,1)

For DFE D= (0,6) F= (2,6) and E= (2,3)

Step-by-step explanation:

I believe tha name of a sequence of transformation will be translation and the Reflection

Hope this Helped

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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:

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

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

6 0

3 years ago

Mental math explanation

Allushta [10]

Mental math is something that can be done i your head.

for example you can do 1+1 in your head and you dont need a calculator or paper to do it so i would be called mental math

Hope this helps

remember if you ever need anything and you dont want to waste points you can just post it on my pfp or message me the question

7 0

3 years ago

Read 2 more answers

Determine whether the relation is a function. find the domain and range

11Alexandr11 [23.1K]

What’s the question

5 0

4 years ago

Help pls and explain !!

Aneli [31]

Answer:

d) 5/3 x