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3 years ago

Constant of proportionality between yyy and xxx of \dfrac{8}{5}

Mathematics

1 answer:

Artist 52 [7]3 years ago

5 0

It’s 8 I think not sure

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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

3. Consider the functions f(x) = x2 + 10x – 5, g(x) = 8x + 1, and h(x) = 3x – 4. List the functions from least to greatest value

weeeeeb [17]

<u>x = 3</u>
f(x) = x²+ 10x - 5
f(3) = (3)² + 10(3) - 5
f(3) = 9 + 30 - 5
f(3) = 39 - 5
f(3) = 34

g(x) = 8x + 1
g(3) = 8(3) + 1
g(3) = 25

h(x) = 3x - 4
h(3) = 3(3) - 4
h(3) = 9 - 4
h(3) = 5

<u>x = 6
</u>f(x) = x² + 10x - 5
f(6) = (6)² + 10(6) - 5
f(6) = 36 + 60 - 5
f(6) = 96 - 5
f(6) = 91

g(x) = 8x + 1
g(6) = 8(6) + 1
g(6) = 48 + 1
g(6) = 49

h(x) = 3x - 4
h(6) = 3(6) - 4
h(6) = 18 - 4
h(6) = 14

I would explain to someone that you don't need to do any calculations to know the order of the functions when x is equal to 15 by knowing that f(x) is equal to 370, g(x) is equal to 121, and h(x) is equal to 41 to know that it is easy finding the function of x without calculating the answer.

4 0

3 years ago

Solve t-v/r=k, for v

Arisa [49]

Answer:

t-v/r = k

t-k = v/r

v = r(t-k)

Hope this helps!

6 0

3 years ago

$6.53 Round to the nearest dollar