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3 years ago

What is the seventh term in the arithmetic sequence? 3, 11, 19, 27,

Mathematics

2 answers:

Sergio [31]3 years ago

4 0

The answer will be 51

Bad White [126]3 years ago

3 0

The seventh term is 51.

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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

2 years ago

Stephanie went to the shoe store and bought two pairs of flip-flops one pair was 19.95 and the other was 23.55 if she had $50.00

Levart [38]

Answer:

$6.9

Step-by-step explanation:

23.55 + 19.95 = 43.1 (total amount spent)

She had $50 and she spent $43.1 on shoes

50 - 43.1 = 6.9

= $6.9

7 0

2 years ago

Laticia randomly selected 25% of the seventh grade students in her school and asked them their favorite season. Of the students

Marysya12 [62]

I would say 75 or 150
Do u know how many students are in her school?

4 0

2 years ago

Can someone plz help me with this question

Nostrana [21]

10^2 is 10×10 which is 100
4^2 is 4×4 which is 16
so it simplifies to 100-16×2. By Order Of Operations, parentheses, exponents, multiply and divide, add and subtract 100-16×2 makes 100-32 which is 68

5 0

2 years ago

Identify whether the given numbers are perfect squares, perfect cubes, or neither.

Anna [14]

8- perfect cube. 2×2×2=8
9-perfect square. 3×3=9
21- neither.
27- perfect cube. 3×3×3=27
1331- perfect cube. 11×11×11=1331
1332- neither
100- perfect square. 10×10=100
1000- perfect cube. 10×10×10=1000
126- neither
125- perfect cube. 5×5×5=125
25-perfect square. 5×5=25
81-perfect square. 9×9=81

5 0

2 years ago