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3 years ago

Find the 123rd term of the sequence: 5, -5, 5, -5,.......

2 answers:

7 0

You see, every even number (2, 4, 6, 8, 10) is going to be negative. Every odd number (1, 3, 5, 7, 9) is going to be positive. Since 123 is odd, the answer is 5.

Lelu [443]3 years ago

4 0

The 123rd term of the sequence is 5

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Please help ASAP

Arisa [49]

Answer:

2 lessons per day for 27 days and 3 lessons a day for 4 days.

Step-by-step explanation:

66/31 = 2 4/31

Since 66 divided by 31 is 2 with remainder 4, you need to do 31 days of 2 lessons per day. To do the extra 4 lessons, do one more lesson per day for 4 of those days.

That results in:

2 lessons per day for 27 days and 3 lessons a day for 4 days.

Check:

2 lessons a day for 27 days = 2 * 27 = 54

3 lessons a day for 4 days = 3 * 4 = 12

54 + 12 = 66

If you do 2 lessons a day for 27 days and 3 lessons a day for 4 days, you will do the 66 lessons.

5 0

3 years ago

A shelf holds 2 cans of tomato soup, 8 cans of vegetable soup, 1 can of chicken noodle soup, and 8 cans of potato soup. Without

yanalaym [24]

You have an 8 out of 19 chance to grab a potato soup.

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

Can someone pleaseeee hellpp??

monitta

Answer:

5.8

Step-by-step explanation:

To get the mean, add up all the numbers

9+4+8+3+5 = 29

Then divide by how many numbers there are (5)

29/5 =5.8

The mean is 5.8

4 0

3 years ago

Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s). Line m has no y

goblinko [34]

If the line does not have a y int, then it means it does not cross the y axis....that means it is a vertical line.....its x int is (3,0)...so this equation for the line is x = 3

if the line does not have an x int, it means it does not cross the x axis...that means it is a horizontal line...its y int is (0,-4)...so the equation for this line is
y = -4

4 0

3 years ago

Read 2 more answers