All categories

2 years ago

25 points+brainliest!

Mathematics

2 answers:

STatiana [176]2 years ago

8 0

Answer:

an= 10n+90

a10= 190

Serggg [28]2 years ago

5 0

Answer:

$\displaystyle a_n=100+10(n-1)$

$\displaystyle a_{10}=190$

Step-by-step explanation:

We have the arithmetic sequence:

100, 110, 120, 130...

And we want to determine the equation for the nth term of the sequence.

The standard form for the nth term of an arithmetic sequence is given by the formula:

$\displaystyle a_n=a+d(n-1)$

Where $a_n$ is the nth term, $a$ is the first term, and $d$ is our common difference.

From the sequence, we can see that the first term $a$ is 100.

Also, each term is +10 the previous term. Hence, our common difference $\displaystyle d$ is +10.

Therefore, our equation is:

$\displaystyle a_n=100+10(n-1)$

To find the 10th term, we can substitute 10 for n and evaluate. Hence:

$\displaystyle a_{10}=100+10(10-1)$

Evaluate:

$\displaystyle a_{10}=100+10(9)=100+90=190$

Therefore, the 10th term is 190.

You might be interested in

What divide by what equals 0.0505........

Oksana_A [137]

$\frac{5}{99} = 0.05050505...$

5 0

2 years ago

Match each vector operation with its resultant vector expressed as a linear combination of the unit vectors i and j.

Cloud [144]

Answer:

3u - 2v + w = 69i + 19j.

8u - 6v = 184i + 60j.

7v - 4w = -128i + 62j.

u - 5w = -9i + 37j.

Step-by-step explanation:

Note that there are multiple ways to denote a vector. For example, vector u can be written either in bold typeface "u" or with an arrow above it $\vec{u}$ . This explanation uses both representations.

$\displaystyle \vec{u} = \langle 11, 12\rangle =\left(\begin{array}{c}11 \\12\end{array}\right)$ .

$\displaystyle \vec{v} = \langle -16, 6\rangle= \left(\begin{array}{c}-16 \\6\end{array}\right)$ .

$\displaystyle \vec{w} = \langle 4, -5\rangle=\left(\begin{array}{c}4 \\-5\end{array}\right)$ .

There are two components in each of the three vectors. For example, in vector u, the first component is 11 and the second is 12. When multiplying a vector with a constant, multiply each component by the constant. For example,

$3\;\vec{v} = 3\;\left(\begin{array}{c}11 \\12\end{array}\right) = \left(\begin{array}{c}3\times 11 \\3 \times 12\end{array}\right) = \left(\begin{array}{c}33 \\36\end{array}\right)$ .

So is the case when the constant is negative:

$-2\;\vec{v} = (-2)\; \left(\begin{array}{c}-16 \\6\end{array}\right) =\left(\begin{array}{c}(-2) \times (-16) \\(-2)\times(-6)\end{array}\right) = \left(\begin{array}{c}32 \\12\end{array}\right)$ .

When adding two vectors, add the corresponding components (this phrase comes from Wolfram Mathworld) of each vector. In other words, add the number on the same row to each other. For example, when adding 3u to (-2)v,

$3\;\vec{u} + (-2)\;\vec{v} = \left(\begin{array}{c}33 \\36\end{array}\right) + \left(\begin{array}{c}32 \\12\end{array}\right) = \left(\begin{array}{c}33 + 32 \\36+12\end{array}\right) = \left(\begin{array}{c}65\\48\end{array}\right)$ .

Apply the two rules for the four vector operations.

$\displaystyle \begin{aligned}3\;\vec{u} - 2\;\vec{v} + \vec{w} &= 3\;\left(\begin{array}{c}11 \\12\end{array}\right) + (-2)\;\left(\begin{array}{c}-16 \\6\end{array}\right) + \left(\begin{array}{c}4 \\-5\end{array}\right)\\&= \left(\begin{array}{c}3\times 11 + (-2)\times (-16) + 4\\ 3\times 12 + (-2)\times 6 + (-5) \end{array}\right)\\&=\left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle\end{aligned}$

Rewrite this vector as a linear combination of two unit vectors. The first component 69 will be the coefficient in front of the first unit vector, i. The second component 19 will be the coefficient in front of the second unit vector, j.

$\displaystyle \left(\begin{array}{c}69\\19\end{array}\right) = \langle 69, 19\rangle = 69\;\vec{i} + 19\;\vec{j}$ .

$\displaystyle \begin{aligned}8\;\vec{u} - 6\;\vec{v} &= 8\;\left(\begin{array}{c}11\\12\end{array}\right) + (-6) \;\left(\begin{array}{c}-16\\6\end{array}\right)\\&=\left(\begin{array}{c}88+96\\96 - 36\end{array}\right)\\&= \left(\begin{array}{c}184\\60\end{array}\right)= \langle 184, 60\rangle\\&=184\;\vec{i} + 60\;\vec{j} \end{aligned}$ .

$\displaystyle \begin{aligned}7\;\vec{v} - 4\;\vec{w} &= 7\;\left(\begin{array}{c}-16\\6\end{array}\right) + (-4) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}-112 - 16\\42+20\end{array}\right)\\&= \left(\begin{array}{c}-128\\62\end{array}\right)= \langle -128, 62\rangle\\&=-128\;\vec{i} + 62\;\vec{j} \end{aligned}$ .

$\displaystyle \begin{aligned}\;\vec{u} - 5\;\vec{w} &= \left(\begin{array}{c}11\\12\end{array}\right) + (-5) \;\left(\begin{array}{c}4\\-5\end{array}\right)\\&=\left(\begin{array}{c}11-20\\12+25\end{array}\right)\\&= \left(\begin{array}{c}-9\\37\end{array}\right)= \langle -9, 37\rangle\\&=-9\;\vec{i} + 37\;\vec{j} \end{aligned}$ .

7 0

2 years ago

The height of a baseball after being hit by a batter is given by h(t) = -16t2 + 60t + 4 (where h(t) is the

lakkis [162]

Hayya please help with mine its right above

3 0

2 years ago

Ship A receives a distress signal from the southwest, and ship B receives a distress from the same vessel from the north. At wha

Fiesta28 [93]

I've attached an image showing the coordinates of Ship A and Ship B.

Answer:

distressed vessel is located at (1, 1.5)

Step-by-step explanation:

From the image attached, we can see that;

Coordinate of ship A is (3, 4)

Coordinate of ship B is (1, 1)

Now, we are told that Ship A receives a distress signal from the southwest, and ship B receives a distress from the same vessel from the north.

This means that the distressed ship is somewhere in between ship A & B.

To solve this, we will extend the arrow line from Ship A in that same SW direction and we will do the same for ship B in the same North direction. Their point of intersection is where the distressed ship is located.

Now, if we do that, we will see that they will intersect at the point; (1, 1.5)

Thus,distressed vessel is located at (1, 1.5)

5 0

2 years ago

Herman already has 1 sticker, and he will get 33 additional stickers from each sticker pack he buys.

adell [148]

Answer:

Herman needs to buy 3 sticker packs to have a total of 100 stickers.

Explanation:

33 x 3 = 99

99 + 1 = 100

4 0

2 years ago