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2 years ago

A(1) = -13 a(n-1) + 4 find the 2nd term in the sequence

Mathematics

1 answer:

kirill115 [55]2 years ago

5 0

Solution

The sequence is an Arithmetic sequence.

Here the 1st term = -13

An = A(n-1) + 4

Therefore, the common difference (d) = 4

2nd term = first term + common difference

2nd term = -13 + 4 = -9

The answer is -9.

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Gennadij [26K]

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2 years ago

42 is 8% of what number

Setler [38]

8%=0.08
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So your answer is 525

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2 years ago

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Write the Number with the same value as 28 tens

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280 because 28 *10 is 280.

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2 years ago

Tickets to a concert cost $2 for children, $3 for teenagers and $5 for adults. When 570 people attended the concert, the total t

-Dominant- [34]

Let

x--------> the number of children

y------> the number of teenagers

z------> the number of adults

we know that

$2x+3y+5z=1,950$ ------> equation A

$x+y+z=570$ -----> equation B

$y=\frac{3}{4}x$

$x=\frac{4}{3}y$ -------> equation C

Substitute equation C in equation A and equation B

$2[\frac{4}{3}y]+3y+5z=1,950$

$\frac{17}{3}y+5z=1,950$ --------> equation D

$[\frac{4}{3}y]+y+z=570$

$\frac{7}{3}y+z=570$ --------> equation E

Multiply equation E by $-5$

$-\frac{35}{3}y-5z=-2,850$ --------> equation F

Adds equation D and equation F

$\frac{17}{3}y+5z=1,950\\\\-\frac{35}{3}y-5z=-2,850\\\\---------\\\\-\frac{18}{3}y=-900\\\\y=3*900/18\\\\y=150\ teenagers$

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the answer is the option

$150\ teenagers$

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2 years ago

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What are the endpoint coordinates for the midsegment of △BCD that is parallel to BC?

natulia [17]

Answer:

The endpoint coordinates for the mid-segment are: $(-2,-1)$ and $(1,0)$

Step-by-step explanation:

According to the given diagram, the coordinates of the vertices of $\triangle BCD$ are: $B(-3,1), C(3,3)$ and $D(-1,-3)$

Now, the endpoints of the mid-segment of $\triangle BCD$ which is parallel to $BC$ will be the mid-points of sides $BD$ and $CD$ .

Formula for the coordinate of mid-point : $(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2})$ , where $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are two endpoints.

So, the mid-point of $BD$ will be: $(\frac{-3-1}{2},\frac{1-3}{2})=(\frac{-4}{2},\frac{-2}{2})=(-2,-1)$

and the mid-point of $CD$ will be: $(\frac{3-1}{2},\frac{3-3}{2})=(\frac{2}{2},\frac{0}{2})=(1,0)$

Thus, the endpoint coordinates for the mid-segment of $\triangle BCD$ that is parallel to $BC$ are $(-2,-1)$ and $(1,0)$

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2 years ago

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