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Svetach [21]
3 years ago
10

PLEASE HELP What is the rule for the sequence with the first four terms below?

Mathematics
1 answer:
ivolga24 [154]3 years ago
3 0

Answer:

A) [First try is correct for all U1 to U4]

Step-by-step explanation:

1. Trying each option

(let x = 1,2,3,4) since it goes from U1 to U4 (Geometric sequence)

A.) 10(-4/5)^x

10(-4/5)^1 = -8

10(-4/5)^2 = 6.4

10(-4/5)^3 = -5.12

10(-4/5)^4 = 4.096

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If the selling price of a home is $505,000 with a 15% down payment, what is the amount of the down payment?$75,000$75,500$75,750
AlladinOne [14]

Answer

the down payment is $75,750

Explanation

The selling price of a house = $505, 000

The person made a 15% down payment

Down payment = 15% x $505, 000

Down payment = 0.15 x $505, 000

Down payment = $75, 750

Therefore, the down payment is $75,750

5 0
1 year ago
X - 7y - 10<br> - x - 7y + 10<br> x - 7y + 10<br> Find the quotient. (5x - 35y + 50) ÷ 5
weqwewe [10]

Answer:

Divide each term by 5, looking at the coefficients. The answer is x - 7y + 10


Step-by-step explanation:


3 0
3 years ago
Read 2 more answers
jeanine Baker makes floral arrangements. She has 13 different cut flowers and plans to use 7 of them. How many different selecti
Ksju [112]

A total of 1,716 selections of the 7 flowers are possible.

Step-by-step explanation:

Step 1:

There are 13 flowers from which Jeanine Baker plans to use 7 of them.

To determine the number of selections that are possible we use combinations.

The formula for combinations is; ^{n} C_{r}=\frac{n !}{(n-r) ! r !}.

Step 2:

In the given formula, n is the total number of options and r is the number of options to be selected.

For this question, n = 13 and r=7.

So ^{13} C_{7}=\frac{13 !}{(13-7) ! 7 !} = \frac{13 !}{(6) ! 7 !} = 1,716.

So a total of 1,716 selections are possible.

3 0
3 years ago
Trevon’s school is selling tickets to a fall musical. On the first day of ticket sales the school sold 13 adult tickets and 12 s
julsineya [31]

9514 1404 393

Answer:

  • adult: $7
  • student: $10

Step-by-step explanation:

Let a and s represent the prices of adult and student tickets, respectively.

  13a +12s = 211 . . . . . . ticket sales the first day

  5a +3s = 65 . . . . . . . ticket sales the second day

Subtracting the first equation from 4 times the second gives ...

  4(5a +3s) -(13a +12s) = 4(65) -(211)

  7a = 49 . . . . . . . simplify

  a = 7 . . . . . . . divide by 7

  5(7) +3s = 65 . . . . substitute into the second equation

  3s = 30 . . . . . . . subtract 35

  s = 10 . . . . . . . divide by 3

The price of one adult ticket is $7; the price of one student ticket is $10.

7 0
3 years ago
I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be
Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: g'(-1)\,(1) since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: g'(-1)= 7 as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

6 0
3 years ago
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