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Vedmedyk [2.9K]
4 years ago
7

When 118 is divided by 11, what is the quotient and the remainder

Mathematics
1 answer:
zlopas [31]4 years ago
4 0
Remainder would be 8 I believe idk
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What is the area of this cross section of this rectangular prism?
tino4ka555 [31]

Answer:

Step-by-step explanation

cross-section is 10 inches by 16 inches

area = 10*16 = 160 square inches

6 0
3 years ago
The graph of a translated exponential function is shown below. Its parent function is y = 4x.
Trava [24]
The function has been translated down 3 units and right 2 units. The equation of the graphed function is
.. y = -3 +4^(x -2)

6 0
4 years ago
-10,20,-40 what are the next two terms
dybincka [34]

Answer:

80

Step-by-step explanation:

10+10= 20

20+20=40

40+40=80

4 0
3 years ago
Read 2 more answers
A brick of mass 4 kg hangs from the end of a spring. When the brick is at rest, the spring is stretched by 3 cm. The spring is t
Zielflug [23.3K]

Answer:

Step-by-step explanation:

In this system we have the force of the spring and the gravitational force. The equation that describes that is

F_{s}+F{g}=ma\\k(y_0+y)-mg=ma

where y0 is the equilibrium position when the string is free and y0+y is the new equilibrium position when the object is hanged of the string. By replacing by derivatives we have

ky_0+ky-mg=ma\\mg+ky-mg=ky=ma\\\\ky=m\frac{d^2y}{dt^2}\\\\my''+ky=0

the solution for this differential equation is (by using the characterisic polynomial)

y(x)=Acos(kt)+Bsin(kt)\\k=\omega^2m

hope this helps!!

7 0
4 years ago
Read 2 more answers
Find the general term of {a_n}
Assoli18 [71]

From the given recurrence, it follows that

a_{n+1} = 2a_n + 1 \\\\ a_{n+1} = 2(2a_{n-1} + 1) = 2^2a_{n-1} + 1 + 2 \\\\ a_{n+1} = 2^2(2a_{n-2}+1) + 1 + 2 = 2^3a_{n-2} + 1 + 2 + 2^2 \\\\ a_{n+1} = 2^3(2a_{n-3} + 1) + 1 + 2 + 2^2 = 2^4a_{n-3} + 1 + 2 + 2^2 + 2^3

and so on down to the first term,

a_{n+1} = 2^na_1 + \displaystyle \sum_{k=0}^{n-1}2^k

(Notice how the exponent on the 2 and the subscript of <em>a</em> in the first term add up to <em>n</em> + 1.)

Denote the remaining sum by <em>S</em> ; then

S = 1 + 2 + 2^2 + \cdots + 2^{n-1}

Multiply both sides by 2 :

2S = 2 + 2^2 + 2^3 + \cdots + 2^n

Subtract 2<em>S</em> from <em>S</em> to get

S - 2S = 1 - 2^n \implies S = 2^n - 1

So, we end up with

a_{n+1} = 4\cdot2^n + S \\\\ a_{n+1} = 2^2\cdot2^n + 2^n-1 \\\\ a_{n+1} = 2^{n+2} + 2^n - 1 \\\\\implies \boxed{a_n = 2^{n+1} + 2^{n-1} - 1}

5 0
3 years ago
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