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Rasek [7]
3 years ago
13

Please help me on this question.

Mathematics
2 answers:
AnnZ [28]3 years ago
5 0
The person above has the right answer I checked
scoundrel [369]3 years ago
4 0

Answer:

f(1)=-48(-\frac{1}{4} )

= 12

f(n)=f(n-1)\:. (-\frac{1}{4} )

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<u>OAmalOHopeO</u>

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Johan's family took him out to dinner to celebrate his amazing 2nd
leonid [27]

Answer:

$190.32

Step-by-step explanation:

$175.00 x 1.08756=$190.32

(1.08756 represents 1 + 8.756%)

8 0
3 years ago
Ava wants to figure out the average speed she is driving. She starts checking her car’s clock at mile marker 0. It takes her 4 m
olga_2 [115]

The average speed is obtained by finding the slope of the function in the given interval:

v =\frac{y2-y1}{x2-x1}

Substituting values we have:

v =\frac{6-3}{8-4}

Rewriting:

v =\frac{3}{4}

v = 0.75

Then, the generic equation of the line is:

 n-n0 = v (t-t0)

Where,

v: average speed (slope of the line)

(t0, n0): ordered pair that belongs to the line.

Substituting values we have:

 n-6 = 0.75 (t-8)

Answer:

the average speed of the car is:

b. 0.75

an equation of the line is:

b. n-6= 0.75(t-8)

8 0
3 years ago
Read 2 more answers
The side length of an equilateral triangle is 4 inches. Find the area of the<br>triangle.​
timofeeve [1]

helppppp

Step-by-step explanation:

5 0
3 years ago
How to rewrite 11/12 in two different ways
KatRina [158]

Answer:

1) 33/36 = 11/12

2) 22/24 = 11/12

Step-by-step explanation:

Hope it helps you! Please give me Brainlest!

3 0
3 years ago
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* Some amount of billiard balls were arranged in an equilateral triangle. And 5 balls were extra. When the same set of billiard
Agata [3.3K]

Lets say billiard balls are arranged in rows to form an equilateral triangle, then the first row consists of 1 ball, second row consists of 2 balls, and third row consists of 3 balls,  and so on. So there must be n balls in the n^{th} row.  

So, the total number of balls that forms the equilateral triangle with n rows is:  

1+2+3+4+5+....+n=\frac{n(n+1)}{2}

Let x_1 and x_2 be the total number of balls in the first and second arrangements respectively.  

Then,

x_1=\frac{n(n+1)}{2} +5

It has been said that there were 11 lesser balls in the second arrangement:  

x_2=\frac{1+(n+1)}{2} \times (n+1)-11=(n+1) \times \frac{(n+2)}{2} -11

Since, x_1=x_2

\frac{(n+1)}{2} \times n+5=\frac{(n+2)}{2} \times(n+1)-11

multiplying both the sides by 2

(n+1)\times n+10=(n+2)(n+1)-22

n+n^2=n^2+n+2n+2-22-10

2n=22+10-2

2n=30

n=15

Therefore,

x_1=\frac{(n+1)}{2}\times n+5=\frac{15+1}{2}  \times 15+5=125

So, there were 125 balls at the set.

4 0
3 years ago
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