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lakkis [162]
3 years ago
15

What fraction would represent the ratio of the cost

Mathematics
1 answer:
Stolb23 [73]3 years ago
3 0
To write 3000 as a fraction you have to write 3000 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.
3000 = 3000/1 = 30000/10

And finally we have:
3000 as a fraction equals 30000/10
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Find the average rate of change for f(x)=9x^2+3 over the interval [x,x+h].
butalik [34]

Answer:

9h +18x

Step-by-step explanation:

\frac{f(x +h) -f(x)}{x+h -x} \\ \frac{9(x +h)^2 +3 -(9x^2 +3)}{h} \\ \frac{9(x^2 +2xh +h^2) +3 -9x^2 -3 }{h} \\ \frac{9x^2 +18xh +9h^2 +3 -9x^2 -3}{h} \\ \frac{9h^2 +18xh}{h} \\ 9h +18x

4 0
3 years ago
The sum of the first 10 terms of an arithmetic series is 100 and the sum of next 10 300 .Find the series.​
Free_Kalibri [48]

Let a be the first term in the sequence, and d the common difference between consecutive terms. If aₙ denotes the n-th term in the sequence, then

a₁ = a

a₂ = a₁ + d = a + d

a₃ = a₂ + d = a + 2d

a₄ = a₃ + d = a + 3d

and so on, up to the n-th term

aₙ = a + (n - 1) d

The sum of the first 10 terms is 100, and so

\displaystyle \sum_{n=1}^{10} a_n = 100 \\ \sum_{n=1}^{10} (a + (n-1)d) = 100 \\ (a-d) \sum_{n=1}^{10} 1 + d \sum_{n=1}^{10} n = 100 \\ 10a+45d = 100

where we use the well-known sum formulas,

\displaystyle \sum_{n=1}^N 1 = 1 + 1 + 1 + \cdots + 1 = N

\displaystyle \sum_{n=1}^N n = 1 + 2 + 3 + \cdots + N = \frac{N(N+1)}2

The sum of the next 10 terms is 300, so

\displaystyle \sum_{n=11}^{20} a_n = 300 \\ (a-d) \sum_{n=11}^{20} 1 + d \sum_{n=11}^{20} n = 300 \\ (a-d) \left(\sum_{n=1}^{20} 1 - \sum_{n=1}^{10} 1\right) 1 + d \left(\sum_{n=1}^{20} n - \sum_{n=1}^{10} n\right) = 300 \\ 10a+145d = 300

Solve for a and d. Eliminating a gives

(10a + 145d) - (10a + 45d) = 300 - 100

100d = 200

d = 2

and solving for a gives

10a + 145×2 = 300

10a = 10

a = 1

So, the given sequence is simply the sequence of positive odd integers,

{1, 3, 5, 7, 9, …}

given recursively by the relation

\begin{cases}a_1 = 1 \\ a_n = a_{n-1} + 2 & \text{for }n>1\end{cases}

and explicitly by

a_n = 1 + 2(n-1) = 2n - 1

for n ≥ 1.

7 0
2 years ago
I need help on these questions who ever answers this question will be the brainliest!!
Nana76 [90]

Answer:

buut

Step-by-step explanation:

hvb hbd

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3 years ago
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