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3 years ago

Solving Exponential and Logarithmic Equations In Exercise, solve for x. 100(1.21)x = 110

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

8 0

Answer:X=110/121

Step-by-step explanation:

Multiply 100 by 1.21 then by x

121x=110

Divide both side by 121

X=110/121

X=0.91

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Helps are really appreciated!!

Nastasia [14]

Answer:

a) 58.24 square centimeters

b) 5cm

Step-by-step explanation:

a) Area of a parallelogram

The formula for the Area of a parallelogram = Base × Height

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Base : 11.2cm

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= 58.24 square centimeters

b) Volume of a cuboid

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3 years ago

Find the number of terms, n, in the arithmetic series whose first term is 13, the common difference is 7, and the sum is 2613.

siniylev [52]

Answer:

Step-by-step explanation:

Recall that the sum of an arithmetic series is given by:

$\displaystyle S = \frac{n}{2}\left(a + x_n\right)$

Where n is the number of terms, a is the first term, and x_n is the last term.

We know that the initial term a is 13, the common difference is 7, and the total sum is 2613. Since we want to find the number of terms, we want to find n.

First, find the last term. Recall that the direct formula for an arithmetic sequence is given by:

$x_n=a+d(n-1)$

Since the initial term is 13 and the common difference is 7:

$x_n=13+7(n-1)$

Substitute:

$\displaystyle S = \frac{n}{2}\left(a + (13+7(n-1)\right)$

We are given that the initial term is 13 and the sum is 2613. Substitute:

$\displaystyle (2613)=\frac{n}{2}((13)+(13+7(n-1)))$

Solve for n. Multiply both sides by two and combine like terms:

$5226 = n(26+7(n-1))$

Distribute:

$5226 = n (26+7n-7)$

Simplify:

$5226 = 7n^2+19n$

Isolate the equation:

$7n^2+19n-5226=0$

We can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 7, b = 19, and c = -5226. Substitute:

$\displaystyle x =\frac{-(19)\pm\sqrt{(19)^2-4(7)(-5226)}}{2(7)}$

Evaluate:

$\displaystyle x = \frac{-19\pm\sqrt{146689}}{14} = \frac{-19\pm 383}{14}$

Evaluate for each case:

$\displaystyle x _ 1 = \frac{-19+383}{14} = 26\text{ or } x _ 2 = \frac{-19-383}{14}=-\frac{201}{7}$

We can ignore the second solution since it is negative and non-natural.

Therefore, there are 26 terms in the arithmetic series.

Our answer is A.

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I need help with this problem from the calculus portion on my ACT prep guide

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Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it:

$\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}$

Now let's re-writte n/(n+1):

$\frac{n}{n+1}=\frac{n}{n\cdot(1+\frac{1}{n})}=\frac{1}{1+\frac{1}{n}}$

Then the limit we have to find is:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{n}{n+1}=\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}$

Note that the limit of 1/n when n tends to infinite is 0 so we get:

$\lim _{n\to\infty}\frac{2}{5}\cdot\frac{1}{1+\frac{1}{n}}=\frac{2}{5}\cdot\frac{1}{1+0}=\frac{2}{5}=0.4$

So from the test ratio r=0.4 and the series converges. Then the answer is the second option.

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1 year ago