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MrRa [10]
4 years ago
10

Help me solve this??? 3p - 1 = 5(p - 1) - 2(7 - 2p)

Mathematics
1 answer:
zvonat [6]4 years ago
5 0
Hey there!

3p - 1 = 5(p-1)-2(7-2p)

Use the distributive property 

3p - 1= (5)(p)+(5)(-1)+(-2)(7)+(-2)(-2p)

3p - 1 = 5p - 5 - 14 + 4p

3p - 1= 9p - 19 

3p - 9p =-19 + 1 

-6p = 18 

Divide both sides by -6 

-6p/-6 = -18/-6

p = 3 

Best of luck with your studies!
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Step-by-step explanation:

an exponential equation has the form y=ar^x where y=final amount, a=initial amount, r=rate, and x=time, in this case

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3 years ago
Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co
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Answer:

<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If \sum a_n is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho

If \rho < 1, the series converges absolutely

If \rho > 1, the series diverges

If \rho = 1, the test fails.

Given the series \sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}

To test for convergence or divergence using ratio test, we will use the condition above.

a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}

\frac{a_n_+_1}{a_n} =  \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\

\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\

aₙ₊₁/aₙ =

\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\

note that any constant dividing infinity is equal to zero

|\frac{1+2/\infty+1/\infty^2}{5}|\\\\

\frac{1+0+0}{5}\\ = 1/5

\rho = 1/5

Since The limit of the sequence given is less than 1, hence the series converges.

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3 years ago
Can you find the sum of this triangle?<br>please ​
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3 years ago
Find the x intercepts of f(x) = –16x^2 + 22x + 3
Helga [31]
That is where the line crosses the x axis or where y=0

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we gon do grouping
ac method
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set eaqual to 0

8x+1=0
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x intercepts ate x=-1/8 and x=3/2
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3 years ago
Given that g(y) = y2 - 1, find g(a - 1).<br>Give your answer in its simplest form.​
AlekseyPX

Answer:

a^2 - 2a.

Step-by-step explanation:

Replace the y in g(y) by a - 1 and simplify.

g(a-1) = (a - 1)^2 - 1

= a^2 - 2a + 1 - 1

= a^2 - 2a.

8 0
4 years ago
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