All categories

3 years ago

Write the explicit rule for the given the sequence: 125, 25, 5, ...

Mathematics

2 answers:

Korvikt [17]3 years ago

6 0

Answer:

An=125(1/5)^n-1

Step-by-step explanation:

r=common ratio

A1= first term in the sequence

I hope this helps some

skelet666 [1.2K]3 years ago

4 0

Answer:

an=a1(r)^(n-1)

a1=first erm

r=common ratio

firs term is -5

we times each term by 5 to get next term so r=5

is the explicit formula

hope this helped

You might be interested in

Math question! <img src="https://tex.z-dn.net/?f=%20%5Cint%5C%20%7B%20%5Csqrt%7Btan%28x%29%7D%20%7D%20%5C%2C%20dx%20" id

stira [4]

Solution: $\int\limits \: \sqrt{tan(x)} \: dx$
1) pass: make the replacement
 $y = \sqrt{tan(x)} = tan(x)^ \frac{1}{2}$
$dy = \frac{1}{2} *(tan\:x)^{- \frac{1}{2}} * sec^2x\:dx$
$2dy = \frac{1}{y}*(tan^2*x+1)dx$
$2ydy = (y^4+1)dx$
$dx = \frac{2y}{y^4+1} dy$

2) pass: We substitute in the integral of the statement
 $I = \int\limits \sqrt{tanx} \: dx$
$I = \int\limits \:y* \frac{2y}{y^4+1} dy$
$I = \int\limits\: \frac{2y^2}{y^4+1} dy$

3) pass: Using the Gauss Lemma, we will factor the polynomial (y⁴ + 1) knowing that there are no real roots, so we will directly try to factorize into two polynomials of degree 2:

$y^4+1 = (y^2+ay+1)(y^2+cy+1)$
$y^4+1 = y^4+(a+c)y^3 + (ac+2)y^2+(a+c)y+1$
Make the system linear, to find the values of "a" and "c".
 $\left \{ {{a+c=0} \atop {ac+2=0}} \right.$
$a+c = 0 \:\to c=-a$
$ac+2=0\to ac = -2\to a*(-a) = -2\to -a^2 = -2\to a =\sqrt{2}$
$a+c = 0 \:\to c=-a\to c = - \sqrt{2}$
We have:
$(y^4+1)= (y^2+ \sqrt{2y} +1)(y^2- \sqrt{2y} +1)$

4) pass: We will use the partial fractions method, using the fraction from within the integral:
 $\frac{2y^2}{y^4+1} = \frac{Ay+B}{y^2+ \sqrt{2y}+1 } + \frac{Cy+D}{y^2- \sqrt{2y} +1}$
$= \frac{(A+C)y^3+(- \sqrt{2}A+B+ \sqrt{2}C+D)y^2+(A- \sqrt{2}B+C+ \sqrt{2}D)y+(B+D) }{y^4+1}$

$\longrightarrow \left \{ {{A+C=0\to A=-C} \atop { -\sqrt{2}(A-C)+(B+D)=2 }} \right.$
$-\sqrt{2}(A-C)+(B+D)=2 \to - \sqrt{2} *2A+0=2\to A = - \frac{1}{ \sqrt{2} }$
$A+C=0\to C=-A\to C = - (- \frac{1}{ \sqrt{2} } )\to C = \frac{1}{ \sqrt{2} }$
$(A+C)+ \sqrt{2} (D-B)=0\to 0+ \sqrt{2} (D-B)=0 \to B=D=0$
$B+D=0$

5)pass: Adopt what was used above:

$I = \int\limits \frac{ -\frac{1}{ \sqrt{2} }y }{y^2+ \sqrt{2}y+1 } dy+ \int\limits \frac{ \frac{1}{ \sqrt{2} }y }{y^2- \sqrt{2}y+1 } dy$

$I = - \frac{1}{ \sqrt{2} } \underbrace{\int\limits \frac{y}{y^2+ \sqrt{2y}+1 }dy }_{I_1}+ \frac{1}{ \sqrt{2} } \underbrace{\int\limits \frac{y}{y^2- \sqrt{2y}+1 }dy }_{I_2}$

6) pass: Now, solve it separately

$I_1 = \int\limits \frac{y}{y^2+ \sqrt{2y}+1 }dy$
$2I_1 = \int\limits\frac{2y}{y^2+ \sqrt{2y}+1 }dy = \int\limits \frac{2y- \sqrt{2}+ \sqrt{2} }{y^2+ \sqrt{2y}+1 } dy$
$2I_1 = \int\limits\frac{2y+ \sqrt{2} }{y^2+ \sqrt{2y}+1 }dy - \int\limits\frac{ \sqrt{2} }{y^2+ \sqrt{2y}+1 }dy$
$2I_1 = ln|y^2+ \sqrt{2}y+1| - \sqrt{2} \int\limits \frac{1}{(y \sqrt{2}+1)^2+1 }dy$
$\boxed{I_1 = \frac{1}{2} ln|y^2+ \sqrt{2} y+1| - \frac{ \sqrt{2} }{2} arctan(y \sqrt{2}+1 )}$

and

$I_2 = \int\limits \frac{y}{y^2- \sqrt{2y}+1 }dy$
$2I_2 = \int\limits\frac{2y}{y^2- \sqrt{2y}+1 }dy = \int\limits \frac{2y- \sqrt{2}+ \sqrt{2} }{y^2+ \sqrt{2y}+1 } dy$
$2I_2 = \int\limits\frac{2y- \sqrt{2} }{y^2- \sqrt{2y}+1 }dy - \int\limits\frac{ \sqrt{2} }{y^2- \sqrt{2y}+1 }dy$
$2I_2 = ln|y^2- \sqrt{2}y+1| + \sqrt{2} \int\limits \frac{1}{(y \sqrt{2}-1)^2+1 }dy$
$\boxed{I_2 = \frac{1}{2} ln|y^2- \sqrt{2} y+1| + \frac{ \sqrt{2} }{2} arctan(y \sqrt{2}-1 )}$

7) pass: Let's use the expression of $I$

$I = - \frac{I_1}{ \sqrt{2} } + \frac{I_2}{ \sqrt{2} }$

$I = - \frac{ \frac{1}{2}ln|y^2+ \sqrt{2}y+1|- \frac{ \sqrt{2} }{2}arctan(y \sqrt{2} + 1) }{ \sqrt{2} } + \frac{ \frac{1}{2}ln|y^2- \sqrt{2}y+1|+ \frac{ \sqrt{2} }{2}arctan(y \sqrt{2} - 1) }{ \sqrt{2} }$
$I = - \frac{1}{2 \sqrt{2} } ln|y^2+ \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2} +1)+$
$+\frac{1}{2 \sqrt{2} } ln|y^2- \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2}-1)$

8) pass: Now, returning to the expression as a function of x, we finally arrive at the final answer:

$I = - \frac{1}{2 \sqrt{2} } ln|y^2+ \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2} +1)+$
$+\frac{1}{2 \sqrt{2} } ln|y^2- \sqrt{2} y +1|+\frac{1}{ \sqrt{2} } arctan(y \sqrt{2}-1)$
$I = - \frac{1}{2 \sqrt{2} } ln|( \sqrt{tan\:x})^2+ \sqrt{2}( \sqrt{tan\:x}) +1| + \frac{1}{ \sqrt{2} } arctan(( \sqrt{tan\:x}) \sqrt{2} +1$
$+\frac{1}{2 \sqrt{2} } ln|( \sqrt{tan\:x})^2- \sqrt{2}( \sqrt{tan\:x}) +1| + \frac{1}{ \sqrt{2} } arctan(( \sqrt{tan\:x}) \sqrt{2} -1$

Answer:

$\boxed{I = - \frac{1}{2 \sqrt{2} } ln | ( \ tan\:x+\sqrt{2\:tan\:x} +1| + \frac{1}{ \sqrt{2} }arctan( \sqrt{2\:tan\:x} +1) +}$
$\boxed{+ \frac{1}{2 \sqrt{2} } ln | ( \ tan\:x-\sqrt{2\:tan\:x} +1| + \frac{1}{ \sqrt{2} }arctan( \sqrt{2\:tan\:x} -1) +C}}$ $\end{array}}\qquad\quad\checkmark$

4 0

4 years ago

Which expressions have a value equal to 22? Select all that apply.

xz_007 [3.2K]

18+4

Step-by-step explanation:

4 0

3 years ago

If the polynomial x^5 − 10^5 can be split as the product of the polynomials

AfilCa [17]

Hello,

x^5-10^5=(x-10)(x^4+10x^3+100x^2+1000x+10000)
=========================================

7 0

3 years ago

Read 2 more answers

Q. Saline IV fluid bags cost $64.20 for 24, 1000 ml bags from one supplier. A new supplier sells 500 ml bags for $2 each

Trava [24]

Answer:

The 1000 ml bags, over $3000

Step-by-step explanation:

First, determine how many bags you need from each vendor. Then find the cost per bag with each vendor. Remember, the bags are reported to be only half full when discarded, therefore you need the equivalent of 5,000 1,000 ml bags and 5,000 500ml bags.

4 0

3 years ago

?????????????????help???

Otrada [13]

They are not equivalent fractions.

3 0

3 years ago

Read 2 more answers