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4 years ago

Enter the explicit rule for the geometric sequence. 60,12,12/5,12/25,12/125,...

Mathematics

2 answers:

Oliga [24]4 years ago

4 0

Answer:

see explanation

Step-by-step explanation:

The n th term (explicit rule) of a geometric sequence is

$a_{n}$ = a₁ $(r)^{n-1}$

where a₁ is the first term and r is the common ratio

here a₁ = 60 and r = $\frac{12}{60}$ = $\frac{1}{5}$ , hence

$a_{n}$ = 60 $(\frac{1}{5}) ^{n-1}$

Tems11 [23]4 years ago

4 0

Answer:

60(1/5)^(n-1).

Step-by-step explanation:

The common ratio r is 12/60 = 12/5 / 12 = 12/25 / 12/5 = 1/5.

The first term a1 = 60 so the explicit rule is

a1 * r^(n-1)

= 60(1/5)^(n-1).

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F(x) = 2x^ 2 - 3x + 6 find f(-7)

aleksandr82 [10.1K]

Answer:

$f(-7) = \bold{125}$

Step-by-step explanation:

Given:

$f(x) = 2x^2 - 3x + 6$

$f(-7) = 2(-7)^2 -3(-7) +6 \\ f(-7) = 2(49) -3(-7) +6 \\ f(-7) = 98 +21 +6 \\ f(-7) = 125$

5 0

3 years ago

Read 2 more answers

Find the quadratic function that has the vertex (4,5) and passes through the point (8,21) If possible, can you explain the work?

Ganezh [65]

Start by looking at the vertex form of a quadratic function, f(x) = a(x - h)^2 + k. The variables h and k are the values of the vertex. Plug those in to get f(x) = a(x - 4)^2 + 5. To find the variable a, plug in the point given for the x and y values. So, you get (21) = a((8) - 4)^2 + 5. Solve for a algebraically, and you get a = 1. Finally, plug everything in and simplify the equation. You should get that the quadratic function is f(x) = x^2 - 8x + 21. Hope this helps!

8 0

4 years ago

Indicate the equation of the given line in standard form. The line that is the perpendicular bisector of the segment whose endpo

noname [10]

Well the line that bisects RS, will cut RS in two equal halves, therefore, that line will cut RS perpendicularly at the midpoint of RS.

now, what the dickens is the midpoint of RS anyway?

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
R&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
S&({{ 5}}\quad ,&{{ 5}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{5-1}{2}~~,~~\cfrac{5+6}{2} \right)\implies \stackrel{midpoint}{\left(2~~,~~\frac{11}{2} \right)}$

so, we know that perpendicular line, will have to go through (2, 11/2)

now, a perpendicular line to RS, will have a negative reciprocal slope to it. Well, what is the slope of RS anyway?

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -1}}\quad ,&{{ 6}})\quad 
% (c,d)
&({{ 5}}\quad ,&{{ 5}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{5-6}{5-(-1)}\implies \cfrac{5-6}{5+1}\implies -\cfrac{1}{6}$

and let's check the reciprocal negative of that,

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad -\cfrac{1}{6}\\\\
slope=-\cfrac{1}{{{ 6}}}\qquad negative\implies +\cfrac{1}{{{ 6}}}\qquad reciprocal\implies + \cfrac{{{ 6}}}{1}\implies 6$

so, then, what's is the equation of a line whose slope is 6, and goes through 2, 11/2?

$\bf \begin{array}{lllll}
&x_1&y_1\\
% (a,b)
&({{ 2}}\quad ,&{{ \frac{11}{2}}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 6
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-\cfrac{11}{2}=6(x-2)
\\\\\\
y-\cfrac{11}{2}=6x-12\implies -6x+y=-12+\cfrac{11}{2}\implies \stackrel{\textit{standard form}}{-6x+y=-\cfrac{13}{2}}$

6 0

3 years ago

Please help ill give brainliest

Naddik [55]

I think it’s option a

8 0

3 years ago

What expressions are equal if both a and b are negative numbers ?

lbvjy [14]

Should be 2 as this makes the numbers positive and the action of subtraction doesn't change

6 0

3 years ago