8) Kelvin is at his house located at (3, 4) on a

Mathematics

1 answer:

dolphi86 [110]3 years ago

4 0

3+1 /2. = 2. The x is 2
4+0 /2=2 the y is 2

The answer is (2,2)

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What is the common ratio and explicit formula for -2,6,-18,54?

Snezhnost [94]

Answer:

The explicit formula for a given geometric sequence is of the form

$a_{4} =\frac{1}{(a_{1})^4}$

Step-by-step explanation:

Given sequence is ${\{-2,6,-18,54}\}$

Let $a_{1}=-2,a_{2}=6,a_{3}=-18,a_{4}=54$

To find the common ratio r:

$r=\frac{a_{2}}{a_{1}}$

$=\frac{6}{-2}$

$=-3$

Therefore r=-3

$r=\frac{a_{3}}{a_{2}}$

$=\frac{-18}{6}$

$=-3$

Therefore r=-3

Therefore common ratio is -3

Therefore given sequence is geometric sequence.

The explicit formula for a geometric sequence is $a_{n}=a_{1}^{r-1}$ , where common ratio is r.

From given sequence we are having r=-3

Therefore the explicit formula for a given geometric sequence is of the form $a_{4} =a_{1}^{-3-1}$

$a_{4} =a_{1}^{-4}$

$a_{4} =\frac{1}{(a_{1})^4}$

6 0

3 years ago

The graph of F(x) shown below has the same shape as the graph of G(x) = x^2 but it is shifted down 5 units and to the left 4 uni

grandymaker [24]

$\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } A\cdot B\\\\ \bullet \textit{ flips it upside-down if } A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis}$

$\bf \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{ C}{ B}\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is negative, to the right}\\\\ ~~~~~~if\ \frac{ C}{ B}\textit{ is positive, to the left}\\\\ \bullet \textit{ vertical shift by } D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{ period of }\frac{2\pi }{ B}$

with that template in mind, let's see

down by 5 units, D = -5

to the left by 4 units, C = +4

$\bf G(x)=x^2\implies G(x)=1(1x+\stackrel{C}{0})^2+\stackrel{D}{0} \\\\\\ \begin{cases} D=-5\\ C=+4 \end{cases}\implies F(x)=1(1x+\stackrel{C}{4})^2\stackrel{D}{-5}\implies F(x)=(x+4)^2-5$